The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
Pauli Exclusion Principle, Electron Spin, and Magnetism
I am teaching a lot of atomic theory (including quite a bit of quantum mechanics) to my 5th grade gifted and talented students. Even though my BA is in English, I take science very seriously (in a fun way) and really strive to get them excited about science. People seem to assume that kids are largely incapable of learning the really cool, deep science.
That being said. Is magnetism caused by the, "uniform motion of electrons," or is it a result of a large number of electrons spinning in the same direction along axis that are parallel? Is either statement close to the truth? Am I right to assume that Pauli's Exclusion Principle is not violated as long as none of the electrons are occupying the same space, even if they are all spiining the same way?
Both can cause magnetism. a large number of electrons spinning in the same direction will give you a permanent magnet. ANd a massed flow of electrons will give you an electromagnet. Even in a magnet many of the electrons are spinning in opposite directions, the but there is a surplus in one direction. For most substances the electrons will pari up, with one spinning one way and one spinning another way. It is only in special elements which have unpaired electrons not involved in a bond that you will have a spare electron to have one spin without being counter balanced. These are likely to be the transition elements or rare earth elements. They have a d-shell or f-shell that is "inside" the atom. Someone else can probably explain this better. Graeme Bartlett (talk) 04:46, 1 October 2009 (UTC)[reply]
(ec)Our magnetism article has a section explaining the sources of magnetism in different types of materials (and there are actually several different types of magnetism, each with different electronic origins; even protons and atomic nuclei have magnetic properties, which your students may have experienced in an MRI scan). Lots of complicated stuff in there! But many of the ideas should make some sense (and of course ask specific questions about it if not!).
The Pauli Exclusion Principle says two electrons cannot be the same in all respects. So if you have two different places you are already "different", and the spin can be "same" and not be a problem. This is the origin of the idea of "2 electrons in each atomic orbital": the pair electrons are the same except for spin; two different spin possibilities, so 2 electrons per orbital. DMacks (talk) 04:51, 1 October 2009 (UTC)[reply]
We need to distinguish three different types of magnetism: ferromagnetism, diamagnetism, and paramagnetism. All of these are essentially modeled by a quantum interaction between an electron's intrinsic magnetic moment and an externally applied magnetic field; in the case of ferromagnetism, the field is due to local interactions with other electrons in a magnetic domain (resulitng in a strong feedback and mutual alignment). In general, it is not a clean description to say that the magnetic moment is caused by "uniform motion" of electrons - this implies bulk motion (net measurable current) which contradicts experimental observation. In general though, there is a quantum-mechanical interaction between an electron's orbit description and its magnetic moment, if a magnetic field is present. If the magnetic field is uniform, the effect on all electrons' bulk motions in the material will be a similar alignment - but this is not the cause of the magnetic moment. Even if it were possible to freeze an electron motionless - (which is not actually possible) - that electron would still have an intrinsic magnetic moment. This is a very non-intuitive, non-classical effect - but it has been repeatedly observed by numerous experimental methods. A "stationary" electron is always "spinning" - it can never be stopped from spinning - it has this property built into the definition of an electron and it cannot be removed. Nimur (talk) 06:11, 1 October 2009 (UTC)[reply]
'Spin' is not actually a rotation of the particles. Electron's don't spin about their axis. (They don't even have an axis, what with lacking any kind of volume or internal structure as far as we know.) Two electrons with the same spin can occupy the same space. Two electrons with the same spin cannot occupy the same orbital - which is not a point in space but an entire pattern of motion (although 'motion' is a somewhat ill-defined concept at that level as well), different orbitals do overlap spatially. --Pykk (talk) 05:58, 2 October 2009 (UTC)[reply]
It may be helpful to think of electrons in their orbitals like a three-dimensional standing wave, though the actual term is more properly called a wave function. Bonding theories like Molecular orbital theory basically work on treating bonding like constructive and destructive interference of these waves. I commend you for taking this on with 5th graders; most adults cannot wrap their heads around quantum theory. You need to completely abandon your intuitive understanding of how physics works to really "get" it... --Jayron3205:26, 3 October 2009 (UTC)[reply]
I wouldn't advise it - the explosion would scatter the bits that didn't get annihilated all over the place. Not very hygienic. --Tango (talk) 03:22, 1 October 2009 (UTC)[reply]
The most hygienic manner in which to handle garbage would undoubtedly be with out direct body contact -- you can use impermeable gloves or chopsticks. The most hygienic fashion in which to dispose of garbage would follow suit, making sure whatever you encase the garbage in does no disintegrate until after the garbage disintegrates -- or you could decontaminate the garbage in an autoclave. DRosenbach(Talk | Contribs)03:42, 1 October 2009 (UTC)[reply]
While not a bad idea on the face, what happens if you have your rocket malfunction and crash, or worse, explode in midair? Then instead of having your nuclear waste contained in a nice small barrel, you scattered it across 500 square miles. Googlemeister (talk) 13:09, 1 October 2009 (UTC)[reply]
I did say a reliable rocket. Also, nuclear waste is regularly enclosed in strong containers that will withstand explosions and falls from great heights, but I agree that there is a small risk. Perhaps it would be safer to transport it in very small quantities up to an orbiting space station just outside the atmosphere, then send the rocket from there. Of course, once we get nuclear fusion going properly, there will be no nuclear waste, and we can use the fusion reactor to get rid on non-nuclear toxic waste. Dbfirs16:25, 1 October 2009 (UTC)[reply]
The Earth is moving sideways at 30 km/s relative to the Sun — it'd need to drop like a stone to fall into the Sun. If ever feasible you'd probably want to do gravity assist(s) to help. Considering that we have what, 70,000 tons of nuclear waste and a thousand tons of the highest grade waste alone, at the current prices (hundreds of millions of dollars per ton) this would be a very expensive method of disposal. Sagittarian Milky Way (talk) 18:59, 1 October 2009 (UTC)[reply]
Let's say you're doing a single particle double-slit experiment with detectors. You shoot the electron at the screen and measure which slit it goes through. Your detector has a unique design. It stores the information somewhere and prevents it from decohering. If you ask it for which slit the particle went through, it tells you. However, you can also tell it that you don't want to know. Then, it makes a conjugate measurement of it's information storage, erasing the original observation. Obviously, whether you see interference patterns depends on what question you ask your detector. The paradox is - what if you ask your detector after you have run all the trials? What if you pick your question to be the one that would cause the set of probabilities that is not the one you observed (ie. see interference + ask which slit or don't see interference + and ask to forget)? 76.67.76.19 (talk) 04:23, 1 October 2009 (UTC)[reply]
The short answer is that the pattern on the screen will be the same (no interference) regardless of what you do to your stored information and when. Your choice of measurement basis affects what you learn from the measurement, but the pattern on the screen is only affected by the initial act of copying. -- BenRG (talk) 12:58, 1 October 2009 (UTC)[reply]
It's not a measurement, but the interference pattern disappears if any information about the path remains anywhere, in quantum or classical form. The path from source to slit A to position X on the screen interferes with the path from source to B to X, but the path from source to A to |A>|a> to |X>|a> doesn't interfere with the path from source to B to |B>|b> to |X>|b>, because the final states are different. (|a> and |b> here being orthogonal states of your stored qubit.) -- BenRG (talk) 23:41, 5 October 2009 (UTC)[reply]
Clearly, it depends on the material. The graph illustrates several volatile fluids which have a dramatic enthalpy-of-vaporization dependence on temperature. Most materials do not have such a large variation. Nimur (talk) 05:56, 1 October 2009 (UTC)[reply]
So for most metals (for phase changes like changing from alpha-manganese to beta-manganese, or from solid to liquid lead) I can assume that the enthalpy of phase change is constant? John Riemann Soong (talk) 07:58, 1 October 2009 (UTC)[reply]
Per the "no homework" rule, I'm posting on here to check that my thinking is correct, not just fishing for answers. Anyway, among others, I was given these two problems for a Physics class:
"You are sitting in a boat in the middle of a lake. Also in the boat with you is an anvil. If you throw the anvil into the lake, what will happen to the level of the water in the lake?"
My thought here is that the level will go down. When the anvil is inside the boat, it's displacing water equal to its weight. When it's thrown out, it's displacing water equal to its volume. Since the anvil is presumably less dense than the lake (barring a wooden anvil on a mercury lake), the water level will go down.
"You are attempting to lift a bar with a man grasping either end. Once off the ground, the men each perform a chin-up. Does this increase the amount of work needed to lift the bar? What if the men were jumping up and down on the bar, instead?"
Newton's third law would mean that as the men perform their chin-ups, the force they're exerting downwards on the bar would equal the force upwards on themselves. As I'm lifting the whole lot, the work required to lift the system wouldn't change. If they're jumping up and down on the bar, the initial jump upwards wouldn't change the work needed for the same reasons, but when they land, they're adding an additional downward force due to gravity. So the work in that case would increase, provided they complete at least one full jump and landing.
I think you meant "more dense" instead of less dense in your answer for the first question. For your second question do they start performing their exercises during or after you have lifted the bar? --antilivedT | C | G07:18, 1 October 2009 (UTC)[reply]
Your first answer looks good, although you made an error here: "the anvil is presumably less dense than the lake". In your second answer, for the bar to remain stationary, you must oppose the downward force on the bar with an equal, opposite force. As the man begins to pull up, he exerts another force on the bar due to his upward acceleration, in addition to his body weight. As he decelerates at the top of his pull up, he exerts less force on the bar than his body weight, so it becomes slightly easier for you to keep the bar still. Axl¤[Talk]07:30, 1 October 2009 (UTC)[reply]
Yes, it's all a matter of timing. In the first example, as you throw the anvil slightly upwards to clear the side of the boat the average water level may rise marginally, then it will fall as the anvil is in the air, then rise again as the anvil enters the water, but the overall effect is a slight fall (too small to measure in a normal-size lake with a liftable anvil). In your second example, it is theoretically possible to reduce the amount of work done in lifting the bar if the chin-up men complete their upward acceleration before you start lifting, and decelerate (slightly pushing the bar upwards) as you lift. They would have to perform very athletic chin-ups to achieve this. With the opposite timing, the work needed would increase. Exactly the same applies to those jumping on the bar except that no upwards force is possible (unless they hook a toe under the bar), so work will nearly always increase. Remember that there is an increased force as they begin their jump, as well as on landing, so the work needed would always increase (over the empty bar) unless you lifted in mid-jump. Such split-second timings are unlikely to be achieved in practice. If the number of chin-ups or jumps is an integer during your lift, then the total work done is the same as it would be if the men were just hanging from or standing on the bar. Remember that if you lift whilst the jumpers are in the air, you are lifting only the bar, but if you lift whilst they are starting the jump, you are lifting the bar, plus their weight, plus an additional force needed to accelerate them upwards. Very hard work! You must be super-human! Dbfirs08:46, 1 October 2009 (UTC)[reply]
The work needed to lift the bar and the gymnasts is equal to their combined weight times the height through which you raise their combined centre of mass. As long as the gymnasts end up in the same position relative to the bar at the end of each lift, the height through which the centre of mass is raised is the same, so the work done is the same. It does not matter whether they do chin-ups, jump up and down, or do a whole Olympic high bar routine while the bar is being lifted - as long as they always end up in the same position relative to the bar at the end of the lift, the work done on the bar and gymnasts is always the same. Gandalf61 (talk) 10:12, 1 October 2009 (UTC)[reply]
True, but only if they end up at rest in that position. Extra work needs to be done if they are still moving at the end of the lift. Also, the two jumpers or chin-uppers might have done some of the work. Dbfirs16:18, 1 October 2009 (UTC)[reply]
If the gymnasts are still moving at the end of the lift, extra or less lifting work will be needed depending on the direction of their combined kinetic energies. Slightly extra work is needed in any case due to the energy they dissipate in causing air turbulence. Cuddlyable3 (talk) 23:38, 1 October 2009 (UTC)[reply]
Good point - I'd forgotten about air turbulance, though the amount of extra work is minimal at the speed most people do chin-ups! Kinetic energy is a scalar, so how can less work be needed if kinetic energy is given to the gymnasts? I'm still thinking about this ... Dbfirs11:49, 2 October 2009 (UTC)[reply]
And to whoever it was that tried to delete the above, he needs to back off. I would, in fact, like to hear the explanation (if any) as to why the item at 4:30 is realistic or fanciful, as I've always wondered about it, and it's within the general realm of the original question here. More importantly, why it's realistic or fanciful. →Baseball BugsWhat's up, Doc?carrots11:04, 1 October 2009 (UTC)[reply]
Well, it pains me to break the news to you - but cartoons aren't real. I understand how this might come as a shock...would this be a good time to break the news about Santa Clause and the Tooth Fairy? SteveBaker (talk) 11:14, 1 October 2009 (UTC)[reply]
I didn't say it was "real", I asked if it was "realistic". You've got a heavy object (again, an anvil) falling from a significant height, and Sam catches it. Assuming all that's possible, does the boat drop in the water somewhat? It sounds to me like a cousin to OP's question. The boat sinking is obviously a gag. But would the boat's waterline drop at all? →Baseball BugsWhat's up, Doc?carrots11:23, 1 October 2009 (UTC)[reply]
Yes - of course it drops...the ship has to displace water equal to it's weight...including that of the anvil. In cartoon-physics, anvils have near-infinite mass - so the ship sinks. SteveBaker (talk) 11:35, 1 October 2009 (UTC)[reply]
If the anvil were heavier than the water displaced by the boat itself, presumably it could sink the ship, right? So if it were dropped from an airplane and had sufficient mass, combined with a sturdy deck and a Sam of super-human strength, it might sink the ship, right? But instead it's dropped from the crows nest - making the situation fairly similar to the OP's question. When the anvil is dropped from the crows nest, would the ship rise slightly? And would it then fall back to its original waterline once Sam catches it, but also drop a little farther due to the added force of the anvil's fall? →Baseball BugsWhat's up, Doc?carrots11:42, 1 October 2009 (UTC)[reply]
But the weight of the anvil is not invariant under situational transformation in comic physics. In this case, this is even true in real physics - the anvil up on the mast is further from the middle of the Earth than down on the deck. I'd guess that the mast is roughly 1.5 times the radius of the Earth, so on the deck it will have near infinite massIn comic physics, mass and weight are interchangable*blush* (per Steven), while in the crows nest it will only have near (infinite/5.25) mass, obviously much less. --Stephan Schulz (talk) 11:51, 1 October 2009 (UTC)[reply]
The mast doesn't look that long, but it could be trick photography. The point being, if the ship were made of balsa, i.e. lighter than the anvil, then the anvil dropping probably wouldn't sink the ship by weighing it down, it would probably sink it by breaking all the way through it. But assuming the deck is strong enough to withstand the anvil hitting it (or to withstand it hitting Sam), would the anvil cause the waterline to at least momentarily drop somewhat lower than it was when the anvil was at rest in the crows nest? I say momentarily because I'm guessing the boat would have to eventually rise back up to its original waterline, with the anvil again being at rest. Right? →Baseball BugsWhat's up, Doc?carrots12:01, 1 October 2009 (UTC)[reply]
You can only gauge the relative mast height from long distance shots showing the curvature of the Earth. When the anvil is released, the ship will start an upward movement, and perform a dampened oscillation around the new waterline. When the anvil hits Sam, a downwards impulse is transmitted to the ship, which leads to a second oscillation. I think those two will simply add up until the ship settles again. --Stephan Schulz (talk) 13:44, 1 October 2009 (UTC)[reply]
That makes sense. And the more I think about it, the more similar this question is to the original, especially as Sam throws the anvil off the ship once it's underwater. It's really not too different from OP's hypothesis, except it's taken to a comical extreme. →Baseball BugsWhat's up, Doc?carrots13:52, 1 October 2009 (UTC)[reply]
The ship would rise when the anvil was released from the top of the crow's nest. Time would elapse before the anvil reconnected with the ship. Upon reconnecting with the ship there would result a lowering of the ship to the level it had been at prior the the anvil being released from the point of the crow's nest. There would be bobbing motion through all these steps, as the ship is free to move to beyond its point of equilibrium, in the process of finding its point of equilibrium. The only factor that I am aware that I am leaving out of this description involves the distance from the center of the Earth as that may be negligible in this instance. I think if the distance from the center of the Earth were factored in the result would be a lower level of the ship in the water with the anvil on the deck than with the anvil in the crow's nest. Bus stop (talk) 14:05, 1 October 2009 (UTC)[reply]
So my problem set isn't making any sense for me. Say that below some phase change temperature T_c (this particular case is 993 K), an alpha phase is favoured over the beta phase. But apparently if I calculate the free energy of each phase (from enthalpy and entropy of each phase), I predict a much lower phase transition temperature. On the other hand, I'm given the enthalpy of the phase transition and the temperature it occurs at (so I could well, predict the entropy change of the phase transition, and I have this value).
I am however given their differential heat capacities ... which happen to be constant (not dependent on temperature). But don't the free energy lines (with dependence on temperature) on the phases cross ultimately because of temperature-differential changes in heat capacity? John Riemann Soong (talk) 11:46, 1 October 2009 (UTC)[reply]
"But apparently if I calculate the free energy of each phase (from enthalpy and entropy of each phase), I predict a much lower phase transition temperature." - as a guess - are you assuming that the entropy enthalpy is constant with T ?
Second part - the free energy lines cross because of the dependence of free energy with temperature ie don't you need to calculate
GT= Gstart/standard - ʃSTdT
Where ST is the entropy as a function of temperature - the relationship includes the heat capacities..
