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Yes, there are applications. For instance, Fermi's theory does not account for neutral currents which allow neutrinos to scatter ellastically off of matter. That fact was skillfuly used at snolab to clinch the proof that solar neutrinos oscillate into different kinds of neutrinos along their path towards the earth. Dauto (talk) 20:52, 29 March 2010 (UTC)[reply]
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Space Radiation and the Earth's Atmosphere —how are high-frequency light rays affected?
I have a question about how exactly the Earth's atmosphere protects us from space radiation?
As far as I'm aware, high-frequency radiation (gamma rays, x-rays, and most ultraviolet rays) are in some way "filtered" by the atmosphere and thus never gets to the surface. Whereas lower-frequency radiation (some ultraviolet rays, visible light, infrared waves, microwaves, and radio waves) penetrate the atmosphere and are felt down here.
My question is twofold:
(1) How does the atmosphere interact with the higher-frequency radiation, exactly? Does it reflect the gamma rays, etc. back into outer space; decrease the frequency of it so as to render it into visible light, etc; or some other process?
and
(2) Is this done by the radiation of the atmosphere, the chemical composition, or something else?
Would exploding a hydrogen bomb on the planet Jupiter or any of the other gas giants turn them into a second sun? And if so, would there be any effects on the Earth?80.1.88.25 (talk) 12:33, 29 March 2010 (UTC)[Trevor Loughlin][reply]
The size and heat of even the largest H-bombs ever set off on Earth is still quite small on a planetary scale even if they are large on a human scale. The amount of energy that is released when, say, an asteroid hits a planet is vastly larger (though not always concentrated in the same area). In any case, even if a significant fusion reaction was set off inside the planet (where the material would be of highest density), without there being more mass and gravity, it would not be self-sustaining like reactions in the Sun. (For the same reason, setting off an H-bomb on Earth does not begin a fusion reaction in the nitrogen in the air or the water in the sea, even though in theory both could undergo fusion under the right conditions. But Earth does not have the right conditions for that.) --Mr.98 (talk) 12:42, 29 March 2010 (UTC)[reply]
Bombs cause explosions, which move things outwards. To turn Jupiter into a second sun you would need to compress it to increase the temperature and density to a point where fusion can occur. However, once that fusion did occur the radiation pressure would cause the planet to expand again and the fusion would stop. You would need some way to compress it and keep it compressed, and I don't think there is any practical way of doing that, even assuming futuristic technology. I think any attempt to compress it would end up using more energy than would be generated. --Tango (talk) 13:07, 29 March 2010 (UTC)[reply]
Agree with Mr.98 and Tango. None of the gas giants are large enough to sustain fusion reactions at their core - otherwise they would be stars already. The minimum theoretical mass for a star with a similar composition to our Sun is about 75 times the mass of Jupiter, and the least massive known star, AB Doradus C, is 93 Jupiter masses - see star. Jupiter became a star at the end of 2010: Odyssey Two, but only with the help of advanced alien technology.
To answer the second part of your question, if Jupiter did somehow become a star, but without changing its size, mass or orbit, I think there would be a minimal impact on the Earth. Jupiter is 1/10 of the diameter of the Sun, and even at its closest, it is four times as far away from Earth as the Sun. This means that its apparent area in the sky is at most about 1/1600 that of the Sun. If we assume a Jupiter-star would emit the same amount of radiation per unit surface area as the Sun, then it would increase the solar radiation on Earth by less than a tenth of 1%. This is the same order of magnitude as the natural solar variation that we experience as the Sun's output changes over its 11-year solar cycle. Gandalf61 (talk) 13:10, 29 March 2010 (UTC)[reply]
To the contrary, I think it would be devastating if Jupiter ever became a sun. Much of Earth's night would be gone, which would wreak havoc on nocturnal animals and mess up the circadian rhythms of the diurnal animals. --206.130.23.67 (talk) 14:39, 29 March 2010 (UTC)[reply]
If Jupiter (at the current size) were burning like a star, it would not eradicate nights on Earth. Your statement appears to be based on the idea that it will shed enough light every night to cause problems. First, it is 1/1047 the volume of the sun. It is more than 4 times further away from Earth than the sun. (Please check my math - I am doing this with a crying baby on my lap, so not concentrating well.) Combined, the luminosity will be minuscule compared with the Sun (or even a full moon). It will be very visible, but mainly as a very bright star, not something that lights up the night. Further, that is only if it is in the night sky. Jupiter is not always in the night sky. -- kainaw™14:50, 29 March 2010 (UTC)[reply]
Since we don't know any way of turning Jupiter into a star, it is difficult to guess what luminosity it would have. A key thing to keep in mind is that human light perception is logarithmic. A full moon doesn't seem that much dimmer than the sun - it's plenty to see by. That is despite the Sun being 449,000 times brighter than a full moon. If Gandalf61 is correct (his guess is as good as anyone's) and Jupiter would be only 1000 times dimmer than the Sun, it would be many many times brighter than the full moon. You wouldn't really notice that it was dimmer than the sun at all. --Tango (talk) 15:09, 29 March 2010 (UTC)[reply]
Using the basic luminosity function L=4πR2σT4, you can see that the radius of Jupiter is directly related to the luminosity. Since Jupiter is 1/10th the radius of the Sun, it has 1/100 less luminosity (squared produces 100 instead of 10). If we assume T to be the same, Jupiter would be 1/100th the luminosity of the Sun. That would be the apparent luminosity if Jupiter were the same distance from Earth as the Sun. It is more than 4x further away, so it would have an apparent luminosity less than 1/100th that of the sun. -- kainaw™15:34, 29 March 2010 (UTC)[reply]
That is consistent with Gandalf's guess (I think you did the same calculation). As I say, being 1/1600 the brightness of the Sun wouldn't be very noticeable to humans (your pupils would dilate a little more, and then everything would seem the same). That formula doesn't really apply, though, since we're assuming something weird is happening to make Jupiter a fusor. --Tango (talk) 15:56, 29 March 2010 (UTC)[reply]
OK - now that I can work out equations and type without little fingers trying to press all the keys on the keyboard... The apparent luminosity (usually referred to as "magnitude") will be approximately 1/1600 that of the Sun if Jupiter has the same temperature as the Sun. But, that will not be a reasonable assumption. Instead, consider a very small star, still larger than Jupiter, but damn small. VB 10 is the best I can find as a reference. It has an absolute luminosity that is 0.0008 that of the Sun. Therefore, the apparent luminosity of Jupiter, if it was as bright as VB 10, would be about 0.00005 that of the sun (1/20000, which is significantly dimmer than the 1/1600 previously quoted). Further, this would not an object the size of the Sun in the sky. It would be a tiny dot that is visibly brighter than the other stars. So, it is not reasonable, in my opinion, to claim that it will disrupt all nightlife on Earth - which is the point I was attempting to make much earlier. The best I can explain it is comparing lighting up a room with a flood light or with a tiny little LED. Even if the tiny little LED is 10 times brighter than the flood light, it won't do a good job of lighting up the room. In our case, the tiny little LED is 20000 times dimmer than the flood light. -- kainaw™17:34, 29 March 2010 (UTC)[reply]
According to apparent magnitude, yes. It is also a given since we know we can see the moon during the day and it is around 450,000 times dimmer than the Sun. So, an object 20,000 times dimmer will be visible. Jupiter would be a tiny little bright dot in the sky. -- kainaw™19:11, 29 March 2010 (UTC)[reply]
The angular diameter is irrelevant, it is brightness (what you are calling apparent luminosity) that matters. How spread out it is will determine whether you can look at Jupiter without blinding yourself, but that's about it. A flood light and an LED of equal brightness will illuminate a room equally well. 1/20,000 times the brightness of the Sun is 22 times brighter than a full moon. Basically, it would like having a full moon much more often than we do now. That may not have too big an impact, but it would have one. --Tango (talk) 19:22, 29 March 2010 (UTC)[reply]
I pity the poor baby having to put up with Kainaw discussing Jupiter sun's when he/she's supposed to be the centre of Kainaw's attention :-P Nil Einne (talk) 06:53, 30 March 2010 (UTC)[reply]
Also, nuking Jupiter will not turn it into a non-star fireball, because to have a fire you need heat (supplied by nuke), fuel (Jupiter has quite a bit of hydrogen), and Oxygen (technically Chlorine might do), which is lacking, so no fire will be maintained. Googlemeister (talk) 13:54, 29 March 2010 (UTC)[reply]
It's also worth pointing out that the fusion matterial in the hydrogen bomb is either a mixture of deuterium and tritium or a lithium deuteride. Hydrogen itself leads to a slow nuclear burn instead of an explosive one. Thanks god, otherwise the sun would have exploded a long time ago :-) Dauto (talk) 14:49, 29 March 2010 (UTC)[reply]
While there are probably good arguments for why it couldn't happen with something as small as Jupiter, I suspect one could imagine a slightly sub-critical mass of hydrogen that might be ignited with a large enough thermonuclear trigger that would cause it to generate enough energy from fusion to keep the reaction going. Sort of like how many substances don't spontaneously burn, but add an initial spark and they will produce enough of their own heat to keep the reaction going. Type Ia supernova are actually a little like this. It is believed that an initial "spark" of carbon fusion generates heat that triggers fusion in surrounding material and the ensuing cascade of fusion rapidly burns through the entire star (causing it to blow apart). It would be interesting to know what would happen if you artificially heated a nearly critical mass of hydrogen. Would it keep producing fusion, or would the extra heat cause it to expand enough that the reaction would snuff itself out. I don't think one can draw an obvious conclusion either way. Dragons flight (talk) 08:56, 31 March 2010 (UTC)[reply]
I've looked and all I keep getting is frosted glass which is not what I mean.I wanted to know why ,when windows frost up,the pattern looks like ferns or flowers? —Preceding unsigned comment added by 88.96.226.6 (talk) 13:47, 29 March 2010 (UTC)[reply]
The article titled (unsurprisingly) Frost contains a section on window frost. Ice forms a a hexagonalcrystals. The frost patterns you see result from a propagation of this hexagonal structure across the surface of the glass, much as in a snowflake. The angles involved in the branching are all generally 120o (the internal angle of a hexagon), which gives the pattern you are familiar with. --Jayron3214:24, 29 March 2010 (UTC)[reply]
Other than the information given in sinusitis, we can't really give any more information specific to your case because we don't offer medical advice. You should consult a medical professional if you need to know more. —Akrabbimtalk20:14, 29 March 2010 (UTC)[reply]
Well, Matt asked for details and I gave it. I just wanted to know if there were any cases where someone infected someone else with sinusitis. --Reticuli88 (talk) 20:30, 29 March 2010 (UTC)[reply]
It was a rhetorical question. You got the infection from somewhere, therefore it is contagious. How contagious, as in how worried should you be if you sneeze, is a more complicated question that depends on a lot of variables that we wouldn't be able to evaluate here. —Akrabbimtalk20:58, 29 March 2010 (UTC)[reply]
Sinusitis and a cold could be the same organism. Or it could be an opportunistic infection with bacteria after a cold (which is a virus). So you can infect other people, but that doesn't mean it will turn into sinusitis for them - it might simply be a cold. A bacteria sinusitis is unlikely to be infectious. If you are contagious you might transmit the germ, but you can't control in which part of the body it will cause illness in the other person. Ariel. (talk) 21:03, 29 March 2010 (UTC)[reply]
The textbook I use derives the relationship τ = dL/dt in the following manner: dL/dt = d/dt(r x p) = dr/dt x p + r x dp/dt = v x mv + r x F = r x F = τ. But doesn't this assume that the position vector for the torque and the particle are the same (ie that the force is being applied directly on the particle)? 173.179.59.66 (talk) 17:30, 29 March 2010 (UTC)[reply]
Whatcha talking about? By definition the coordinate of the point of action of force on a point particle (and the torque as well) is the coordinate of the particle itself. Dauto (talk) 17:53, 29 March 2010 (UTC)[reply]
Well, what if we were looking at a massless lever with a large mass somewhere in the middle. The the force can be applied at the end of the lever, while the particle would have a different r than the force. 173.179.59.66 (talk) 21:41, 29 March 2010 (UTC)[reply]
No, a lever is made of many particles, the external force acts on a massless particle at one end of the massless lever and the action gets transmited to the other end by internal forces between the massless particles used to make the lever. Lots of forces and each one of them act exactly at the coordinate of the particle they are acting on. Dauto (talk) 22:16, 29 March 2010 (UTC)[reply]
Agreed, but then how would you calculate the torque on the massive particle? I would imagine that you would have to use τ = rF x F, and this would equal d/dt(Lm) = d/dt(rm x p), with rF not equal to rm. So what gives? 173.179.59.66 (talk) 01:12, 30 March 2010 (UTC)[reply]
I guess I'm missing something lol, maybe an example with numbers would help me. Imagine we have a 2 metre rod of negligible mass with a small rock with a mass of 10 kg attached at the middle (the rod will be rotated about one end). A force of 3 Newtons is applied to the end of the rod. What's the angular acceleration? τ = 6Nm, and τ = d/dt(rmmv) = d/dt(Iω) = (mrm^2)α = (100kg.m^2)α, so α = 0.06rads/s. Here, rF didn't equal rm. I agree that the internal forces of the rod will ensure that a force with a different magnitude than F will act on the particel at r = rm, but it remains to be shown that dL/dt still equals rF x F, which I don't think my book adequately proves. If I'm not making sense, let me know. 173.179.59.66 (talk) 02:38, 30 March 2010 (UTC)[reply]
Everything you said above is correct except for "it remains to be shown that dL/dt still equals rF x F". Yes, for that example rF and rm are different but that's only possible because you are dealing with an extended object. The book's proof is intended for a set of point particles. But since extended objects are made of point particles, at the end of the day the proof also applies for extended objects. The internal forces will be whatever they have to be in order for dL/dt to be equal to rF x F. Either that happens, or the massless bar will bend and brake. Dauto (talk) 04:49, 30 March 2010 (UTC)[reply]
How do you know the internal forces will arrange themselves so that dL/dt = rF x F? I would think it would have to be justified on the grounds of mechanical advantage or something, which my book doesn't seem to address. It seems to me that the proof doesn't apply to extended objects for that reason. 173.179.59.66 (talk) 06:41, 30 March 2010 (UTC)[reply]
Yes, there are some missing steps in the proof but they are almost trivial. Denoting the extended object angular momentum by and its external torque by , we want to prove that . But we already know that . So the only step missing in the proof is to show that . This comes about because the internal forces come in action-reaction pairs of oposing forces acting along a common axis which leads to pairs of oposing torques that cancel pairwise identically. Dauto (talk) 15:12, 30 March 2010 (UTC)[reply]
I don't know what you mean by a central torque. But I know that the reason that they cancel pairwise is that the pair of forces act along the line connecting the particles. May be that's what you mean by central torque? That's true because this is the only way to preserve the axial symmetry. There is only one important exception. Electric charge particles in motion produce magnetic fields that generate forces on othe moving charges. The magnetic force depends on the velocity of the particles as well as on their positions and can generate non-axial forces. Heck, The magnetic forces may not even opose each other, violating Newton's third law. But all is well since in those situations there will be electromagnetic radiation being produced and if the energy, momentum, and angular momentum of the radiation is also included in the calculation than the usual laws (including ) will still apply. This situation is clearly outside of the intended scope of the proof presented by your book. Dauto (talk) 19:52, 30 March 2010 (UTC)[reply]
If a product has a caducity of some days in the fridge (7 C) or several months frozen (-15 C), what is the problem of defrosting and re-freezing it? You buy it frozen, let it defrost 1 day and then re-freeze it and defrost it again in another further day. That just makes 2-3 days spoiling, is that a huge deal?
