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October 23

Women & Yeast Infections

What causes some women to get a yeast infection after going swimming? Why swimming? If a pool has chlorine in it, shouldn't it help prevent it? Why does it happen and why are some women more prone to it? — Preceding unsigned comment added by 76.173.30.220 (talk) 00:40, 23 October 2011 (UTC)[reply]

First of all, public swimming pools are notorious exchanges of microorganisms -- because so many people swim in them, it's a near-certainty that at least some of them will introduce harmful microorganisms into the pool, which can then spread very efficiently through the water. And while chlorine can reduce the sheer amount of pathogens in the water (ironically, by oxidizing them), it cannot ensure the elimination of all the nasty bugs in the water (especially since there are just so many of them, and since some kinds are immune to chlorine altogether). As for why some women are more prone to it, some of the reasons may be: differences in the woman's vaginal environment (more acidic = less susceptible), different patterns of sexual activity (promiscuous sex makes the woman more prone to infection, while frequent monogamous sex may decrease her susceptibility), immune system health (obviously, immunocompromised = MUCH more susceptible), stress levels (more stressed = more prone to infection in general), and just the woman's general health level (healthy = more resistant to infection, not feeling well = more susceptible). This list is by no means all-inclusive. 67.169.177.176 (talk) 01:16, 23 October 2011 (UTC)[reply]

You claimed "promiscuous sex makes the woman more prone to infection, while frequent monogamous sex may decrease her susceptibility." There is a need for a reliable source to substantiate this puritanic claim, unless it just means that more partners means a higher likelihood of one having a STD. Edison (talk) 05:00, 23 October 2011 (UTC)[reply]

Quote from the article Vaginitis: "Further, either a change in pH balance or introduction of foreign bacteria in the vagina can lead to infection known as vaginitis." The first condition is always the case during unprotected sex, while the second is much more likely with multiple sex partners, because more sex partners = more varieties of bacteria, funguses, etc., that are likely to be introduced into the vagina. Therefore, promiscuous unprotected sex is definitely a risk factor in developing vaginal infections, even if venereal disease is not a factor. As for frequent monogamous sex, although it does change the pH balance as well, there is circumstantial evidence that receiving semen from one man only on a regular basis could improve the woman's immune response through repeated exposure to pretty much the same antigens all the time; however, at this time it is not certain that this significantly improves resistance to yeast infections, which is why I put "may" before this claim, to show that it's not proven. I also should have qualified these claims as applying only to unprotected sex; obviously, protected sex (using a condom) is not a risk factor for developing vaginal infections, even if promiscuous. 67.169.177.176 (talk) 20:55, 23 October 2011 (UTC)[reply]

With your later admissions, it seems clear your original statement is dubious. In fact it seems likely someone who is not 'promiscious' but in a stable committed relationship is less likely to use a condom. If that is they case, they could easily be at higher risk (compared to both the 'promiscious' woman who uses condoms all or nearly the time and the woman who rarely has sex with any partner) despite only having one partner but frequent sex, even with an unproven reduction in susceptibility. In other words, if you're talking averages, the info you have provided does not establish your original claims or at least suggests they could be misleading as originally presented, a 'promiscious' woman may be less likely to get an infection then a woman having frequent monogamous sex (with the close to celibate woman possibly being the lowest risk). Unless you can provide some actual sources showing otherwise of course.... Nil Einne (talk) 00:07, 25 October 2011 (UTC)[reply]

Your statement ignores the effects of the antigens found in the semen on the woman's immune system, which, in the stable committed relationship scenario, would tend to counteract the effects of the pH changes inevitable during unprotected sex. This is the same mechanism that greatly reduces the risk of miscarriage when the woman has frequent, monogamous, unprotected sex during pregnancy -- I just don't remember the scientific name for this effect, or else I'd have given you the article link by now. 67.169.177.176 (talk) 04:52, 25 October 2011 (UTC)[reply]