You can visualize them as Photons, traveling through space at the speed of light. If you want to think of it as a wave, then consider it as electric and magnetic fields oscillating perpendicular to each other traveling at the speed of light. Look up Light. Rkr1991(Wanna chat?)12:44, 1 October 2009 (UTC)[reply]
And if you want to think of light as a wave - the varying magnetic field produc es and electric field (at right angles), and the varying electric field produces a magnetic field. These fields exist OK in a vacuum. Graeme Bartlett (talk) 14:03, 1 October 2009 (UTC)[reply]
The electromagnetic field progates by medium of photons, the electromagnetic fields gauge boson. Photons are the field, and the field is photons, the two 'classical style' parts of the wave-prticle duality as described by QED. The field as such is not smooth as is taught in classical dynamics, but lumpy and made up of phtons as in QED. However for most purposes the smooth classical model suffices, and is useful for being drastically simpler in most cases. Elocute (talk) 17:37, 1 October 2009 (UTC)[reply]
Magnets work fine in a vacuum; they attract and repel just as in air. Electrostatic attraction and repulsion works fine in a vacuum. Thus there is no reason that electromagnetic waves would not propagate in a vacuum. Edison (talk) 04:47, 2 October 2009 (UTC)[reply]
Followup (Photon)
OK, here's a spinoff question, which I'm not so sure has an answer. Supposedly an object's mass increases exponentially as it accelerates toward the speed of light, right? So that, in theory, a massive object can never get very close to the speed of light, and certainly not equal to speed of light, because its mass would become infinite, assuming the math about light-speed is correct. Or that's what I recall from 9th grade science. However, an extremely small particle can be accelerated nearly to the speed of light. The built-in assumption is that photons are traveling at the speed of light, i.e. at this assumed upper-bound of speed. But are they really? Do photons have literally 0 mass, or do they merely have a mass that's too small to measure? In the latter case, are they really traveling at that upper bound of speed? Or are they just short of it? →Baseball BugsWhat's up, Doc?carrots12:52, 1 October 2009 (UTC)[reply]
Well, I'll give the annoying answer. Yes - the photons are moving at the speed of light, because their speed is what we know, call, and have measured to be the speed of light since, by definition, light moves at the speed of light (well, usually). ~ Amory (user • talk • contribs)13:00, 1 October 2009 (UTC)[reply]
As far as we know, photons have a rest mass of zero (or, more preciely, the concept of rest mass does not apply to them since they can never be at rest). If they had a non-zero rest mass then the speed of light would depend on its energy, and hence on its frequency - but observations of gamma-ray bursts in distant galaxies show to a high degree of accuracy that the speed of light is independent of its frequency [2]. Usenet Physics FAQ lists some other theoretical and experimental evidence for zero rest mass. Gandalf61 (talk) 13:10, 1 October 2009 (UTC)[reply]
By definition, the speed of light is the speed that photons travel. That follows. The speed of light also varies depending on what it's traveling through. A partial or spinoff answer to the OP's question, ironically, is that the presence of a "medium", e.g. clear glass, actually slows down the speed of the photons traveling through it, if I recall correctly. When "the speed of light" is said by itself, it usually implies "in a vacuum". But I wonder if the speed of light itself, in a vaccuum, is not the actual upper bound of speed, or whether light-speed is just short of the actual upper bound - maybe so slightly short of it that it's unmeasurable. →Baseball BugsWhat's up, Doc?carrots13:18, 1 October 2009 (UTC)[reply]
Gandalf's above bit about photons having zero rest mass is how we assert that c is in fact the upper bound. We have no evidence that speeds above that exist. Of course, once you get to "so slightly short that it's unmeasurable"... well, that's why 0.999...=1 — Lomn13:25, 1 October 2009 (UTC)[reply]
Photons always (under every conceivable circumstance) travel at c, this is in all meaningful sense a tautology, as when we measure c, we measure the speed of photons. However the speed of light is a different issue for reasons relating to the difference between physical and effective (optical) path lengths. That is, a photon going through a medium doesnt go slower, it just travels further at the same speed. It is correctly noted that it cannot be taken as given that all photons (and thus all light) travels at the same speed, no purely logical reasoning will deduce this, but that as Gandalf said, the evidence to support it is strong. A side note that the mass increas is not exponential, it has a pole at c.Elocute (talk) 17:33, 1 October 2009 (UTC)[reply]
There's not really such a moment since there's an uncertainty relation between the energy of the photon and the point in time it hits the mirror. That aside, there's no reason the velocity of a photon has to be continuous in time (unlike for a massive particle). In other words, it can go from c to -c without ever being in between and in fact it's required to. You could call the value at that point of discontinuity zero, but what would that mean? Rckrone (talk) 00:07, 2 October 2009 (UTC)[reply]
OK, so it's always either headed toward the mirror or headed away from the mirror. The instant it touches the mirror is essentially 0 time, hence it doesn't count, i.e. it doesn't "compress" and bounce off like a spaldeen or something. So let's say the room is full of mirrors. The photons would bounce all over the place, forever, wouldn't they? Or if not, then where would they go? →Baseball BugsWhat's up, Doc?carrots02:16, 2 October 2009 (UTC)[reply]
Yeah, in the ideal case with perfect mirrors they would bounce around forever. What happens in practice is that mirrors absorb some of the photons that hit them and the photons' energy becomes heat (or gives some electrons enough energy to make a break for it). The other issue is that photons have momentum, so when they bounce of a stationary object (like an electron of some atom), they have to transfer some energy to the object for momentum to be conserved. The effect is pretty minimal for visible light but a comes into play with high energy photons like X-rays (see Compton scattering). Rckrone (talk) 02:57, 2 October 2009 (UTC)[reply]
OK, that was the missing piece. The photon travels at light speed while it's a photon, but it can be transformed into another state, yes? I think this exhausts my questions on this topic. I probably knew all this stuff in 9th grade, but that was some time back. I appreciate your help and your patience. →Baseball BugsWhat's up, Doc?carrots03:17, 2 October 2009 (UTC)[reply]
Here's a paper that goes through a lot of experiments that have been done to show that photons have zero rest mass [3] to high accuracy. As others have mentioned, this isn't something scientists can take for granted, despite the fact that the universal constant c that crops up in relativity is also called "the speed of light." They're only the same under the assumption that photons have zero rest mass, although that seems to be a pretty good assumption. A story that worked out differently was neutrinos which were first assumed to be massless but experiments showed this to be wrong. Rckrone (talk) 19:25, 1 October 2009 (UTC)[reply]
If I take an airstream which is at 500K and vaporize 1 gallon of water to cool it to 400K, will evaporating the same amount of water in the same volume of airstream which is at 600K cool it to 500K, or would it be more like 480K? Googlemeister (talk) 15:14, 1 October 2009 (UTC)[reply]
It depends how exact you want to be. See specific heat capacity. Evaporating the same amount of water requires the same amount of energy - but that assumes that the only relevant energy transfer is into the latent heat of vaporization. It also assumes a constant heat-capacity of the air stream with respect to temperature - again, this is probably a fairly invalid assumption, because at higher temperatures, the density may be lower due to thermal expansion - so there may be less mass present. Heat capacity is usually quantified in terms of mass, so a moving airstream needs some way to parameterize its mass flux. The intricacies are going to depend on what approximations you want to make - but since you've only specified so few parameters, it might be safe to use a very simple model. Nimur (talk) 16:35, 1 October 2009 (UTC)[reply]
Name of integrated foreign DNA?
Is there a technical name for foreign DNA (from some kind of pathogen, for example) that has been integrated into the DNA of the host? I'm not asking about integrons or retroviral proviruses, but rather some non-directed integration. Thanks. —Preceding unsigned comment added by 129.49.7.150 (talk) 15:31, 1 October 2009 (UTC)[reply]
Sure you can. It might take them a little while to get adjusted to their new territory, but they will. It depends on their disposition. Some will hide for awhile, others will adjust quickly. Just be as good to them as you can, and over time they'll get used to their new surroundings. →Baseball BugsWhat's up, Doc?carrots17:40, 1 October 2009 (UTC)[reply]
I've moved house with a cat. It is important to keep them indoors for a week or two so they learn that this new place is home, otherwise they may try to go home to the old house. We shut the cat in the bathroom for the first day or two so as not to expose her to the chaos of moving. It all went very smoothly. --Tango (talk) 19:00, 1 October 2009 (UTC)[reply]
Indoors vs. indoors/outdoors vs. outdoors cat is a key issue. And "sealing them off" in a room with everything they need will likely ease their discomfort (along with lessening the likelihood of them running away). →Baseball BugsWhat's up, Doc?carrots21:05, 1 October 2009 (UTC)[reply]
We had a cat that used to jump up on the back shelf of the car to come with us when we went off on weekends. Didn't half surprise a person who was admiring our nice soft toy at the back when he stretched and yawned. Dmcq (talk) 22:49, 1 October 2009 (UTC)[reply]
Turns out that we had heavy rain the day after moving day and our outdoor cat sat like a soldier waiting to be let in, Our old house had "cat doors" but I hadn't yet installed them in our new house. Anecdotally, they seem to tolerate the move better than →hydnjo (talk) 22:59, 1 October 2009 (UTC)[reply]
While I don't suggest that particular model of cat carrier, it is important to have one, as the average cat will go nuts inside a moving car. For long trips with some difficult cats, you may need to get tranquilizers from the vet. Once you arrive, I agree to lock them in a room with familiar items, including food, water, a blanket to sleep on, and a cat pan they've used before (it will have "their smell" on it). Some outdoor cats may need to give up that lifestyle, if the outdoors are not as safe in the new location. StuRat (talk) 01:23, 5 October 2009 (UTC)[reply]
I strongly advise you to keep the cat locked up at the new place for a week or so to make it clear that this is now "home". Otherwise, there's a reasonable chance it will attempt an "incredible journey" to its old place, as our cat did after a move. --Sean14:54, 2 October 2009 (UTC)[reply]
I've been thinking a lot about clouds recently (as one does when trying to write software to make pretty pictures of them) - and it occurs to me that it's a bit odd that clouds form such dense, compact 'lumps'. The basic idea is that water evaporates off the ocean or something - the water vapor moves up to some altitude where it condenses into droplets...but why isn't there just a more or less uniform sheet of grey 'stuff'? Why does it form clumps that look like bunny rabbits? (Except that one over there that looks like a cartoon fish). Some of them are uniform-ish sheets of grey - but some of them are really dense - with relatively hard edges. Mutual gravitation of the water droplets? Nah...something like surface tension? Van-der-waal's forces? None of these seems plausible. SteveBaker (talk) 21:50, 1 October 2009 (UTC)[reply]
My daughter asked me this a little while ago. I think it is because (1) when warm and cold fronts of air mix you get edges between air of differing temp and humidity (2) when one side of the interface starts to form water droplets it rapidly changes its absorption and radiation of radiation which extenuates the difference between it and the other side of the boundary. I don't know though it is a guess, but some feedback seems needed to give the clean edge versus natural mixing, and the upper side of clouds are often sharper which fits with radiative heat loss once droplet form. I have also noticed that sometimes jet trails of planes disappear but sometimes they just spread out and out thicken. Again, my assumption is that in the right marginal conditions the heat loss from radiating droplets can be a positive feedback cooling air and increasing droplets but generally the mixing of the water and heat just reverts the trail to the surrounding conditions. --BozMotalk22:07, 1 October 2009 (UTC)[reply]
"compact lump" sounds to me like some mechanism is minimising surface area, that would tend to result in a sphere (like surface tension, but it isn't a liquid so I guess it isn't actually surface tension, I'm not sure what it would be [certainly not gravity, it will be something based on electromagnetism]). I imagine it is flattened by atmospheric effects - the atmosphere is full of layers with different winds, temperatures, humidities, etc., the cloud is restricted to one layer (if it crossed the boundary between two layers with different winds it would get ripped apart). That flattening restricts your degrees of freedom, which increases the chance of spheres meeting and becoming bunny rabbits. I'm not sure what stops those bunny rabbits becoming a larger sphere... --Tango (talk) 22:10, 1 October 2009 (UTC)[reply]
While you are looking at the topic Steve the WP article on Double diffusive convection could do with some work. There was a whole graduate course on this when I was a lad, including atmospherics...I don't think there is any attraction involved, only feedback but I could be wrong. --BozMotalk22:14, 1 October 2009 (UTC)[reply]
There are essentially two types of cloud (well there are in climate models, and probably in reality): convective and "large scale". Large scale clouds are when the entire atmospheric layer becomes saturated (typically because the whole lot cools, either in place or because of forced ascent) and clouds form - these are the grey slabs. Convective ones are the bunny rabbits. In this case, the entire layer isn't supersaturated, but the column is unstable with respect to overturning, once you take into account heat of condensation. But in that case the ascent is organised, and this is where the clouds form. So ascending air cools, hence vapour condenses, hence the air warms and ascends, and so on (i.e. since the air is being heated as it ascends the moist adiabat is different to the dry, so the plume stays buoyant). And the plume is what makes the clouds lumpy (it can be a tall plume if its a thundercloud or a rather squat one if only a little puffy cloud) William M. Connolley (talk) 22:46, 1 October 2009 (UTC)[reply]
Here's a not-too-incorrect explanation of why daytime clouds are lumpy.[4] The lumpiness is fundamentally caused by atmospheric turbulence. There's a lot going on in clouds -- rumor has it there are people who make their living trying to predict them. Short Brigade Harvester Boris (talk) 00:16, 2 October 2009 (UTC)[reply]
I think there's really another question here. Besides being "lumpy", some clouds, including the standard cumulus humilis cloud have an oddly discrete outline to them, which I think adds to the impression of lumpiness. "Non-lumpy" clouds like nimbostratus, or even fog obviously also have all kinds of structural variation due to slight differences in temperature and pressure, we just don't think of them as lumpy because they're indistinct. So maybe a better question would be to ask why some clouds, like cumulus, are so sharply delineated, thus letting us see their lumpiness. I have a couple of hypotheses of my own on that, but no evidence to back them up and so will keep them to myself in hopes somebody else can come forward with something more concrete. I would also like to record for posterity that this is easily the most times I've typed the word "lump" in a single paragraph! Matt Deres (talk) 03:39, 2 October 2009 (UTC)[reply]
The pedant in me wished to point out that you didn't use the word "lump" even once in the preceding paragraph. "lumpy"=3, "lumpiness"=2. It's OK though - I have my 'inner pedant' under control again now. SteveBaker (talk) 20:34, 2 October 2009 (UTC)[reply]
Cumuliform clouds are buoyant parcels that are distinct from the air around them. They are the same as "thermals" in the lower atmosphere that as birds and glider pilots use. If the thermal ascends so high that it cools enough for the water vapor in it to condense, it becomes visible as a cloud. (This height is called the "lifting condensation level" and corresponds to the flat bottoms of these clouds.) The distinct boundary of the cloud is essentially the interface between the rising parcel and the outside environment. Stratiform (layered) clouds have less distinct boundaries because they are large masses of air that get lifted rather than distinct blobs, as William explains above. The high-level wispy clouds are made of ice and have fuzzy boundaries because ice crystals doesn't evaporate as readily as water droplets. Short Brigade Harvester Boris (talk) 13:10, 2 October 2009 (UTC)[reply]
October 2
cancerous lump?
Hi, im 15 years old, and i recently found a lump on my left arm, near the tricep i guess, and i think i have a similar one on my right arm, my dad said its just muscle, but i dont think so, is it natural or cancerous? —Preceding unsigned comment added by Joejoe94 (talk • contribs) 01:53, 2 October 2009 (UTC)[reply]
If you are concerned, you and your parents should consult a doctor. Wikipedia does not provide medical advice. — Lomn02:09, 2 October 2009 (UTC)[reply]
The rain after earthquake
Did anyone notice this? Or is it just me, I remembered or notice it's almost always raining after the quake. Or maybe it's just a matter of geographical location or coincident or false observation. roscoe_x (talk) 02:10, 2 October 2009 (UTC)[reply]
It's not quite what you're looking for, but we have articles on earthquake cloud and earthquake weather (and see also here. Those are phenomena said to presage the earthquake, though, not come after it. You'll note that our article asserts that "Geologists maintain that there is no connection between weather and earthquakes. " but it is uncited and, again, is talking about pre-earthquake effects. Matt Deres (talk) 03:44, 2 October 2009 (UTC)[reply]
Would it be possible to shatter a toilet by alighting on it?
In Guinness a team of kung fu masters demolished a real house without tools so it's very probably humanly possible. Whether without injury I dunno, they could've just pulled it out and thrown it off the second floor (when the floor still existed). Could an amount of jumping-on force reasonable for a person merely wanting to get on (say, to change the lightbulb or close the upper window) break it and cause grave injury? Corner case of course, (i.e. temperature unfavorable, the dude is the heaviest person that has the athletic capability to want to do such a thing, he does this all the time for decades, and the workers who brought it up the stairs ages ago banged it pretty good) Sagittarian Milky Way (talk) 05:54, 2 October 2009 (UTC)[reply]
Your question reminded me of the Great John Toilet Company, but sadly, their site appears to be gone. Their products still seem to be available, and a google search will turn up pictures and brochures. One reason cited for making these XXL johns was just what you are talking about: safety. Presumably, a standard toilet might break under a very heavy person, if there was some kind of defect. And it's easy to imagine very nasty damage: split femoral arteries and the like.--Rallette (talk) 10:24, 2 October 2009 (UTC)[reply]
Well, it'd be entirely reasonable for a person jumping up on a toilet to break cheap plastic hinges holding the seat together. It's then entirely reasonable that serious injury could result when the now-detached seat slides out from under someone. It's not as dramatic as the porcelain suddenly shattering, but it's a lot more likely to be called "plausible" by the Mythbusters. — Lomn13:14, 2 October 2009 (UTC)[reply]
Corner case is a concept usually applied to more complex systems than a toilet but I understand what the OP is getting at. It would take a kung fu master to answer how best to shatter a toilet deliberately but the OP has described a feasible way of doing it without martial art. In the obviously unlikely event of a toilet shattering accidentally merely by someone standing on it, and that causing a serious injury, the cause of the breakage would become a legal question - who, if anyone, can be held negligent? We can't predict a legal outcome. Cuddlyable3 (talk) 14:34, 2 October 2009 (UTC)[reply]
Our Dawn Eden article describes an incident where a toilet "collapsed", causing injury, but I can't find the original reference, so it's not clear if the shitter shattered. --Sean15:05, 2 October 2009 (UTC)[reply]
I'm currently doing research on the Tanager Expedition and I'm trying to get past the physical sickness and vomiting reflex I experience each time I read about the "sampling" of the species they encounter. Granted, this is 1923, but I'm curious about the ethical guidelines of scientific expeditions today in comparison to the ones conducted in the early 20th century. My question is this: If and when a biologist encounters a new species, or even a previously known species that is undergoing current study, what are the guidelines about killing that species in the wild to bring back to the laboratory? In other words, how does science approach the study of living organisms, while at the same time, acknowledging their right to exist in spite of our efforts to study them? Reading the notes of the primary biologist for Tanager, it seems like there were no ethical guidelines, and anything they encountered that was of any interest was immediately killed on the spot. Viriditas (talk) 07:44, 2 October 2009 (UTC)[reply]
I don't think science is practiced differently around the world. My understanding is that scientific methodology is a shared value, so we should find a common procedure for studying captured organisms. So, what is it? Catch and release? Kill and dissect? Study from afar and don't interfere? It sounds like it depends on the situation rather than our values. I doubt there are that many different scenarios so it should be easy to describe. I'm just curious what they are and how they differ from those used in the past, or even if they do. Viriditas (talk) 11:08, 2 October 2009 (UTC)[reply]
Conservation is recognized as an important, actually "vital" is the appropriate adjective, issue today. The links Scray provided demonstrate this. The situation was different in 1923 in many ways: the diversity of species was less well known and the rate at which they were being lost due to human interaction was little known (and do we really know today?). It is only recently that the ethic of conserving biodiversity has been recognized internationally following advances in understanding of genetics, atmospheric pollution and the potential for discovering medicines in nature. At the time of the Tanager Expedition to Hawaii, not yet a US state but already the site of American political and military involvement for which the US would later apologise, Hawaii was with its many islands an ecological terra incognita. That gave the motivation for the expedition. It had some ethically laudable goals that included halting the extinction of local species by feral rabbits and investigating hitherto unknown species. Then as now, one has to discover a species before one can take specific steps to conserve it. That research usually requires obtaining "samples" in some way, dead or alive. We should not protest against this or this or doing warning: image that may upset this because these are the only ways to understand the natural world. The OP is clearly upset by this aspect of Biology and is curious about how the ethical issues are debated today. However the Ref. Desk is not meant to be a debating forum so if you need advice or opinions, it's better to ask elsewhere. Cuddlyable3 (talk) 14:14, 2 October 2009 (UTC)[reply]
I'm not upset by biology in and of itself nor am I trying to debate, but you seem to have reinforced my original point by saying "these are the only ways to understand the natural world". Obviously, such an absolutist statement can only be wrong. While a tricorder is still science fiction, there may come a time when biologists of the 19th and 20th century are looked at in the same way we look at primitive societies who practiced human sacrifice to appease their Gods. Surely for them, it was the "only" way. Viriditas (talk) 19:28, 2 October 2009 (UTC)[reply]
Your absolutist statement that an absolutist statement can only be wrong is wrong. Statements that use the present tense of "to be" are correct or otherwise at the time they are stated. You oppose my statement by posing a "there may come a time" speculation but that does not deny its present truth. It was once semantically correct to say "We must sacrifice to appease God(s)". That perception has changed just as mine is open to change. But that hasn't happened. Frankly, it is immature to insert a fictional tricorder gadget into discussion of real ethical issues. Cuddlyable3 (talk) 00:52, 4 October 2009 (UTC)[reply]
This is the second response you've given that has nothing to do with my question. The first was when you outright dismissed and ignored my simple and direct question about ethical guidelines in the field today in comparison with those in the past by distracting away from it and bizarrely claiming that I was "upset" by biology and I was trying to start a debate. Looking at my question above, there's no sign of that at all. Meanwhile, you said that the only way we can understand the natural world is to kill and dissect, when clearly there is a lot to understand in the context of the Tanager Expedition, that involves merely observing wildlife. You obviously did not take the time to even read the primary notes by the wildlife biologist that I was referring to in the first place, (Olson 1996) so you were offering an uninformed opinion. We already have non-invasive ways of understanding the natural world, but they are not yet as portable as they could be, but are well on their way. (See also Goldstein & Pinshow, "Taking Physiology to the Field, 2002) In other cases, wildlife biologists are carrying portable X-ray machines (MinXray) into the field. So we already see alternatives in use. (Henen 2002) In any case, Mr. 98 tried to answer my question without your distraction, so it is strange to see you appeal to a discussion of real ethical issues without ever addressing them. Viriditas (talk) 08:15, 5 October 2009 (UTC)[reply]
It's of note, of course, that Gray didn't kill anyone to make his drawings. It makes him somewhat different than the frog bits. One can, of course, object to frog dissection on the grounds that when done by elementary school students, its educational import is rather low and could be easily simulated. It seems to be more a rite of passage than anything else. Which, depending on how you feel about killing frogs, may or may not be justified. (And just speaking historically, the Conservation movement was actually quite prominent in America in 1923 and had existed for quite some time by that point.) --Mr.98 (talk) 14:59, 2 October 2009 (UTC)[reply]
This is a wee bit tangential, but you might take a look at Gregg Mitman's Reel Nature: America's Romance with Wildlife on Film. One of the main narrative arcs, if I remember correctly, is how wildlife photography was initially developed as a way of documenting the hunting of big game (for science!) but over time, the actual killing of the game became frowned upon, and less invasive forms of observation were adapted and valued (hence the difference between Jane Goodall and, say, a primatologist of the 1920s). --Mr.98 (talk) 14:59, 2 October 2009 (UTC)[reply]
Particles from space time traveling thanks to special relativity...
I can not think of this to save my life, hopefully someone could help me out...
What is the name of the particles that enter the earth's atmosphere due to solar flares and make it to the earths surface alive, even though they do not live long enough to make the journy from space to the surface of the earth. But what keeps them alive is that their timeline has slowed down due to special relativity.
great scientists at the other end ...
please help me on a small ... question..