--Quest09 (talk) 17:49, 29 March 2010 (UTC)[reply]
If you are defrosting it in the fridge, you should be fine (although you may damage the food, particularly the texture). If you are defrosting it to room temperature, it could be a problem. --Tango (talk) 19:23, 29 March 2010 (UTC)[reply]
When a product is frozen commercially they (often) use flash freezing, i.e. freeze it so fast that there is no time of ice crystals to form. This preserves the texture of the product. If you refreeze it, you will allow ice crystals to form, and burst the cells of the food, so it will be limp and have a poor appearance. Ariel. (talk) 21:15, 29 March 2010 (UTC)[reply]
This is an issue that often comes up with frost-free freezers, which do, at times, melt the food then refreeze it. This absolutely ruins some types of foods, such as bread, by driving all the moisture to one end. Thus, you get part that's hard as a rock, and part that's slime. Meat can suffer from "freezer burn" for a similar reason. Ice cream also forms large ice crystals when it refreezes, making it no longer smooth and creamy. Some foods do quite well when melted and refrozen, like juices and soups, just as long as they are fully thawed and stirred before you consume them. StuRat (talk) 22:11, 29 March 2010 (UTC)[reply]
Grasping the EPT
Can I jump from a helicopter and safely grasp the electric power transmission line to hang for a few seconds, like birds that commonly sit onto it? 213.154.8.70 (talk) 21:33, 29 March 2010 (UTC)[reply]
(ec) There are at least two cables in a power line, one taking the electricity away, the other bringing it back. If a circuit is made, and you are a part of it, you will be fried to a crisp. With low power cables, they are insulated, but with high power cables, the bare wire is usually exposed, because insulation wouldn't do anything anyways. As long as only one cable is touched, you should be fine. If two uninsulated cables, however, are touched simultaneously, a circuit will be formed, and you will be turned into a crispy critter. --The HighFinSpermWhale21:40, 29 March 2010 (UTC)[reply]
Original poster: Note that the above answer is not to be interpreted as some sort of guarantee. That is, don't jump from a helicopter and say Wikipedia said you could. Gabbe (talk) 21:45, 29 March 2010 (UTC)[reply]
I would advice against the jumping part (It would likely be to high up to survive a fall and to close to the ground to be able to open a parachute), But they DO, on a routine basis, use helicopters as a working platform for work on LIVE extra high voltage power grid lines. (i.e. without turning off the power!). So if what you are yearning for is to touch a live wire (and survive), then that experience is definitely within reach (pun intended!)(Of course Provided that you have sufficient money to bribe the working crew :-) Unfortunately I am not able to remember what kind of protection, if any, one needs to have on the helicopter for it to fly safely in the (I presume) unusual ammount of static electricity. --Seren-dipper (talk) 22:15, 29 March 2010 (UTC)[reply]
You missed one. It's "advise" (rhymes with eyes) for the verb - "I advise you", "I would advise" - and "advice" (rhymes with ice) for the noun "I provide good advice", "the advice was wrong". Conveniently advice the noun ends in 'ice', it doesn't sound particularly wrong when you drop 'ice' into a sentence in place of 'advice': "I (the ice maker) provide good ice", "the ice was wrong (too yellow)". The verb doesn't have ice at the end. --Polysylabic Pseudonym (talk) 06:03, 1 April 2010 (UTC)[reply]
I assume no liability for the following statement. It would be doubtful you could jump from a helicopter and grab even a non-electric rope without falling to your death. Utilities commonly have "live line work" done by personnel carried by helicopters. The personnel wear a metallic garment, and a probe absorbs the arc as the helicopter assumes the potential of the high voltage line (345 kv or 765 kv, for example). Then the worker installs a line separator or whatever, at the line potential, but isolated from ground potential. So for information only, and not proffered as advice, I assume that if I jumped from a helicopter and somehow grasped a high voltage line, without falling to my death, there would be a surge of electric current when I passed near the wire, causing an electric arc which might cause injury or death, which might cause my hands to grasp the wire, or to be unable to grasp the wire. I have seen photos of high voltage line maintenance in China in which the worker shot a nonconductive line across a transmission wire, then pulled herself up to the conductor, then attached a little wheeled cart which she used to travel to the line separator needing service. (By no means should you attempt this). The trick is to be at the line potential, without a path to ground or a different phase. Once there, corona discharge might be painful or disabling, even without arcing to ground or a different phase. The higher the voltage, the more difficult it would be to hand on. Birds land and perch on distribution lines, but not on transmission lines. Edison (talk) 01:59, 30 March 2010 (UTC)[reply]
There was a programme on the BBC nor long ago - Richard Hammond's Invisible Worlds - which showed an engineer sitting on a power line. I don't think it said whether it was a distribution or transmission line, but they used a special camera to show the electric field around him. It has to be said he was wearing a special metal mesh suit as well! --TammyMoet (talk) 08:19, 30 March 2010 (UTC)[reply]
The question is unclear and difficult to answer. If I touch my toes with my fingers, then yes, there is a circuit through the body, and some current could flow, if something is creating a voltage difference in the body. It is not clear why there would be a difference of voltage. Don't try such an experiment with household or utility electricity. There is the likelihood of electrocution due to touching two conductors or a conductor and ground (the wood of a utility pole counts as ground), and arcing can occur across some distance of air, perhaps several feet, depending on the voltage. At high voltages, there is corona discharge even without a connection to ground or a different phase. This is due to the air being ionized near the sharper points connected to the conductor. The shape of a conductor affects the location and amount of corona discharge. Edison (talk) 16:25, 30 March 2010 (UTC)[reply]
But would there be a corona discharge between hands or legs? Is there a way to prevent it (maybe by placing a non-conductive material under me or knowing the upper limit of discharge's improbability)? 213.154.0.237 (talk) 06:08, 31 March 2010 (UTC)[reply]
Probably if you took a chopper beside a wire, allowed yourself to be charged by contacting the helicopter to the wire so you wouldn't get a static shock, then climbed out and hung on the wire, nothing would happen. The two wires close to each other do not carry opposite charges. --Cheminterest (talk) 20:46, 31 March 2010 (UTC)[reply]
What is the psychological term denoting the strong reaction (weeping, sobbing) when divulging a traumatic experience?
Resolved
When a patient for the first time divulges an until then repressed traumatic event, then many people (patients) experiences a (to them) surprisingly, and sometimes even frighteningly strong reaction of crying, sobbing, shaking etc. What is the specific psychological term for this (strong) reaction? --Seren-dipper (talk) 21:36, 29 March 2010 (UTC)[reply]
Most residential houses in the US appear to be supplied with 220-240 voltage which is split into 2 legs with each leg carrying 110-120 volts AC. Both legs seem to be connected by a neutral that is sometimes referred to as ground. Does this mean that one of the wires in a wall outlet has not difference in voltage that the ground outside or the floors in the house and that the bottom of the sin or cos wave for each leg is where neutral is actually at rather than at the zero point of the wave? 71.100.3.207 (talk) 02:24, 30 March 2010 (UTC)[reply]
To your first question: yes, one of the prongs is at ground potential.
To your second: it's hard to describe without drawing a picture. Each of the "hot" "phases" oscillates to a positive and negative voltage about that neutral. The scheme is described in our Split-phase electric power article. (Too bad there's not a picture on that page, either.) —Steve Summit (talk) 02:30, 30 March 2010 (UTC)[reply]
voltages of the two "hot" legs in Split-phase electric power Okay, here's a simple illustration. The black wire alternates from +120V to -120V with respect to the neutral (white) wire, which stays at ground potential (0V). Meanwhile, the red wire is alternating from -120V to +120V, 180 degrees out of phase. The voltage from either the black wire or the red wire with respect to the white wire is 120VAC. But if you connect a load across the black and red wires, you get 240VAC. (As you can see from the figure, at any given instant, the distance from the black curve to the red curve varies from +240 to -240.) This is how you can get 120V for ordinary household appliances, or 240V for high-power appliances such as electric stoves, ovens, and clothes dryers. (These are of course the voltages for North America; YVMV.) —Steve Summit (talk) 03:26, 30 March 2010 (UTC)[reply]
Please provide an English translation for "has not difference in voltage that the ground outside or the floors in the house and that the bottom of the sin or cos wave for each leg is where neutral is actually at rather than at the zero point of the wave?" Edison (talk) 05:09, 30 March 2010 (UTC)[reply]
I believe the first part has to do with the distinction between a floating ground and an earthed ground. That is, is neutral (zero) voltage in the electrical distribution system 0V with respect to a metal rod sunk into the ground. The answer to that, at least at the household level, is yes. The neutral bus bar in the electrical distribution panel in most US homes is directly tied to a metal rod sunk into the soil. The second part is asking about where the zero reference point is on the sinusoidal wave of the live conductor - does the alternating voltage swing from -120V to +120V (or whatever the numbers happen to be) or from 0V to 240V (with respect to the given neutral "zero"). In US household current it's the former. -- 174.31.194.126 (talk) 06:15, 30 March 2010 (UTC)[reply]
A U.S. utility is typically allowed to let the voltage (at the meter) vary +/- 6 volts from the nominal 120. With voltage drop along the branch wiring in the house, this typically still supplies at least 110 volts at the outlet in the house. During peak loads there may be a "brownout" which causes voltage too low for some equipment to operate. The peak of a 120 RMS voltage is higher than 120 volts. To get the peak value, divide the RMS of a sinusoid by .707 and the result is 169.7 volts. A cheap (non-RMS) voltmeter will report a voltage other than the RMS value if the waveform is distorted, but will correspond to the RMS value for a pure sinusoid. So the voltage to neutral/earth from a "hot" wire varies typically from about -170 to about +170, 60 times per second (in the U.S.). If there is distortion (harmonics) in the waveform, the peak may vary substantially. The 240 volts in household supply is the RMS measurement between the two "hot" conductors. In 120/240 supply, the transformer low voltage winding is center-tapped for the neutral, so the two hot supply lines measure 120 from the neutral. The neutral is typically grounded to a substantial metallic low resistance earth ground at the transformer and at the neutral bus of the electric panel. The earth ground in the house might be a metallic cold water pipe coming in from the street, or it might be a driven ground rod. If the neutral were ungrounded, such as in the case of the ground at the utility pole not being connected and the neutral at the house power panel not being grounded, there would still be 240 between the hot wires and 120 from either of them to the neutral wire, but with no earth ground there could be very high voltage, up to the primary voltage, between any of the supply wires and ground, a very hazardous situation. Edison (talk) 16:04, 30 March 2010 (UTC)[reply]
Role of the Pudendal Nerve in male sexual function?
I've been studying the role of the Sympathetic Ganglion in sexual intercourse, and I am stumped when it comes to the role of specific neurons in the male ejaculatory function.
As far as I can tell, the Prostatic Plexus and the Pelvic Splanchnic Nerves relay the sensory function —as well as increased heartbeat, blood pressure, body temperature, etc.— and the Pundendal Nerve relays the motor function (enabling ejaculation).
To wit, the Sympathetic Ganglion conveys the relexive arc from the genitals to the spinal cord (and back again), whereas the sensation of arousal/orgasm is received (some time later) by the brain.
My question, however, relates to the Pudendal nerve's function apropos ejaculation and the (subsequent) termination of sexual excitement —and the start of the refractatory period.
If —during the act of coitus— the Pudendal nerve were numbed or incapacitated, how would that affect a male's sexual performance? Would a man become able to feel the thrill of ejaculation for an extended period of time, without actually ejaculating (or losing his arousal)?
There is a technique, taught to couples where the men suffers premature ejaculation, where the perineum is pressed and held in just before ejaculation: this is said to delay ejaculation. I'm not an expert on human anatomy, but is this where this particular nerve is situated? --TammyMoet (talk) 08:16, 30 March 2010 (UTC)[reply]
To save further responders a little time, let me provide the link Pudendal nerve. (Yes, I could have linked one of its previous mentions, but I didn't want to bring down the Wrath of Steve :-) .) 87.81.230.195 (talk) 10:20, 31 March 2010 (UTC)[reply]
PAC MAN MOON
I seen a quick article on the PAC MAN MOON shape (Saturn i think) It seemed glaringly obvious to me that the with the giant impact area that the energy transfered from the impact through the planet to other side and this was some sort of risidual engery left over from this. Ok maybe i am completly wrong but thats what struck me looking at it? So the question is i suppose could this be the case? —Preceding unsigned comment added by Chromagnum (talk • contribs) 05:18, 30 March 2010 (UTC)[reply]
Your description in terms of energy sounds pretty ropey. However, the idea that the creation of the Herschel crater could be linked to the variations in the observed temperature is suggested in the article you provided:
"The Cassini team says the creation of the crater itself might have played a key role in changing conditions across extensive regions of the moon's surface."
The Cassini-Huygens probe has an official website from NASA. Here's a little more technical information from the NASA/JPL Press Release: 1980s Video Icon Glows on Saturn Moon. The early assumption is that the temperature difference is due to surface texture changes, which affect the heating and cooling rate. "Even if surface texture variations are to blame, scientists are still trying to figure out why there are such sharp boundaries between the regions, Spencer said. It is possible that the impact that created Herschel Crater melted surface ice and spread water across the moon. That liquid may have flash-frozen into a hard surface. But it is hard to understand why this dense top layer would remain intact when meteorites and other space debris should have pulverized it by now, Spencer said."[3]. Nimur (talk) 14:48, 30 March 2010 (UTC)[reply]
Energy/Heat yes bad wording on my part wanted a quick answer just seemed blatanly apparent the two would be linked thanks for your answersChromagnum (talk) 17:07, 31 March 2010 (UTC)[reply]
Number of organs
I've read that humans have 22 or 23 organs depending on how you count.
How do you define an organ? Are cerebrum and cerebellum seperate organs? What about hypophysis? Is it an organ on its own or a mere extension of the brain? The eyes? If the stomach is granted the title of organ, what prevents the poor duodenum from being called one? Maybe the entire gastrointestinal tract is a single organ, then. Last thing I heard, salivary glands have been lobbying for independence. Seriously, unless the IAU decides what is an organ and what is a dwarf organ, I don't think it matters much. 88.242.231.192 (talk) 10:28, 30 March 2010 (UTC)[reply]
Maybe he was thinking of chromosomes? Humans have 23 pairs of chromosomes, often split into the 22 pairs of autosomes and one pair of sex chromosomes. His numbers match this fairly precisely. If he meant chromosomes, the answer is that the number of chromosomes varies wildly from species to species (note that the preceding link gives total number, not number of pairs, so humans show up with 46). —ShadowRanger(talk|stalk)12:21, 30 March 2010 (UTC)[reply]
No. I mean organs. Yes, I realise it's hard to quantify natural phenomena that fall on a gradient such as languages, planets, species, and organs. That does not mean that questions regarding them are not interesting or important. -Craig Pemberton13:22, 30 March 2010 (UTC)[reply]
Well, anyways, other animals, even other mammals, have organs that we do not share. Ruminants have a much more complex digestive system, for example. And I have it on good authority that many politicians manage without a neocortex. --Stephan Schulz (talk) 14:08, 30 March 2010 (UTC)[reply]
(ec)If you can't rigidly define an organ, then your question is meaningless. I have found at least one source that supports your "22 or 23" contention (where the 23rd is the skin, though it neglects to enumerate the rest). I can guarantee you that the number of organs in other animals would not match those of a human though, except by coincidence or in cases of extremely recent speciation (many of the apes may match us though). According to the page I'm referring to, the definition is "a relatively independent part of the body that carries out 1 or more special functions." Given that the appendix would probably not count as an organ under that definition, it would logically follow that any mammal with a functioning appendix (that actually serves the purpose of digesting cellulose) would not have a matching number of organs. Most herbivores have one, so the number wouldn't match. Similarly, many herbivores have multiple stomachs, which would screw up the calculation. And we haven't even left mammals yet. Birds, reptiles, etc., all have substantially different digestive tracts, reptiles, being cold blooded, don't have a liver in the same sense as humans, non-mammals wouldn't have a direct analogue to the mammalian reproductive system, etc. —ShadowRanger(talk|stalk)14:14, 30 March 2010 (UTC)[reply]
Similar to variations in the digestive system, how about variations in the scent glands of animals, such as skunks ? Venom producing/delivering organs would be another. Then some animals have other sensory organs, like extra UV and infrared-sensitive eyes and pits, and the echo-location "melon" on a porpoise or dolphin. Also, some animals have organs which provide the ability to sense weak, or create strong, electrical fields. Only (female) mammals have (milk producing) breasts and a uterus. StuRat (talk) 14:21, 30 March 2010 (UTC)[reply]
Evolution - E. coli
Greetings! Two questions:
How old - evolutionwise - is E. coli? (~ 300-400 Mill. years?)