Follow-up: the phenomenon I was thinking of (and vainly trying to find a link to for 2 days) is paternal tolerance. This has been proven to prevent miscarriages, and is hypothesized (but not proven at this time) to also reduce the risk of vaginitis (all forms, including yeast infections). 67.169.177.176 (talk) 05:09, 25 October 2011 (UTC)[reply]

Not I did not. Read my comment carefully. I said 'even with an unproven reduction in susceptibility'. In other words, even if there is this purported reduction in susceptibility, it doesn't automatically follow that the woman in the stable committed relationship who does not use condoms will have a lower risk of yeast infections or vaginitis then either the woman who rarely has sex or the woman who has sex with multiple partners but nearly always uses a condom since the pH changes and introduction of a great number of yeast, bacteria etc (despite the lower variety present on the sexual organ of her partner then would be present on multiple different partners) could easily be greater given the lack of protection. (Note I did not say then and I'm not saying now it's definitely the case, I'm just saying from the evidence presented thus far, both scenarios seems equally plausible as well as a whole host of other scenarios so your original statement was misleading.) You cannot look at factors in isolation if you are talking about overall risk. And unfortunately without the source you are thinking of we cannot even know if the source said the woman may have a lower susceptibility (as you said at first) or a lower risk (as you suggest now).

I'm aware of the miscarriage theory but that's a somewhat different thing, the hypothesis is that womans immune system develops tolerance due to frequent exposure to paternal antigens and this reduces the risk her body will respond in an unwanted way to said antigens coming from the fetus. This it would seem to suggest it would lead to a greater susceptibility to yeast infections from her partner because she develops some tolerance to the antigens including those present on external bodies like yeast her partner introduces. However the immune system is a complex thing so I'm not denying it is possible she would develop some sort of improved immune response to the antigens on external bodies her partner introduces simply that isn't so much paternal tolerance. Of course it's also possible a woman who has frequent unprotected sex with multiple partners may in fact have the lowest susceptibility (but perhaps not the lowest risk) because the frequent introduction of a variety of different antigens best primes the immune system to respond.

Nil Einne (talk) 09:53, 26 October 2011 (UTC)[reply]

In other words, nobody can tell for sure which woman will have the lowest risk/susceptibility -- the slut, the nun, or the ardent lover. I will not dispute this statement, especially since there are so many other confounding factors as well. 67.169.177.176 (talk) 05:55, 27 October 2011 (UTC)[reply]

ice age

is it true we're entering an ice age — Preceding unsigned comment added by 69.5.89.125 (talk) 02:13, 23 October 2011 (UTC)[reply]

Read further up the page, a related question is posted under 19 October. Plasmic Physics (talk) 02:20, 23 October 2011 (UTC)[reply]

See global cooling, global warming and Milankovich cycles. ~AH1 ^(discuss!) 15:02, 24 October 2011 (UTC)[reply]

Neutrino interactions

Hi, I watched a program recently which, as I understood it, seemed to say that neutrinos interact with matter only when they hit an atomic nucleus, and that the great expanses of empty space around atomic nuclei were the reason why such interactions were so rare, and why, for example, neutrinos can easily fly straight through the Earth. However, according to my calculations based on the separation of atoms and the sizes of atomic nuclei, it seems to me that, barring unfortunate exact alignment with a huge perfectly regular crystal, a particle of negligible size could expect to hit a nucleus after only a few tens of centimetres, or at most a few metres, of some reasonably dense material like rock. So, what gives? Have my calculations gone wrong, or can neutrinos also fly straight through atomic nuclei? 86.179.116.118 (talk) 02:53, 23 October 2011 (UTC)[reply]

That program was wrong in two accounts. First, neutrinos can interact with nuclei but they also can interact with electrons which are found outside of the nuclei. Second, the reason they can go through the earth is because they interact very weakly. Dauto (talk) 06:20, 23 October 2011 (UTC)[reply]