..can we have a plot of change of acidic character of an acid when we keep on adding water to it.
if not .. then i would love to know the trend of acidity in above case.
thankyou
SCI-hunter (talk) —Preceding undated comment added 13:17, 2 October 2009 (UTC).[reply]
You yourself can intuitively guess that "acidity" will decrease when you keep adding water to an "acid",ie, the substance will become less acidic. The rest depends on how you scale acidity. Rkr1991(Wanna chat?)13:36, 2 October 2009 (UTC)[reply]
thanks a lot for your quick attention ...
what i actually want to know is :((pH)) of solution
as we go on adding water h3o+concentration in acid goes on increasing.so pH of acid must decrease.but it is diluted.kindly elaborate with plot.
thanks
SCI-hunter (talk) —Preceding undated comment added 13:43, 2 October 2009 (UTC).[reply]
Adding more H2O decreases the concentration of H3O+ in it. As you said, it's diluting the same amount of acid in a larger amount of solution. DMacks (talk) 16:26, 2 October 2009 (UTC)[reply]
A favorite exam question deals with the pH of a water solution with an H+ concentration of 10–7 M. What happens if it's diluted? DMacks (talk) 19:28, 2 October 2009 (UTC)[reply]
i think .. i got that ..
water is diluting the same amount of acid as that required by the acid in a balanced chemical equation.and the rest of water remains idle.so amount(concentration) of h30+ in water goes on lessening with increasing amount of water.right??
one more misconception::can i say pH decreases for a while and then increases.because h3o+concentration must first increase untill equlibrium is reached and then its concentration decreases to increase pH.(le chatiliers principle of equlibrium.)
SCI-hunter (talk)
A chemistry professor said it was common to try and trip up a grad student at oral exams by asking how much the pH of a very strong acid changed if it was diluted with an equal volume of water. (There was a standard rule of thumb change I do not recall). That was fine. Then they would ask how about if the result were diluted again with an equal volume of water? the nervous student might be so unfortunate as to repeat the previous answer in terms of numerical change in pH. In the extreme case, he would keep giving the same answer for each dilution, until he had agreed that a strong acid could change into a strong base by repeated dilution with water. Oops. Edison (talk) 01:36, 4 October 2009 (UTC)[reply]
Human intelligence
Will prolonged ice age cause human intelligence to devolve? If a community could not operate any 'schools' say starting today for say the next 10 / 100 / 1000 / 10,000 .... years and most energies of adults are focused on survival, can we truly expect the children to be more intelligent or less? To me it seems like we will become less intelligent until the environmental conditions change and allow us back to have educational focus and less survival focus.
Intelligence increased in the past when there was no schools and there were ice ages. What makes you think that another ice age and no schools would reverse that? Dmcq (talk) 16:20, 2 October 2009 (UTC)[reply]
"Intelligence" is such a poorly define concept that it is impossible to say. Knowledge would certainly decrease without education. --Tango (talk) 16:59, 2 October 2009 (UTC)[reply]
It wouldn't negatively affect human intelligence in an evolutionary sense, or at least there's no obvious mechanism to me that would cause that. Even if their knowledge would suffer, I don't think "devolve" is the right word. Rckrone (talk) 18:16, 2 October 2009 (UTC)[reply]
Indeed, I can't see the genetics of intelligence changing. If civilisation completely fell and we lost (almost) all technology, then evolution might move us towards a genetic makeup better suited to the environment, but even that seems unlikely (there wasn't a great deal of evolutionary change during the last glacial period). If we maintain a decent level of technology, then I don't expect much evolution at all - we'll handle everything technologically, like we do now, so almost all selection pressure will be removed. So, nothing interesting for genetics - memetics on the other hand... --Tango (talk) 21:29, 2 October 2009 (UTC)[reply]
"Intelligence" is best defined as "the traits and abilities measured by intelligence tests." 10,000 years into an ice age, with no schools, people might not be able to read or write or solve math problems beyond basic arithmetic. How would intelligence be measured? Book learning might matter less than survival skills and interpersonal skills. (Bargh say "We killed 16 seals. There were 4 of us. Each should get 4 seals." "Huh, huh. Bargh stupid. He give Clargh 4 seals when Clargh didn't kill any in the hunt, just came along. Look Bargh step in hole in ice. Now I take his 4 seals. I smart." "Dargh figure out how to make good seal skin kayak. He smart, too." "Fnargh get everyone to follow him as leader and lead neighboring tribe into ambush. kill'em and take all their bear fat. He very smart." "Gargh can take malted barley and hops and make yummy fizz drink. He smartest." Edison (talk) 00:09, 4 October 2009 (UTC)[reply]
I would suggest that there is a difference between "intellegence" and "education." Historically, Aboriginal peoples (and other visible minorities) were considered by Europeans as "unintellegent savages." However, in retrospect, Aboriginals and visible minorities are every bit as intellegent as Europeans although their intellegence may be a different kind (eg. "city smart" vs "wilderness smart;" there are different kinds of intellegence). The difference is education. I am probably "better educated" than many historical figures (as measured by access to formal academic education) but certainaly would not consider myself as smart as many of them (e.g. Plato, Michaelangelo, etc.). Another example of this would be my father who dropped out of school in grade 10. I am better educated than he is (having 3 university degrees including post-graduate) but when it comes to solving mechanical problems, I am as dumb as a stump in comparrison. So, in response to the original query, I would argue that we (as a species) would continue to be every bit as intellegent as we now are but would end up being less educated (due to lack of opportunity). —Preceding unsigned comment added by 69.77.185.91 (talk) 16:05, 5 October 2009 (UTC)[reply]
Architecture, High Ceilings (United States)
I wasn't sure if Science was the correct category or not...my question is: Why do rooms in buildings built prior to the 1920s/30s in the United States, particularly residential apartments, have higher ceilings than many modern ones? It would've made sense to prefer lower ceilings in that era, when heating technology was less effective. . .Thanks --68.175.44.30 (talk) 16:46, 2 October 2009 (UTC)[reply]
I don't know why but a few ideas occur to me:
Your observation may be selection in that only the better apartments have survived.
High ceilings allowed better circulation of air and that's a good idea with a bad fire particularly if one can afford the fuel.
I don't have a definite answear, but I do know that the advent of electric lighting changed architecture. When the sun is high in the sky, you need a tall window for the light to reach deep into the room. Dmcq may also have a point in that upper-class buildings may have had higher ceilings and these are the ones that survive. I think I see this in the old upper and lower-class buildings in my city. The reason may be architectural style, light or simply more room for the chandeliers. EverGreg (talk) 18:11, 2 October 2009 (UTC)[reply]
This reference[5] bemoans the "disregard for...aesthetic dimension of a ceiling" compared with 100 years ago and concludes "Ceiling[s] in houses and apartments usually range from about eight feet to an expansive twelve feet high. Many New York apartments are prewar buildings with ceiling averaging nine feet six inches high; most newer buildings have lower ceilings. Ideally, the ceiling height should be in proportion to the size of the room. The larger the room, the higher the ceiling can be and still please the eye." Cuddlyable3 (talk) 19:03, 2 October 2009 (UTC)[reply]
Fire places are esthetically pleasing, but wretched ways of heating a room. Freeze on one side while toasting on the other, with masses of hot air going up the chimney. Stoves beat them hands down. Edison (talk) 00:23, 4 October 2009 (UTC)[reply]
Perhaps it had more to do with staying cool in summer? Hot air rises - so a nice high ceiling allows heat to stay out of your way? That's a guess though. SteveBaker (talk) 20:28, 2 October 2009 (UTC)[reply]
Not USA, but just to comment that the older part of the house where I was born and brought up had ceilings just over six feet high, and one doorway was only five foot seven high. It dated back to the early 1600s, but many houses in the area where I live have similar low ceilings. When an extension was built in 1835, they made the new doorways and ceilings much higher. Dbfirs01:53, 3 October 2009 (UTC)[reply]
I think that the rise both in standards of living (thus the price of labor) and in safety standards since the 1920s/30s has meant that building construction is more expensive in real terms than it was then. So builders may be looking more for ways to cut costs now, particularly in apartment buildings where any extra cost is replicated for each unit. But this is just my theory, I don't have a cite and could be quite wrong. --Anonymous, 04:12 UTC, October 3, 2009.
A reason given for high ceilings in 19th century homes in the U.S. was to let the hot air rise in the summer so it was cooler nearer the floor. I agree that homes in the 1600's, especially in the Northern U.S. had low ceilings, which kept heat down where it was useful. They generally seem mean and peasant-hovel-like. The larger a room was, the higher a ceiling needed to be so it seemed in proportion and gracious. A high ceiling also enclosed a larger air volume, so that if the room were packed with people at a party, it did not seem "close" or lacking in breathable air. Grand furniture, such as 4 poster beds with canopies or tall posts, or wardrobes 8 feet tall looked in proportion, and tall paintings could be hung on walls. Elaborate molding could grace the top of the walls. In a 19th century southern ancestral home the 10 foot ceiling in the most-used bedroom was lowered to 8 feet in the 20th century for easier heating, and an oil heater replaced the blocked off fireplace in that room. I see more than one reason for 10 foot or higher ceilings. Edison (talk) 00:23, 4 October 2009 (UTC)[reply]
To me a better solution than one floor with extremely high ceilings is two floors with low ceilings. The hot air will still rise up the stairs to the 2nd floor. When it gets too hot upstairs during summer days, stay downstairs. When it gets too cold downstairs during winter nights, stay upstairs. StuRat (talk) 00:53, 5 October 2009 (UTC)[reply]
No it's not true. The misconception is derived from the fact that a small number of infants seem to be slow to take their first air breaths, and it is difficult for the attendant adults to restrain themselves from administering some form of tactile stimulation. However, actual spanking is not what is done. alteripse (talk) 17:52, 2 October 2009 (UTC)[reply]
I'm pretty sure this is an old fashioned thing that has survived in movies and television because it makes for good story-telling short-hand. (The entire birth can be presented to the audience as a light slap and a loud cry.) APL (talk) 18:25, 2 October 2009 (UTC)[reply]
The article Midwife notes that in ancient Rome a midwife "inspected the newborn for congenital deformities and testing[sic] its cry to hear whether or not it was robust and hearty." Her method of testing its cry is not specified. Those who promote Unassisted childbirth have produced videos[6][7] of their deliveries where spanking the infant is usually unnecessary. (The boy in the second video typically takes his time to arrive but is vociferous enough once he does.) Cuddlyable3 (talk) 18:30, 2 October 2009 (UTC)[reply]
What they actually do in most parts of the world is some variation on the Apgar test. There is a whole procedure associated with this which assesses how the baby is coping and whether medical intervention is needed. They do test for reflexes (typically by stimulating the babies feet) - and they expect to hear some sort of a cry when they do that - or the baby doesn't score a point for that. Read the article - it explains it pretty well. SteveBaker (talk) 20:16, 2 October 2009 (UTC)[reply]
Quick silly question hopfully some one can answer for me...
If your in a car, traveling at the speed of light, then you turn on the headlights...what happens? Does the light still project due to the velocity already being at the speed of light, then the light particles just travel at double the speed?
74.218.50.226 (talk) 19:44, 2 October 2009 (UTC)[reply]
So someone will point out that it is not possible for objects with nonzero rest mass (presumably this applies to your car) to travel at the speed of light, because it would take an infinite amount of energy to accelerate them to that speed.
That's true, but shouldn't stop us from examining your question; Einstein thought about just such questions, and they helped lead him to the special theory.
One answer is that, for an object (however that word be interpreted) traveling at the speed of light, the Lorentz contraction of time goes all the way to zero. So you don't have time to turn on the switch. This is how it was originally determined that neutrinos have nonzero rest mass. If they had zero rest mass, they would have to travel at the speed of light. But then there would be no way for them to switch between electron, muon, and tau neutrinos, because there would be no time for that to happen in. But we know this switching happens (because there are too few observed electron neutrinos from the Sun). --Trovatore (talk) 20:05, 2 October 2009 (UTC)[reply]
I don't think that's true about neutrino oscillation; I think it's only necessary that the masses not be all the same. The observed oscillation implies that there are three different masses, but one of the three could still be zero, in principle. -- BenRG (talk) 22:53, 2 October 2009 (UTC)[reply]
No, it's correct. If they were massless they would get from the sun to our detectors in zero proper time (to the extent proper time is even defined in that case), so there would be no time for them to oscillate. I guess it's possible that one or both or the flavours the sun doesn't emit could be massless, but then I would expect we would detect almost all them since once a neutrino became that flavour it would be stuck there, and that isn't what we see. --Tango (talk) 23:48, 2 October 2009 (UTC)[reply]
The e, mu and tau neutrinos don't have well-defined masses, zero or nonzero. There are three definite neutrino masses, but they are masses of mixtures of the three neutrino flavors. Conversely, each flavor of neutrino is a mixture of all three mass eigenstates. As far as I know it doesn't matter if one of the eigenvalues is zero. A zero-mass neutrino won't oscillate, but neither will a neutrino of any other definite mass. -- BenRG (talk) 17:45, 3 October 2009 (UTC)[reply]
Well, you can't travel AT the speed of light - only a little below it. But let's suppose you're travelling at 99.9999999999999999999999999999999999999999999999999999999999999999999999999% of the speed of light - is that OK? Well, the weird thing about light - and the entire reason we have their weird relativity stuff - is because when you're travelling at that ungodly speed and you turn on your headlights - the light shoots off away from you at EXACTLY the speed of light. In fact - you can't really tell that you're moving at all. Now - there is a problem with that - which is if someone is watching you (flagrently going WAY over the speed limit!) you seem to be moving at ALMOST the speed of light - and the beams from your headlights are travelling only just a fraction faster. That's odd because you seem to think the light is racing away very quickly. What's really going on is that time has compressed for you. SteveBaker (talk) 20:09, 2 October 2009 (UTC)[reply]
(EC) Your car can't go that fast, so let's say instead that it's going 1% of the speed of light. You shoot a bullet out the front window at 2000 feet per second. From your perspective, the bullet is moving away from you at 2000fps. From someone "standing still" as you whiz by at 1%c, the bullet is going 1%c + 2000fps -- faster! Now, you turn on the headlights. From your perspective, the light shoots away from you ... at the speed of light, of course. From the stationary observer's perspective, the light from your headlights is going ... the speed of light! Sorry, I know it doesn't make sense, and it seems like it should be going 1.01c, but the speed of light is constant in all reference frames. --Sean20:15, 2 October 2009 (UTC)[reply]
These all do make sense to me and thanks for the answers...one last quick question for all of you physicsheads 8')
Sorry again if these questions seem silly, but I love physics...but I think I'm missing half of my brain...
Why is it that light can only travel at the speed of light? —Preceding unsigned comment added by 74.218.50.226 (talk) 20:55, 2 October 2009 (UTC)[reply]
I don't think there really is a "why", it's just the way it is. It falls out of the mathematics without too much effort. --Tango (talk) 21:32, 2 October 2009 (UTC)[reply]
An "enlightening" discussion that answers a question taken straight from a Steven Wright joke, and which I also wondered about when I first heard it years ago.[11] The second question, about why the speed of light is what it is, I once heard posed differently: "When God created the universe, and when it came to setting the speed of light, did He have any choice?" To put it another way, what is it about the universe that compels light to travel at the speed it does? Maybe if that could be determined, it could answer a lot of other questions about the nature of the universe. →Baseball BugsWhat's up, Doc?carrots05:33, 3 October 2009 (UTC)[reply]
Many different factors can occur depending on the situation. For example, when I am at my Mother-in-Law's house, the major effect on me is stress. 68.245.48.172 (talk) 21:26, 2 October 2009 (UTC)[reply]
No, not tropism. I mean the general effects that an environment would have on an organism. i.e. why twins can grow up with different personalities. --Glaesisvellir (talk) 03:11, 3 October 2009 (UTC)[reply]
Actually the standard term for the effects that environment has on an organism is "environment". Sort of a trick question, not that I think it was intended that way. Regards, Looie496 (talk) 23:23, 3 October 2009 (UTC)[reply]
October 3
Is there really meaning to the speed of light?
This shoots off from the most recent comment from the last speed of light question a couple posts above: When God chose the speed of light, did he have a choice? Anyway, given that essentially every measurable quantity is dependent to an extent on the speed of light, is there meaning to its value? For example, all elementary forces are transmitted at the speed of light; the speed of light is the maximum speed of transfer of information; a change in the speed of light would influence the energies of atomic orbitals. So my question really is, if the speed of light were to be changed, universe-wide, how would all of these things it influences scale to eachother? Basically, would everything get smaller at the same proportion, or would the size of things like different atoms, or different bond lengths, scale disproportionately, as I suspect? Someguy1221 (talk) 05:56, 3 October 2009 (UTC)[reply]
Unfortunately there's no well defined answer to this question; it depends on arbitrary notational choices. For example, the energy of a photon is E = hν = hc/λ where ν is the frequency and λ is the wavelength, so changing c won't change the energy-frequency relationship but it will change the energy-wavelength relationship. But Planck could just as well have chosen a constant k with the property that E = k/λ = kν/c, and in that case it would be the other way around, even though the physics is the same. As another example, particle physicists quote masses in kg but also in MeV/c². If you change c then those no longer agree, so the theory isn't even internally consistent without further changes. It is possible to choose new values of the constants and get a new theory that works, but it's not well defined whether you've changed "just one" or "several". It's also possible to change the numeric value of c without changing anything of physical importance. For example, you can just redefine the meter such that c = 123456789 m/s and express all other constants in terms of the new meter. -- BenRG (talk) 09:34, 3 October 2009 (UTC)[reply]
You are right in some cases. All the physical constants are related to each other. You can't change one without changing some of the others, but you can choose which ones to change and how. For example, . So if you double the speed of light, you need to change one or both of and , but you can choose the details. There are ways to do it where everything would change in such a way that the physics is identical and you wouldn't be able to tell. --Tango (talk) 18:09, 3 October 2009 (UTC)[reply]
If you use Planck units, all the universal constants are one, but all the masses, charges, etc. are arbitrary numbers. In a sense, this means that all the universal constants are meaningless, and it's just the attributes of elementary particles that matter. On the other hand, if you make all your definitions based on elementary particles, it's the universal constants that matter. I think the best you could say is that any given measurement is meaningless, and what matters is when you compare different measurements to each other. — DanielLC18:31, 3 October 2009 (UTC)[reply]
I like to concentrate on dimensionless constants. It doesn't matter what the masses of electrons and protons are in kilograms, what matters is the ratio between the two, which is dimensionless and thus not arbitrary. --Tango (talk) 19:45, 3 October 2009 (UTC)[reply]
I would like to know what will happen if we try to magnetise a magnetic material beyond its saturation,and what behavioral change we get if the materaial back to below saturation.Roon93 (talk) 06:45, 3 October 2009 (UTC)
Thanks[reply]
In the simplest, where a body is stretched (in tension) or squeezed (in compression) by a single force and an equal and opposite reaction, the (average) stress is simply the force divided by the cross-sectional area perpendicular to the direction of the force. In more complex cases, where shear stresses are involved, stress has to be treated as a tensor. See stress (mechanics) for more derails. Gandalf61 (talk) 10:23, 3 October 2009 (UTC)[reply]
That article is rather math-heavy, unfortunately. The physical reaction is called the strain or deformation (mechanics). In addition to tension and compression, there's also bending and twisting (torsion). Heat and noise can be generated, too, and eventually a phase change or fracture may occur. StuRat (talk) 00:32, 5 October 2009 (UTC)[reply]
The easiest physical interpretation of a stress that I can think of is that if I push with a force F directly on a sheet of metal with area A, then I am exerting a normal stress of magnitude F/A. If I push parallel to the surface of the metal plate instead of directly into it (e.g., placing your hands on a wall and pushing upwards), then I am exerting a shear stress (or shear traction) of the same magnitude. You'll have to clarify before I can give a meaningful answer to, "How can we realize stress in a body?", Awickert (talk) 16:59, 5 October 2009 (UTC)[reply]
Speed of light between Casimir plates?