As it is not found in the guts of cold-blooded animals (molluscs, fish) what ist the main type of gut bacteria in these species? Thanks Grey Geezer 07:26, 30 March 2010 (UTC) —Preceding unsigned comment added by Grey Geezer (talk • contribs)
Thanks! Problem seems to be the poor fossil record, so only genetics and mutation rates are used for time estimates. Grey Geezer 07:41, 31 March 2010 (UTC) —Preceding unsigned comment added by Grey Geezer (talk • contribs)
why pumice stone
i have a practical in course its aim is to determine the boiling point of the given sample of water.
in the setup of apparatus we are said to put some pumice stone in the boiling tube containing water.
tell me why that pumice stone is to be kept? {its not a HW question its indeed a doubt}
in precaution it is clearly mentioned that "the bulb of thermometer should not dip in liquid. it should be about 4-5 cm above the liquid surface."
You can measure the boiling point of water by measuring the temperature at which steam condenses (on the thermometer, for instance). Therefore the thermometer does not need to be immersed in the water. As for the pumice, its purpose is to assure that the water boils at its boiling point instead of superheating. If the water is quite pure, and the vessel in which it is heated is quite clean and smooth, then the water is liable to be heated above its boiling point and then violently erupt, yielding a false measurement (and possibly giving you a nasty burn as well). The porous pumice provides a rough surface on which the water will readily vaporize into bubbles. You can read more here, in considerable detail.--Rallette (talk) 09:27, 30 March 2010 (UTC)[reply]
See also boiling chip, if you ever need to buy some from a catalog. As Rallette notes, small, inert, irregularly-shaped chunks with a large surface area (for their volume) are used to discourage superheating and violent bubbling. TenOfAllTrades(talk) 13:07, 30 March 2010 (UTC)[reply]
transpiration
is it so that a tree on high altitudes will have more or less transpiration , assuming that temperature is same for it in comparison to another tree at lower altitude...thanx--Myownid420 (talk) 07:37, 30 March 2010 (UTC)[reply]
There seem to be quite a few studies on this. The abstract of the first one in my search says: "An important factor at high elevations is the reduced barometric pressure. It has long been recognized that the reduction of air pressure brings about an increase of evaporation, and the increase of plant transpiration in response to low air pressure has also been demonstrated under laboratory conditions." Other factors also seem to influence the rates of transpiration at different elevations; your best bet is to read some of the articles for yourself. Deor (talk) 11:52, 30 March 2010 (UTC)[reply]
Note that at that site, they mention the World Trade Center, Iraq conspiracies, and conspiracies around Tesla (this one actually has a bit of truth to it). It's surprising they don't mention the JFK assassination and the rise of Hitler. StuRat (talk) 15:47, 30 March 2010 (UTC)[reply]
I don't blame you for not liking that source, as it has all the hallmarks of crackpot pseudoscience.
Whenever someone says they've got an extraordinary new theory, which is unbeknownst to (and is perhaps being suppressed by) mainstream science, it's possible in some theoretical sense that they're right -- science does always consider the possibility that it's wrong, and it does always have more to learn -- but it's spectacularly, ludicrously, vanishingly improbable. Electromagnetic waves aren't an energy source anyway, so to suppose that these hidden, different kind of waves could somehow provide limitless "free" energy is just, I'm sorry, nonsense. —Steve Summit (talk) 13:14, 30 March 2010 (UTC)[reply]
According to the feedback column of New Scientist the "scalar energy" notion is an increasingly prevalent brand of what it calls fruitloopery. Consensus on Wikipedia, for what it's worth, said pretty much the same thing: Wikipedia:Articles for deletion/Scalar field theory (pseudoscience). However, unless I am missing something, I disagree with Steve Summit's assertion that electromagnetic waves aren't an energy source. See for example solar power. Steve - perhaps you mean that they aren't an "ultimate" energy source, in that they must have been generated by something else such as nuclear reactions within the sun, but under that interpretation kinetic energy wouldn't be an energy source in lots of real world situations either. Equisetum (talk) 14:01, 30 March 2010 (UTC)[reply]
I made the point a bit too hastily, but Stu caught my meaning. I think of electromagnetic fields and waves as being a means of transmitting energy, but not of storing it, let alone being a source of it. (Although it'd probably be hard, I concede, to rigorously distinguish between a "source" and a "storage", if it came to that.) —Steve Summit (talk) 16:57, 30 March 2010 (UTC)[reply]
Just curious; wouldn't it be more useful to have an article on scalar energy that says it's crap rather than just not have one at all? I'm not familiar with the WP deletion policies but common sense says that having information telling you it's highly improbable leads to less confusion than no information at all. I'll read into the guidelines further and perhaps I can answer my own question. -Pete5x5 (talk) 14:47, 30 March 2010 (UTC)[reply]
For very common ideas, like perpetual motion, we have an article explaining the common fallacies. Unfortunately, there is no limit to the number of incorrect theories - it's hard to have an article on all of them. We reserve our articles for only those most important pseudoscientific theories - the ones which are notable. Nimur (talk) 15:07, 30 March 2010 (UTC)[reply]
So that web page links to http://Cheniere.org - which will be instantly familiar to old hands here at the science ref desk. Not that we've necessarily seen that page before. You don't even have to read it. It has lots of
Centered text
In lots of different fonts'
AND many different sizes
For some reason, this style of web page is completely unique and almost universal amongst pseudo-science nut-jobs. I have no idea why - but it's almost 100% reliably so. SteveBaker (talk) 00:47, 31 March 2010 (UTC)[reply]
My last molar on the upper left is giving me trouble. I had a crown placed on it a few years ago, and several months ago, ¼ of it broke off. I haven't had the money or the insurance to get it fixed until now, and it hasn't given me much discomfort until the past week or so anyway. (This is not a request for medical advice, because I *know* what I need to do-- have the dentist fix my tooth. I have the appointment. It's in May.) The associated toothache comes and goes, but when it comes, it sometimes is blindingly bad.
I'm taking Ibuprofen and treating topically with Orajel, which helps some. However, I have also found that swishing slightly-below-room-temperature water around my mouth also helps ease the pain some.
Cooling the nerve may be helping you a bit. Cold brings a certain amount of numbness, and if your tooth is damaged, the nerve is exposed more directly to the cool water. If the tooth is infected, the cool water may also be reducing swelling a bit. You may also be reducing the degree of infection a bit (by washing away some of the pathogens), but that isn't likely to have a measurable effect on your pain, and they reproduce quickly enough and enough of them are likely unreachable by the water that the long term effect is negligible. —ShadowRanger(talk|stalk)14:18, 30 March 2010 (UTC)[reply]
Thanks. However, I had some slushy (partially frozen) lemonade yesterday, and five minutes later, I thought I was going to die. So, apparently not TOO cold? Kingsfold (talk) 14:40, 30 March 2010 (UTC)[reply]
Too cold is going to be a problem, but lemonade (regardless of temperature) would probably make it worse. Lemonade is quite acidic, and contains a lot of sugar (which bacteria in your mouth process, producing more acids). —ShadowRanger(talk|stalk)15:19, 30 March 2010 (UTC)[reply]
Might be the acid in the lemonade, but it was Crystal Light, so... no sugar. I'll try another flavor today (not frozen), and see if it still causes trouble. Kingsfold (talk) 15:48, 30 March 2010 (UTC)[reply]
May seems like a long time to wait while you are in such distress, can't you get an appointment with a dentist sooner ? StuRat (talk) 15:41, 30 March 2010 (UTC)[reply]
Great idea. Wish I had thought of that. (Wink.) I'm on the "call if someone cancels" list. I'm considering changing dentists.... Kingsfold (talk) 15:48, 30 March 2010 (UTC)[reply]
I had toothache that was relieved by swirling water. It was due to a large cavity between my teeth - due to moving town I had not had regular check-ups for some time - my fault for not signing up with another dentist and not the NHS's. Coincidence - it was the last tooth on my upper left as well! Luckily it was just a wisdom tooth, so it was removed without any cosmetic or functional effects. Am I right in guessing from the OP saying that they could not afford treatment that OP is from the USA? (Why dont Americans revolt and demand free health care from their government?) 78.146.180.118 (talk) 20:17, 30 March 2010 (UTC)[reply]
Did you read the post? He does have an appointment to get it fixed, and presumably can afford treatment. But his regular dentist is backed up, so he has to wait a little while (or find another dentist, which he has not yet tried to do). —ShadowRanger(talk|stalk)20:22, 30 March 2010 (UTC)[reply]
But I though this kind of delay was what Americans imagined happened with the UK NHS, and not what is supposed to happen in the US? I mean, May and he's in pain! Over a month away! This side of the Atlantic you'd probably get it done as an emergency appointment the same day, or the next day at most. And if you had a low income it would be completely free, otherwise a nominal charge of a few pounds. 78.146.180.118 (talk) 20:26, 30 March 2010 (UTC)[reply]
I feared that this discussion would eventually turn into a back-and-forth about Nationalized Health Care. I'm not convinced that I'd get in any sooner were the United States to adopt a system like the one in the UK. If anything, I'd think it would be *more* difficult to get in to secure an appointment....
The other chapter of the story (that I didn't particularly care to get into on the Science reference desk) is that it was the crown that broke-- not my tooth. As such, I am only being charged 25% of the usual cost of a crown. (I didn't realize this until the hygenist notified me of this at a recent checkup.) So, I can stick it out for a month and maybe get in sooner on "standby;" or I can go to the hassle of changing dentists, probably not get in any sooner, AND pay full price. My vote: stick it out with my usual, albeit busy, dentist. (Unfortunately for my tooth.) Kingsfold (talk) 02:09, 31 March 2010 (UTC)[reply]
As described above, you'd probably get emergency treatment for pain the same day in the UK. So you would definately have quicker treatment. I'm shocked that you consider such very bad treatment and totally unacceptable delay, not to mention not getting treatment merely because you cannot afford it, normal. Why don't Americans have a second Revolution against getting such third-world health care? If the same thing happened in the UK the dentists would probably get sacked. 78.147.25.63 (talk) 16:34, 31 March 2010 (UTC)[reply]
It's a weird part of the U.S. view of health care. For some reason, historically dentistry wasn't considered health care. Same goes for "basic" vision and hearing problems; one of my mostly deaf relatives pays about $5000 out of pocket every few years when he needs a new hearing aid. So a lot of people who have "health care" don't actually have dental coverage. If a dental problem spreads (blood infection or the like), and you lack insurance and the ability to pay, you can visit an ER for treatment. The new health care reform bill takes some steps to address this, particularly in the case of children (though most of the reforms won't kick in until 2014). It's bad, but at least it is getting better. Single-payer would probably have been more effective at controlling costs and better at achieving universal coverage, but it was politically impossible. —ShadowRanger(talk|stalk)19:34, 31 March 2010 (UTC)[reply]
People in the US don't revolt about inadequate health care because so many of us are completely deluded about US's health care being the best in the world. Afterall this is America! Dauto (talk) 07:12, 1 April 2010 (UTC)[reply]
Yes, and being cancer, there are a nearly infinite number of possible symptoms. That said, we cannot give medical advice at this reference desk. You should contact a doctor for specific advice. —ShadowRanger(talk|stalk)15:24, 30 March 2010 (UTC)[reply]
Here is a PDF of Canadian cancer statistics for 2008. It also has a special section of Childhood (Ages 0-14) cancers beginning on page 60. You can use that to look for articles on specific cancer types. -- Flyguy649talk15:50, 30 March 2010 (UTC)[reply]
Cancer being cancer, it will probably manifest with similar symptoms as it does in adults; however, in children, some cancers may manifest differently, and there are a whole slew of cancers that occur mostly during childhood. It depends on the cancer. 70.179.127.14 (talk) 20:38, 30 March 2010 (UTC)[reply]
Some cancers, such as leukaemia, have greater prevelence in children and teenagers than other forms (such as carcinomas and melanomas). You may be interested to read this link. Symptoms for cancer will depend highly upon where the cancer is and at what stage it's at. Please seek medical advice if you are concerned about specific, personal health matter. Regards, --—Cyclonenim | Chat 00:59, 31 March 2010 (UTC)[reply]
number of sites
If Iran packed every square inch of land under its control with nuclear sites is there a limit in the amount of electricity it could sell to other countries without fear of invasion if Iran cut them off? 71.100.3.207 (talk) 17:24, 30 March 2010 (UTC)[reply]
You're going to have to clarify what you mean. How would Iran's nuclear sites affect its ability to sell electricity (do you mean nuclear power plants, as opposed to nuclear weapons or nuclear refining sites?), and how would selling electricity have anything to do with fear of invasion? Another country could theoretically become dependent on foreign electrical power, but they usually maintain enough homegrown generation capacity to keep their military functional. —ShadowRanger(talk|stalk)17:43, 30 March 2010 (UTC)[reply]
Lets see... after a certain distance the cost of transmission might become too high for electricity to sell. with that restriction what is the limit of power Iran could reasonably expect to sell, use or store without unreasonable risk of invasion by others to remedy an energy crisis?. 71.100.3.207 (talk) 18:16, 30 March 2010 (UTC)[reply]
The range for transmission would likely not be much of an issue. You do lose power over long distances, but given that there are serious (if unlikely to be fulfilled) proposals to use solar power in the American Southwest to power the whole country, it's clear that the losses aren't enough to make it completely impractical. Answering the question of how much power is "too" much, in the sense of attracting the attention of foreign powers looking to remedy an energy crisis, that's asking us to predict the future and human political psychology all in one. According to this chart of power generation capabilities, the U.S. produces as much power as China, Japan and Russia put together. Iran is producing roughly 10% of what Japan uses, so unless you assume that Iran's neighbors are more likely to launch an invasion than Japan's neighbors, Iran should be able to multiply it's power generation capabilities by 10x without risk. —ShadowRanger(talk|stalk)18:43, 30 March 2010 (UTC)[reply]
Land isn't the limiting factor. I would expect them to run out of nuclear fuel long before they ran out of space. --Tango (talk) 17:50, 30 March 2010 (UTC)[reply]
Breeder reactors aren't magic. They can turn certain thorium isotopes into usable reaction material, and they can use the initial uranium fuel more thoroughly, but they're still limited by the need for fairly uncommon materials. —ShadowRanger(talk|stalk)18:43, 30 March 2010 (UTC)[reply]
The transmission cost for electric power is mainly the one-time investment cost in infrastructure i.e. lines, pylons, switchgear and transformer stations. The selling price for electricity varies with market demand, which varies with season. Availability of cheap and reliable nuclear power will tend to stifle use of other forms of power, particularly if it is perceived as less polluting. There is no economical way to store electric power so it makes more sense to run nuclear stations at partial capacity and adjust output to suit demand. Neighbouring countries that might invade Iran for reasons that have nothing to do with electric power would seem less likely to do so if they depended on Iran for cheap power. But the OP makes a big assumption that the world powers will allow Iran to develop nuclear capabilities unchecked. Cuddlyable3 (talk) 19:17, 30 March 2010 (UTC)[reply]
And the capital costs of such an endeavor are probably beyond them and everyone else. Nuclear power plants cost a non-trivial amount to build and take a long time to pay for themselves. --Mr.98 (talk) 01:01, 1 April 2010 (UTC)[reply]
Some things are clearly beyond anyone's (and the whole world's) capacity. Iran cannot acquire such potential overnight or in a year or in ten years. There's nothing to discuss here. They can build the system organically, one radius at a time, but it's a whole different scenario. NVO (talk) 19:21, 30 March 2010 (UTC)[reply]
Putting it another way... what are the primary limiting factors in the number of sites Iran should build to produce only fuel for the purpose of generating electricity (to sell, use or store - zero degree inductors) to minimize risk of invasion to stop production of WMD? ShadowRange says 10 times its current electricity generating capacity which seems well beyond the amount the world would be comfortable with in absence of direct International supervision.71.100.3.207 (talk) 20:53, 30 March 2010 (UTC)[reply]