Expanding on Dauto's excellent but brief explanation: the nucleus of an atom is not a solid object. It consists of neutrons and protons, which in turn are made up of quarks. The figures usually given for the "size" of an atomic nucleus - around 1 to 10 femtometres - are a measure of the range of the strong force, which keeps the neutrons and protons (or, more precisely, their constituent quarks) bound together in the nucleus. But neutrinos do not feel the strong force (or, indeed, the electromagnetic force which keeps electrons in orbit around the nucleus). They only interact via the weak force, which has a range of about 1/100 of a femtometre. So to a neutrino even an atomic nucleus is mostly empty space - the neutrino has to come very close indeed to a quark (or an orbiting electron) before it will interact with it via the weak force. Gandalf61 (talk) 10:06, 23 October 2011 (UTC)[reply]

Thanks guys. 86.179.116.3 (talk) 13:06, 23 October 2011 (UTC)[reply]

Why not ? (RDX)

Why they don't use RDX in gun-cartridges ? Jon Ascton (talk) 06:47, 23 October 2011 (UTC)[reply]

As a propellant, you mean? It's because RDX "burns" extremely fast so that the pressure would shatter the weapon, instead of propelling the bullet through the barrel. For a gun propellant you want something that "burns" slower than that.Sjö (talk) 09:04, 23 October 2011 (UTC)[reply]

(EC) For one thing, RDX would be too brisant. Gunpowder works well as a propellant because it's a low explosive that deflagrates at subsonic speeds, so it has low brisance. In contrast, RDX is a high explosive that detonates, producing a supersonic shock wave which would likely rupture the gun barrel. Red Act (talk) 09:17, 23 October 2011 (UTC)[reply]

Molecular bonding

What does the MO diagram for monomeric beryllium hydride look like? Is there sp hybridisation? This my idea:

If this is true, then why is there no sp mixing to allow for the diberyllium molecule? Plasmic Physics (talk) 09:15, 23 October 2011 (UTC)[reply]

Well, one thing is not to merge hybridization and MO, two very separate approaches. MO is more straightforward. Start with symmetrical H1s+H1s and antisymmetric H1s-H1s linear combinations. Match these with atomic orbitals on Be. The two symmetry-suitable atomic orbitals on Be are 2s and the 2pz (z defining the molecular axis). Add 4 valence electrons and you are there.--Smokefoot (talk) 12:23, 23 October 2011 (UTC)[reply]

How is it possible to separate the two theories. Hybridisation explains the two different MO diagrams above. Left: with sp hybridisation, the 1σ* MO is unoccupied, because it is higher in energy than the 2σ MOs. Right: without hybridisation, the 1σ* MO is occupied, because it is lower in energy than the 2σ MOs. I'm not sure what you mean by your combinations, the s orbitals are all spherical and therefore symmetrical. Plasmic Physics (talk) 14:04, 23 October 2011 (UTC)[reply]

When building molecular-orbital electronic diagrams, it's common to consider the symmetry with respect to the combinations--symmetry-adapted linear combinations of atomic orbitals--not each just each AO's symmetry itself or hybridization on individual atoms first. The antisymmetric H1s-H1s combination is the case where the two H 1s AO are of opposite phase compared to each other. There is a node between them, and the +z and –z directions from the Be center have opposite phase. That's the same symmetry pattern as the Be 2pz. So "Be 2pz has same symmetry as the antisymmetric 1Hs-1Hs" and therefore is a good combination for a MO. Likewise, the 1Hs+1Hs combination is the case where the two H 1s AO are of the same phase, which means "no node between them", so using the Be2s continues that pattern of having no node in the middle of the molecule, and therefore is another good combination for a MO. DMacks (talk) 14:22, 23 October 2011 (UTC)[reply]

OK, then which MO diagram is correct, if neither, what is it? Initially, I chose left. Plasmic Physics (talk) 22:49, 23 October 2011 (UTC)[reply]