When light moves through a dense medium such as glass or water it slows down. By the same logic, since the gap between Casimir plates is less "dense" in terms of energy than the normal space outside, would for example, a pulse of laser light shone between such plates arrive slightly faster than it would in "normal" space, and (assuming this was the case) could such a phenomenon be exploited for intelligible superluminal signaling?Trevor Loughlin (talk) 10:47, 3 October 2009 (UTC)[reply]
If you read further down in the casimir effect article, under "wormholes", you'll find reference to a paper in the journal Physical Review, which discuss what you propose! :-D The discussion is rather complicated I'm sure, as the speed of light enters as a constant in Casimir's own derivation of the effect. Not to mention that c is related to other constants like the Fine-structure constant and the planck constant. How these would change with a changing c is not obvious. EverGreg (talk) 11:04, 3 October 2009 (UTC)[reply]
There's an article on the Scharnhorst effect. I find the whole thing implausible because light speed is built into relativistic QFT at such a fundamental level. If you have Lorentz invariance and you can solve the initial value problem then you don't have faster-than-light signaling. So I suspect this is just a miscalculation, such as the first term of a series approximation that ends up getting canceled out by later terms. Or the photon might really acquire an effective tachyonic mass, but that doesn't imply superluminal propagation (as explained here). Most speculation about the Casimir effect is fueled by the idea that it has some special relationship with vacuum energy, which is not true, as explained in this paper. There's nothing wrong with invoking vacuum fluctuations in deriving the Casimir effect, but you can derive anything in QFT with or without reference to vacuum fluctuations. -- BenRG (talk) 12:36, 3 October 2009 (UTC)[reply]
I have heard through talk, and read in Wiseman's SAS Survival Handbook, that the only remote possibility of a poor hapless soul surviving an oncoming wall of superhot gas and rock is to get into water and dive down for as deep and as long as possible. Considering that one may hold one's breath for about two minutes, and also taking into account that the water probably will get very very hot, would this method actually see you live? Lady BlahDeBlah (talk) 13:39, 3 October 2009 (UTC)[reply]
See: Time (min), Distance (Max), and Shielding (Max) for the protection from threats, real or percieved. These proven parameter adjustments can be applied to Items as varied as Mothers-in-law to hand grenades, and yes even to 'pyroclastic' flows. 68.245.155.69 (talk) 16:09, 3 October 2009 (UTC)[reply]
If you could dive down just a few feet in a body of water, the top foot might boil off, and thus not contribute to heating the rest, while the next foot might get up to near boiling, but then that heat would be mixed with the coolth (my word) of the rest of the water to make a warm, but not deadly hot, body of water. I would think all the ash would be the real threat, as a several foot thick layer of ash could bury you in the bottom of the water. StuRat (talk) 00:25, 5 October 2009 (UTC)[reply]
The water is probably the best option. The speed at which the clasts in the flow are moving is enough to cause significant injury, the temperature will certainly kill you, and the velocity of the flow makes running away rather implausible. But I don't think that the flow would pass before one would need a breath of air, so the true survivalist needs a mini airtank :). Another option I just thought of would be to try to run to the top of a nearby high point. The density-driven pyroclastic flow will converge towards topographic lows. Awickert (talk) 17:03, 5 October 2009 (UTC)[reply]
Visualizing magnetic fields in 3d
It's easy to visualize magnetic field lines in two dimensions by putting iron filings on a flat surface, as at right. Is there any such way to visualize the third dimension? I'm thinking of something like a flow of ferrous mist or an electron gun and special glasses or something. Thanks. --Sean14:47, 3 October 2009 (UTC)[reply]
a Fluidized bed would allow you to suspend fine magnetic particles in a 3 dimensional arrangement. An electromagnet held in place and energized, would, conditions taylored, allow you to have a 3d view similar to the 2d example shown. 72.58.186.50 (talk) 16:14, 3 October 2009 (UTC)[reply]
Well, you could mount your bar magnet on some kind of non-magnetic axle that would allow it to rotate about its long axis. Then you could use the paper-and-iron-filings trick to visualise the field - take a photo - rotate the magnet (say) 10 degrees - tap the paper to let the iron filings rearrange themselves - take another photo - repeat 18 times so you have a photograph of the field lines every ten degrees over the entire volume around the magnet. Then you could load the pictures into a computer - use GIMP or Photoshop to turn the white parts of the paper transparent then apply the photos as texture maps on a 3D model using Blender, 3DStudio or Maya or something. My slight concern is that you can't use the paper-and-iron-filings approach to get a proper cross-section of the field because it's off to the side of the centerline of the magnet - but so long as the 3D model you build in the computer correctly reflects the position of the paper when each 'slice' of the field was imaged - that shouldn't matter...but for complicated shapes of magnet - you might not be able to get the paper close enough to image the field in detail close-up to the magnet. SteveBaker (talk) 16:00, 3 October 2009 (UTC)[reply]
Sometimes iron filings are suspended in some sort of clear viscous liquid or gel. Here's an example that showed up on google [12]. Electron guns are used to visualize how magnetic fields affect moving charged particles, but they don't show the shape of the magnetic field lines the way iron filings do. The electrons' path is made visible by firing them through a gas that's ionized by the electrons and glows. Rckrone (talk) 16:03, 3 October 2009 (UTC)[reply]
Iron filings arrange themselves into only an approximation to the magnetic Field lines. The density of true field lines at any point is proportional to the field strength at that point while the lines of iron filings are equispaced. They also do not show the direction of the field vector. Cuddlyable3 (talk) 00:03, 4 October 2009 (UTC)[reply]
Mobile phone charging disabling vibration function
Is there a reason why when I charge my mobile phone (and all others I've had in the past) that the phone won't vibrate as it normally does when someone rings? Smartse (talk) 15:33, 3 October 2009 (UTC)[reply]
See this. as far as a workaround to the issue (I am assuming that you would not need to feel the vibration, but only need to hear the phone 'vibrate'). As to why the feature is disabled, I would need to know the model to research it. Speculation on my part includes:
Charging/power consumption circuit limitations
Developer seeing no need to vibrate when unit is (presumably off the body) and charging.
Developer's lack of consideration for the mildly obsessed.
It's not really a problem but the three phones I've had: Philips Savvy, Sharp gx10i and Nokia 6300 have all done it. Considering this it seems reasonable to guess that there is a specific reason why beyond all three manufacturers wanting to make me confused! Smartse (talk) 16:25, 3 October 2009 (UTC)[reply]
I've only had two mobile phones. I always kept them on vibrate and rarely removed them from the charger. Both vibrated while charging. Currently, I have an LG Voyager that vibrates while charging without a problem. Therefore, this is not an industry standard. It just happens that the phones you've selected contain this feature. -- kainaw™21:04, 3 October 2009 (UTC)[reply]
Go with #2 above, Developer may have intended for a charging phone on the bathroom sink not to vibrate into the toilet, but plain omission of feature is most likely. Oh and if your phone will load the Vibrating wave file linked above...Go for it, and set the phone for Vibe & Ring. It will stay out of the crapper while charging and still make that buzz sound you NEED to here. 68.244.20.23 (talk) 21:39, 3 October 2009 (UTC)[reply]
I've always assumed it was so you could charge it in a docking station kind of charger (for example, a hands-free set in a car) without it breaking. --Tango (talk) 21:42, 3 October 2009 (UTC)[reply]
Perhaps they seek to avoid damaging the charging connector - so they avoid imposing vibration stress on it? Seems the most likely explanation to me. SteveBaker (talk) 23:14, 3 October 2009 (UTC)[reply]
I think Tango and SB have points here. Also remember depending on the type of connection it may actually come off enough to stop charging which many would likely find annoying Nil Einne (talk) 23:34, 3 October 2009 (UTC)[reply]
With all the phone's I've owned (all Motorola) the phone stores separate settings for both on and off the charger, and you cannot edit the other settings while not in that mode. Thus, to have it on vibrate at all times, you need to unplug it, set it to vibrate, then plug it in (says switching to loud mode or something of the sort) and you need to set it to vibrate yet again. The difficulty is that you can't feel it vibrate when it's three rooms away, but your call. 76.228.199.201 (talk) 08:32, 6 October 2009 (UTC)Ehryk[reply]
That "deer caught in headlights" reaction - what is it?
I was reading the fight-or-flight response article to see if there was some way of explaining the reactin when a deer that freee when it's caught in headlights - or any creature (including humans) freezes. I'm thinking it's too much informtion causing an overload in the neural network. but, I'm not really sure. I suppose a severe enough situation might even be considered a transient ischemic attack, but I doubt it; just because blood vessels constrict and muscles tense doens't mean a clot is actually forming, which is what happens ina TIA, right? (Of course, I'm sure not many vets have tried to diagnose TIAs in deer :-) )So, is it just too much information having to be processed at once, as I suspect? What articl4e would point me in the right direction? Thanks in advance.4.68.248.130 (talk) 16:16, 3 October 2009 (UTC)[reply]
"Too much information causing an overload" is a reasonable first pass at this. What you're seeing is probably a combination of three things: (1) a very strong form of the orienting response that causes an animal to direct full attention to anything that is intense and surprising, (2) a freezing response that many camouflaged prey species are "programmed" to emit in response to threat, (3) conflict between multiple possible motor responses (go forward, go back, etc) that results in none of them being executed. The intense activation can sometimes cause a TIA (i.e., fainting) in humans, but the basic freezing-in-the-headlights reaction doesn't depend on that. Looie496 (talk) 17:05, 3 October 2009 (UTC)[reply]
See Collision. From a different frame of reference: It is the last thing that a person may see before the impact. It will be the rut season here soon, and many a driver at the body repair shop may reflect on that image as they are having the front end damage appraised. 68.244.20.23 (talk) 21:30, 3 October 2009 (UTC)[reply]
As far as I was aware, animals freeze in headlights purely because in the wild they'd do the same when they saw a predator, in order to stay camouflaged and be less noticeable. I'm not a biologist, but I didn't think "overload" came into it. Chase me ladies, I'm the Cavalry (talk) 23:27, 3 October 2009 (UTC)[reply]
To clarify something in the original question, a TIA is generally caused by a constriction of blood vessels or by a cardiovascular problem that causes blood pressure to drop, not by a clot. In most cases a clot that lodges in the brain will cause a stroke, which is like a TIA except that it isn't transient -- clots don't go away quickly. Looie496 (talk) 00:44, 4 October 2009 (UTC)[reply]
I agree that it's the "freeze so they don't see me" reaction (they don't know that our headlights make them glow brightly against the black night). They have judged that the car is faster than them, so running from it will do no good. Thus, they think their only hope is if they aren't spotted. In their mind anything as large as a car always wants to eat them, they just can't imagine that we are driving down a road and want to avoid them as much as they want to avoid us. I wonder if African deer (or whatever they call them there) react more like we'd expect, by running out of the road, due to being familiar with elephants and such. Millions of years of evolution there may have taught them that large moving objects aren't hunting them, but still need to be avoided. StuRat (talk) 00:15, 5 October 2009 (UTC)[reply]
Although actually the deer has no idea how large the car/predator is. For that matter, neither do you late at night, you only make assumptions based on your prior experience with vehicle lighting, which the deer has none of. It could be a child on a bicycle with balsawood light mountings for all you know - but your human instinct/knowledge will cause you to flee anyway (cue Nelson saying "ha ha"). I'd also agree that it's a freeze-response - which is why hunting deer by night using intense lights is forbidden in many areas. (I'd link nightlighting, but it doesn't dab to a relevant page). Franamax (talk) 00:36, 5 October 2009 (UTC)[reply]
You're just talking about the headlights, which is all you detect from a distance when in another car. But when outside and standing on the ground, the sound and especially bass sound and vibrations give a clue that the approaching object is large. Deer, being particularly keen on observing for large moving objects they would think of as predators, would be well able to read all these signs to determine that the object approaching is much larger than them. StuRat (talk) 01:32, 5 October 2009 (UTC)[reply]
As a former resident of semi-rural Scotland and more recently as a regular driver through the New Forest at night, I can say from first hand experience that deer don't just freeze in one's headlights, they are totally unable to judge a car's speed and predict its (straight line) direction, and are liable to dash from out of the darkness off the road into one's path at the last second, particularly if one or a few are separated by said road from the bulk of their herd on the other side. This is unnerving, because apart from one's not wanting to hurt any animal unnecessarily, a collision with an animal in the New Forest is (1) an offense and (2) automatically deemed the driver's fault. The numerous free-roaming cattle and ponies are actually less of a hazard because they rarely startle or move quickly, having learned that traffic gives them the right of way. 87.81.230.195 (talk) 02:55, 5 October 2009 (UTC)[reply]
As a regular (ex-)driver of rural and wilderness Ontario, I can tell you for sure that the minute you see deer anywhere even close to the road, you slow down as quickly and smoothly as reasonably possible. Deer might decide the best way out is right through your windscreen. As I alluded in my reference to nightlighting (which is hunters driving slowly and quietly along back roads using a spotlight to illuminate the forest edge where deer come out to graze, so no Stu, sound has nothing to do with it, it's the dazzling light), deer freeze then suddenly make an entirely unpredictable escape response. The second rule of seeing deer near the road you're driving on (after slowing down right away) is to keep driving slow, 'cause where there's one deer, there are more somewhere in the darkness. They will be making sudden bolting movements too, and they can leap onto the roadway in a matter of seconds. Sorry, no links for this, just original and quite thorough research! Franamax (talk) 03:16, 5 October 2009 (UTC)[reply]
When your car gets within a certain distance, the deer may well figure it has been spotted by a predator, and dart off in a random direction. The goal here is to confuse and outmaneuver the predator. It has no concept that the car is headed in a fixed direction, as this is alien to it's experience (or the experience of it's ancestors when they evolved their reactions). StuRat (talk) 02:02, 6 October 2009 (UTC)[reply]
The traditional explanation I grew up with was that the shadows adjacent to the road moved in frightening ways as the car approached and deterred the animals from leaving the road. Polypipe Wrangler (talk) 11:53, 6 October 2009 (UTC)[reply]
Atomic cannon
I'm not talking about this, M65 Atomic Cannon, but actually using a nuclear device as the propellant to drive a solid slug (tungsten probably) at high velocity. Lets say we use a nuclear device of about 50 kilotons or so, and a tungsten slug of 50 kg. We put the cannon in orbit, aim it at the ground, and fire it. Assuming the cannon can actually survive this, would the slug impact the ground with energy greater than the 50 kt nuclear charge? What about if the cannon was aimed at a target in space, and didn't have gravity to assist the slug? ScienceApe (talk) 17:17, 3 October 2009 (UTC)[reply]
I'd have to do some calculations to determine if the slug would vaporize, melt, or merely splinter into fragments. But even in the ideal case of merely splintering, the fragments heading toward the earth would be limited by terminal velocity in the atmosphere. So probably more energy would actually be imparted to a target in space (assuming the target is at rest relative to the cannon) than to a target on the ground. Ironically, the slug would impact the ground with greater energy if it were just pushed out of orbit at relatively low speed rather than being fired by a nuclear cannon, since the air resistance is less if the slug remains in one piece. Red Act (talk) 18:11, 3 October 2009 (UTC)[reply]
Well terminal velocity only applies if the object is in freefall. Furthermore, it actually doesn't matter if the remains in one piece or not since the conservation of momentum states that the fragments will still impact at the same velocity. But Tungsten is very hard so I think it would survive atmospheric entry. There are real-life examples of kinetic bombardment using just tungsten rods. See, Kinetic bombardment. ScienceApe (talk) 18:37, 3 October 2009 (UTC)[reply]
An object 66 tons (Hoba meteorite) survived speeds of 11-72 km/s, far above current kinetic energy rounds, with only a thin melted crust, even though iron-nickel is less strong and tungsten is 2.5 times denser. It is supposed to be almost the largest meteorite that the atmosphere can slow down from space to terminal velocity, so even though tungsten's extra heft over iron will help 50 kg won't do it. All guns deflagrate inside instead of explode like a bomb, would it even be possible to make something like that that could survive being fired? Sagittarian Milky Way (talk) 19:07, 3 October 2009 (UTC)[reply]
I'm not really sure what your point is. Our article on kinetic bombardment indicates that tungsten projectiles would be able to survive atmospheric entry. The US Government has had projects (Project Thor) which involved firing tungsten projectiles from orbit to targets on the surface of our planet. In any case, I'm mostly just interested in the kinetic energy of the projectile, and how it compares to the energy of the nuclear charge. ScienceApe (talk) 19:25, 3 October 2009 (UTC)[reply]
I didn't reread the article but didn't remember the tungsten telephone pole/crowbar.
Fine. Proof of concept. Just use more usual means for acceleration. I doubt you'd would want to use a nuclear weapon, and just for fun, an equal mass of powder shoots a bullet about bullet speeds, so a 50 kiloton nuclear weapon should shoot a 50kg slug.. insanely fast. Large fractions of c fast.. Gravity would barely touch it. A projectile more on the order of the kilotonnage of the bomb in mass would be needed. Sagittarian Milky Way (talk) 20:12, 3 October 2009 (UTC)[reply]
The object is in freefall after it leaves the cannon. After it leave the cannon, there is no further accelleration except for gravity and air resistance.
Staying in one piece does matter, because air resistance is higher on a lot of little pieces spread about than on one big piece. When thinking about conservation of momentum in this problem, it's important to not neglect momentum that gets transferred to air molecules due to air resistance.
But the terminal velocity is different for an object fired from a cannon. Considering the transit time from orbit to the surface of the planet could be as low as just a few seconds, the projectile would not encounter air resistance long enough for it to be a significant factor.
Not at the extreme velocities that would be involved in this example though. The amount of energy absorbed by the atmosphere might just be a tiny fraction of the total kinetic energy of the fragments. But, I don't think tungsten would break up.
No, it's a real concept. Do a search on Project Thor, or Rods from God. In any case, I feel like this discussion is more about whether orbital bombardment is possible, rather than if the projectile's kinetic energy would surpass the energy released by the nuclear fuse used to propel the projectile. ScienceApe (talk) 19:43, 3 October 2009 (UTC)[reply]
I wonder how much of the energy of the nuke would go into the kinetic energy of the cannon and escape as heat/whatever? My elementary physics tell that should be less than what gravity adds, ie 50 kg * 10 m / s^2 * (some thousands of kilometers? idk) = much less than a gigajoule (this is an absolute limit as gravity is already weaker than 10 m / s^2 at sea level no matter where you are and air resistance is ignored). A kiloton is ~ 4 terajoules. The answer to the original question is no, but it would be nice to have some numbers to support. --194.197.235.240 (talk) 20:54, 3 October 2009 (UTC)[reply]
Kinetic energy is given by . Thus, velocity .
According to our article about TNT equivalent, 1 kiloton of TNT is equal to 4.184 x 1012 joules.
So, for 50 kilotons of TNT with a 50 kg slug, and assuming that 10% of the energy from the explosion is converted to kinetic energy in the projectile, numbers become
.
This works out to a muzzle velocity of approximately 914000 meters per second.
The troposphere contains approximately 75% of the Earth's atmospheric mass, and is 15km deep. To determine how long the projectile will be in contact with the troposphere, the relevant equation is
where is the height of the target above the ground, is the initial height of the projectile (in this case, zero), is the initial velocity of the projectile, is the acceleration due to gravity, and is the time required for the projectile to reach the target.