Ah, see now you're talking about something different. Your original question, at least as I interpreted it, was what level of power generation would make its neighbors get greedy enough to risk war. That said, the calculus involved in what would trigger international sanctions, or even a U.N. sponsored invasion is at least as complicated; the neighboring states may be more volatile, and therefore unpredictable, but worldwide politics is at least as complicated, and much harder to read. What would it take for Russia or China to not veto a use of force resolution? If that doesn't happen, what would it take for another country to become sufficiently worried as to attack without U.N. approval? Neither question can be easily answered. The Iraq invasion was launched with little evidence; with Iran, we *know* they are further along than Iraq had been in the last 15 years, but the U.S. hasn't invaded. Some reticence after the Iraq debacle is expected, but where the new line is drawn is not clear. —ShadowRanger(talk|stalk)21:00, 30 March 2010 (UTC)[reply]
Misunderstanding can often require that a question be rephrased. To be more precise let me rephrase by asking then is there any correlation between number of sites and Iran's power generation capability versus Iran's WMD capability and if so what number of site would trigger world wide consternation or would that be triggered only by Iran's refusal in absence of limiting the number of sites or WMD correlated activity to guarantee the security of Israel from intent of use it has already expressed? 71.100.3.207 (talk) 01:39, 31 March 2010 (UTC)[reply]
WMD production capabilities aren't directly correlated to power generation. As noted in the article on Breeder reactors, a single breeder reactor can consume its initial fuel (and in fact produce fuel from other elements, such as thorium-232) with far greater efficiency than the typical non-breeder light water reactor, but they also produce nuclear weapon fuel (specifically plutonium) in significant quantities as a consequence of operation. Therefore, the most efficient means of producing power is also the one that gives the international community the most sleepless nights. Failing to reprocess the fuel is a waste, which Iran understandably wants to avoid. Proposals to ship the partially used fuel to Russia for reprocessing have not progressed (last I checked). The point is, trying to infer Iran's nuclear weapons capability from their power production isn't possible unless you know their power generation technique. And it still won't tell you if they have nuclear weapon capabilities, because you can breed plutonium and enrich uranium without hooking it up to the power grid. After all, if the whole base is a secret, it's handy to have it off the main grid, so an in-house generator/breeder is quite handy. —ShadowRanger(talk|stalk)19:26, 31 March 2010 (UTC)[reply]
Thorium reactors produce Uranium-233, which is similar in its properties to plutonium (fissile, high background neutron rate, difficult to handle), but not quite the same thing. Just a small caveat. --Mr.98 (talk) 01:01, 1 April 2010 (UTC)[reply]
I'm not sure there's any way to gauge in real terms your proposal. What you have proposed is not an economic question but a political one. There is unlikely to be any price that Iran can put on 3% enriched uranium that would make it significantly more politically palatable. Or put another way, the reason people don't invade the Netherlands is not because they produce a lot of the world's nuclear fuel. The economic output is simply not all that important in the vast scheme of their economy, or in the world's political balance. If an invasion will happen, it won't be because the economic aspects weren't important enough; if an invasion doesn't happen, it won't be because the economics of uranium was cheap. --Mr.98 (talk) 01:01, 1 April 2010 (UTC)[reply]
calorie consumption
Is there a formula to calculate the number of food calories needed to fulfill both the BMR requirements and number of calories needed to ride a bicycle different distances and at different speeds? 71.100.3.207 (talk) 17:39, 30 March 2010 (UTC)[reply]
There are a million different calorie calculators online, just Google (or Bing) for them. This is one. They're all going to have a very limited amount of accuracy, due to the fairly substantial differences in metabolic rates between people. —ShadowRanger(talk|stalk)17:47, 30 March 2010 (UTC)[reply]
You'd have the same problem with a formula. Unless you spend a lot of time to determine your particular basal metabolic rate, you will just have to settle for an average, which won't be accurate enough to be of much value, unless you just happen to be a throughly average person. See basal_metabolic_rate#BMR_estimation_formulas. Also, the air temperature, humidity, hilliness of the path, clothes you wear, whether it's sunny or cloudy, etc., would all have an effect on the calories you use biking. But, if you want to ignore all that, here's a site which has the formula to calculate calories burnt bicycling: [5]. StuRat (talk) 18:14, 30 March 2010 (UTC)[reply]
Yes, that is more or less the function I'm looking for. However, the exact formulas I'm looking for will show calorie consumption based on: miles and duration, weight of bicycle and rider, bicycle resistance (due to condition or need for repair), wind and hill resistance and urban versus rural traffic conditions. I'm open to all other significant conditions that might effect calorie consumption to a great extent. 71.100.3.207 (talk) 17:45, 31 March 2010 (UTC)[reply]
A mathematical formula with all of those variables (and some you might have neglected such as average altitude, riders clothing, tire pressure etc) would probably be too complicated for the average Joe to use. Also, it would surprise me if such a formula would still be all that accurate, since person A and person B could both do the same task with all of the quantifiable variables the same, but the non-quantifiable variables will give different results. If person A was in great shape, and person B was in poor shape, all else being equal, person B would burn more calories because being in better shape also means using energy more efficiently, but how do you apply a number for that into your equation? Googlemeister (talk) 19:22, 31 March 2010 (UTC)[reply]
Formulas are generally created to replace searching tables of data. Some table lockup routines merely give you the closest answer. For instance a table for age based on height and weight and gender will start by including ages for only the selected gender. Next from the previous list only ages with weights equal or near to the weight entered will be retained, Next only ages with heights that are equal or close will be retained from the previous list. So long as you end up with more than one age you can keep adding variables like hair color to eliminate ages you don't need. In this case we are talking about calories and not age. 71.100.3.207 (talk) 19:24, 2 April 2010 (UTC)[reply]
Presumably a sufficiently precise set of machinery would leave minimal traces. Finishing the job with an automated round of sanding would remove what remains. —ShadowRanger(talk|stalk)21:04, 30 March 2010 (UTC)[reply]
From Google search, most now appear to be made in China. They do not include a description of the manufacturing process. The wood is variously listed as birch, pine, fir, or poplar. Strangely, wood dowels, which are basically the same thing, are sometimes made in the U.S.. A broomstick or dowel could be made on a lathe from a piece square in cross section sawed out of a larger seasoned timber, as suggested by [6]. Rather than placing a dowel in a lathe and spinning it around its axis, it could also be made by feeding it lengthwise through a milling machine in which cutters of half-circle profile form it from above and below. This should leave some longitudinal ridges or grooves, in contrast to lathe manufacture, which should leave annular ridges or grooves. Edison (talk) 21:07, 30 March 2010 (UTC)[reply]
(ec)In my (admittedly limited) experience, they rarely have grooves at all. So they could be using either technique, then sanding it down to the perfectly smooth cylinder. Same results either way. Old school operations presumably use whichever technique they were using originally (since upgrading machinery costs money up front), while modern operations use whichever technique wastes less wood (since wood costs money over time, and a new operation may as well go for the efficient approach). Per Sean's comment below, the rate of mishaps associated with a particular technique may also guide the decision. Lathe clearly indicates that it is used for tool handles of all kinds, so presumably *some* broomsticks have been made using that technique. —ShadowRanger(talk|stalk)21:21, 30 March 2010 (UTC)[reply]
Here's a description of making a dowel with a router. I find it hard to imagine lathing something as long and thin as a broomstick without it flying off, but I guess it's possible. --Sean21:16, 30 March 2010 (UTC)[reply]
Google Book search does not provide a view of the text other than this snippet: Wood working digest, Volume 65 - Page 59, Hitchcock Publishing Company, Wheaton, Ill - Business & Economics - 1963: "DAH, British Guiana Broom Stick Machine We have a prospect for semi-automatic equipment for high speed production (turning and sanding) of broom sticks, ..." So some modern ones were made by turning. In an 1895 book there is a description of a lathe machine with a hollow mandrel and internal curtters which rounded the broomstick as it passed through the machine [7]. As ShadowRangerRIT noted a piece of wood as long and thin as a broomstick would not be easy to turn at high speed. Cutters could spiral around the piece of wood, which might leave spiral groves. Something like a pencil sharpener, but with cutters that make a cylinder rather than a conical point at the end. Sanding could smooth away the ridges. Edison (talk) 22:04, 30 March 2010 (UTC)[reply]
My broomstick does not appear to have been sanded (ie grain is not filled with sawdust) neither does it it have longitudinal grooves: so hows it done?--79.76.190.44 (talk) 22:33, 30 March 2010 (UTC)[reply]
Sawdust can be removed by a number of methods, from simply falling out during the course of transport to active washing to remove the dust. It's all speculation though; unless you visit (or contact) one of the Chinese broom manufacturers you probably won't know for sure. —ShadowRanger(talk|stalk)19:13, 31 March 2010 (UTC)[reply]
March 31
Dying from "old age"/"natural causes"
This usually means heart problems, a stroke, cancer or something like that. Supposing you remain perfectly healthy and fit your entire life, and you never get sick or have any health problems, would you just die from nothing at a certain age or would you be able to live indefinitely?--92.251.191.108 (talk) 00:32, 31 March 2010 (UTC)[reply]
No you would not die of "nothing". Imagine an old car wearing out-- eventually something fails that is fatal. "Natural causes" is simply a legal category that means "not trauma, not at the hands of someone else, not drowning or lightning or asphyxia, or poisoning...". In some other contexts, "natural causes" = "we dont know the cause of death but we didnt think anyone killed him and doubt that an autopsy would show anything interesting". alteripse (talk) 00:40, 31 March 2010 (UTC)[reply]
Even if you lived your entire life without a single illness, disease or trauma, there is still evidence to suggest aging still occurs and that aging as an effect on your bodies ability to repair itself. Therefore, even after living a perfectly healthy life, your body will ultimately become unable to repair or replace its damaged cells sufficiently to continue living.
The way I see it, the human body is like a pair of pants. When it gets worn-out, it will start to tear - say, for instance, at the knees. And when the pants themselves are strong, you can sew it up. But the more you wear said pants, the harder it will be to patch a hole, because the fabric just isn't strong enough to hold string and will unravel when pulled on. You can put as big a patch on as you want, but it's not going to be holding onto anything if the rest of the pants are the consistency of paper. Now in the world of trousers, you could (in theory) have pants made entirely of patches, but that doesn't really work for human beings... ZigSaw01:47, 31 March 2010 (UTC)[reply]
Interestingly enough that is one of the major interests in stem cell research. In theory you could patch everything that wears out instead of reproducing but unless you hate sex you will probably not opt for stem cell renewal as reproduction versus reproducing in the old fashioned way. 71.100.3.207 (talk) 02:16, 31 March 2010 (UTC)[reply]
Convieniently, we have an article on aging, and a more biologically oriented article Senescence. These both have a lot of other good articles linked, depending on how interested you are. There are some species that effectively don't age (like Hydras, maybe), but humans and almost everything else cannot live indefinately. See Biological Immortality and Maximum life span. Buddy431 (talk) 04:45, 31 March 2010 (UTC)[reply]
Well, there are some people where it does seem like they "died from nothing". That is, their heart just stops beating. There may be an underlying cause other than old age, but it isn't known. StuRat (talk) 04:48, 31 March 2010 (UTC)[reply]
This is supposed to be a maelstrom. That is to say, a huge destructive whirlpool of chaos and death. I look at this picture and I see calm water. Am I missing something really important, or did someone post a picture of something entirely not a maelstrom? ZigSaw01:44, 31 March 2010 (UTC)[reply]
"The Moskstraumen (popularly known as the Maelstrom) is a system of tidal eddies and whirlpools, one of the strongest in the world, that forms in a strait..." (The sky view is more illustrative of the point.)71.100.3.207 (talk) 02:19, 31 March 2010 (UTC)[reply]
In addition, our article doesn't say anything about 'chaos' or 'death'. It's better to find out what is actually meant by a maelstrom (or whatever you're thinking of) rather then rely on any pre-existing ideas you may have Nil Einne (talk) 06:22, 31 March 2010 (UTC)[reply]
I have to say — I too was surprised how greatly exaggerated was the power of the great Maelstrom; all I knew of it was from reading Twenty Thousand Leagues Under the Sea, and while I had assumed that the book exaggerated a bit, I didn't know until giving my answer earlier how greatly it exaggerated. Nyttend (talk) 12:14, 31 March 2010 (UTC)[reply]
There is a powerful whirlpool in the Fraser Canyon in BC that is more dramatic, and not something you'd want to get caught in. It's at the Hell's Gate feature that was created when blasting dropped a huge amount of rock into the river, making it much narrower and deeper than it was naturally. I couldn't find a great picture of that whirlpool, but this gives you a good idea of the tumult involved. Vranak (talk) 20:03, 31 March 2010 (UTC)[reply]
Light Cones in GR
I've read that light cones in GR are tilted, however, I thought that light cones were traced out by light emitted from a spacetime point in time. So, shouldn't a gravitational field not only tilt a light cone, but curve it? Thanks:) 66.202.66.78 (talk) 04:34, 31 March 2010 (UTC)[reply]
Yes, in general the path of a light ray in a gravitational field is curved. When articles talk about (and sometimes illustrate) light cones being tilted, they use "light cone" as a short-hand for the "direction" in spacetime in which a light ray would initially set off if emitted from a given point. You can think of these light cones as first order approximations or linearisations - they tell you about the direction of the tangent to a light ray passing through a given point in spacetime. It is analogous to illustrating a vector field by drawing lots of little arrows representing the direction and magnitude of the field at various points. Gandalf61 (talk) 08:47, 31 March 2010 (UTC)[reply]
Also, it's not really that the light cones tilt. When that seems to happen in illustrations, it just means that the coordinate system chosen happens to be skewed with respect to the real structure of spacetime. It is a true effect of GR that one often cannot choose a coordinate system that aligns with the light cones everythere. But at any particular event you're interested in, you can get the light cone to stand nicely upright, simply by choosing coordinates appropriately. –Henning Makholm (talk) 08:54, 31 March 2010 (UTC)[reply]
Thank you:) I have another, related question: suppose we are in 2+1 dimensions. If this space is flat and we emit out a pulse of light(light going in all directions out from us) it will make a sequence of circles in time. Supposing we are near a massive object, how do those circles deform? To be honest, I don't really know much about GR and am not sure how much sense this question makes. Thank you for any help; if I am going terribly wrong in my thinking, could you point me in the right direction:) Am I making the mistake of thinking in terms of a given set of coords? If so, could you consider how the light propogates in each coordinates, then consider "deformations" in terms of how the light pulse transforms between them? Again, I may be talking nonsense, most of what I work on is far removed from diffgeo and GR 66.202.66.78 (talk) 09:05, 31 March 2010 (UTC)[reply]
My question might be better summed up by asking if each for each null geodesic will be there a light ray in the pulse that will follow it? Essentially, do the circles of light map out these paths? If so, is it only in an open set around the event that the pulse is emitted from? Thanks:) Sorry if I am being confusing/senseless. 66.202.66.78 (talk) 09:37, 31 March 2010 (UTC)[reply]
Not to yammer on; the more I think about it, I'm guessing that the circles would trace out something like the field lines for the electric fields, and that this would end up being the same as the vector field above in terms of information about the gravitational field in the area. I'll stop posting now:) 66.202.66.78 (talk) 09:57, 31 March 2010 (UTC)[reply]
Yes, each null geodetic represents a possible light ray. Assuming, of course, that it does not go through a region of spacetime filled with opaque matter, in which case you need to hypothesize some non-interacting massless particle to travel along the geodesic instead. Or just stipulate that null geodesics are much more interesting than something as annoyingly material as light. :-)
A couple of possible obstacles here. First, an interesting question whose answer I don't know: Does it make sense to do GR in 2+1 spacetime? Certainly one can write down the tensor equations in 2+1 dimensions and look for solutions, but I suspect you will not get an inverse-square law for gravity out of it. The usual simplification here is to say that we're really working in 3+1 dimensions but assume enough symmetry of the situation that we can learn something by studying what happens within a 2+1 (or 1+1) dimensional symmetry surface of the entire spacetime.