Neither scheme looks correct. I would avoid the term sigma bond (a term mainly from hybridization approach). You already know that there are 2e in the 1s level of Be. We do not include those core electrons and core orbital in our scheme, by convention. So that leaves 4e for the scheme. You put 2e in an MO that is Be 2s combined with the (symmetric) H2 SALC and another pair in the Be 2pz combined with the (unsymmetric) H2 SALC. Both of these bonding MO's have cylindrical symmetry, so you could call them sigma, but better would be the irreducible representation from group theory. As DMacks mentioned, a key step that takes some serious pondering to understand, is the factoring the symmetry equivalent H atoms (always 1s AOs) into symmetry adapted linear combinations. Once you understand SALC's, you can do CH₄, H₂O, NH₃. Each bonding scheme harbors its own surprise relative to the hybridization approach, which can be misleading (e.g. that all eight bonding electrons are equivalent in methane or that water has "rabbit ears" for its lone pairs, etc). There are a finite number of SALCs, practically speaking, which is why teachers can recite them quickly. --Smokefoot (talk) 17:13, 24 October 2011 (UTC)[reply]

People have already said some really good stuff. One article not mentioned is Linear combination of atomic orbitals. That provides a decent outline of using MO theory qualitatively, as is being done here. --Jayron32 17:59, 24 October 2011 (UTC)[reply]

So, the mistake in the diagram is that both bonding orbials should be degenerate? Plasmic Physics (talk) 22:21, 24 October 2011 (UTC)[reply]

  __    2σ*
__ __  2×1σ*
  __    2σ
↑↓ ↑↓  2×1σ = 2s¹(Be) + 2p_z¹(Be) + 1s¹(H1) + 1s¹(H2)

  ↑↓    n = 1s²(Be)

I have not heard of group theory or SALCs before in any of my chemistry papers. Plasmic Physics (talk) 10:48, 25 October 2011 (UTC)[reply]

Hello? Aanybody there, or am I just talking with myself again?

Key is stuck in lock; how to remove anyhow?

When my sister came home at about 4:30, she woke me up from the noise of trying to get the key out of the lock. I looked, and the key very much appeared stuck. I tried turning it, and it seems like I was just bending the metal of the key, and she told me not to mess with it until the morning.

I have a bad feeling for what would happen if I reported this to the landlady, so how do I remove a defective key on my own?

By the way, this was the same key that she got from Menards when she got mine duplicated; apparently Menards has poor quality service. Or did my sister just choose the cheapest/least-reliable type of key?

I hope to get this solved soon before they find out. Also, if I hire a locksmith, will they remove the mess from the keyhole, or will they be forced to change the cylinders and get me a new set of keys? (If they change the cylinders, the maintenance people would need a new set of keys too, so I hope it doesn't come to that.) Thanks, --70.179.174.63 (talk) 09:57, 23 October 2011 (UTC)[reply]

Can you hold the key with needle nose pliers? It seems the best way to remove it, if it's at least protruding a little. Nail clippers will probably work too. Otherwise, try to put some machine oil into the lock with a needle and pull the key a little bit out with the same needle. 2.138.250.205 (talk) 10:28, 23 October 2011 (UTC)[reply]

If it is a Yale night latch, then if you remove the latch (inside part), then the barrel (its attached to a plate behind the latch), you can then push the stub of the key with a sewing needle. For thin-door double-sided locks, then you can normally push a needle through them as well, if the key hasn't been turned. CS Miller (talk) 11:18, 23 October 2011 (UTC)[reply]

Incidentally, this does just sometimes happen. I had a key get stuck in a lock that was just impossible to get out. I ended up just replacing the lock. It wasn't very hard and not very expensive. We somehow got the locksmith to make it so our old keys worked in the new door. --Mr.98 (talk) 13:14, 23 October 2011 (UTC)[reply]

No guarantee, but it often works to squirt some WD-40 into the slot around the key and jiggle it back and forth for a while until the WD-40 works its way into the mechanism. (The two universal solutions are WD-40 and duct tape. When two things move indpendently and you need them to move together, use duct tape. When two things move together and you need them to move independently, use WD-40.) Looie496 (talk) 14:24, 23 October 2011 (UTC)[reply]

Locks are easy to remove. Why not just take it apart, take out the key, and see what the problem is. It is likely that one of the pins bent or something got in the lock and jammed the pins. -- kainaw ™ 14:31, 23 October 2011 (UTC)[reply]

A locksmith or hardware store can create a new cylinder keyed to match the old one, so the old keys work. Rekeying a lock should cost under $8 (US) if you do the work of removing and installing, in addition to the cost of the new lock. Edison (talk) 19:33, 23 October 2011 (UTC)[reply]

Try WD-40. Really. It's easy and cheap and often works in cases that seem hopeless. Looie496 (talk) 23:00, 23 October 2011 (UTC)[reply]

Center of mass of a van

Is there a practical way for me to locate the center of mass of an old Ford Econoline van?