I do not remember what equations are used to determine the amount of energy frictional heating can impart into an object, but considering that the projectile will be in contact with troposphere (and hence 75% of the mass available to heat it up) for just 1.5 hundredths of a second, I cannot imagine that any reasonably aerodynamically-shaped projectile would break up in flight. It simply would have no time to heat up and start to ablate. J.delanoygabsadds22:04, 3 October 2009 (UTC)[reply]
Just a note on the math here: it looks like you've got the sign wrong for acceleration due to gravity, it will be opposite to that of the initial velocity. By my Casio fx-350D with the keys that don't always work anymore, that would actually be 0.224 sec in the troposphere. None of which really matters, 'cause at that velocity the result would likely be (as we all watched the object leave the solar system) "oops, let's try aiming at the boss's car in the next test". :) Franamax (talk) 01:00, 5 October 2009 (UTC)[reply]
Air resistance is proportional to velocity squared, so would be enormous at those velocities. The time being so short means there wouldn't be time to dissipate any heat, so I can imagine the object would get very hot. I can't see why the time being short would stop it having time to heat up - that's like the old joke "I have to write this letter quickly before my pen runs out." For future reference, units in TeX need to be written inside a \mathrm{...} command, otherwise you've actually written the product of m, e, t, e, r and s. --Tango (talk) 23:59, 3 October 2009 (UTC)[reply]
Energy lost due to air resistance is of order:
Taking A as 200 cm^2 (for a 50 kg slug of tungsten), Cd as 0.3 (somewhat rounded), and density of air as 0.5 kg / m^3 and your numbers for velocity and time gives: 1.9×1013 J or roughly 90% of the energy you presumed it started with. It wouldn't so much get hot as it would explode on impact with the atmosphere. And that's neglecting the hypersonic aspects of the problem. Air resistance gets worse, not better, when one tries to travel at very high velocities. Dragons flight (talk) 01:35, 4 October 2009 (UTC)[reply]
If it's losing that much energy then you can't assume constant velocity, as you seem to be doing. That means the energy loss will actually be less than you say. It will still be a very significant amount, though - I would expect enough to vaporise it before it hit the ground (I can't be bothered to calculate it...). --Tango (talk) 02:16, 4 October 2009 (UTC)[reply]
The maximum amount of speed that can be imparted by Earth's gravity = 11 km/s, Earth's escape velocity. Gravity trails off quickly with distance so you could impart almost as much without having to go too far from Earth. You could also use much of your already existing orbital speed (8 km/s) by bleeding off a little velocity, so coming in at an angle and do this from low earth orbit. The effects of nuclear explosions article says 40-50% of the energy is commonly in the blast wave. If you want Earth to be hit by more projectile energy than the bomb put in simply make it heavy enough that it can't go too fast. This would go hand in hand with the whole not destroy cannon thing but remember the gravitational part of the energy was put in my you so isn't free. Sagittarian Milky Way (talk) 22:16, 3 October 2009 (UTC)[reply]
What do you consider to be "too far"? Gravity in Low Earth Orbit is only a few percent lower than on the Earth's surface. While gravity does drop off quite quickly (with the square of distance), the altitude of LEO is very small compared to the radius of the Earth. If you go to Geostationary Orbit, then gravity is quite a lot lower (about 36 times lower, I think). If I've done the calculation correctly (somebody please check me!) if you fall from 200km straight down (so that's LEO but without the horizontal velocity) you will hit the ground at only 2km/s, quite a lot less than escape velocity. --Tango (talk) 23:59, 3 October 2009 (UTC)[reply]
Okay, subjective. But compared to the infinity actual distance.. And if measured in time and delta-v it isn't much more than what we do to reach objects very near us now (ISS, 2 days, which is routine). If dropping straight down is discarded you could just redirect your LEO which is already there so it intersects with your target, getting much speed without having to give much speed or start up high. (of course you gave that speed when you launched it) It'd come in at an angle, which has air resistance implications. Sagittarian Milky Way (talk) 03:04, 4 October 2009 (UTC)[reply]
The specific heat capacity of tungsten actually varies considerably with phase and temperature, but for just a rough order-of-magnitude calculation, I'll use a constant value of the 24 J mol-1 K-1 that's listed in the tungsten article. Given also the atomic weight, boiling point, heat of fusion and heat of vaporization listed in that article, I calculate that 50Kg of tungsten at 0 K can absorb on the order of 272 MJ of heat before it vaporizes. In comparison, a 50 kiloton bomb releases close to a million times as much energy. So it's very hard to see how you could reasonably propel a 50 Kg tungsten slug with a canon powered by a 50 kiloton atomic blast without the slug vaporizing. Red Act (talk) 23:11, 3 October 2009 (UTC)[reply]
Your calculations aren't necessarily correct depending on the way the device is designed. Certain designs of Project Orion (nuclear propulsion) allow for a pusher plate that doesn't ablate much if at all, which shows that specific heat of a substance doesn't dictate whether it will "vaporize" or not if propelled by a nuclear explosion. And also, just because it vaporizes doesn't mean the matter that makes it up disappears. Conservation of mass applies here, so the particles that make it up will still be given tremendous kinetic energy. In any case, you aren't really understanding the point of the question. I'm simply asking if the kinetic energy of the mass that the cannon propels will be equal or greater than the energy released by the nuclear charge used to propel the mass. ScienceApe (talk) 15:41, 4 October 2009 (UTC)[reply]
In both cases, the energy the slug has would be far less than the energy of the nuclear blast. This is because most of the energy released by the blast would be in the form of heat and radiation, with only a small portion being kinetic energy, and only a small portion of that would then go into the slug. As for the case of the slug fired toward Earth, the acceleration due to gravity would be totally insignificant, while the air resistance would be overwhelming. Much like the Tunguska Event, the object would vaporize in the high atmosphere, creating heat, light, sound, possibly an EMP, and an atmospheric shock wave aimed at the ground. Since it would be spread out over a very large area, though, the shock wave might not even be noticeable, from such a small slug, when it hit the ground.
You don't seem to be getting the way air resistance works, it's the total air you are displacing and the speed with which it is moved that determines how much energy is robbed from the projectile. A faster object moves just as much air, and accelerates it to an even faster speed, so loses even more energy. As a thought experiment, imagine making the same ground trip in your car at 60 MPH or at 120 MPH. Which would use more energy (gasoline) ? The 120 MPH trip, of course, even though the time the trip takes is half as much. The reason, just as in your Q, is all the air the car moves along with it, at either 60 or 120 MPH. StuRat (talk) 23:57, 4 October 2009 (UTC)[reply]
No, if the blast is in space, most of the energy would be in the form of X-rays. If it were a pure fusion bomb catalyzed by perhaps antimatter as is discussed in this report, http://arxiv.org/PS_cache/physics/pdf/0510/0510071v5.pdf most of the energy would be in highly energetic neutrons.
I'm not really interested in debating whether or not the slug would vaporize in an atmosphere, I simply used the example of the weapon being fired from space because I thought it might be slightly more realistic to use such a weapon in orbit, rather than in a terrestrial setting. But if you are still curious about the feasibility of kinetic bombardment, I suggest you do some searches on "Project Thor" and "Rods from God", both of which involve using tungsten slugs fired from orbit to targets on the ground. ScienceApe (talk) 01:52, 5 October 2009 (UTC)[reply]
The idea behind those projects is to use gravity to accelerate the slug. You would fire some kind of rocket to knock it out of orbit and it would then hit the ground at 11km/s (minus a bit for air resistance, which wouldn't be anywhere near as significant as for your design since the speed is about 100 times less and velocity is squared and cubed in the relevant formulae, and minus a bit since you aren't dropping from infinity, but you would probably drop from pretty high since the orbital velocity you have to shed would be less). A 50kg tungsten slug hitting the ground at 5km/s (a number I've made up, but it should be in the right order of magnitude) has an energy of 625 MJ or about 150kg of TNT. That's not on the scale of a nuclear explosion, but it's a big bang. --Tango (talk) 02:11, 5 October 2009 (UTC)[reply]
"No, if the blast is in space, most of the energy would be in the form of X-rays." I said radiation, which includes X-rays. And you keep assuming that faster travel through the atmosphere means less heating, when everyone here has told you it's just the opposite. At the speeds you're talking about, the slug would be vaporized in the high atmosphere. Yes, if just dropped, it would survive, but that's due to the much lower speeds and therefore air resistance. StuRat (talk) 14:24, 5 October 2009 (UTC)[reply]
StuRat, as I already stated, I am not interested in debating whether or not the slug would vaporize in an atmosphere, but in any case I never said that faster travel through an atmosphere means less heating. Tango, what I'm interested in is the kinetic energy of a tungsten slug fired from an atomic cannon that I described in my first comment. Whether for the slug or the sum of its fragments if it doesn't survive the explosion. ScienceApe (talk) 20:09, 5 October 2009 (UTC)[reply]
I'm not sure exactly what question about kinetic energy hasn't already been answered. The original question was "would the slug impact the ground with energy greater than the 50 kt nuclear charge?" The answer to that question is definitely not, regardless of whether or not the slug vaporizes. If the slug vaporizes, the vast majority of whatever kinetic energy the nuclear explosion imparted to the slug will ultimately go into heating the air, over a large area around where the slug exploded. Very little of the kinetic energy the slug had will make it all the way to the ground. And the effect of gravity on the tungsten vapor will be negligible. If the slug doesn't vaporize in the atmosphere, that implies that as of when the slug left the cannon, it was so slow and so cool that only a tiny fraction of the energy of the nuclear explosion ever got transferred to the slug. The slug will gain some additional kinetic energy due to gravity, but nowhere near enough to compensate for all the energy of the nuclear explosion that didn't get transferred to the slug.
Later on, your question became "if the kinetic energy of the mass that the cannon propels will be equal or greater than the energy released by the nuclear charge used to propel the mass". Correct me if i'm wrong, but it sounds like at that point you're just asking about events in the near vicinity of the canon, correct? If that's the question, the answer again is definitely not. The nuclear explosion will be considerably less than 100% efficient in transferring its energy to the slug. And if you're only asking about events in the near vicinity of the canon, the effect of gravity is negligible.
If the above doesn't answer your question, could you please be more precise about exactly what it is that you're asking? Like it sounds like maybe you want us to solve the problem while pretending that the atmosphere didn't exist, and pretending that tungsten has an infinite melting point? Pretend problems like that are fine, but if that's the case, what efficiency would you like us to pretend the cannon has? Red Act (talk) 22:00, 5 October 2009 (UTC)[reply]
Actually, the problem can basically be answered "definitely not", even under the assumptions that the earth doesn't have an atmosphere, and the melting point of tungsten is infinite. Even without any atmosphere, the energy gain due to gravity would be less than the energy of a 50kg mass at the escape velocity of 11km/s, which is 3GJ or 723kg of TNT. So the impact energy would have to be less than the 50kt nuclear charge, unless the canon was considerably more than 98.5% efficient, which it realistically can't be. It'd definitely be more efficient to just drop the nuke. Red Act (talk) 23:20, 5 October 2009 (UTC)[reply]
I'm interested in knowing what the kinetic energy of the slug would be (a rough estimate is fine), and/or the muzzle velocity of the slug. For comparison sake, I would also like to know the kinetic energy of a bullet fired from an M-16 compared to the energy released from igniting the powder used to propel the bullet. How efficient is the powder at transferring kinetic energy to the bullet? Assume all takes place in a vacuum. ScienceApe (talk) 19:40, 6 October 2009 (UTC)[reply]
Yes. You don't even need a black hole. Any gravity field will accelerate light. Amazingly light speed won't change when it is accelerated though its velocity, momentum, and energy will change. @tango: remember that speed and velocity are two different things Dauto (talk) 18:16, 3 October 2009 (UTC)[reply]
Good point. I interpreted the question as referring to a change of speed, and I expect that is how it was meant, but it isn't actually what was said. --Tango (talk) 18:25, 3 October 2009 (UTC)[reply]
Actually I'm going to raise some issues with that point. Your interpretation of lensing in a gravitational field is only valid for wave-treatment of light. Since we don't have a theory of quantum gravity we can't comment of the behavior of a photon in a gravitational field, other than to say in the classical limit, the path of a beam of photons will be bent. Elocute (talk) 19:30, 3 October 2009 (UTC)[reply]
We know empirically what light does in a gravitational field. It makes no difference how you interpret light, we know what we see. --Tango (talk) 19:39, 3 October 2009 (UTC)[reply]
Ah it does make a difference, because quantum particles don't accelerate per se, their vectors just in an exchange of what in this case would be the hypothetical graviton. I suppose if you consider an instantaneous change of velocity acceleration then they do accelerate, but they spend zero time actually accelerating. I did chuckle a bit at "We know what we see" as an approach to an inherently quantum problem. Elocute (talk) 19:48, 3 October 2009 (UTC)[reply]
Well I am sorry to have offended but all I was trying to add was that when using the more traditional definition of acceleration as dv/dt you will find that a=0 for all regions where a is well defined. Elocute (talk) 20:15, 3 October 2009 (UTC)[reply]
Obviously you need to average the change in velocity over some reasonable length of time. This isn't an inherently quantum problem. What you say holds just as well for billiard balls. They don't accelerate, the atoms sub-atomic particles (I suppose I should be precise, since you seem to want to be excessively pedantic) in them just exchange virtual photons, but obviously that is not a useful interpretation when playing snooker. It also isn't a useful interpretation in astrophysics. --Tango (talk) 20:17, 3 October 2009 (UTC)[reply]
The difficulty here is not quantum but general-relativistic. Dauto says "I define the bending of the path as an acceleration". Fine, except in what sense does the path "bend"? Locally, it does not bend; the light follows a geodesic in spacetime, which is a straight path, as best a straight path can be defined on a manifold of nonzero curvature.
If you refer everything to asymptotically flat spacetime removed from the gravity well, you can talk about the path "bending" in the sense that if you take a vector representing the photon going in to the well, and subject it to parallel transport through flat spacetime to where the photon comes out, you'll find they're no longer parallel. It may be possible to make a reasonable definition of "acceleration" from that; I'm not sure. But the conceptual and definitional problems are not trivial. --Trovatore (talk) 20:25, 3 October 2009 (UTC)[reply]
You're making things more complicated than they have to be. The geodesic path followed by the light beam will only be perceived as a straight line by an observer in free fall. But why should we use an observer in free fall? The reasonable choice of observer is one that is inside of a rocket which has its engines on and poiting downwards in such a way that it is hovering above the black hole at a constant distance. That observer will see the light beam follow a curved path from which he can inferer that the light is indeed subjected to an acceleration. Dauto (talk) 21:52, 3 October 2009 (UTC)[reply]
The answer is "not really", in that the light doesn't undergo any proper acceleration. It could be said to undergo a "coordinate acceleration", but that really just reflects an arbitrary choice of coordinate system. The light's proper velocity does not change, and the light travels along a perfectly straight line (a geodesic). Whether the light's energy and momentum appear to change depends on what choice of coordinate systems you choose to use at the beginning and at the end, which is arbitrary. Red Act (talk) 20:07, 3 October 2009 (UTC)[reply]
That's a property of gravity, rather than light, though. Nothing in free-fall experiences any proper acceleration. I think it is clear that we are talking about acceleration in some observer's frame, not the light's own frame. --Tango (talk) 20:14, 3 October 2009 (UTC)[reply]
I was about to make that point as well. Thank you Tango. In other words, massive particles also follow geodesics (though a different family of geodesics) and that does not keep us from saying that they do accelerate under gravity. Dauto (talk) 20:17, 3 October 2009 (UTC)[reply]
For simple problems, it can be convenient to think of a massive object as accelerating due to gravity. But that’s really treating the problem loosely. In a more rigorous sense, what’s really happening is that the massive object is moving at a constant velocity. It just appears to be accelerating, when viewed in an accelerating coordinate system, such as an accelerating coordinate system that’s stationary with respect to the surface of the earth. Red Act (talk) 20:47, 3 October 2009 (UTC)[reply]
What do you mean by "some observer's frame"? There is no inertial frame of reference that encompasses the entire spacetime involved. "Some observer's frame" isn't a well-defined concept in a gravitational problem like this. Red Act (talk) 20:32, 3 October 2009 (UTC)[reply]
Of course it isn't well defined, I haven't specified the observer, but that doesn't really matter. "Inertial reference frame" is a concept from Special Relativity, it doesn't really apply to General Relativity, where all frames are equivalent. Obviously we can't encompass the entire spacetime since there is a singularity in it, but we can choose a useful co-ordinate system that works away from the black hole. --Tango (talk) 20:43, 3 October 2009 (UTC)[reply]
It’s not just the singularity of the black hole that causes a problem with “acceleration in some observer's frame, not the light's own frame”. There would still be a problem with that idea even if it was, say, a star that was affecting the light’s path. There is no global coordinate system that behaves “nicely”, i.e., has spatial and temporal intervals that correspond to the coordinates in all directions, except in the region of spacetime that’s far enough from the gravitating object that it’s essentially flat. In other words, the “nice” coordinate chart has to leave out the interesting part of spacetime, in which light is affected by the gravitating object. Sure, if you only look at the light’s path before it enters the interesting part of spacetime, and after it leaves the interesting part of spacetime, the angles and positions are as if the interesting part of spacetime was flat, and the light’s path was bent, i.e., the light was accelerated. But that’s not what really happened. What really happened was that the light travelled in a straight line, on a curved spacetime.
If the coordinate system doesn’t ignore the interesting part of spacetime, then it’s necessary to make rather arbitrary choices as to what coordinate system you use, since due to the curvature of spacetime there is no coordinate system that has all of the features one might hope for. Schwarzschild coordinates are fairly popular, but there are some disadvantages to those coordinates, and various advantages to other coordinate choices, such as isotropic coordinates, Eddington coordinates, Gaussian polar coordinates, Kruskal–Szekeres coordinates, or Eddington–Finkelstein coordinates. Since you’re forced to make an arbitrary choice of coordinates anyway, you might as well choose a coordinate system in which along the light’s path in question, only real (i.e., proper) acceleration appears as an acceleration, instead of dealing with a coordinate system in which there is an additional, unreal “acceleration” that’s really just an artifact of the choice of coordinates. Red Act (talk) 04:05, 4 October 2009 (UTC)[reply]
Red Act, what you are saying is true, but the exactly same points are also true for the trajectory of a massive particle and that does not keep us from describing that as an acceleration. For consistency we should also describe what's happening to the light beam as a acceleration, even though the proper acceleration vanishes. Dauto (talk) 20:54, 3 October 2009 (UTC)[reply]
It’s more important to bring up the distinction between proper acceleration and mere coordinate acceleration in this problem than in a problem in which only massive objects are involved, because the constancy of the speed of light is such an important principle. I suppose if the OP is just wondering whether or not light is affected by gravity at all, then bringing up the distinction is just unnecessary pedantry. But my impression is that the OP might be wondering why acceleration of light near a gravitating object doesn’t wind up violating light always traveling at exactly the speed of light. If that’s what the OP is really wondering, then pointing out that light doesn’t really accelerate at all is the most accurate way of pointing out why the speed of light doesn’t get violated. Red Act (talk) 04:05, 4 October 2009 (UTC)[reply]
about photosynthesis,
I'm a grade 11 student from sri lanka. i faced a question in a paper and i wasnt able to find the answer for it. it asked whats the source for O2 which is produced in the process of photosynthesis. the answers were
H2O
CO2
NO3-
CO3-
please help me to solve this question. —Preceding unsigned comment added by 112.135.199.4 (talk) 18:50, 3 October 2009 (UTC)[reply]
Actually, the help desk is a "school help place". The fact that a question arises from something studied in school in no way invalidates the question. Our "we don't do homework" policy must be confusing someone. Asking for clarification of something you've read or been tested on in school is a good use of the reference desk and should be encouraged, not discouraged. That said, if our original questioner will look at the photosynthesis article, under the section on Light reactions, he (or she) will see that the overall equation for the portion of photosynthesis that produces oxygen is:
Regarding the other choices, the oxygen in CO2 (carbon dioxide) is incorporated into the sugar produced by photosynthesis in the light-independent or dark reaction of photosynthesis, but this reaction does not produce oxygen. While NO3 (nitrate) metabolism is important in plants, it has nothing to do with what the question is asking. And carbon trioxide (CO3) is an unstable molecule; even if the bicarbonate ion was meant, it also has nothing to do with photosynthesis. - Nunh-huh19:44, 3 October 2009 (UTC)[reply]
Quantifying Color Blindness
Has their ever been a method devised for ascertaining the degree of a subject's color blindness?—so, for example, the color blind individual could then correct the saturation of certain hues in an image they've generated, allowing people with full color vision to see it correctly?
There are varying degrees of Color blindness, as noted in the article. There are various tests for color blindness. One I've seen involves a field of dots with some of them colored differently to form the shape of a numeric digit. If you can't see the number, you have that type of color blindness. →Baseball BugsWhat's up, Doc?carrots20:36, 3 October 2009 (UTC)[reply]
The question is about the degree of colorblindness. The Ishihara test doesn't really do that with much precision. I don't know of a test that does this precisely. SteveBaker (talk) 23:11, 3 October 2009 (UTC)[reply]
A color blind artist creates an image that looks ok to them. That means it resembles what they see in the world around them. If part of their system of 3 types of color sensitive cone cells is weak or anomalous they do not see, and can therefore hardly be expected to reproduce, some colored features. However there is no way to process the image to generate missing detail, nor any certain way to make it "more real" than the artist was able to make it. Compare this with the difficulty of reproducing for normal-sighted people what a color blind person actually perceives. Cuddlyable3 (talk) 23:24, 3 October 2009 (UTC)[reply]
er...the 2nd link in the .23 reply describes how the severity can be determined as the test progresses. It should be a good start for the OP with a couple of good internal and external links if they need to delve in further. Its about the journey, not the destination for me. So for the OP, I just point in the direction of TYBR. 173.100.133.44 (talk) 00:25, 4 October 2009 (UTC)[reply]
Why can't you treat carbon monoxide poisoning with a blood transfusion?