Second: Yes, I think you're possibly by assuming that you have a nice enough coordinate system at your disposal. What you do in SR corresponds to intersecting the light front with a succession of spacelike Euclidean hypersurfaces and look at the shape this intersection makes in each "instant" surface. As long as all of your surfaces are parallel, you'll get a nice and meaningful picture that way. However, in GR spacetime does not foliate nicely into a succession of spacelike cuts. Once you choose a coordinate system, you can of course use it to slice up spacetime, but it will be difficult to distinguish whether the answers you get tell you more about your particular choice of coordinates than about anything physically significant. –Henning Makholm (talk) 18:11, 31 March 2010 (UTC)[reply]
You can do GR in 2+1 dimensions, but it's weird. In n spacetime dimensions, the stress-energy tensor has n (n + 1) / 2 components and the gravitational field (Riemann curvature) has n² (n² − 1) / 12 components. The GR field equation equates n (n + 1) / 2 components of the gravitational field to the components of the stress-energy tensor. In 3+1 dimensions, that's 10 of 20, and the other 10 describe the free field (the part that reaches out into the vacuum). In 2+1 dimensions, that's 6 of 6, so there's no free field and no gravitational attraction in the usual sense. There is still a gravitational interaction, but it's purely topological, depending only on how the worldlines are braided together in spacetime. -- BenRG (talk) 01:47, 2 April 2010 (UTC)[reply]
The current bit at In The News says that the collisions between two 3.5 TeV proton beams in the Large Hadron Collider are a world's record for the highest energy man-made particle collisions. Seeing that Electronvolt says one 100W lightbulb takes 8000 TeV to work for one second, why is it amazing that these beams have such a comparatively small amount of energy? Is it simply that protons are very small and thus don't need much energy to go extremely fast, or am I missing something else? Nyttend (talk) 12:49, 31 March 2010 (UTC)[reply]
It's a matter of scale (and comparing a composite total to an individual value). A lightbulb may use 8000 tera-eV, but that translates to around 8 trillion electrons each at 110 volts. Each individual electron, then, is at about 110 eV (assuming US mains voltage) The LHC is colliding protons that are individually at several TeV. — Lomn13:08, 31 March 2010 (UTC)[reply]
Could I use Galileo and GPS at the same time? Could I use Galileo with normal (Same than I use now) GPS-locator? And, I know Galileo be finished (Maybe) at 2014. Aku506 (talk) 12:55, 31 March 2010 (UTC)[reply]
Galileo and GPS use different frequencies. Asking if you can use both at the same time is the same as asking if you can use two cell phones at the same time. There is no problem. As for using Galileo with a GPS unit, the GPS unit must have Galileo compatibility - which some do, but it is clearly noted on the unit. -- kainaw™13:11, 31 March 2010 (UTC)[reply]
Yes, it would be possible to combine Galileo and GPS, just as any two navigational aids can be combined to improve their total performance. However, Galileo isn't the same protocol as GPS. While I'm sure plenty of dual-system receivers will be manufactured, an arbitrary present-day GPS receiver won't use Galileo, just as it won't use digital TV, AM radio, or WiFi. Note that while Galileo claims to have better resolution than GPS, it also has (just as GPS does) a rough-accuracy basic version and a higher-accuracy military (and for Galileo, paid commercial) version. It's not clear that the "better accuracy" claim in our article is comparing real-world basic user performance. We also have an article on GNSS Augmentation (with GNSS being Global Navigation Satellite System, the general term for this sort of thing). — Lomn13:21, 31 March 2010 (UTC)[reply]
LEFT OVER ONIONS ARE POISONOUS
someone sent this info in an email...would this be correct?
Ed says that when food poisoning is reported, the first thing the officials look for is when the 'victim' last ate ONIONS and where those onions came from (in the potato salad?). Ed says it's not the mayonnaise (as long as it's not homemade Mayo) that spoils in the outdoors. It's probably the onions, and if not the onions, it's the POTATOES.
Extended content
I have used an onion which has been left in the fridge, and sometimes I don't use a whole one at one time, so save the other half for later.
Now with this info, I have changed my mind....will buy smaller onions in the future.
I had the wonderful privilege of touring Mullins Food Products, Makers of mayonnaise. Mullins is huge, and is owned by 11 brothers and sisters in the Mullins family. My friend, Jeanne, is the CEO.
Questions about food poisoning came up, and I wanted to share what I learned from a chemist.
The guy who gave us our tour is named Ed. He's one of the brothers Ed is a chemistry expert and is involved in developing most of the sauce formula. He's even developed sauce formula for McDonald's.
Keep in mind that Ed is a food chemistry whiz. During the tour, someone asked if we really needed to worry about mayonnaise. People are always worried that mayonnaise will spoil. Ed's answer will surprise you. Ed said that all commercially- made Mayo is completely safe.
"It doesn't even have to be refrigerated. No harm in refrigerating it, but it's not really necessary." He explained that the pH in mayonnaise is set at a point that bacteria could not survive in that environment. He then talked about the quaint essential picnic, with the bowl of potato salad sitting on the table and how everyone blames the mayonnaise when someone gets sick.
Ed says that when food poisoning is reported, the first thing the officials look for is when the 'victim' last ate ONIONS and where those onions came from (in the potato salad?). Ed says it's not the mayonnaise (as long as it's not homemade Mayo) that spoils in the outdoors. It's probably the onions, and if not the onions, it's the POTATOES.
He explained, onions are a huge magnet for bacteria, especially uncooked onions. You should never plan to keep a portion of a sliced onion.. He says it's not even safe if you put it in a zip-lock bag and put it in your refrigerator.
It's already contaminated enough just by being cut open and out for a bit, that it can be a danger to you (and doubly watch out for those onions you put in your hotdogs at the baseball park!)
Ed says if you take the leftover onion and cook it like crazy you'll probably be okay, but if you slice that left-over onion and put on your sandwich, you're asking for trouble. Both the onions and the moist potato in a potato salad, will attract and grow bacteria faster than any commercial mayonnaise will even begin to break down.
So, how's that for news? Take it for what you will. I (the author) am going to be very careful about my onions from now on. For some reason, I see a lot of credibility coming from a chemist and a company that produces millions of pounds of mayonnaise every year.'
Also, dogs should never eat onions. Their stomachs cannot metabolize onions .Please remember it is dangerous to cut onions and try to use it to cook the next day, it becomes highly poisonous for even a single night and creates Toxic bacteria which may cause Adverse Stomach infections because of excess Bile secretions and even Food poisoning.
I've summarized the relevant question while collapsing the long email, as I found the distinction between you and the email you're quoting confusing. Anyway, as with most circulating emails, Snopes has an entry. Some mayo inhibits bacterial growth, and certainly many places around the world get by on far less refrigeration than the US. Onions, though, aren't attested as some sort of food poisoning magnet. — Lomn15:12, 31 March 2010 (UTC)[reply]
Dandelion bulbs look a lot like onions and are poisonous. A few people a year get poisoned by making that mistake. Onions themselves aren't known for food poisoning to the best of my knowledge. --Tango (talk) 15:45, 31 March 2010 (UTC)[reply]
I got the first two letters right! Yes, I meant daffodil. I had an image of a daffodil in my mind, I just attached the wrong word to it. Thanks! --Tango (talk) 21:22, 31 March 2010 (UTC)[reply]
I don't think the flesh of onions contracts bacteria any quicker than a hard-boiled egg would. And commercial mayonnaise has preservatives in it. --Cheminterest (talk) 21:02, 31 March 2010 (UTC)[reply]
I often make spaghetti for myself for lunch, and I'll toss little bits of onions into the sauce; I take a few weeks per onion, and I don't do anything more than warm the sauce, and I've never had trouble. Of course, anecdotes don't prove it, but they prove that you won't always be poisoned. Nyttend (talk) 02:11, 1 April 2010 (UTC)[reply]
melting ice
Which would melt faster, a 4' deep swimming pool that is filled with a solid block of ice, or an 8' deep swimming pool (same area as pool A) which has a 4' thick solid block of ice floating on 4' of near freezing water if they are in equal conditions for sunlight temperature etc...? I am trying to decide if a pond that freezes in winter will thaw faster if it freezes solid then a deeper pond that forms a thick layer of ice on top, all else being equal. Googlemeister (talk) 19:27, 31 March 2010 (UTC)[reply]
This is not a definitive answer, but I believe the answer is "it depends". If the ground underneath the ice/water is warmer or colder than freezing, you'd get a different result. If the water is insulating the ice from the warmer ground (or warming ground, if the heat is being conveyed from the surface via conduction), then presumably the added bulk of cold water that must be warmed would delay melting. For a pond, the water underneath is presumably flowing to a greater or lesser extent, which would probably aid the melting process by eroding the ice from underneath. —ShadowRanger(talk|stalk)19:43, 31 March 2010 (UTC)[reply]
Yes. But given that (to my knowledge) production of anti-matter creates a matching amount of matter (since matter can be neither created nor destroyed in the grand scheme of things), you'd be producing a black hole's worth of matter in order to annihilate the original black hole. Unless you spread it out really thinly, you'd have another black hole in short order. —ShadowRanger(talk|stalk)19:38, 31 March 2010 (UTC)[reply]
Antimatter doesn't have "antimass", and so there's no expectation I can see that you produce hydrogen for a given quantity of antihydrogen. For the specific question, though, the answer appears to be "no". A matter-antimatter reaction converts the rest mass into kinetic energy. However, this is where mass–energy equivalence kicks in. Per the nature of black holes, the resultant energy can't escape. As such, the effective mass can't change. See, from m-e equivalence, "if a body gives off the energy L in the form of radiation, its mass diminishes by L/c2". Note that this is consistent with the nature of an event horizon -- once the antimatter passes within the black hole's event horizon, what happens to it is unknowable and irrelevant. — Lomn19:55, 31 March 2010 (UTC)[reply]
Hmm... How does that work with the description of Hawking Radiation then? My understanding is that the universe regularly creates and annihilates matter and anti-matter "virtual particles". When this happens at the edge of a black hole, the anti-particle gets sucked in to the black hole, while the paired particle escapes as Hawking Radiation. To satisfy conservation of energy/mass, the anti-particle annihilates it's equivalent particle in the black hole, thereby reducing the black hole's mass by the same amount as that of the particle that escaped. I'm not a physicist, so perhaps the distinction between virtual antimatter and real antimatter is messing me up, but wouldn't supplying an enormous amount of antimatter produce a similar reduction in mass? —ShadowRanger(talk|stalk)21:12, 31 March 2010 (UTC)[reply]
I don't really understand it myself, but virtual particles behave differently to real ones. I think that is why your argument doesn't hold. I'm not sure what the correct argument is, though. --Tango (talk) 21:26, 31 March 2010 (UTC)[reply]
Re: Hawking radiation: the virtual particle thing is that energy is transformed into particles, which typically annihilate back into energy. For HR, one of the two particles (and it doesn't matter whether it's the electron or the positron) escapes, reducing the mass of the black hole by whatever the correct conversion amount is. What happens to the other particle (again, positron or electron) doesn't matter, as it remains hidden behind the event horizon. The reduction in mass results from the loss of a particle but isn't contingent on whether that particle is matter or antimatter. — Lomn21:34, 31 March 2010 (UTC)[reply]
By the way, according to Antimatter#Fuel, "Known methods of producing antimatter from energy also produce an equal amount of normal matter, so the theoretical limit is that half of the input energy is converted to antimatter." So at least according to the current understanding, creating enough antimatter to match the mass of a black hole would intrinsically produce enough matter to make a black hole as well. And use a god-awful amount of energy to boot. —ShadowRanger(talk|stalk)23:44, 31 March 2010 (UTC)[reply]
Antimatter has positive mass, so the black hole would get even heavier (and bigger) instead of lighter (and smaller). Dauto (talk) 20:42, 31 March 2010 (UTC)[reply]
No, because energy has mass - see mass-energy equivalence. The no hair theorem says that a black hole made from matter is indistinguishable from one made from the same mass of antimatter with the same charge and angular momentum - or, indeed, from a black hole made from 50% matter and 50% antimatter. Gandalf61 (talk) 22:44, 31 March 2010 (UTC)[reply]
I'm more than willing to admit to a weakness in my physics on this matter. Although I'm frankly perplexed by the concept of a 50/50 matter/anti-matter black hole. Presumably compressed that densely, a matter/anti-matter reaction would be inevitable? It would produce energy, which presumably could not escape (assuming the mix wasn't actually 50/50 and what remained could maintain the black hole's integrity), but I'm frankly perplexed as to what would happen with that energy. I realize matter and anti-matter both have mass, but I thought the annihilation reaction between them eliminated the mass entirely, in favor of energy (though what form that energy takes is beyond me). —ShadowRanger(talk|stalk)22:54, 31 March 2010 (UTC)[reply]
Reading further, the energy apparently takes the form of either high energy photons (which have no rest mass) or a new particle-antiparticle pair. So I suppose a matter/antimatter composite black hole would be possible if all particle-antiparticle reactions produced new pairs of particle-antiparticle. It seems like the collision of a matter black hole and an antimatter black hole (of sufficiently similar mass), or an exactly 50/50 black hole might have a different result though. If the annihilation is sufficiently energetic, and the vast majority of the colliding holes is transformed into gamma rays, might it be possible that the instantaneous reduction in mass would allow for the whole thing to self-destruct, leading to the release of some matter and/or antimatter propelled by the energy of the gamma rays? This doesn't work if the gamma rays are considered to have mass for the purposes of holding the black hole together, but I'm stuck in a conundrum here. The gamma rays only have mass when moving, so if they are moving they might produce the mass necessary to keep the black hole intact. But if the black hole remains intact, the gamma rays can't escape, i.e. they cannot move, which means they have no mass and the black hole wouldn't have the necessary mass to stay together. Gah! Driving myself insane. I'm either one step away from figuring out the cause of the Big Bang, or (more likely) I really suck at advanced theoretical physics. —ShadowRanger(talk|stalk)23:13, 31 March 2010 (UTC)[reply]
The products of a matter / antimatter reaction have exactly the same effective mass as the reactants. That's true even if you products are gamma rays. Gamma rays have no rest mass but their relativistic mass matches exactly what you'd predict from E = mc2, and that's what counts for determining gravity (more or less). Mass is energy, energy is mass. The annihilation of matter and anti-matter has no effect on the size of the black hole. The advantage of gamma rays is that they are easily converted to other forms of energy, whereas particles are usually hard to convert to useful energy. So in principle, matter / anti-matter reaction can produce lots of useful energy, but we don't know any easy sources of anti-matter. Dragons flight (talk) 00:02, 1 April 2010 (UTC)[reply]
Like most people here (evidently), I find the Hawking radiation thing perplexing. If you smack a kilogram of hydrogen into a kilo of antihydrogen, you certainly get 2kg times the speed of light squared joules of energy. If that happens inside the event horizon of a black hole, the energy can't escape so as far as an outside observer is concerned - it's overall mass/energy (and therefore it's gravity and general blackholeyness) doesn't change. The explanation of how Hawkins radiation works isn't so simple. Since the virtual particle and virtual anti-particle both have positive mass, it takes energy to create them - and energy is produced when they are destroyed. The "popular science" explanation is that this energy is "temporarily borrowed" from the universe and has to be "repaid" shortly afterwards. That is (to say the least) an unsatisfying explanation - but it's the only one I've heard that I can understand! Anyway - following that rather tricky explanation to it's conclusion - when one of the virtual particles vanishes into the black hole, never to return and the other one doesn't - the annihilation of the original pair of virtual particles doesn't happen. So the energy that was 'borrowed' from "the universe" gets "paid back" inside the black hole...which loses energy as a result. This is a really unsatisfying answer...I wish we had a better one that was reasonably comprehensible. However, what we CAN say for sure (and understand easily) is that it's not that the antiparticle has negative mass or turns into negative energy somehow...that's definitely not what happens because we can easily imagine that the 'normal' matter particle of the pair falls into the hole about 50% of the time and if antiparticles shrank the black hole due to "negative mass", it would all be cancelled out by the normal particles falling in the other half of the time. So the average black holes wouldn't evaporate. Hence, we know the explanation is more tricky than that...even if those here on this reference desk can't explain why that is. SteveBaker (talk) 00:04, 1 April 2010 (UTC)[reply]
I don't purport to understand Hawking radiation either, but: Instead of the virtual-twins-separated-at-birth story popularized by Hawking himself, we could consider a story where a particle inside the black hole tunnels across the horizon. The event horizon is (by one definition) an "outward-moving light front that doesn't get anywhere", so all one needs to escape is to overtake it -- and moving faster than light is one of the things a tunnelling/virtual particle can do, at the cost of having imaginary momentum and a chance of pulling it off that decreases exponentially. This is not extraordinarily fanciful; in quantum electrodynamics you routinely have to consider contributions from worlds where a particle moves from event A to event B faster than c.