I have access to ordinary tools, but not to anything hard-core, like a lift or a winch. I ask because I'm thinking that's the best place to install a mast, if I want to turn the van into a sailing craft. So my second question is: Am I correct, that the center of mass would be the best place to install a sail mast?

Thanks in advance for any help. -GTBacchus^(talk) 20:12, 23 October 2011 (UTC)[reply]

Good luck! I assume you have seen the top gear episode. With regards to your second question I think that most boats I have sailed on have the mainmast somewhat forwards of the centre of mass. In fact I think the most important thing is where the centre of effort ends up in relation to the centre of lateral resistance (this website explains it a lot better than I could, it's written from a model yachting perspective but the principles are the same). The idea is that you want just a little weather helm, as then the boat will luff up and depower itself automatically in a gust. You certainly don't want Lee helm as that can lead to repeated uncontrolled gybes if you get overpowered. In terms of determining the centre of lateral resistance, it is usually approximated as the geometrical centre of the below waterline hull (by the way, you may want to add some sort of keel, centerboard or even a leeboard to the van, which you will need to take account of in the calculations, as otherwise you will go predominantly sideways at anything tighter than a beam reach). The centre of effort is estimated as the geometrical centre of the sail plan. You probably don't need to be too precise about this, a sailing van will probably handle like a recalcitrant hippopotamus however well it is designed. P.S. That all being said you do of course want to make sure that the mast is not so far off the COM that it will cause the van to tip over in the water. Equisetum (talk|contributions) 22:18, 23 October 2011 (UTC)[reply]

I don't watch television, so I don't know what you mean by the "Top Gear" episode. Did they outfit a car with a sail? Regarding handling and tipping, I forgot to mention - I imagine outfitting the van with (wheeled) pontoons on either side, like a land-catamaran. Otherwise, tipping to the side would be an inevitability, I think. This would probably only be practical to use somewhere very flat, like the Nevada desert or something.

Thanks for the link and the info! :D -GTBacchus^(talk) 23:19, 23 October 2011 (UTC)[reply]

Oh, you want to make a land yacht, not a boat! (far to sensible for us Brits), I'm not sure how much my advice applies in this case as the steering/lateral movement is somewhat different. Watch the clips below for a look at what happens if you try to make real boats out of land vehicles! Equisetum (talk|contributions) 09:37, 24 October 2011 (UTC)[reply]

Ah, I see that my original question wasn't clear on this point. By "sailing craft", I just meant something powered by a sail. I'd definitely stay away from water. I guess I might start with models, and see what I can learn that way. A Ford land-yacht with pontoons and a mast would probably go over quite well at Burning Man or some similar venue. -GTBacchus^(talk) 17:21, 24 October 2011 (UTC)[reply]

Hello my good man - couple of links for the Top Gear segments in question - first failed attempt is at [1] and the second somewhat more successful one is at [2]. Enjoy old chap! Quintessential British Gentleman (talk) 00:26, 24 October 2011 (UTC)[reply]

Thank you! I'll watch these when I get home where my computer has sound. :) -GTBacchus^(talk) 17:21, 24 October 2011 (UTC)[reply]

sleep

why do we sleep? — Preceding unsigned comment added by 203.112.82.128 (talk) 21:05, 23 October 2011 (UTC)[reply]

Sleep's functions aren't really well understood, but for some of the proposed theories, see Sleep#Functions. Red Act (talk) 21:13, 23 October 2011 (UTC)[reply]

And for what happens if you can't sleep, which might imply why you need to sleep, read about fatal familial insomnia. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:33, 23 October 2011 (UTC)[reply]

It has to be a very good idea, because it seems to exist in one form or another in almost every species. Imagine Reason (talk) 12:14, 24 October 2011 (UTC)[reply]

3 CRT questions

1) Are there any qualitative differences between the CRTs in televisions and those in computer monitors?