As far as I understand part of the problem during CO poisoning is that the CO binds to the hemoglobin preventing oxygen from doing so. I have often wondered whether this could be partially alleviated by replacing a proportion of the patient's blood with tranfused blood, thereby replacing the compromised hemoglobin with fresh hemoglobin. Having read Carbon_monoxide_poisoning#Treatment there is no mention of any attempt to do this. I wonder, is it the case that doing so would not be useful because CO causes other problems (including binding to myoglobin) so that solving the hemoglobin issue alone would not be beneficial. Or is it that it could, in theory, be useful but has practical problems preventing its use. Thanks Illegal Cat (talk) 21:00, 3 October 2009 (UTC)[reply]
See Emergency! as Wikipedia cannot give medical advice. As far as the TV show goes, both Early and Brackett would issue out the perscription for 100% Oxygen (along with the generic: administer D5W (or Ringers'), monitor the patient and transport immediately). 72.58.9.205 (talk) 21:15, 3 October 2009 (UTC)[reply]
The practical problem, to which you alluded, is nothing more than "blood transfusion" is not a particularly speedy process. Factor that into your thinking? --DaHorsesMouth (talk) 22:38, 3 October 2009 (UTC)[reply]
OK, but would it even be effective? It might be slow, but if it was in some way effective, then perhaps it would be used in bad cases, even if it was slow. After all, it seems as though the half-life of the bound CO is 80 minutes on 100% oxygen, so that is hardly speedy. It seems unlikely that speed is the only issue to not doing a transfusion. Perhaps the problem with clotting factor when too much blood is replaced would be a more substantial issue. Illegal Cat (talk) 23:04, 3 October 2009 (UTC)[reply]
Sorry, you've lost me. Are you saying that you feel my question does fall under the "no medical advice" guidelines? As for the article's talk page, I am not suggesting an article improvement. I have, it appears, made up a treatment and I want to understand why it wouldn't work. Is that not a reasonable question to ask here, and one that does not fall it the realm of asking for medical advice? Illegal Cat (talk) 23:04, 3 October 2009 (UTC)[reply]
I just did a bit of research -- apparently the story is that although this idea has been suggested many times dating back to the nineteenth century, it doesn't actually work very well, because transfusing a patient with severe CO intoxication leads to a set of pathological reactions known collectively as "reperfusion syndrome" (see PMID12453699), which are greatly reduced by hyperbaric oxygen. Looie496 (talk) 23:16, 3 October 2009 (UTC)[reply]
Thanks, I will read up on this effect. I suspected that there would be some kind of side effect, well either that or the fact that fixing the hemoglobin alone had little benefit. Illegal Cat (talk) 23:19, 3 October 2009 (UTC)[reply]
C3, the .205 answer was to what happened on the TV show when a CO case came up. I think I made that clear that I was reciting the plot not passing out triage chits. I'm not a Doctor, but I have watched an actor play one on TV. The episode (from Emergency!) I remember most was a knucklehead who wandered into the ER with his son after installing a new furnace. I think it was Early on the horn with Roy(Back at the subjects house, with the rest of the family) on that one. But it is interesting that he was spot on 35 years back w/ the current research. Never did see them bring the poor dog out of the house, Hope the little guy made it. 70.4.152.122 (talk) 00:06, 4 October 2009 (UTC)[reply]
The procedure you are suggesting is called an exchange transfusion. Your question is good, and has been asked before, the latest I found was this letter to the editor of Chest, which advocates for further studies. At the time the letter was written (2002), the author was able to find only one published case report. --NorwegianBluetalk08:01, 4 October 2009 (UTC)[reply]
Because carbon monoxide doesn't just inhibit hemoglobin oxygen binding, but also oxygen binding in cytochrome c oxidase and other enzymes, leading to cellular asphyxiation. It's not just the oxygen transport that gets interrupted. --Pykk (talk) 17:34, 4 October 2009 (UTC)[reply]
I suspect that the timing is the critical factor. That is, if the patient had bad enough CO poisoning to need a transfusion, they would be dead before it could be set up. On the other hand, if the carbon monoxide poisoning is at a low enough level that there's time to set up a transfusion, then giving them pure oxygen is enough to save them, and transfusions, with all their added risks, are not needed. StuRat (talk) 23:33, 4 October 2009 (UTC)[reply]
October 4
Help with clarifying what something I've added an article is....
I've just been doing an un-stubbing of the article for Departure Bay, British Columbia (Canada). I've added this (sourced and all). Below is a cut / paste from my additions to the article. I'm too tired to figure this out myself now; I've searched all over Wikipedia and can't find it (though I don't have or know of any good sources on chemistry / explosives). Can anyone out there help me clarify what "duraline stumping powder" is or could be? The original source I have consulted does not clarify. I'm not sure, but maybe it's designed for blowing up stumps. What gets me, though, is that it's included in a list of very common / general explosives (i.e., black powder and dynamite). Any help would be appreciated....
In 1892—after realizing that irrespective of the new plant in the Northfield area, demand was still increasing—the Hamilton Powder Company built an explosives (mainly black powder, dynamite, and duraline stumping powder) manufacturing plant on the shore of Departure Bay. The production of black powder was relatively dangerous, and the death of employees was frequently the result of accidents.
Duraline is a brand name. For info about stumping powder, see Tree stump#Stump removal. It makes sense that a company that manufactured black powder would also manufacture stumping powder. The two products are similar in that they are both powdered mixtures containing potassium nitrate. Red Act (talk) 06:22, 4 October 2009 (UTC)[reply]
Wikipedia has an article on Tree Stumps? That's just awesome. Anyway, I thought "duraline" sounded like a brand name, but I decided it must not be given the context and the fact that "Duraline" just somehow sounds like a modern brand name, not a 1892 brand name from, relatively speaking, the middle of nowhere! But you must be right; maybe I'll just rephrase the paragraph to say solely "stumping powder" so the context is consistent. Thanks for your quick help,
Hold on, I've just read the article and I'm not sure I completely understand. Do you mean to say that stumping powder is not an explosive, it's rather a rotting agent? So my whole phrasing of the sentence needs to be restructured? I just found a quote later on in the reference I'm using saying it's "less dangerous than either black powder or dynamite and was used mainly by farmers to blow up stumps." Now I'm confused, I have two ambiguous hints, seemingly contradictory! What should I do, lol ;) ...
Whoops, it appears I've lead you astray! "Stumping powder" is for sure the term for an explosive used historically to blast stumps away, as a google search on the term will readily show. The modern chemical stump removal products (commonly potassium nitrate) used to accelerate the decay of the stump, as described in the tree stump article, is different. The modern products to accelerate decay of a stump are not called "stumping powder". Sorry! Red Act (talk) 19:08, 4 October 2009 (UTC)[reply]
Well, now we've figured it out. Thanks for adding it to the tree stump page; I'll just remove the "duraline" on the Departure Bay page and link it to tree stumps (it's a bit of a poor WP:EASTEREGG link, but it'll have to do).
Read pH. As you will see, it is the negative logarithm of the hydrogen ion concentration. This means that a smaller concentration of hydrogen (hydronium) ions results in a large value in pH. Keeping this in mind, let us analyse this problem. HCl is a strong electrolyte, ie it dissociates completely to give Hydrogen cations and Chlorine anions. So, it gives a relatively higher concentration of Hydrogen ions. However, Acetic acid, being a weaker electrolyte, only partially dissociates into its respective ions. In fact, there exists an equilibrium between the ions and the acid itself. Hence, the concentration of hydrogen ions it can supply is considerably lower, and hence it has a higher pH. Rkr1991(Wanna chat?)09:23, 4 October 2009 (UTC)[reply]
I've read about some quite interesting contraptions used in a 16th century battle. In the 1552 siege of Eger, the greatly outnumbered defenders constructed a giant wooden wheel, filled it with explosives, and rolled it down the hill to break the ranks of their enemy. It's a pity no contemporary drawings exist, so we will never know how exactly it could have looked like, but based on what I've read, they had hundreds of "obsolete" guns, and they filled the wheel with them, setting the fuses so that they will be set off sequentially. (as in 1552 in Eastern Europe they still used bows and arrows and the most advanced gun of the time was probably just an arquebus, it think obsolete might mean handegonne). They filled some of them with sulfur, so it had some kind of flame thrower as well.
So, my question would be:
Can this be considered an early form of volley gun?
Why was it never standard issued, if it was so effective, at least against enemy morale? Surely it would be limited to those on higher ground, but I never heard it being more than an improvised device. Or were there other uses of similar devices in other times and places I don't know about?
I found a movie on youtube [13] (at 6:30 to 7:30) containing it, however, it is just an artistic impression, it might or might not have been similar to it. --91.8.225.151 (talk) 11:20, 4 October 2009 (UTC)[reply]
I don't think this could be considered a volley gun: I understand that term to mean a device firing multiple shots in the same direction in a controlled manner.
It seems to me that such a device would only be effective in very limited circumstances, be too uncontrollable and subject to chance variations in terrain, shifting battlefield conditions or other factors, and so be more trouble than it was worth. A similar device called the Great Panjandrum was invented by the engineer (and novelist) Nevil Shute Norway during the Second World War, but in trials proved unsuccessful, as did a fictionalized radio-(un)controlled version of it in the TV show Dad's Army. 87.81.230.195 (talk) 14:22, 4 October 2009 (UTC)[reply]
The description sounds fanciful, like conjecture by someone of a later generation. People under seige usually do not have time to build something novel which uses up time, weapons, and gunpowder. How could they possible have timed the fuses so that the thing fired when it was aimed at the enemy as it rolled down the hill? What does a contemporaneous account say, exactly? Edison (talk) 23:29, 4 October 2009 (UTC)[reply]
It seems really unlikely to me. If these "obsolete" guns still fired - you can bet they'd have had a use for them up on the battlements. Our article: Siege of Eger says "...a water-mill wheel packed with gunpowder which he rolled into the Ottoman ranks. His secret lied in the gunpowder not simply exploding but sparking even more fire." - which sounds much more likely. They wouldn't have had to "construct" anything if the water-mill wheel already existed - and this description suggests that they just stuffed it full of explosives - no fancy triggering mechanism - no obsolete weapons - no "volley gun" - just a big rolling bomb. Some sites say that rusty gun-barrels (not entire guns) were packed with sulpher, flints and other 'junk' to add to the shrapnel. One presumes that the idea was that the force of the main explosion would force this junk outwards and cause more horrific injuries. Still no "volley gun" though. SteveBaker (talk) 02:10, 5 October 2009 (UTC)[reply]
When these tree seed fall in the wild, the way they end up is probably random, so it's unlikely that it matters much. Plants are sensitive to gravity, so that roots and shoots quickly adopt the appropriate directions - see Gravitropism. You could try a small experiment by dropping some of them many times and seeing if they end up a particular way more often: if they do, that's the way to go.
I'm presuming, by the way, that by "horse nuts" you mean the seeds or 'conkers' of the Horse chestnut tree: which hasn't been much modified by human cultivation; seeds of plants that have been more heavily modified by selection, cross-breeding and other techniques may be less robust, so that the sowing orientation matters more.
However, there may be other factors than orientation that affect whether or not a particular tree's seeds start to grow, like whether they're buried or not (and how deep, and in what), or the temperatures they're exposed to - the article on Germination mentions some of these factors. Some seeds will only grow if they've been exposed to frost, or to fire, or even if they've passed through the digestive tract of a particular animal or bird! I don't think any of these apply in this case, but I haven't found any reference that says so definitely. Googling on 'Horse chestnut germination' brings up quite a few relevant references and discussions (do explore them yourself), but overall conkers seem to be unfussy and I don't see any critical factors. Be aware though, that Horse Chestnut trees are quite slow growing (and can live for centuries), so you're not going to have a spectacular, flowering tree anytime soon. 87.81.230.195 (talk) 14:04, 4 October 2009 (UTC)[reply]
Built in GPS Mobile Phones
Lately I was searching for a mobile that serves all important features and couldn't find any matching device.
I was wondering for the reasons, companies don't provide such services together in one device:
Dual mode :GSM/CDMA
GPS asisted and built-in.
OS: Windows Mobile
Camera
SD card
Is it because there is no much request for such gadgets (in my opinion these aren't gadgets), or complications of manufacturing both networks? Perhaps, the question is about technology rather than science. --Email4mobile (talk) 15:08, 4 October 2009 (UTC)[reply]
A website like CNET allows you to add requirements to narrow down a list of phones that match your criteria. I couldn't find any such phones on that site, but other comparison-shopping sites have similar feature filters. Nimur (talk) 16:31, 4 October 2009 (UTC)[reply]
How hard and where did you look? Without intending to disrespect Nimur, I'm somewhat doubtful cnet is particularly complete when it comes to this sort of thing. I doubt it is even mostly complete with the major international brands. And there are a whole bunch of Chinese phones that it definitely doesn't contain. There are quite a few Chinese dual mode phones and also some smart phones usually with Windows Mobile (probably because until the launch of Google Android there was little other ready made option and companies are still getting their heads around Google AndroidO). It wouldn't surprise me if there's one that combines both. I did find [14] although I'm not sure if they're a Chinese company. You're generally unlikely to hear about these phones as they're not sold (or approved by regulators) in most developed countries, in fact China is usually their primary market. In fact it may be difficult to find anything about some of these phones sometimes from my experience only the cheap phones with MTK chipsets and Nucleus Plus OS and a cloned iPhone (particularly) or Android or whatever interface get much attention (and no I don't consider these smartphones because they usually aren't powerful or featured enough to be called smart phones IMHO). Nil Einne (talk) 17:13, 5 October 2009 (UTC)[reply]
There are not many dual mode GSM/CDMA phones at all. (If CDMA was supposed to be CDMA 2000 or similar earlier standart). -Yyy (talk) 07:30, 6 October 2009 (UTC)[reply]
Calculating electrical resistance
Calculating the electrical resistance of a body such as a wire, with a length much greater than its diameter, is straightforward, knowing the resistivity of the material, e.g. 1.59×10−8 ohm-metres for silver. Equally easy is the resistance of several wires in parallel, even if they're uninsulated and touching - they can be regarded as separate, or as a single wire with the combined CSA. But how would the resistance be determined of something more complicated, say a thin square lamina between opposite corners, or a thin circular one between points diametrically opposite? Each shape could be considered as consisting of many wires of different lengths in parallel, but it all looks nastily infinite and bound to require integral calculus.→86.148.186.208 (talk) 15:32, 4 October 2009 (UTC)[reply]
You might find sheet resistance helpful. The correct unit to use here is "ohms per square" (Ω/☐) (That is "ohms per square", not "ohms per square inch", not "ohms per square meter" - it has units of ohms). This has huge application in ASIC design - in semiconductor processing, there is always a need to calculate the effective areal resistance between two ports, and even with CAD it is not usually suitable to perform a full nasty integral every time. For a non-rectangular area (which is very uncommon in semiconductor processing), you can approximate as a sum of rectangular regions. Two or three regions can form a pretty reasonable polygon for estimation purposes. Keep in mind that geometric tolerances in silicon manufacture are often on the order of 10-20% - so very accurate estimation of the resistance is less important than designing for a wide range of acceptable values! Nimur (talk) 16:20, 4 October 2009 (UTC)[reply]
The integrals don't seem that difficult, and I can't imagine how to get the resistance without doing them; for the circular one the resistance would be muʃt r2 sin2(a) da (from a = 0 to pi) (where mu is the resistance in omhs per unit length, t is the thickness, r is ...) - just by summing the cross sectional areas from one side to another. The resistance accross a diagonal of rectangle is also doable.83.100.251.196 (talk) 17:48, 4 October 2009 (UTC)[reply]
Correction for the disc - using R=muL/A R=muʃr sin(a) / t r sin(a) dx = muʃdx/t (between 0 and pi) = pi mu/t - interestingly the resistance accross the disc is unrelated to its radius! that is for edge to edge resistance between opposite points, not inner points...in a linear field..83.100.251.196 (talk) 18:48, 4 October 2009 (UTC)[reply]
I see what you're trying to do with that integral, but I'm fairly certain it's incorrect. You're basically making the assumption that the voltage is constant along a line perpendicular to the line going through the two end points. To see that it's incorrect, think about what equipotential lines should really look like in an infinitesimal region near one of the end points. Red Act (talk) 22:20, 4 October 2009 (UTC)[reply]
Yes I aggree (quote me "for a linear field") to get this I'd need the disc sandwiched between two capacitive plates.. Not really what was originally asked.83.100.251.196 (talk) 23:54, 4 October 2009 (UTC)[reply]
Another thing to bear in mind is that the resistance between two points is theoretically infinite because the contact areas are theoretically zero. In practice, the answer depends on the size of the contact areas. To get a meaningful answer you need to measure the resistance between two well-defined contact areas, such as two parallel wires bonded to opposite sides of a square sheet, or two metal discs bonded to the ends of a cylinder. I don't see how you're going to get a meaningful answer to the case involving the opposite corners of a rectangle. --Heron (talk) 18:46, 4 October 2009 (UTC)[reply]
Interesting links, Nimur, but I'm not convinced. I think you're all answering the wrong question. The 4-point probe that you refer to is just a Kelvin probe. It solves the practical problem of contact resistance, but no amount of experimental ingeniousness will get around the underlying geometrical fact that you can't measure the resistance of an object that terminates in a point, which is what the OP wanted to do. If a material has finite conductivity then a perfectly sharp corner of the stuff must have infinite resistance. And if the corner is not a mathematical point but a real-life slightly blunt one, then its resistance depends entirely on its imperfect geometry. You may be able to measure resistivity using your devilish 4-point method, but you can't measure resistance, because the latter is determined by the geometry of the object. All your calculations are doomed to fail if you ignore this distinction. --Heron (talk) 20:04, 5 October 2009 (UTC)[reply]
Okay, but there is no better real-world approximation to an infinitesimal-point-of-contact than a needle-point. That is why it is used in industrial processes - to measure resistance and resistivity. Nimur (talk) 00:09, 6 October 2009 (UTC)[reply]
To accurately calculate the resistance of a complicated object (as opposed to measuring that resistance), you'd generally need to use the finite element method, except in cases where symmetry enables you to simplify the problem enough to solve it, or at least closely approximate it, analytically. Red Act (talk) 22:40, 4 October 2009 (UTC)[reply]
I have a wire Frame cube. The resistance of each edge of the cube is 1 ohm. What is the resistance from diagonally opposite corners of the cube? This is not home work but something been puzzling me for years. How does one work it out using normal circuit theory?--SpiceJar (talk) 22:33, 4 October 2009 (UTC)[reply]
As far as I know temperature is the numerical expression of average kinetic energy of motion of molecules.So what will the temperature of space where there are no molecules? —Preceding unsigned comment added by 113.199.141.152 (talk) 16:28, 4 October 2009 (UTC)[reply]
Temperature is not simply the average kinetic energy of motion of molecules -- for a precise definition you would have to read the Entropy article, but it isn't easy going. Basically temperature is a number that determines which way the net flow of energy will go when two systems are brought into contact with each other -- it always goes from higher temperature to lower temperature. In outer space with no massive particles, the temperature is set by the cosmic microwave background radiation. If you shield a portion of space so that no particles whatsoever can penetrate, the temperature is undefined. Looie496 (talk) 16:44, 4 October 2009 (UTC)[reply]
Also, there is a small amount of stuff out there. The average is about one atom per cubic centimeter, I believe. --Sean16:57, 4 October 2009 (UTC)[reply]
The mass-density and energy-density depend where you are, of course. A temperature for a perfect vacuum can be defined, but it is quite different from the conventional usage of "temperature" (which is exactly as the OP stated - the average kinetic energy of the molecules). Nimur (talk) 18:37, 4 October 2009 (UTC)[reply]
I am a body of mass 1kg floating in deep space. My surface area is 1 m^2. My surface emissivity is 0.95. I have been here many millions of years. What is my temperature? (Not homework just curiosity) —Preceding unsigned comment added by SpiceJar (talk • contribs) 22:49, 4 October 2009 (UTC)[reply]
Assuming that by deep space you mean away from any potential source of radiation such as stars, your temperature is the same as the cosmic microwave background which is about 2.7 K. all the info you gave about size mass, and composition is unecessary. Dauto (talk) 00:35, 5 October 2009 (UTC)[reply]
It is necessary to work out how long it would take to reach thermal equilibrium, but it is not sufficient - the initial temperature is obviously needed. I would expect something that small to have reached equilibrium after millions of years even if it started of extremely hot, though - the square-cube law is important here, small objects exchange heat very quickly. --Tango (talk) 02:22, 5 October 2009 (UTC)[reply]
Indeed. A white dwarf is just such an object, only a bit heavier than 1kg. However, none has reached thermal equilibrium yet - the universe is too young (and "too young" at 14 billion years, not 6000 ;-). --Stephan Schulz (talk) 14:39, 5 October 2009 (UTC)[reply]
Um, our white dwarf article gives a range of 0.17-1.33 solar masses, average of 0.6, this corresponds to an estimate of 1kg being off by about 30 orders of magnitude. Googlemeister (talk) 18:18, 5 October 2009 (UTC)[reply]
Yes, a white dwarf is rather big so takes a long time to cool down. A 1kg mass with a surface area of 1m2 would cool down much faster. --Tango (talk) 22:24, 5 October 2009 (UTC)[reply]
Energy transformation and space-time
I went through some interesting subjects today as in "speed of photons/light" and the "current source voltage source" discussions. I was wondering of the answers; though, accidentally I had a relevant question to ask. Does energy require time and/or space to be transformed from one form to another or can this transformation reduce to power terms at all cases?--Email4mobile (talk) 17:13, 4 October 2009 (UTC)[reply]
I think you are asking whether any such form of energy transfer can occur instantaneously. The most complete answer is that all physical interactions appear to be quantized on some level. This means that the smallest/fastest/most-infinitesimal version of any particular interaction must occur on a discrete time scale. Notably, though, we have not fully explained the quantum nature of gravity, for example; there are other complex interactions which are not necessarily best-modeled with a quantum theory. But, it is probably safe to say that there is no process which can occur faster than one interaction per Planck time. Nimur (talk) 18:48, 4 October 2009 (UTC)[reply]
Indeed. Time periods less than the Planck time are essentially meaningless within the framework of our current theories. It is possible (I'd even say it's likely) that we'll come up with a theory in the future that has a better explanation of what happens on small time-scales, but even then it is likely to still be a quantum theory, so there will be a minimum amount of time that any process must take. --Tango (talk) 19:06, 4 October 2009 (UTC)[reply]
International Space Station Experiments
Is there a comprehensive fully itemised list of all the experiments that have been and are currently being conducted, and how and where can I access it? —Preceding unsigned comment added by Hardy6273 (talk • contribs) 18:56, 4 October 2009 (UTC)[reply]
Not only is there a comprehensive list - the data collected resides in the public domain (with few caveats and exceptions). You can access the data, perform original research, and submit your results for peer review. It's surprising to me that few "crank scientists" take advantage of this - instead, they seem to prefer to speculate about perpetual motion machines instead of freely accessing the best scientific data this side of the Van Allen belts. Nimur (talk) 19:05, 4 October 2009 (UTC)[reply]
I hope not. For those not sufficiently familiar with cranks to understand, allow me to explain: The data is only useful to people with significant scientific knowledge and understanding, which is precisely what cranks lack (and often pride themselves in lacking). --Tango (talk) 21:43, 4 October 2009 (UTC)[reply]
Sclerophyllous plant communities
Besides the Mediterranean biome and in Oceania, is there any other region of the world that presents a mostly sclerophyllous vegetation cover, or one that this vegetation can be considered to play a significant part in? —Preceding unsigned comment added by 201.21.180.57 (talk) 23:58, 4 October 2009 (UTC)[reply]
Hi. I'm wondering if it is possible for the H1N1 virus to produce brain-related symptoms, for example: insomnia, mild depression, psychomotor retardation, prolonged shallow or manual breathing, fast and variable heartrate (OK, that's not entirely brain-related, but still), headache, irritability, numbness, and lack of desire for certain things? I also wonder if it could produce other symptoms such as irregular temperature in the extremeties (ears, hands, knees, feet; some hot, others cold), itching in the face and hands/feet, nosebleeds, dehydrated/cracked lips, low levels of urine, asthma-like symptoms, arrythmia, pain in the abdominal organs (kidneys, spleen, pancreas, gallbladder, etc.), drowsiness, and partial paralysis while in a half-sleeping state. I do know however that it can cause nausea, vomitting, diarrhea, fatigue, dizziness, fainting spells, and even seizures. Also, would having both the seasonal flu and swine flu make the symptoms worse? Would an altered sleep schedule? I am in no way asking for medical advice or a diagnosis. I am not trying to diagnose any of my own symptoms, nor those of someone else I know, nor do I have an illness that would make me lie about this. Please do not provide any information that pertains or alludes to a diagnosis or medical advice. Please also do not interperet this as a medical advice question. Thanks. ~AH1(TCU)00:10, 5 October 2009 (UTC)[reply]
Diseases and people are sufficiently individual and variable that almost any symptoms or signs might, in some circumstances, be caused in someone by almost any disease; when considered in permutations the possibilities would be endless. Most of the things you list are, in my entirely untrained non-medical lay opinion based on very limited knowledge, not particularly characteristic of Swine flu, but both individually and even more so in some permutations might be possible signs/symptoms of many conditions as serious or more serious, ranging up to imminently lethal.