It feels plausible to me that tunnelling could actually be mathematically equivalent to the standard story. It would also "explain", heuristically, why a smaller black hole emits more Hawking radiation: As the distance to tunnel through decreases, the probability of success increases exponentially. –Henning Makholm (talk) 01:58, 1 April 2010 (UTC)[reply]
Let me try to clear up a couple of common misconceptions that may be hampering people's understanding.
First it is important to understand the relationship between energy of a particle , its momentum and its rest mass ---- That relationship is given by the equation . If the particle is at rest than that equation reduces to the notorious . Note that the last equation is not telling us that a particle of mass can be converted to energy . It is simply telling us that a particle of mass has energy . The energy is already there but not in a form that can be detected by more conventional means. When a particle meets its antiparticle and they annihilate each other, photons get released. The energy of the photons will be the same energy that was already there in the form of mass-energy. If that happens inside a black hole, the energy of the photons will have as much of a gravitational effect as the particles that were there before them and the total mass of the black hole won't change. Simply, mass creates gravity because it contains energy and all forms of energy do create gravity.
Second it is important to understand what makes virtual particles different from real particles and how does that difference allow for Hawking radiation to leave the black hole ---- Virtual particles do not have to obey the relationship . That means that, unlike real particles, they are allowed for a brief moment to exist and have any energy, unrestricted by the need to satisfy a relationship with a rest mass and a momentum. They are even allowed to have negative energy. Oddly, the technical speak for that is to say that real particles must be on their mass shell while virtual particles may be off shell. Because of that, pairs of particles and antiparticles pop out of the vacuum all the time, one with a positive energy and the other one with a negative energy so that energy conservation is preserved ---- energy conservation is always preserved. There is no need to borrow any energy. The borrowing of energy is one of the worst analogies I've ever seen. It produces much more confusion than any benefit that it might have. These virtual particles live only long enough to ask themselves one single question : "Am I on my mass shell?" For at least one of them the answer is always "no" because it has negative energy and therefore it must vanish back into the oblivion whence it came from dragging with it the other particle. But near a black hole things are different. If the particle with negative energy falls in the black hole than it doesn't need to answer that question anymore because it's already gone to a different kind of oblivion. It has been absorbed by the black hole and since it had negative energy, the black hole becomes a little lighter. The other particle, the one with positive energy, might also fall in the black hole, in which case nothing was accomplished. But if it happens to be on shell and if it happens to have enough energy to climb out of the hole than it may leave the black hole as Hawking radiation. Dauto (talk) 06:03, 1 April 2010 (UTC)[reply]
I would like to add that individual virtual particles are - by definition - impossible to observe. If you ever observe a particle directly, then by definition it must be real. The notion of virtual particles as individual entities is really part of the interpretation of quantum mechanics, rather than part of physics itself. They emerge as intermediate states in Feynman diagrams. Feynman diagrams describe a process of doing mathematical calculations such that one can predict the probability of reaching a specific quantum mechanical end state given a specific initial state. Those probabilities are real and observable. The calculations however require one to do integrations across intermediate states that are assumed to contain unobservable (i.e. virtual) particles which can have arbitrary energy not conforming to the particles' normal momentum-energy relationship. In fact, to predict any physically observable quantity, one doesn't just consider a single possible virtual particle, but one must integrate over all conceivable particle energies and momenta (including negative energy states). Whether one considers a virtual particle to be a real entity or merely some form of mathematical convenience for accounting purposes is a really a philosophical question rather than a physics question. Many physicists do view them as real entities, but personally, I tend to see their strange properties as too bizarre. Hence I prefer to think of them as a mathematical contrivance. In the case of Hawking radiation, the mathematics of virtual particles falling into the black hole can be shown to be the same as the mathematics of real particles quantum tunneling out of the black hole. So again, whether you choose to believe in virtual particles as real and individualized things boils down to a question of philosophical interpretation, rather than one that can be settled by any direct measurement. Dragons flight (talk) 08:03, 1 April 2010 (UTC)[reply]
I find it interesting that you dislike virtual particles because you "tend to see their strange properties as too bizarre" and than you suggest as an alternative 'real' particles tunelling out of the black whole. I guess particles moving faster than the speed of light (or backwards in time if you will) are not too bizarre. What I tell my students about quantum weirdness is "Embrace the weirdness instead on trying to 'understand' it of explain it away as a mathematical artifact. The same thing applies here. Virtual particles have real observable effects and should be considered as real as anything else. There is a long history in physics of second guesses about how real our theoretical constructs really are. Electromagnetic fields, electromagnetic potentials Wave function, quarks, quasi-particles, etc, all were at some point considred by some as simple mathematical artifacts. Ultimately I think this is a pointless philosophical discussion. I say: "If it works, than it's real" Dauto (talk) 17:46, 1 April 2010 (UTC)[reply]
Yep, I do think superluminal quantum tunneling seems more natural / intuitive than talking about one half of a virtual particle pair falling into a black hole and contributing negative mass. Is that so crazy? ;-) Newtonian gravity and the Bohr atom are also useful theoretical constructs, but if you interpret them literally you end up with incorrect models of reality. Eventually, science led us to better theories and new perceptions. Personally, I suspect that the conception of virtual particles as real individualized things will also end up getting displaced by a better understanding eventually. You're right that the math works of course, and that's good enough for now, but a theory that requires the existence of ephemeral, untouchable particles with strange variations in energy and momentum is enough for me to believe that something is missing in our understanding. Given that interpreting the mathematics is more philosophy than science, I choose to adhere to a point of view that feels more natural until there is evidence to the contrary. Dragons flight (talk) 23:36, 1 April 2010 (UTC)[reply]
I think they are both bizarre but I also think that doesn't matter. Being bizarre is not a reason to drop a theory (or an interpretation of a theory). What matters is whether it is consistent and how well it works to explain the experiments, observations and, lets not forget, the gedanken experiments. Dauto (talk) 02:09, 2 April 2010 (UTC)[reply]
The thing that exists is the math describing how these processes work. Virtual particles and quantum tunneling rates are both approximations that hold in many circumstances, but neither is "really" going on. As far as we know, what is happening is that a (wave)function over a quantum mechanical Hilbert space is changing over time in respect to certain rules. Virtual particles are just a way of putting some of these rules into easy boxes for simplified human calculation and tunneling is more the wavefunction slowly leaking out of a black holes than individual particles escaping one at a time. 99.238.19.115 (talk) 03:52, 2 April 2010 (UTC)[reply]
The source is the difference in temperature between the water and the (high altitude) air. Warmer water makes for a more powerful hurricane. A hurricane is basically a heat engine. Water can store a lot more heat than land, that's the main reason hurricanes die when they go over land. Ariel. (talk) 01:01, 1 April 2010 (UTC)[reply]
April 1
TEMPERATURE TIME GRAPH
i got this question in my exams . may you answer and explain
Four students set up the experiment on plotting the temperature time graph of hot body as it cools to room temperature. They all perform the experiment using identical apparatus, under identical conditions and plot their graphs on similar graph papers using similar scales. The teacher asks them to observe their graphs carefully and list down the(approximate) values of the angles, made with time axis, by the tangents to their graphs, at the beginning and towards the end of their observations. The values, noted by them, were as follows:
i dont know how to put the image.Well TEMP is on y axis and TIME is on x axis
Student A : 60o; 30o
Student B : 65o; 22o
Student C : 75o; 14o
Student D : 85o; 5o
Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. ~ Amory (u • t • c)06:26, 1 April 2010 (UTC)[reply]
Another relevant article is Newton's law of cooling. Heat transfer in these sorts of problems describes an exponential decay of the temperature. I'm not sure what to make of the measurement of the angles, because it depends on the temperatures, time scales, and the amount to which steady-state has been achieved. All we can say is that the measurements are all plausible. It's possible that you have graphs as well as angle measurements; if so, try to see which graphs look like an exponential temperature decay toward room temperature. For comparison, here's the omnipresent Science Reference Desk Reference Coffee/Milk Cooling Curve Simulator - the red curve shows a standard cooling curve, which should look like your hypothetical measurements. I'm going to go out on a limb and say that every one of those angle-measurements could be valid, given a particular decay time-constant. If we had enough information to calculate the decay constant, we could narrow down some more. As an engineer, I'm going out on a limb to say "steady state for decay is approximately 5 time constants"; and then, "14° is approximately 1/5 of 75°", and then, "x ≈ tan x", so let's go with C; but to be honest, unless you can personally justify any of those approximations in your specific case, I'd recommend asking your teacher for clarification. Nimur (talk) 14:03, 1 April 2010 (UTC)[reply]
Myownid420 says that he has images, but doesn't know how to put it up here. If that's the case, then the question is to basically estimate a couple of angles on a picture. We don't have the picture, so we can't possibly estimate the angles. Buddy431 (talk) 14:53, 1 April 2010 (UTC)[reply]
Nimur, all those are not plausible temperatures. If you warm up some water on your kitchen stove to 85C and then set it on the counter, it will never cool to 5C unless you keep your kitchen inside a refrigerator. Googlemeister (talk) 15:31, 1 April 2010 (UTC)[reply]
Those are not temperatures, they are angles of the tangent line of the temperature-vs.-time curve. It's slightly confusing because both angle and temperature are measured in degrees ° - but they are different things entirely. I suspect this might have been an intentional trick question. Nimur (talk) 15:45, 1 April 2010 (UTC)[reply]
Note that you can't calculate this angle, not even in principle, unless you have the graph paper. Temperature and time have different units, and only if you kow the scale from the graph paper can you convert both time and temperature into distance, after which you can calculate the angle. Might the questioner want you to use a heuristic such as "the value closest to the average of all values is most likely to be correct"? 83.134.157.144 (talk) 16:43, 1 April 2010 (UTC)[reply]
Right, but we can calculate the ratio between initial- and final- values, if we make suitable assumptions as I outlined above. We can do this because exponential functions have special properties - the tangent line is defined by the derivative, which is linearly proportional to the function value. Nimur (talk) 19:49, 1 April 2010 (UTC)[reply]
On the basis of his second reading, you'd suppose that 'D' was most likely to be correct. The temperature should be asymptoting down to room temperature - and therefore the shallower the angle for that second number, the better. But that's guessing - we know that they all stopped the experiment before the liquid reached room temperature because (strictly) it takes an infinite amount of time to do that. If they stopped it sooner then maybe 'C' got the measurements most accurately. But if some of them are sloppy experimenters (which they evidently are) then maybe the most average value is the best to take - but that's 'C' if you take the best average of the first number and 'B' if you take the best for the second. If they used 'log' graph paper then you'd expect the two angles to be the same - so now 'A' got it closest to being right. So I can make the case for any of the four experimenters to have been "the best". This is a REALLY bogus question. SteveBaker (talk) 20:01, 1 April 2010 (UTC)[reply]
I suppose so. If we could see that the graph was plotted on log graphpaper - then the answer is very different than if it's on linear graphpaper. Perhaps that's what the original questioner was trying to suggest. If it's log graph paper then I think we know that 'A' is the best answer because the slope of the graph is predicted by Newton's law of cooling and it's the same at the start and end of the experiment. Hence whichever person had first and last numbers sufficiently close together has the most accurate data. SteveBaker (talk) 20:34, 1 April 2010 (UTC)[reply]
well my question is not home work. i am a tenth standard student and my exams are over this question was in my last exam ie of practical science skills, sorry i still cant upload the photo.
let me tell you how to make it
1. make x axis and y axis of definite length
2. mar a point O(x,y) such that x=y
3. from point O draw an arc of radius less than x (slightly)
4. let this arc touch the line (0,y) and (x,o)
If, in a reference frame O, two clocks seperated by a distance L are synchronized, then the two clocks may not be synchronized in another reference frame moving at a relative speed v. What, then, is the difference in timing between the two clocks as observed in the moving reference frame? Using length contraction and such (as per http://galileo.phys.virginia.edu/classes/252/synchronizing.html), I get Δt = vL/c^2. However, using Lorentz transformations gives me a different result: t' = γ(t - vx/c^2), t'2 - t'1 = γ(t2 - vx2/c^2) - γ(t1 - vx1/c^2) = γ(t2 - t1 +v/c^2(x1 - x2) = γvL/c^2. Why the discrepancy? 173.179.59.66 (talk) 05:49, 1 April 2010 (UTC)[reply]
Does the material the drinkware is made of (metal, glass, paper, plastic) really have any effect on the quality and taste of wine? Or is it just snobery?