2) I'm aware that CROs use electrostatic deflection rather than magnetic, whilst TVs and computer monitors use magnetic deflection. I'm aware of the merits of using electrostatic deflection, so why don't televisions and computer monitors use electrostatic deflection also?

3) Is there any "principal" limit to the resolution of a monochrome CRT display? And if not, why isn't there a market for cheap, ultra-high resolution monochrome displays?--Leon (talk) 23:24, 23 October 2011 (UTC)[reply]

A1 According to the page on Cathode ray tube, there does not seem to be a significant difference between those used in TVs and those used in monitors, althogh the inductance of the monitor deflection coils may need to be lower to allow faster scanning than the 62us (or thaereabouts) per line as in Tvs.

A2 TVs and computer monitors dont use electrostatic deflection because, I believe, this method has difficulty in scanning over a wide angle to produce a large picture as required in TV sets and monitors

A3 I cant see any limit to the resolution of a monochrome CRT apart fron the video bandwidth limitation in driving the grid of the tube and the granularity of the phosphor. --92.25.104.152 (talk) 13:52, 24 October 2011 (UTC) Oh yeah, I suppose everyone wants color now?--92.25.104.152 (talk) 13:58, 24 October 2011 (UTC)[reply]

A magnetic field will cause deflection perpendicular to the field, so the deflection doesn't change the distance between electron and magnet. It's more or less a linear relation between the magnetic field and the angle of deflection. With electrostatic deflection, the small distance between electron and electrode at maximum deflection would make it very sensitive to small fluctuations. Like trying to deflect a metal ball rolling from a slope with a strong magnet without the ball getting too close and stick to it. Metal ball in a magnetic field is (more or less) the equivalent of an electron in an electric field. DS Belgium (talk) 01:38, 25 October 2011 (UTC)[reply]

Seismic wave polarisation

For S-waves, what is the difference between horizontal and vertical polarisation? I mean, I understand polarisation in general, but apparently SV waves are able to be reflected/refracted to P waves, and SH waves are not. I'm not sure why the difference isn't just arbitrary. Why doesn't it just depend on the orientation of your coordinate system? --130.216.172.64 (talk) 23:38, 23 October 2011 (UTC)[reply]

I'm not an expert, but it seems to me that this has to do with the wave's orientation relative to the Earth's surface. 67.169.177.176 (talk) 00:21, 24 October 2011 (UTC)[reply]

The waves are reflecting and refracting off of discontinuities on the earth. Those discontinuities are caused by temperature and pressure gradients which are vertical. The discontinuity surfaces, being perpendicular to those gradients, are horizontal. Those planes set the orientation of the coordinate system. The oscillation direction for a SH-wave is horizontal and will be on the plane of discontinuity. On the other hand, the oscillation direction for a SV-wave is not horizontal (it's not vertical either unless the wave is propagating horizontally). That means that those waves interact differently with the discontinuity surface. The SV-waves hit the surface like a drumstick hits a drum, creating sound waves (p-waves). The SH-waves skim the surface and create no sound waves. Dauto (talk) 05:12, 24 October 2011 (UTC)[reply]

October 24

Reptile heat source

Do I really need to buy the $12.95 pet shop light bulb for my ball python or can I just get a Shoprite incandescent 4-pack for $1.99? UV-A and UV-B ehh... DRosenbach ^{(Talk | Contribs)} 00:13, 24 October 2011 (UTC)[reply]

Hello old chap - this link seems to have a lot of information about lighting, heating and different bulb types - [3]. Quintessential British Gentleman (talk) 00:20, 24 October 2011 (UTC)[reply]