Were I myself to want an answer to the question with any urgency, I would be inclined to ask someone definitely knowlegeable about such matters face-to-face - a GP springs immediately to mind (I happen to know one socially). If I wanted to use such information to, say, work up ideas for a novel, I might hang around common rooms in a medical teaching establishment or otherwise get to know advanced medical students socially, and invite their brainstorming over drinks. A large Science Fiction Convention would be another good venue to pursue such ideas, as there are likely to be student and qualified medical and or biology-discipline personnel as well as laypersons with relevant knowledge among the attendees (again, I speak from personal knowledge).
Our article on influenza notes the common symptoms, and our article on the 2009 flu pandemic notes that symptoms are generally in keeping with standard seasonal flus. We note that neurological symptoms have been noted in some cases; you'll want to check those references for further details. — Lomn02:24, 5 October 2009 (UTC)[reply]
Check also the article Fever since it is a common symptom of influenza and it is generally a sign of the body correctly coping with the infection. A question like "Would an altered sleep schedule...make the symptoms worse?" is asking for medical advice. Cuddlyable3 (talk) 13:07, 5 October 2009 (UTC)[reply]
Headache is common. Fatigue is common, but not drowsiness per se. Dehydration and subsequent low levels of urine could result from the nausea, or from simple irritability with poor feeding in children. The other symptoms you mention would be uncommon, but theoretically possible as the IP above describes. Sleep deprivation depresses the immune system, but merely an altered sleep schedule (such as working the graveyard shift) would not have any effect to my knowledge. - Draeco (talk) 18:47, 5 October 2009 (UTC)[reply]
I assume you mean hail. Hail forms in high-altitude clouds, which can be cold regardless of surface temperature beneath them. Sufficiently large bits are able to reach the ground still frozen, even on hot days, because air is a poor thermal conductor. — Lomn12:47, 5 October 2009 (UTC)[reply]
Carbon Monoxide Levels Given Off By Older Vehicles
Let me begin by saying that this question is in no way a roundabout way of obtaining medical or legal advice. That said,
the other day, I noticed that I was able to smell the exhaust of a car idling outside of myself, which made me wonder exactly how much co is in exhaust fumes, and how many ppm would be in the air as a function of time (odd thing to wonder about, I know?) Thus, lacking the means to figure this out myself, I wanted to see if anyone here had even a rough idea of how many ppm of co would be in the air after a vehicle was idling outside for t minutes(namely, t = 15 minutes, 30 minutes, 1 hour) For simplicity, assume we are talking a room of average size, that the vehicle is older and ill configured, and that the only major source of ventillation in the room is the window outside of which the vehicle is idling. [Again I want to stress that this not a roundabout way to obtain legal/medical advice, obviously the above data would in no way imply anything relating to either types] —Preceding unsigned comment added by 71.61.48.205 (talk) 12:52, 5 October 2009 (UTC)[reply]
Presumably the OP did not pass out from inhaling those fumes. The article does not quite come out and say, so I have to ask: Once CO is exposed to the open air, does it quickly join with O2 molecules and become pairs of CO2's? Or does it simply dissipate? Not that you want to be surrounded by CO2, either. →Baseball BugsWhat's up, Doc?carrots13:03, 5 October 2009 (UTC)[reply]
I'm aware that you can't smell co, nonetheless, it is present in exhaust fumes. Or rather, any car giving off extremely odorus exhaust fumes is likely improperly calibrated, and thus, also giving off carbon monoxide. Thank you though:) Given that most naturally occuring oxygen is O2, it is unlikely that it bonds with CO; though other reactions might take place [though they couldn't be overly significant, else co poisoning would be much more rare.] [this is op by the way] —Preceding unsigned comment added by 71.61.48.205 (talk) 13:07, 5 October 2009 (UTC)[reply]
There is far too little information to make a meaningful scientific estimate of "ppm emissions per unit of time". If the vehicle is outdoors, wind will be a major factor in controlling the local particle levels. A simpler starter-question would be - "Can you find or measure the net CO emissions from a vehicle in number-of-particles-per-minute?" The answer to this is of course "yes" - in fact, it is required in many states - California's mandatory measurements include Carbon Monoxide (and Carbon Dioxide, nitrous oxide, unburned hydrocarbon, and other contaminants). How this net flux of contaminants disperses is a very hard problem to define. If you were in a tiny enclosed room running the engine and assuming uniform dispersal, you could count the net particles released and divide by the volume of the room. (Note: it is very dangerous to run a car in a closed environment - you can suffocate and die before you ever smell the contaminants). But outdoors, all bets are off. Local and prevailing winds will disperse the particles pretty well - so even if you correctly know the net contaminant released (by mass), it's hard to know what your vehicle's contribution is to contaminant density. Nimur (talk) 15:26, 5 October 2009 (UTC)[reply]
The issue is not so much about how well set-up the car is (although that does have an impact) - but whether it has modern catalytic converters. These are required by law on vehicles less than 25 years old - but those older than that are exempt in most parts of the world - so the biggest risks are with cars older than 25 years. There are far too many unknowns to be able to calculate a direct answer to the OP's question though: How far was the car from the window? What was the wind speed and direction? What engine capacity does the car have? ...I'm sure there are others. But since any one of those things would affect the answer (possibly by a couple of orders of magnitude each) - our error bars are far too wide. SteveBaker (talk) 15:21, 5 October 2009 (UTC)[reply]
I figured that this was too broad of a question to realy get a specific answer. I guess my major curiosity is how large an impact sources of co outside a building can have on the co levels inside a building; I guess asking for an approx ppm value was a little absurd, I was still a little sleepy. In other words, can you ever get an appreciable co level from running a vehicle outside of a window, how long would it take, what kind of conditions would there need to be, etc? This would probably be more in line with what I was wondering; sorry for the confusion...71.61.48.205 (talk) 16:28, 5 October 2009 (UTC)[reply]
CO does not react with normal oxygen at room temperature. Oxygen is a diradical -- and this makes it only reactive at higher temperatures where other radicals are produced -- hence why combustion takes place at high temperatures. Diradicals tend to react only with other radicals since if a diradical reacted with most "normal" molecules (that have filled orbitals), a monoradical would be produced (which is higher in energy). You could probably quench it with a flood of reactive singlet oxygen, but you'll also destroy whatever organic material is with the CO (read: the tissues of most living things) in the process. John Riemann Soong (talk) 22:14, 5 October 2009 (UTC)[reply]
OK, so the answer to my question would essentially be that it dissipates, i.e. it spreads out and the potential harm is significantly reduced - unlike as with a closed environment, such as a sealed house with a malfunctioning stove or furnace, or a sealed garage where someone is trying to snuff it by running their car. →Baseball BugsWhat's up, Doc?carrots23:55, 5 October 2009 (UTC)[reply]
According to the weather channel, there is a 30% chance of isolated thunderstorms. No severe weather warnings at this time. I would thus say there is a probability of less then 30% of severe weather in Fort Worth Texas today. Googlemeister (talk) 13:48, 5 October 2009 (UTC)[reply]
The National Weather Service is the authoritative government bureau for these kinds of questions. They are the original source for nearly all redistributed commercial forecasts (as seen on the Weather Channel or local news); and they collaborate very closely with military and civil aviation weather bureaus to establish uniform estimates of weather conditions. You can see their forecasts directly, as well as most of the raw data and RADAR imagery that is used to make these forecasts, at nws.noaa.gov. Fort Worth Weather Forecast Office is a regional center, so there is no shortage of accurate information for you. Nimur (talk) 14:36, 5 October 2009 (UTC)[reply]
TLC: detecting 0.5% yield?
Suppose by nature of your laboratory (a kitchen) and the nature of your reagents/catalysts (phosphoric and acetic acid), your desired experiment gives you like a yield of 0.5%, but you want to prove the yield exists. What's a practical way of showing a new ester has formed, besides the subjective means of smell and taste? Would 0.5% product show up against the starting material on TLC? John Riemann Soong (talk) 15:22, 5 October 2009 (UTC)[reply]
Depends on what you are detecting - for visual inspection using fluorescent backed tlc plates aromatics>>non-aromatics. For aromatics I think a 1:200 ratio is just detectable by eye (spot size ~1:30 ?). With a color based detection agent it might be easier.83.100.251.196 (talk) 19:26, 5 October 2009 (UTC)[reply]
Mostly a bunch of aliphatic esters, though I may decide to use various aromatic acids (or possibly their anhydrides) or something like that. Actually because of the nature of some of my reagents (they are food-based) I may get a spread of products (and food-wise this is desirable). John Riemann Soong (talk) 22:05, 5 October 2009 (UTC)[reply]
Theobromine tastes bitter as the free base... It's clear that making the salt of an alkaloid increases it's solibility and hence tastability - see quinine tannate as an example of an insoluble quinine salt with little or no taste.83.100.251.196 (talk) 20:14, 5 October 2009 (UTC)[reply]
Seeking list of mercury content of fish species
I have not succeeded in my lengthy Google search for a list of the mercury content of a variety of fish species, including things like sardines and the different types of mackeral. I'm looking for the numeric quantitative amounts of mercury, preferably in metric, not just something categorised as low medium or high. Maybe someone could try a different approach to me and find the information please? 78.149.191.96 (talk) 17:37, 5 October 2009 (UTC)[reply]
Of course, mercury content will vary by geographic location, especially for freshwater species. Here's a quantitative analysis of a few different fish "types" (well, fish and crustaceans): [15]Franamax (talk) 18:20, 5 October 2009 (UTC)[reply]
Also, I nabbed a copy of "Survey of total mercury in some edible fish and shellfish species collected in Canada in 2002" which is the source Health Canada used. You would need to email me to get a copy for your personal use only. Franamax (talk) 18:38, 5 October 2009 (UTC)[reply]
And some quant stuff from the US FDA: [16] (eww, catfish ;) I'm just using the gsearch term "mercury content in fish". Also, note the distinction between total mercury and methylmercury when you read this stuff. Methylmercury is way the more dangerous form than inorganic variants. Franamax (talk) 18:49, 5 October 2009 (UTC)[reply]
[17] "Many researchers have reported geographic variability in Hg concentrations among commercially important fish and shellfish species. "
That would depend a lot on when you were born, I suppose. Mars can see the same constellations as we can, and Mars' axis is roughly parallel to our own. I imagine the signs would be the same, though the Sun Signs wouldn't line up with the same months as they do on Earth. (In fact, the martian year is a different length, earth months wouldn't line up with the zodiac at all!) APL (talk) 22:12, 5 October 2009 (UTC)[reply]
The axial tilt isn't what's important. You want the orbital inclination (which is only about 2 degrees different from Earth's) - that's what determines the path of the Sun against the background stars. Astrology is in no way scientific, though, so this is the wrong reference desk to discuss it on - try Miscellaneous. --Tango (talk) 22:27, 5 October 2009 (UTC)[reply]
(edit conflict) Actually, the axial tilt plays an important role in determining which constellations the Sun appears in front. If the Earth were 0˚ or 45˚, we'd see a different range of stars at a given latitude and date. The precession of the axis is not, as I said, taken into account - since the Astrological system was first "defined" the axis has rotated roughly one sign. And I think the question was placed appropriately given the answers have to come from a scientific reason, even if the topic isn't. ~ Amory (u • t • c)22:41, 5 October 2009 (UTC)[reply]
Yes, you would see the stars in different places, but you also see the sun in a different place. It is the sun relative to the stars that determines a star sign, and that is dependant on the planet's orbit around the sun only. The ecliptic (the path the sun takes through Earth's sky) is the project of the Earth's orbit onto the celestial sphere. It depends solely on the orbit. --Tango (talk) 22:44, 5 October 2009 (UTC)[reply]
Which constellations the sun crosses your planet's celestial equator in is determined by the heliocentric ecliptic longitude of the point it's axis points to. This determines the origin (the closest point the planet's ecliptic approaches the north celestial pole is 0° mars ecliptic latitude, 90° mars ecliptic longitude) The 0, 0 point is the sun equator crossing that moves north. Whether the pole that's north is the one pointing closest to Earth's north ecliptic pole or the one pointing closest to Earth's North Celestial Pole or the one that rotates counterclockwise when viewed from above.. is ambiguous even in astronomy. So far this is astronomy. I don't know what they'd pick. For Mars this is unambiguous. But not for every planet in the solar system.
They attach great and utter significance to ecliptic longitude twelfths measured prograde from this point, as if these gores eminated mystical powers, oooh. It is one of the three pillars (signs, aspects and houses, which is like the hour of culmination or something) So does this mean rotating an axis will render 2000 years of astrological development obsolete? (haha) They'll have to find out everything all over again: Being born 60 to 90 degrees after new equinox makes one grumpy, clinical research studies have increased predictive skill for trait [male libido] due to new finding of a sine distribution centered on 28°16 (+/-2)″ Aries Makemake sign with a peak of 0.0011813796363706844 correlation with secondary peak 0.0008/12°2(+/-3)″Gem/*all95%cnfdncelvl/blablabla superimposed
The axial tilt is least important to them becase it just determines the angle it crosses at. And they don't care about that (you know, changing it even 10% only what like starts ice ages or something?) Sagittarian Milky Way (talk) 04:49, 6 October 2009 (UTC)[reply]
The origin is just a coordinate thing, it doesn't affect which constellation the Sun is in, which is (nominally) what star signs are all about. --Tango (talk) 14:34, 6 October 2009 (UTC)[reply]
This is really not a science question. Astrologers really don't give a damn about science - and they are just as likely to make up some kind of crazy answer as they do on any other day. The answer is whatever the nut-jobs say the answer is. Go ask an astrologer...no - better still - go ask three different astrologers and be impressed about how none of their answers will agree with any of the others! SteveBaker (talk) 22:37, 5 October 2009 (UTC)[reply]
To amplify Steve's comment here, The sun (from Earth) is very clearly in Virgo right now. Pretty much right in the middle of it. But for whatever reason, September 23 - October 22 is considered Libra's territory. It makes you wonder what planet astrologers are on.
So we can tell you what constellation the sun is in, as seen from Mars, on any particular day. (Right now it's in Taurus.) But there's no telling how that'll line up with whatever standards astrologers come up with.
It's funny you should mention all that. And it's ironic the OP should ask about the procession of the equinoxes - because, as I understand it, that phenomenon is the reason that the constellations no longer physically align with the astrology charts. It's like refusing to accept the Gregorian calendar, only much worse in terms of the amount of error. So things don't even line up anymore, yet astrologers pretend that they do, which goes to show how bogus it all is. →Baseball BugsWhat's up, Doc?carrots03:15, 6 October 2009 (UTC)[reply]
Just to put Steve's comment into perspective here, I have two medical doctors in my immediate family and I have long amused myself by asking them the exact same medical question and comparing the answers. When I'm feeling especially evil, I ask them both while they're in the same room. It's not just the nut-jobs who often disagree. Franamax (talk) 05:12, 6 October 2009 (UTC)[reply]
Astrology, and in particular the western astrology you're asking about, is essentially a religion. As such, I think the humanities ref desk would be the correct place to ask this question, since that's the category that covers religion. However, I think SteveBaker's answer hits the nail on the head. What your sign would be depends on which person making stuff up you choose to listen to. There is no scientifically meaningful answer. Red Act (talk) 23:38, 5 October 2009 (UTC)[reply]
Yep. The problem with the OP's question is that it's based on a flawed premise. Keep in mind, aside from the fact that astrology is hogwash, that it also dates to a time when people thought the earth was the center of the universe, and that everything else (including Mars) orbited around it. Trying to fit astrology to Mars "does not compute". →Baseball BugsWhat's up, Doc?carrots02:27, 6 October 2009 (UTC)[reply]
First off, axial tilt has nothing to do with what the OP is trying to find out; it is, indeed, the plane of Mars' orbit about the Sun that is the determining factor, just as it is the Plane of the Ecliptic (the plane of Earth's orbit about the Sun) that determines which constellations the Sun appears to be in front of at any particular time. As it turns out, the Ecliptic cuts through 13 constellations, the twelve of the standard Western zodiac plus a corner of Ophiuchus (which the vast majority of astrologers ignore, if, indeed, they've ever heard of it). As there is only a small angle between the orbital planes of Earth and Mars, the Martian zodiac would be very similar, if not identical, to Earth's. (I have not taken the trouble to either research the exact details or figured it out for myself.)
As mentioned by others already, astrologers ignore Precession, discovered by Hipparchus in the 2nd century BC. The dates of the "Sun Signs" that you see in your newspaper were correct in approximately 100 AD, and are now almost a month off. Astrologers also ignore the fact that the constellations are not all the same size and that the segments of the Ecliptic are of different lengths when crossing them. The Sun spends more time "in" some constellations than others. Astrologers ignore this and say that the Sun spends equal time in the various "Signs" or "Houses," and thus where the Sun is in relations the actual constellations doesn't matter. Of course, if one thinks about it, this constitutes an admission that astrology isn't based on the real motions of the heavenly bodies, which was supposed to be the point of it all. As most of us know, there is no scientific validity to Astrology, but by ignoring the actual location of the Sun, planets, and stars, its claims are seen to be empty even by its own standards.
Okay, now, if you still care, here's what to do. Mars' period (year) is almost exactly 687 Earth-days long. Whenever you decide to determine the answer to your question, find out where Mars is in its orbit (not as seen from Earth, but from the Sun). Based on its longitude see what "sign" the Sun is "in" now. Well then, 687 days before then, it was in the same place. You know your age - I don't. Calculate your age in days. Determine what that comes to modulo 687 (just keep subtracting 687s and deal with the remainder). Every 57 1/4 days in the remainder pushes your answer back one Sign.