(im looking for science based answers) —Preceding unsigned comment added by 71.98.64.15 (talk) 06:21, 1 April 2010 (UTC)[reply]
Lots of materials can react with the wine to put off-tastes into it. Plastics, for example, can attract hydrophobic compounds (basically fats) which don't really ever wash off completely (if you have ever microwaved tomato sauce in a plastic dish, you know what I mean). This could redissolve in small amounts into the wine. Metals can be reactive to acids in the wine, while the wine could also dissolve flavors out of paper. Glass is the only truly non-reactive material out of which to drink. The shape of the glass also matters; remember that most of what you associate with taste is actually smell, and depending on the type of drink, the shape of the glass can be used to direct smells either to your nose or away from it. Different shaped glasses also allow differing surface area on the liquid, affecting how the wine will "breathe" (i.e. oxidize in air) which explains the shape of different stemware. Of course, for most people this will make little difference. For people that care about wine, these are big things. --Jayron3206:31, 1 April 2010 (UTC)[reply]
For most people this make little difference only because they are not paying attention. The difference is noticible. Same thing applies to beer too BTW. Dauto (talk) 06:52, 1 April 2010 (UTC)[reply]
How noticeable are these differences? And how long does it take the wine to react to the materials? Is the wine even in the drink ware long enough to make any noticeable difference? —Preceding unsigned comment added by 71.98.64.15 (talk) 08:23, 1 April 2010 (UTC)[reply]
For comparison, pewter tankards are traditional drinking vessels in the UK, and some people keep their own tankard behind the bar at their local pub (another tradition, not of course restricted to pewterware) and/or take one along to beer festivals. However, although there is no problem with beer, which is not markedly acidic, it is now illegal for bar staff to serve cider, which is more acidic (and in the UK always what USAians call "hard", i.e. alcoholic) into pewter mugs made before 1974 when a new low-lead standard was set for the alloy, because the acidity of the cider can leach significant levels of lead from the vessel. Obviously one such drink will likely be harmless in itself, but with habitual use the effect will be cumulative. I'm not sure of the relative acidities of beer, wine and cider, but this demonstrates that the potential for reaction with wine is there.
Speaking as a regular beer, occasional cider and rare wine bibber, I myself notice a marked difference to tastes and other sensory perceptions (e.g. smell, lip feel, etc.) made by various vessel materials, and will by choice only drink from glass or porcelain vessels. 87.81.230.195 (talk) 11:24, 1 April 2010 (UTC)[reply]
Thermal conductivity of the vessel will also have an effect, both on how quickly the liquid warms/cools to room temperature, and on the mouth feel (e.g. a cold glass may distract attention from the contents). --Normansmithy (talk) 15:46, 1 April 2010 (UTC)[reply]
Have there been any double blind studies on rather the taste difference exists or is even different enough for the human tongue to distinguish? I understand the some materials such as pewter can react with certain liquids but im look more for the taste factor. So far the answers (pertaining to taste) are based on anecdotal evidence. —Preceding unsigned comment added by 71.98.64.15 (talk) 17:03, 1 April 2010 (UTC)[reply]
You seem to be ignoring the smell answer. You can smell plastic, paper, and metal (but not glass) even with no wine in it. You can try that yourself, with your eyes closed. StuRat (talk) 17:13, 1 April 2010 (UTC)[reply]
The shape of the container is also very important as is the use of correct technique for pooring the liquid if you're talking about beer. Dauto (talk) 17:28, 1 April 2010 (UTC)[reply]
Some would assert that wine can be pretty good right out of the bottle, although proper glassware can allow better development of the olfactory aspects. Edison (talk) 04:21, 2 April 2010 (UTC)[reply]
A healthy diet must have at least a certain amount of protein,carbonhidrat and fat. For the concreteness let we assume 2000 kcal diet. He must have at least ?? kcal come from protein,?? kcal come from carbonhidrat,?? kcal come from fat. Is these amounts known? Murat Umut. —Preceding unsigned comment added by 85.102.198.65 (talk) 08:09, 1 April 2010 (UTC)[reply]
I am getting the sense in this question that every day one will eat the same thing. You know, one day you could eat a lot of protein, and the next hardly any, and that would be perfectly fine for your health. Vranak (talk) 14:12, 1 April 2010 (UTC)[reply]
This 1,357 page book from the United States Department of Agriculture has an entire chapter on protein intake recommendations, carbohydrates, fats. The recommendations are nuanced by details. So, there is more information in there than can be contained in any reasonable response; and of course, if you have special medical or dietary needs, you should consult a physician. You can get specific numbers for an average, healthy adult American male or female; for example, the USDA recommends 0.80 grams of protein per kilogram of body weight every day[9]. There is additional breakdown of recommendations in this book for other macronutrients. You will probably be interested in the Summary Tables, which specify recommendations for the nutrients you asked about; you can locate whichever age/demographic group is a close match to you, and see what is recommended. Keep in mind that your dietary needs may differ from these general guidelines. Nimur (talk) 14:43, 1 April 2010 (UTC)[reply]
A significantly less technical book is also available from the same source, for free: Dietary Guidelines for Americans, 2005. This has nice summary tables as well as description of nutrients in language that might be more accessible to a non-scientist. "Table 2" in this book breaks down recommended values for all the nutrients you asked about, with amounts specified for two recommended dietary plans. Nimur (talk) 14:49, 1 April 2010 (UTC)[reply]
Side question, for anybody knowledgeable: How corrupted are these USDA guidelines by the US food industry? Did the US food industry basically write them? Comet Tuttle (talk) 17:34, 1 April 2010 (UTC)[reply]
No, these recommendations were not written by the food industry, they were written by employees of the Federal Government: specifically, scientists, medical doctors, and nutritionists at the United States Department of Agriculture. There is a huge quantity of research that was conducted independently by federal agencies, including the USDA and the FDA; and there was independent review by scholars in academia and industry. The book I linked above has literally thousands of citations to internal studies, external publications, international and third-party research, including dissenting-opinions. It also has an entire chapter outlining motivations of the research. It's not productive to promote an unfounded stereotype about conflict of interest. It is probably true that the food industry applies pressure to shape public perception about nutrition; but the Federal Government is a pretty large organization, and to suggest that its research is invalid, or is easily pushed around by outside interests, without specifying a reason or citing any specific grievance, doesn't accomplish anything productive. Considering the relative scaleof the involved parties, it is probably equally valid to insinuate that the American food industry gets pushed around on account of the recommendations of the Federal Government. Nimur (talk) 19:44, 1 April 2010 (UTC)[reply]
I haven't investigated the books or their references yet, so can't say anything to that; but I don't think you can just blow off the concern that the dietary guidelines were influenced by the food industries in the US; here's a NY Times story about the pyramid graphic being delayed a year because (allegedly) of pressure from the meat and dairy industries. The Center for Science in the Public Interestslammed the food pyramid back in 1999, and continued to complain in 2005 that the USDA has a mission of promoting US food industries that is incompatible with the mission of promoting healthy food consumption. Again, this may be giving undue weight to a couple of critics, but I think it's not something to summarily brush off. Comet Tuttle (talk) 21:22, 1 April 2010 (UTC)[reply]
While I have no idea what, if any influence the food industry has on the USA, I disagree that there's any evidence there that the USDA is 'run' or even unduly influenced by the food industry. The article states:
The USDA's stated position was that allowing any meatpacking company to test every cow would undermine the agency's official position that random testing was scientifically adequate to assure safety. The USDA also claims that testing does not ensure food safety because the disease is difficult to detect in younger animals
If a government agency (or whatever) is concerned the actions of some organisation is going to undermine their position or the consumers trust in them, cause undue worry to consumers or worse, mislead consumers it is understandable they may wish to prevent such actions. I'm not saying I agree with the USDA position (I'm unconvinced, but it's irrelevant anyway since this is the RD) or that their stated position is definitely the reason why they are doing what they are doing, but simply that it's entirely plausible it is their reason therefore that case doesn't in itself show that the USDA is unduly influenced by the food industry.
I find that case troubling, too. A regulatory agency trying to limit safety testing at a private company is bizarre behavior. It's difficult to see how limiting testing will improve food safety, which is supposedly the goal of the regulatory agency. On the other hand, limiting testing can lower prices, but that's not supposed to be the goal of regulatory agencies. Note that there are many other decisions by the FDA which also imply a tilt towards industry and against consumers. Just in the food area, we also have the decision not to require labels on genetically modified foods, the ban (until recently) on selling natural stevia as a sweetener while simultaneously allowing the sale of much more dodgy artificial sweeteners pushed by chemical companies, failure to require nutrition labels on fast food items, failure to ban added trans-fats, and having their inspectors only perform visual checks on meat. There are too many instances to list in the drugs area. While any individual instance could be debated, the total makes a strong case for a regulatory agency that was, indeed, controlled by the industry they were supposed to regulate. StuRat (talk) 13:36, 2 April 2010 (UTC)[reply]
Thank you Nimur, this is what I was looking to find. I hope that it will help me convincing my friend that his diet is unhealthy. Murat Umut.85.102.198.65 (talk) 15:29, 2 April 2010 (UTC)[reply]
Phosphate
Unlike other polyatomic ions, such as carbonate and sulfate, the phosphate ion can form two partial ions, H2PO4- and HPO42-. What are their names? Are they called biphosphate and triphosphate, or something else entirely? --Natrium-23 (talk) 15:37, 1 April 2010 (UTC)[reply]
You will frequently see them called "Monobasic" and "dibasic" on the label of the dry chemical, for example, NaH2PO4 = "sodium phosphate, monobasic".[10] This is obviously not the IUPAC nomenclature. -- Flyguy649talk16:30, 1 April 2010 (UTC)[reply]
Actually, the "bi" in "bicarbonate" and "bisulfate" DOES mean 2, and not hydrogen. It is the monovalent form of a divalent ion, so, for example, sodium bicarbonate contains twice as many anions as does sodium carbonate. The name was devised before the structure was fully understood, so when named people noticed that it would take twice as much bicarbonate to precipitate the same amount of a cation as an equivalent amount of carbonate, hence the name "bicarbonate". As far as the phosphate series goes, generally the names are "Dihydrogen phosphate" and "hydrogen phosphate" OR "phosphate, monobasic" and "phosphate, dibasic", as already noted. --Jayron3221:02, 1 April 2010 (UTC)[reply]
Who was the guy who did the first CO2 readings?
There was a guy, an American scientist, who took carbon dioxide readings starting in the 1930s or 40s and did these readings every day until his death in the 1990s, which were then taken over by his son. Ronald Reagan employed a group of scientists to dispute this scientist's findings, but the scientist group found the same thing that carbon dioxide atmospheric levels were indeed rising. The question is, who is this scientist guy and do we have an article about him? I just got told a story about this by someone.--I have winnie the poo tatoo and im 8 (talk) 19:15, 1 April 2010 (UTC)[reply]
I believe you are thinking of Charles Keeling who began his nearly daily Keeling curve in 1958. There are some direct measurements of atmospheric CO2 before that, but they are discontinuous and tend to have much greater uncertainties due to less accurate techniques. Most CO2 records of periods before 1958 are based on studying gas bubbles trapped in ice cores. Dragons flight (talk) 19:21, 1 April 2010 (UTC)[reply]
Improvised teleprompter - looking for one-way mirror in UK
I want to make my ownteleprompter and need to get the one way mirror glass. I live in the UK. My ebay searches aren't yielding anything except this privacy film - would that be likely to work, stretched over some clear acrylic? Proper teleprompter mirrors have an antireflective coating on the camera-side of the mirror.. not sure whether I can prepare such a surface? --78.151.30.114 (talk) 00:55, 2 April 2010 (UTC)[reply]
While I don't know whether you're right or wrong about the anti-reflective coating, I have a suspicion that you may have a misunderstanding of how a two-way mirror works (the proper name is two-way mirror, and that's also the name that correctly describes how it works). A two-way mirror reflects from both sides, and it lets light through from both sides. The reason you perceive it as a mirror from one side and a window from the other has to do with the relative intensities of the transmitted signal versus the reflected one. If you're on the bright side of the mirror, then the reflected signal is brighter (simply because there was more light to start with), and in your eye and/or your brain, it drowns out the signal from the dark side so that you are unable to perceive it.
Both one-way mirror and two-way mirror are commonly used names, and they both make sense, so let's not be saying that only one of them is "proper". "One-way" makes sense because you can only see one way through it. "Two-way" makes sense because it behaves in two ways when seen from different sides. --Anonymous, 04:10 UTC, April 2, 2010.
No, one-way mirror is wrong, period. It should never be used. It encourages people to think that there is some directionality to the mirror. --Trovatore (talk) 07:46, 2 April 2010 (UTC)[reply]
There is directionality. The lighted side and the dark side. But you knew that. --Anon, 10:43 UTC, April 2, 2010.
And it does not "behave in two ways when seen from different sides". It behaves in exactly the same way. (Well, at least in principle -- in practice of course one side could be dirtier than the other or something.) This is a very key point -- if you don't understand that, then you don't understand the gadget at all. --Trovatore (talk) 07:49, 2 April 2010 (UTC)[reply]
It does. It allows you to see what is on the same side when you look into it from the lighted side, and allows you to see what is on the other side when you look into it from the dark side. The symmetry of the underlying mechanism is irrelevant to that fact. And you knew that too. Now please stop telling people what not to say. --Anon, 10:43 UTC, April 2, 2010.