It's depressing to realize how much effort has been wasted over the centuries on this sort of "utter and compounded nonsense." B00P (talk) 05:42, 6 October 2009 (UTC)[reply]
It can be seen as depressing, or it can be seen as funny, as well as giving you a natural feeling of superiority over others, which can be useful. The Martian day is roughly 24 hours, as ours is, but as you suggest, if you want to have a "month" that relates to the 12 signs, then each month is over 50 days long, which probably wreaks havoc with the astrology charts which would expect roughly a 30 day month. For example, Libra would then run from roughly September 42nd through October 40th. Yikes. And to top it off, life expectancy in Martian years would be about 37. Ironically, on Mars, Jack Benny wouldn't have made it much past 39. →Baseball BugsWhat's up, Doc?carrots06:08, 6 October 2009 (UTC)[reply]
Some people above seem to be assuming that Martian astrologers will use the tropical zodiac that's currently the most widely used by Western astrologers. The topical zodiac is the version of the zodiac that's tied to the seasons, and hence has shifted from the Hellenistic zodiac due to axial precession. But the sidereal zodiac, which still matches up with the Hellenistic zodiac in terms of the sun's position relative to the constellations, is also in widespread current use, being used by all Jyotish astrologers, and also some Western astrologers. It seems likely that the sidereal zodiac will become relatively more popular on Mars than it is on Earth, as Martian astrologers will be more likely to view the Earth's axial precession as being less relevant to their lives.
Or maybe the increasingly popular 13-zodiac astrology used by Walter Berg will take over by the time there are colonies on Mars, or maybe some zodiac system will be developed by then that hasn't even been thought of yet. After all, Western astrology seems to be a relatively rapidly evolving set of beliefs. The whole modern sun sign astrology, in which only a person's sun sign is taken into account, was invented only 79 years ago (in 1930), for convenience in publishing horoscopes in newspapers. Westerners who aren't heavily into astrology tend to think of astrology as just being the simple new sun sign astrology that's printed in newspapers, but Western astrology is much more complicated than that to the people who are into it heavily. It seems like a good guess that the relative popularity of sun sign astrology will decrease in the future, as printed newspapers continue to decline, and astrology software makes it easy to get predictions that are based on full-fledged Western astrology, which cares about location of birth, and exact time of birth down to the minute.
One reason to guess that the most popular Martian zodiac might wind up being different from the tropical zodiac is because Martian astrology will have to mutate substantially from Earth-bound astrology, anyway. The position of Mars is important to Earth-bound astrologers, but will become irrelevant in Martian astrology. The position of the Earth, which is meaningless in Earth-bound astrology, will presumably take on some significance to Martian astrologers. The position of Earth's moon is very important in Earth-bound astrology, and indeed is the main thing that supposed causes the day-to-day changes in a person's horoscope in sun sign astrology. But the position of the Earth's moon presumably would be given little or perhaps no importance by Martian astrologers.
In short, it's difficult to try to predict the relative popularity of whatever various zodiac systems will be used by Martian astrologers.
And it still seems to me like this was really a question for the Humanities ref desk. Giving astrology any serious discussion here on the Science ref desk sort of feels similar to us doing an in-depth analysis of the aerodynamics of Superman's cape. Red Act (talk) 09:21, 6 October 2009 (UTC)[reply]
The irony of switching from sun-based astrology to planet-based is that it could be argued that there is actually a tiny grain of possible truth in the sun-based variety because your star sign correlates to the time of year you were born in. There is indeed some evidence of variation in humans dependent on the season they were born in ([19], for example). It's not hard to imagine that in primitive societies, the diet of the mother during pregnancy would be dramatically different for babies born in summer versus winter...that the amount of vitamin D the baby would produce in the early months of life would depend on the amount of sunlight available in summer versus winter...or that children are more likely to be born in one season than another if their parents are poor versus rich. So it's remotely possible that some vestige of these differences could remain with you throughout your life and perhaps have an impact on you that would be in some way predictable by your star sign. However, if this were truly what the astrologers were doing - their first question should be: "Were you born in the northern or southern hemisphere?"...and sadly, that does not figure into their patter.
However, the Astrologers are clearly not producing their results that way - so they don't get to take any credit for this - and worse still, the more serious amongst them generally prefer the more utterly ridiculous planetary-based version of the pseudo-science over the more plausible sun-based variety anyway.
Generally speaking, at what distances is it safe to be near a nuclear reactor? In other words, are the radiation levels (rems/sV) 10 meters from a reactor under normal conditions dangerous? What is the typical distance that workers are from the reactor core itself? I ask because pictures like this make me curious (this one's even better since it's obviously taken by a person). According to this diagram there isn't much between the core and the containing pool. Shadowjams (talk) 23:15, 5 October 2009 (UTC)[reply]
I have visited the PULSTAR reactor at North Carolina State University (in the photo you linked) ; we never wore counters or radiation detectors, let alone PPE. It should be noted that it is a very small reactor (50 kW, I think). Surrounding the chamber was a wall made of paraffin brick - the idea being that hydrogen in the long hydrocarbon chains of paraffin would stop neutrons. I have also visited the operational reactor at MIT - again, no counters or protective equipment. In fact, one of the applications at MIT's nuclear reactor is a medical research program for tumor-ablation - they can open a window to the core and expose slow-neutron radiation on to a live human test-subject to irradiate harmful tumors. You can read about their NRL nuclear medicine program at their website (including radiation levels. (That one has a thermal-neutron source flux; you can find data for other types of nuclear radiation elsewhere on the site). Realistically, on a day-to-day basis, inside a nuclear plant there is more hazard from electrocution, steam explosion, or pump malfunction than from nuclear radiation. Even these risks are properly controlled via effective operational procedure, maintenance, and design. Nimur (talk) 23:58, 5 October 2009 (UTC)[reply]
There are significant neutrons produced in these research reactors. BUT, with enough water, most of the neutrons will scatter, lose energy, be reflected in, and not be a significant problem, for low-power reactors. This is, of course, assuming no meltdowns and etc. (which in the case of many research reactors is just physically not possible anyway). (Neutrons are scattered very easily by light nuclei—so regular water, light water, or paraffin, end up slowing down neutrons significantly.)
At the MIT Nuclear Research Reactor (which you can tour without any dosimeters, though you do have to be screened before you leave), they tell a story about how they used to use solid tools to change the core configuration once, and the neutrons used it as an easy pathway out (and set off all the radiation detectors). Now they use hollow tools that fill up with water. It sounds counterintuitive that water is a better neutron reflector than steel, but it has to do with the mechanism of the neutrons being scattered (it's not the same thing as radiation penetration, e.g. trying to go through lead).
It's of note that comparing research reactors to the PWR (a power reactor) is not necessarily useful... the sizes are quite different, and the cores are quite different in radiation output. I don't imagine you'd stand that close to a commercial reactor core. Size does matter! --Mr.98 (talk) 00:26, 6 October 2009 (UTC)[reply]
Speaking of nuclear power, as opposed to ittybitty reactors at colleges, during normal operation, I would expect high radiation levels near the actual reactor inside the containment building. Someone who works in the field could come up with actual radiation levels. But when it is shut down for maintenance, workers have to be able to go in and physically work on valves, switches, instruments, pipes, wiring, etc. They are very complex systems and require much maintenance during refueling or maintenance outages. When there is a malfunction, like Chernobyl or Three Mile Island, lethal levels of radiation are found inside the containment (like Three Mile Island) or in the surrounding countryside (as at Chernobyl, which had no containment building). In the plant, around the grounds, and near the plant the radiation levels are supposed to be quite low, since people work there. Edison (talk) 16:17, 6 October 2009 (UTC)[reply]
Well, vacancy concentration and vacancy diffusion would increase. Higher heat would theoretically stabilise the enthalpy required to form larger pores, yes. Actually I'm kind of curious how important membrane vacancies are, compared to vacuole formation and channel proteins (active or passive). John Riemann Soong (talk) 03:28, 6 October 2009 (UTC)[reply]
I'm not going to do your homework for you but will help you a little. If by "pores" you mean membrane transport proteins then heat isn't going to make them bigger - it will of course speed up diffusion so molecules will be transported more quickly. In general higher temperatures will make membranes more "leaky" - the lipid tails of the component phospholipids will not be held together by Van der Waals forces so strongly as before. This can increase the permeability greatly, for monosaccharides as well as everything else. You probably want to look at lipid bilayer phase behavior too as this explains this. Smartse (talk) 16:46, 6 October 2009 (UTC)[reply]
Name that substance...
I'm trying to recall the chemical which is transported through the human body so fast and efficiently that if you touch it with your finger, you can taste it in your mouth shortly after. SteveBaker (talk) 04:12, 6 October 2009 (UTC)[reply]
Um, I don't understand. Why would skin contact result in taste? Does the substance travel through blood to the taste receptors or something? Do the receptors accept blood contents then? 66.65.140.116 (talk) 02:33, 7 October 2009 (UTC)[reply]
I don't understand the distinction you're making. Banked turns help compensate for inertia, thereby making it less necessary to depend on friction to overcome the inertia.
The OP refers to the fact that friction need not be present as much while turning on a banked road, ie it "helps overcome the dependence on friction". In fact, it is possible to turn at a specific velocity even if no friction is present. Rkr1991(Wanna chat?)08:00, 6 October 2009 (UTC)[reply]
Easy. A banked corner stops you from flying off the road, ordinarily you depend on road-tire friction for that. APL (talk) 19:45, 6 October 2009 (UTC)[reply]
Cooling towers
If I were to have an aerial view of a nuclear reactor's cooling towers, what would I see? Is most of it empty space? I can't seem to find a decent image from straight down. Dismas|(talk)04:38, 6 October 2009 (UTC)[reply]
I've been looking at nuclear power plants on google earth (hope I don't get arrested by homeland security); if you get a photo of the cooling towers while not in use, all you see is a big shadow inside. Someguy1221 (talk) 06:15, 6 October 2009 (UTC)[reply]
I did some Google searches on "cooling tower" and "diagram". Apparently in at least one design the bottom part of the space has equipment in it while the middle and upper part is indeed empty. For example, see this pageand this diagram. Some other designs are shown on this page. --Anonymous, 09:03 UTC, October 6, 2009.
For the most part, they're just water tanks. There's some inflow and outflow plumbing, and some heavy-duty pumps; and the walls might be reinforced with scaffolding in some designs (though most are freestanding); and they are filled with the heat-exchange water which is cooling to a temperature safe to re-release into the environment without catastrophic thermal pollution. The water is clean and generally speaking, radiation-levels are low (hopefully at about ambient-level) - though this last point is subject to some pseudoscientific debate, the water should not be carrying any radioactive contaminants. Nuclear reactor coolant explains double- or multi-heat-exchange systems - any water that ever actually contacted the core is not usually released into the environment.
The iconic convex shape that has come to symbolize nuclear energy is really just an effective way to build extremely large (by volume) water-reservoirs while using less land-area and less construction material. For the geometrically inclined - the curved shape contributes to structural stability by distributing the load uniformly - but the towers are specifically designed to be ruled surfaces - so they can be built out of conventional construction beams. Nimur (talk) 07:53, 6 October 2009 (UTC)[reply]
So, if I understand you correctly, besides some pumps and associated plumbing at the base, the rest of the tower is filled with water and the top is open to the environment. So the concave design is just a stronger way of making a big water bowl? After EC, you seem to have answered my question. Thanks! Dismas|(talk)08:01, 6 October 2009 (UTC)[reply]
I don't believe this is at all correct. As has been stated elsewhere, cooling towers are mostly empty space. The hot water from the power station is pumped to the top of the tower and falls down through the tower in small drops, cooling as it does so. It is collected at the bottom and returns to the power station as cool water. So the only thing in te tower is equipment to convert the water to drops, drops of water, and air. --Phil Holmes (talk) 09:35, 6 October 2009 (UTC)[reply]
I went on a tour of Three Mile Island when I was young, and we went inside one of the disused cooling towers. Your description is basically correct, but I don't think the water falls from the top of the tower to the bottom. It falls from a much lower height... in the picture above, you can see a dark ring making up the bottom 10% or so of the tower. I believe that's where the water falls, from a height of one story or so. Outside air flows in through that ring (which is mostly open), through the waterfall, and then up the cooling tower. The 90% of the cooling tower that's above the waterfall is there to act as a chimney. -- Coneslayer (talk) 12:28, 6 October 2009 (UTC)[reply]
Plant design, including cooling, can vary from installation to installation. One of the complaints levied against the American nuclear power establishment has been the failure to have a uniform design - every plant is uniquely engineered (at great additional cost), rather than using a tried and tested reference design. (This site has some nice overview and references). Hopefully this situation is changing with the new AP1000 design - which will be installed, among other places, at Shearon Harris in my hometown outside Raleigh. Needless to say, there is no shortage of real debate (and pseudoscientific accusatory politics) associated with this sort of major policy-change - but the hope is that by establishing a single core design, the rest of the construction costs for the plant can also be consolidated and reduced through economy of scale. Nimur (talk) 15:39, 6 October 2009 (UTC)[reply]
inventor of American sealed hard plastic cases used in late 90's & early 00's?
Forgive me if they were found elsewhere, I did not travel abroad extensively during that time period. I'm sure most people here are familiar with those clear plastic cases for things like small electronic items that were essentially welded shut at the edges and required some seriously sharp implements to open, after which they could easily cut your hand/finger/whatever. Is there any info as to who actually invented that method of packaging? I consider it the worst invention in widespread use in the last 20 years... who gets the blame? 61.189.63.208 (talk) 11:12, 6 October 2009 (UTC)[reply]
Playing around with Google Patents, a possible candidate is Cornelius M. Phipps. Phipps was making an improvement over the card-based blister-pack, where a plastic blister was attached a cardboard card, by having his be all plastic. It doesn't look like it is heat-sealed, though, which is part of what makes blister-packs so irritating. Hmm. Anyway, I found this by searching for patents regarding blister packs on Google Patents, and then seeing what patents they referenced, moving back in time in the process. As I would have guessed with something like this, there are a lot possible candidates, depending on what one defines as the essence of the modern plastic case (there are a LOT of patents on the subject). --Mr.98 (talk) 15:31, 6 October 2009 (UTC)[reply]
It's also a matter of choice of plastic - I think the dangerously sharp ones you refer to are usually made of HDPE - which is recyclable! That may have been a motivation for switching to this more irritating form of packaging material. Nimur (talk) 16:08, 6 October 2009 (UTC)[reply]
Because you are seeking fictional creatures, there's no way to complete an authoritative list. You might enjoy reading cryptozoology, but if you are considering adding content about fictional lifeforms, please be sure to establish their notability per our relevant policy. Nimur (talk) 15:43, 6 October 2009 (UTC)[reply]
Depending on your exact criteria, (fictional) Robots and Androids might fit. Note also that, according to a recent BBC TV programme (sorry I can't remember more details: it was on the iPlayer last week but has now gone) the laboratory creation of an entirely artificial living cell (i.e. built from scratch out of chemicals) is expected within the next 2 years. 87.81.230.195 (talk) 16:01, 6 October 2009 (UTC)[reply]
Indeed. The five things on our OP's list are entirely fictional - and that list would be a very long one indeed...but kinda useless for most practical purposes. How about Frankenstein - Asimov's robotics stories - 'Nanites' in StarTrek...there are an awful lot of science fiction/fantasy works that contain artificial life-forms of one kind or another. However, the 'real world' work on creating artificial life from (essentially) raw chemical feedstocks is an amazing thing. If they manage it - it'll be right up there with the top ten things that humanity has achieved. However, it hasn't been completely solved yet. Meanwhile, the vast majority of Artificial Life research has been done in the realms of software - and if you grant the possibility of a digital life-form living in it's own little digital world - then there are hundreds - if not thousands - of examples. Do a Google search on "Alife" and you'll find plenty - many of which you can download and try for yourself. SteveBaker (talk) 18:59, 6 October 2009 (UTC)[reply]
Not so sure about the top ten things list myself. No doubt very tricky but there are lots of very tricky things and perhaps it is only a top ten if you somehow find the concept mind-blowing rather than on pure technical merit. Philosophically isn't it just a form of Procreation? A tighter challenge might be recreating something with extinct DNA perhaps, so you have to learn to code the DNA to match the properties. --BozMotalk19:17, 6 October 2009 (UTC)[reply]
I'm wondering if I should buy the same brand (Giant) and model of bicycle with either hub gears or derailleur gears. My simple preferance is for hub gears; I used them long ago as a teenager: they are reliable - they always work, any adjustment required is simple. However, having used derailleur gears since then (as hub gears were not available) I'm wondering if I've got used to riding with a higher gearing than the 3-speed hub is going to give me. It would be a disapointment to order a hub-gear bike and find that the top gear of the three-speed hub is less than what I prefer. (Pre-order road-testing is not possible). I cannot afford a 7-speed hub as they are much more expensive. The thing about derailleur gears is that in my experience after a few weeks or months they are only useable in one gear - so I just use one gear all the time, and I have never ever managed to get them running properly despite trying. They are also more prone to damage and probably friction. Can anyone tell me please the amount of gearing in a 3-speed hub compared to a derailleur? Thanks 78.146.29.77 (talk) 20:05, 6 October 2009 (UTC)[reply]
Bike wheel gearing is typically measured in inches (being the wheel size equivalent to the gearing). Derailleur gears (which have lower friction loss than hub gears and are less inclined to break under heaby load which is why professionals etc always use them) can range from about 23 inches to 104 inches with a "typical" set up being say 32-96 inch range. There is no standard range and the person selling the gears should tell you the range in inches. Hub gears also vary considerably and of course the range depends on the front cogs (by the pedals) as well as the rear ones. You could move the hub gearing up or down without problem by changing the front cog but you will not get the kind of range a typical derailleur offers. --BozMotalk20:12, 6 October 2009 (UTC)[reply]
Also note that there's no reason that derailleur gear shouldn't continue to work for years, if properly maintained. You may need occasionally to replace the chain and the sprockets, but you'd similarly need to do that with a hub gear. I've ridden thousands of miles without needing to replace any of the components of a derailleur, and it has still been able to use all the gears avaiable. --Phil Holmes (talk) 21:24, 6 October 2009 (UTC)[reply]
Hub gears are pretty much maintenance free, derailleur gears need occasional maintenance but should work for years as well. You'll get away with a 3-speed hub if you live somewhere fairly flat. If money *and* spread of gears is an issue, I would get somebody to show you how to do basic work on derailleur gears - its not all that complicated.195.128.251.43 (talk) 21:55, 6 October 2009 (UTC)[reply]
A few questions
1. If a virus's host dies, in most cases the virus will die out as well, due to its need of the host cells to reproduce. Why hasn't selective pressure led to all viruses becoming non-lethal?
To one: In most cases the pressure of selection has indeed prompted viruses to adopt certain strategies. One of them is to be so virulent as to spread over a whole population quickly, also called virulency, (reducing the need of living hosts) and mutate easily (therefore avoiding the hosts immune system) or to adapt to certain hosts as to not kill them. Also: most viruses who are quite lethal are relatively new to the human population and have not started to adapt to the new host. For an example of the first kind see Rhinovirus, for an eample of the second kind see HIV. An answer to your second question might be found here --217.84.60.60 (talk) 20:46, 6 October 2009 (UTC)[reply]
It is not just a matter of lethality. HIV is extremely lethal, but as it takes years to kill the victim, it has plenty of time to spread to another host. The ones that don't always get to spread far are lethal and fast acting. Googlemeister (talk) 20:55, 6 October 2009 (UTC)[reply]
Have a look at impedance matching. If the impedance of the speaker doesn't match the impedance of the amplifier, significant power is reflected back - leading to potentially significant permanent damage. Nimur (talk) 00:51, 7 October 2009 (UTC)[reply]
Hubble space telescope
I am having a doubt that how the hubble space telescope takes the picture of the galaxies because galaxies size itself in lightyears and the light emitted from a star are any other object will travel in straight line. If this is true then the light that is received by the hubble from distant object will be a small proprotion of the light emitted from them with that light received how they are developing the image of the galaxies and many huge objects. —Preceding unsigned comment added by Dineshbkumar06 (talk • contribs) 01:53, 7 October 2009 (UTC)[reply]
A very good observation. A galaxy has 100,000,000,000 stars. Sometimes more. Each one produces an average 1,000,000,000 times the lighting of all the daylight falling on all the fields, mountains and oceans in the entire world. Even so, it still takes a light gathering area 100,000 times your eyes' to make those pictures. Sagittarian Milky Way (talk) 02:09, 7 October 2009 (UTC)[reply]
Simple Chem Question
Assuming complete combustion:
C2H6O + 2 O2 → 2 CO2 + 3 H2O
If I am given the number of grams of water produced, can I simply perform a stoichiometric conversion to find the number of moles of ethanol in the initial reactants? Acceptable (talk) 02:18, 7 October 2009 (UTC)[reply]