I will not. People should not say one-way mirror. It makes people think that the device is directional, which it is not. On the bright side, you see a roughly equal mix of the bright image and the dark image. And on the dark side, you see a roughly equal mix of the bright image and the dark image. --Trovatore (talk) 17:35, 2 April 2010 (UTC)[reply]
Yeah, but when you construct a box around the camera, there is little light on that side of the mirror, so it works (text is reflected from the screen while the camera receives light from the other side of the mirror). --78.151.30.114 (talk) 01:30, 2 April 2010 (UTC)[reply]
Actually from looking at the diagrams in the teleprompter article, I think the reason the audience (or camera) doesn't see the text has less to do with the relative illumination than with the angle of the mirror. A two-way mirror is essentially a beam splitter. The text is projected up at the mirror from below. Part of the image is reflected towards the speaker; the other half (or so) of it continues up to the ceiling. Essentially none of it goes to the audience, just because the audience is not in the right place to see it. --Trovatore (talk) 01:45, 2 April 2010 (UTC)[reply]
It is also called a "Half-silvered mirrot." Edmund Scientific always had them for sale. Random local glass stores also offered half silvered mirrors for sale. A simple piece of glass would also work nicely, as in the old "Pepper's Ghost" effect. Edison (talk) 04:19, 2 April 2010 (UTC)[reply]
I prefer the name "half-silvered mirror". It makes it clear that it reflects half the light and allows half of it through (roughly). Once you understand that, the behaviour of the mirror is obvious. --Tango (talk) 13:47, 2 April 2010 (UTC)[reply]
I think there is some confusion here over the type of teleprompter being used. The OP wants a teleprompter that allows someone to look into a camera while reading. I think Trovatore is thinking of the teleprompters people use when they want to read while looking at a live audience (the type Obama is (in)famous for using). I think they both use half-silvered mirrors, but in slightly different ways (although I suspect the second type may be less-than-half-silvered, since that will look better from the audience side and you can compensate by just using a brighter projector, the first type could also use a less-than-half-silvered mirror but there is less benefit since the two sides are lit different amounts). --Tango (talk) 13:47, 2 April 2010 (UTC)[reply]
Also, the large planets all have many moons, and the gravitational interaction between them and the planet seems to prevent the rings from coalescing into additional moons. There are smaller planets with multiple moons (Mars has at least 2 and Pluto has at least 3), so perhaps there could be tiny rings there, but I'm not aware of any having been found. StuRat (talk) 13:10, 2 April 2010 (UTC)[reply]
Carbon dioxide accumulation
In various places it seems that it is the short duration of and accumulation or flow of enough CO2 in low spots or ditches or even down a hillside slop and even flat ground to make impossible a breath of fresh air after 2 or 3 breaths that any air breathing life form has collapsed and within seconds to minutes succumbed and died even if air returned within 5 minutes and even if air was in reach of the life form if it stood on its toes or raised its head. This attests to the power of CO2 to knock an animal out quickly but my question is if instead of Nuclear Winter caused by an asteroid striking Earth and killing everything an accumulation of CO2 could be enough from burning matter and release by shock to have killed off most of the dinosaurs? 71.100.3.207 (talk) 02:30, 2 April 2010 (UTC)[reply]
Seems like it would take a lot of carbon dioxide to give enough blanket to enough area. "Everything" on the terrain can't be ditches:) More than just filling low areas and sweeping down hills, you have to get over hills to spread over a continent-sized region.. But certainly the situation has occurred on town-size areas. See for example Lake Nyos. DMacks (talk) 02:46, 2 April 2010 (UTC)[reply]
Agreed. My understanding though is that in the Permian-Triassic extinction event 251 million years ago, which was even worse than the mass extinction that killed the dinosaurs and is thought most likely to have been caused by enormous volcanic eruptions, there is evidence that high CO2 concentrations played an important role in killing large animals. Looie496 (talk) 03:19, 2 April 2010 (UTC)[reply]
The phenomenon which raises this question is the strange accumulation of bone in some places as if they were all living int the same town and the dam broke sweeping them down river to accumulate in eddies and other spots. 71.100.3.207 (talk) 04:28, 2 April 2010 (UTC)[reply]
That's probably closer to the true explanation than carbon dioxide poisoning. After all, there are natural dams. We have beaver dams now, and, although they aren't large enough to cause that type of damage, perhaps some ancient creature did the same thing on a massive scale ? If not, there's still ice dams/glacial dams. You can also get flash floods from sudden rains, without the need for a dam to burst. Another source of floods is when a lake or sea is held back by rock, and that rock collapses. This can happen in a lake in a caldera, for example, and may also have allowed the Mediterranean Sea to catastrophically flood into what became the Black Sea. See Black Sea deluge theory. StuRat (talk) 12:53, 2 April 2010 (UTC)[reply]
Juicy nose
Why is my dog's nose so juicy? I would like to know where all this juice comes from, because sometimes it is so wet I am tempted to get a towel and dry it off. What causes all this juice? I'm not at all concerned about my dogs health; she is completely healthy, she just has a juicy, juicy nose! It's not dripping, like with snot it's just always cold and totally wet. I am wondering *where* this juice comes from. I'm not the only person in the world that has a dog with a juicy nose. I see them all the time. Thanks.70.245.24.14 (talk) 04:51, 2 April 2010 (UTC)[reply]
If you are concerned about your dog's health, you should take them to a veterenarian. Medical advice, even regarding animals, isn't reliable when it comes from random strangers in teh interwbz. --Jayron3204:58, 2 April 2010 (UTC)[reply]
Did you miss the words "not at all"? --Anon, 10:45 UTC, April 2, 2010.
The High Fin Sperm Whale, thank you but where does the juice originate from? That is the nature of my question. Are there some sort of nose glands that produce this juice. 70.245.24.14 (talk) 05:26, 2 April 2010 (UTC)[reply]
Google-ing "dog wet nose" gets tons of hits. One that I found interesting says that the role of the moistness is to help in evaporative cooling. Following some links, the main organ involved in this process seems to be the lateral nose gland. If you want the source for this info, try this article from Science. Other links say that the wetness comes from the dog licking their nose, which I can certainly believe, having observed my own dogs, but I couldn't find any primary sources on that one (not to mention the question of whether the licking is a cause or effect of a wet nose). --- Medical geneticist (talk) 11:46, 2 April 2010 (UTC)[reply]
I've always thought of most animal noses as inside-out relative to humans. That is, the portion that's inside and always kept moist on humans is outside (and inside as well), on dogs. This is likely part of their much better senses of smell. The glands that keep it moist are similar to our sweat glands. StuRat (talk) 12:37, 2 April 2010 (UTC)[reply]
Yes - and part of the reason for that appears to be to give them a better directional sense of smell. When all of the little detectors are up inside your nose, all you can really tell is which nostril the smell came through - but dogs (in particular) have a completely directional sense. SteveBaker (talk) 14:21, 2 April 2010 (UTC)[reply]
Electra hydraulics
For those who have a working knowledge of historic aircraft (especially those made by Lockheed): did the Electra 10-E have a hydraulic system (and I don't mean just the oleo-struts), and if so, what components did it actuate? I'm pretty sure the primary flight controls were all mechanical, as on all other aircraft at the time (AFAIK hydraulic boosters did not begin to be fitted to aircraft until after World War 2, and even then only to the highest-performance jet planes at first); my question is, was there anything else vital to the plane's safety (like flap controls, or landing gear retraction) on the Electra that was hydraulically powered? And no, in case you're suspicious, this has nothing to do with any fringe theories about Amelia Earhart's disappearance; it's for a novel that I plan to write some time in the future. Thanks in advance! 24.23.197.43 (talk) 06:31, 2 April 2010 (UTC)[reply]
drive
Sir,
please send me the drive internal connection diagrams ie, IGBT, RECTIFIER,IVERTER etc
From my office window, I look out over an apartment complex which contains dozens of large, mature, deciduous trees. These appear to be of identical species (sorry - I'm clueless as to what species they are - but I don't think that matters). They are all more or less the same size (40 feet maybe). Presumably they were all bought from the same tree supplier on the same day and planted when the apartment complex was first built. So they ought to be pretty close to identical...possibly even genetically similar or perhaps identical if they were grown from cuttings rather than seed. I'd guess that they are of some fairly fast-growing species.
But they are all slightly different shades of green - some slightly yellowish, some subtly brownish, some are a really vivid almost "lime" green. They all seem pretty healthy though - the color differences are pretty subtle - they aren't brown as in "Autumnal", it's spring here in Texas and the leaves are all pretty new.
Given that everything else is the same (species, age, etc) I presume that this must be due to differences in soil chemistry, available sunlight and amount of water available.
It's also noticable that the color of all of the leaves of a particular tree are identical in color - even though some parts of the tree are in more or less permenant shadow between these large buildings and other parts are getting plenty of light because they are ten feet above the roof-line.
But (chemically) what is changing in the leaves? Is it just a matter of whether there is more or less chlorophyll - or is it the presence of different amounts of some other colorant?
I'm curious because I do computer graphics and I'm wondering about what might constitute a realistic range of randomized colors for otherwise identical trees - I'll probably end up just having one of our artists go and make an artistic choice - but it would be nice to actually understand what's going on. For example, if the differences are only in chlorophyll levels - then it's likely that only the amount of green is changing from tree to tree...but if it were wildly varying amounts of some reddish compound then the answer is different.
I know that in aerial overflight imagery, the most common cause of coloration difference in vegetation is amount of water. This is especially detectable in infrared channels (also picking up amount of chlorophyll). You may be seeing variations of the water table or phreatic surface on a meters-scale level. Whether the quantity of water is a biological cause for a variation in chlorophyll quantity/quality is a question for a botanist... Nimur (talk) 15:33, 2 April 2010 (UTC)[reply]
(ec) The colour of leaves is determined by a combination of various pigments (such as carotenoids), including chlorophyll (those other pigments are what give the autumnal colours). There are also different types of chlorophyll, with slightly different colours (see Chlorophyll). Since it is currently spring, the difference could be due to differing ages of the leaves (young leaves are often yellower than old ones). If the trees get different amount of sunlight, they may be a few days ahead/behind each other. If that is the case, I would expect the differences to reduce over time - I suggest you take daily photographs and compare them. I would expect the soil to be very similar over such a small area (unless different plants had been grown there previously, I suppose). There could be differences in water, though. --Tango (talk) 15:35, 2 April 2010 (UTC)[reply]
(ec) I'd go with the amount of chlorophyll. Note that new leaves are often also a very different shade of green than last year's growth, so the ratio of new leaves to old would affect the overall color. And the percentage of new leaves could depend on water, sunlight, temperature, how many old leaves are retained, and many other factors. StuRat (talk) 15:38, 2 April 2010 (UTC)[reply]
Power lines compared to Overhead cables on train lines
Hi. I am trying to understand what differences there are between power transmission lines that are part of the power grid and power lines that you would find above an electrified train track. I imagine that the power lines are usually almost continuously transmitting electricity, but that overhead train lines are only doing so, when trains pass, and also that far power power passes down power transmission lines than overhead train line. I would also like to understand if the electromagnetic radiation from overhead train lines is at all comparable to the radiation from power transmission lines. Any help would be appreciated. Thanks GregB1968 (talk) 15:47, 2 April 2010 (UTC)[reply]
For the train, they use two power lines, one positively charged and one negative, so that no electricity should flow until the train completes the circuit. I suppose regular power lines are similar, with the appliances at the end being what completes the circuit. However, unlike the train case, the circuit is always completed by many thousands of appliances, so electricity is always flowing. StuRat (talk) 15:57, 2 April 2010 (UTC)[reply]
Our article on overhead lines explains their use for public transport / train systems. They may use mechanical schemes to keep them from breaking under the constant attachment of the train's electrical paddle or connector. Typically, they are higher voltage than the residential circulation lines; voltages vary anywhere from ~ 1000 volts (e.g., San Francisco Muni) to much higher (~ 10 kV in Germany. Sometimes, they also operate at a different AC frequency. Consequently, they have different power distribution and substation implementations; but otherwise, they are not very different from standard transmission lines. Nimur (talk) 16:13, 2 April 2010 (UTC)[reply]
Railway electrification systems usually use the tracks as the one "line" and the overhead as the other. Therefore, only when a train makes a circuit is power transmitted. Power for trams and light rails are typically 750 V DC, while mainline trains may be either DC (typically 1500 V or 3000 V) or AC (15 kV or 25 kV are the most common). With so high voltage, it is often not necessary to have a parallel feeding system, although unless the trains use a different frequency than the power grid (50 Hz in Europe; 60 Hz in North America), they can simply feed from the local grip at regular intervals. Arsenikk(talk)17:11, 2 April 2010 (UTC)[reply]
Thanks for all the input, so far. There is a rail line near to my place of work (Hertford Loop Line) which seems to be 25 kV AC line. The power transmission article seems to suggest that power lines are usually greater than 110 kV. I am wondering what impact that has on electromagnetic radiation. Does the much lower voltage mean that there is much less radiation, and does the fact that there is only intermittent power transmission on the rail line, mean that there is only intermittent electromagnetic radiation (i.e. only during actual power transmission)? Thanks again for the help so far. GregB1968 (talk) 17:38, 2 April 2010 (UTC)[reply]
Stu also even if there no appliances or anything in use, most power lines are at such high voltages that if you touch just the positively charged line, electricity will arc across from the negatively charged one and through your body.--92.251.179.38 (talk) 18:31, 2 April 2010 (UTC)[reply]
One additional question on railway power. I once slept in room in Bremen an in the dark I could see a Cold cathode flicker each time a tram was passing. Next morning I checked and I saw that a steel cable, holding the isolated power cable, was attached to the wall in the room were I was staying. How does enough energy get into my room that a cold cathode is giving light. --Stone (talk) 18:58, 2 April 2010 (UTC)[reply]
If the power is DC there is almost no electromagnetic (radio wave) leak. But there is a (stationary) electric and magnetic field around it. Are you referring to electromagnetic sensitivity? If you are, since it's fictional, I can't tell you if a railway power line is better or worse. Ariel. (talk) 19:11, 2 April 2010 (UTC)[reply]
Are you referring to the medical degree? Your IP address says you are in Pakistan, is that where you want to study? --Tango (talk) 17:02, 2 April 2010 (UTC)[reply]
All medical schools in Pakistan use the Bachelor of Medicine, Bachelor of Surgery (MBBS) degree. Not sure how you would obtain a scholorship but usually such schemes are done directly by the medical schools (e.g. for very good grades at pre-medical level) or they are subsidised by a company in return for you working for them after the degree. Though the latter is rare for being a doctor, you may be able to find a research institution who could fund you if you find that you have exceptional ability and you sign an agreement to work for them post-qualification. Regards, --—Cyclonenim | Chat 21:43, 2 April 2010 (UTC)[reply]
How are proton beams created?
I've always wondered how proton beams for colliders, like the Large Headron Collider, are created. Are the beams exclusively protons? If so, what element is used for the protons? And how are the protons "separated out"?
I don't know for sure, but I would assume that they just ionize hydrogen using an electric field, which also serves to separate the protons and electrons. -- Coneslayer (talk) 18:11, 2 April 2010 (UTC)[reply]
What would happen if someone built an engine that propel a spacecraft at 1,079,252,851 km/h? Supposing we found some method of propulsion that could potentially provide the necessary power, and lets say it's travelling through a complete vacuum so it doesn't get blown all to hell by the friction, is there any reason why the spacecraft couldn't achieve that speed?--92.251.179.38 (talk) 18:25, 2 April 2010 (UTC)[reply]
You see this is what I was expecting. But obviously there's nothing stopping us from building such a propulsion system. What would happen as the craft neared the speed of light?--92.251.179.38 (talk) 18:33, 2 April 2010 (UTC)[reply]
As you accelerate the craft its energy increases which by the Einstein's formula E=mc2 means its mass also increase making it harder and harder to accelerate (More inertia). To reach the speed of light you would need an infinite amount of energy which is not possible. Dauto (talk) 19:00, 2 April 2010 (UTC)[reply]
"obviously" ? Energy (speed) has mass (weight). So as you speed up the craft it gets heavier. In order to speed it up some more, not only do you have to speed up the original weight of the craft (called the rest mass), but also all the additional mass from the energy (called the relativistic mass, and don't for a second think this mass is fictional, or just a mathematical concept - it's totally real). The more you speed it up, the heavier it gets, the heavier it gets the harder you have to push to speed it up - but that very hard "push" uses a lot of energy, so it gets even MORE heavy from all that additional energy. The result is that when you hit the speed of light the mass becomes infinite, so no matter how hard to push, you just can't move it.
Well, you can't hit the speed of light, so it doesn't mean much to say the mass is infinite at the speed of light. The energy required to reach the speed of light (which you calculate as an integral) is infinite. --Tango (talk) 19:31, 2 April 2010 (UTC)[reply]
Special relativity gives a good explanation of what happens—the mass of the craft increases, it starts to perceive time as being slower than other reference frames, its size gets smaller as viewed from other reference frames, and the energy needed to keep it going increases. You can't hit the speed of light this way, or go beyond it—the physics just doesn't work out. If you could get to it (which you can't), the end result would be that its size would be nothing, its mass would be infinite, and its perception of time would stop (if I understand it correctly—I am sure there is a more subtle way of saying those things). The faster-than-light article is a good discussion of why current science says it won't work, and gives some problems with making loopholes in current understandings as a way to get around this constraint. Our theories would have to be substantially wrong (or missing something huge) for this to work out. (Which is always possible, but you'd have to get your new theory to jibe with the rest of the observed data, which would be a non-trivial task! You probably aren't going to manage it!) --Mr.98 (talk) 22:05, 2 April 2010 (UTC)[reply]
is it possible to get pregnant 6 days after (onset of) menstruation?
is it possible to get pregant six days after the onset of mensturation (ie first day being last sunday 1, mon 2, tue 3, wed 4, thur 5, fri - today- 6). thank you. 82.113.106.100 (talk) 22:37, 2 April 2010 (UTC)[reply]
PREGNANCY
When is a woman ready to concieve after menstruation period?#REDIR]]#REDIRECT [[reproduction in human being#REDIRECT Target page name]]
Resolved
Please do your own homework.