(ec) If you need UVA and UVB, your standard incandescent bulb (even "full spectrum" ones) will not be sufficient. Here is a nice page that goes through all of the different lighting options and what they do and don't produce for your pet. It strikes me that trying to be cheap in this department will negatively affect your pet. --Mr.98 (talk) 00:21, 24 October 2011 (UTC)[reply]

Please make sure to use the correct light, no matter what you pay. I work with shelter/rescue groups that often take in various reptiles that haven't been cared for correctly, including the wrong or no light source. Bad things can happen for the animal if something as seemingly simple as light is not given to the animals. Dismas|^(talk) 12:56, 24 October 2011 (UTC)[reply]

But it was explained to me at the pet shop that I don't even need a light -- I could have chosen a heat pad instead of a warm bulb, and that doesn't provide any UV light. DRosenbach ^{(Talk | Contribs)} 01:52, 25 October 2011 (UTC)[reply]

Giant human skeleton

Are there any reliable sources for those pictures [4], [5]?--82.114.165.50 (talk) 02:00, 24 October 2011 (UTC)[reply]

No. LANTZY ^TALK 02:30, 24 October 2011 (UTC)[reply]

Snopes and the National Geographic cover some similar photos. Please note the reference to the square-cube law at the bottom of the Snopes article, and the links in the square-cube law article that show that if there ever would be giants like that then they wouldn't just look like scaled-up versions of humans.Sjö (talk) 05:17, 24 October 2011 (UTC)[reply]

Thanks a lot. What I want to understand is if such photos/articles are logical or just hoaxes as usual. Is there any Wiki article expressing similar material? To be more specific, I'm eager to know about human height history since their ancestors.--Almuhammedi (talk) 12:54, 24 October 2011 (UTC)[reply]

The human height article may be of interest to you. Dismas|^(talk) 12:58, 24 October 2011 (UTC)[reply]

Thanks Dismas. That article seems to be focusing just on the history of past few years. If you look at this site, [6], you will find some exaggerated figures and I need to know if it were true and what reliable sources can also show this.--Almuhammedi (talk) 13:41, 24 October 2011 (UTC)[reply]

See also gigantopithecus. ~AH1 ^(discuss!) 14:37, 24 October 2011 (UTC)[reply]

The site you linked to is unusually bizarre Creationist nonsense. It has nothing to do with actual science. --Mr.98 (talk) 02:22, 25 October 2011 (UTC)[reply]

breast cancer risk

Celibate and childless women are thought to be at greater risk of breast cancer. Is this because childbirth boosts the mother's immune system, and if so what is the mechanism?Pensioner.bsc (talk) 02:01, 24 October 2011 (UTC)[reply]

Does lactation change the susceptibility of breast tissue to cancer? 67.6.179.27 (talk) 02:38, 24 October 2011 (UTC)[reply]

There is an apparent correlation between menstrual cycles and breast cancer. Having children reduces the total number of menstrual cycles, which reduces the risk of breast cancer due to the correlation. -- kainaw ™ 02:44, 24 October 2011 (UTC)[reply]

As for the question by 67IP: yes, lactation triggers changes in the breast tissue that make it less susceptible to cancer. 67.169.177.176 (talk) 05:48, 24 October 2011 (UTC)[reply]

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.

→Σ τ c. 04:59, 25 October 2011 (UTC)[reply]

A censor has removed the discussion on this subject, it was meant to be a scientific question, not a personal medical question. I don't suffer from breast cancer myself, as it is not common in men, but many women do, and sensible discussion on how to minimise the risk should be encouraged, not censored. Pensioner.bsc (talk) 14:24, 25 October 2011 (UTC)[reply]

I restored the top part of the discussion. Once it went into a person asking specifically how to control his/her (but likely her) risk of breast cancer, that is medical advice. It should not be asked here and cannot be answered. -- kainaw ™ 14:46, 25 October 2011 (UTC)[reply]

Despite the phrasing of that question "Is that what women doctors advise?", it is actually not a request for medical advice to ask what doctors advise as an announcement to the general population. We see those all the time - cut cholesterol, keep blood sugar low, etcetera. But to avoid further debate I'll rephrase the last question in academic terms:

Do pills for menstrual suppression or for artificially induced lactation reduce breast cancer risk? (Pensioner.bsc (talk) 21:12, 24 October 2011 (UTC) as rephrased by Wnt (talk) 15:06, 25 October 2011 (UTC))[reply]

That's OK, perhaps I shouldn't have mentioned the word "doctor", which triggered alarm bells!Pensioner.bsc (talk) 17:54, 25 October 2011 (UTC)[reply]

This is a huge topic and I'm just poking a pile of literature and reading what falls off the top. But for example, PMID 1622631 speculated in 1992 that a birth control regimen might be devised to reduce breast cancer along with ovarian and cervical cancer; nonetheless estrogen and progesterone (in older women) can generally have an effect of increasing breast cancer risk (PMID 1622631) and is "positively dependent on oral contraceptive use" (PMID 21995151). While the intended effect of contraceptives raises breast cancer risk, the fact that they increase it in older women makes me think it's more than that. After all, breast cancer has a strong association with estrogen in general. Wnt (talk) 15:18, 25 October 2011 (UTC)[reply]

I've looked a little into the other question about induced lactation, but it's hard to find much about it in the scientific literature. If someone has done a study of breast cancer in wet nurses or women who practice erotic lactation in the long term, I didn't see it. I didn't see anything about domperidone and breast cancer. Bitter melon (Momordica charantia) is reputed to help lactation and might help against breast cancer ([7]) but the evidence is very sketchy and the mechanism might not be this. I'm afraid I don't have a proper answer here. Wnt (talk) 17:18, 26 October 2011 (UTC)[reply]

To get some references down here, I made the quickest possible riffle through PubMed, to the beginning of page 5 of 12 out of "breast cancer childbirth". Late age at first full term birth (>30 yr) increased odds of lobular breast cancer in one study by 2.4-fold, and other types by 1.2, 1.3, and 1.7 fold. (PMID 21509772 and also PMID 19569233). Another study found association of childbirth with breast cancer but not significant evidence that breastfeeding time was correlated (PMID 20851603) Perhaps mammary differentiation during pregnancy reduces breast cancer risk - but in mice, blocking that with dioxin slowed cancer formation after induction with a standard lab carcinogen (PMID 20521247) (I wonder how all this reconciles with a statement that abortion does not increase risk of breast cancer... PMID 21557713) Note also that there is a higher transient risk of breast cancer for a few years after pregnancy, which correlates a little bit with IGF-1 level and size of the infant (see [8] and some references I didn't bother to paste). Wnt (talk) 15:21, 25 October 2011 (UTC)[reply]

Thanks for the refs, as you say a huge topic, and a lot of people working on it. As I saw somewhere in the refs, historically women would be free from menstruation for long periods while pregnant or breastfeeding, and did not live so long. So if too many menstruations in modern women does increase the risk, it would seem sensible to space them out to quarterly instead of monthly. Complete suppression might be a step too far with unknown side effects.Pensioner.bsc (talk) 18:21, 25 October 2011 (UTC)[reply]

What color would the moon be

if it was seen from space, under pure white light, and the reflected light was averaged? With what shade does it tint the sunlight it reflects, and how strongly? Card Zero (talk) 12:22, 24 October 2011 (UTC)[reply]

Under visible light, there isn't really a preferred color, or specific wavelength where the Moon exhibits an unusually high reflectivity. The surface is pretty much shades of gray, just as it appears in photographs taken from space. This article about the color of the moon addresses the point directly; our article on the geology of the Moon will tell you what the surface rocks are made of. TenOfAllTrades(talk) 13:21, 24 October 2011 (UTC)[reply]

I found that Universe Today article already, but didn't have complete faith in it because it links to a page which it says "explains how to get the right color of the Moon in Photoshop". That page in fact is about how to get the wrong color of the moon in Photoshop, by adjusting levels so that the moon appears on average white, and then increasing the saturation of what's left - it's about making a kind of false color image. However, if you agree that it's gray, too, I'll go along with that part. Card Zero (talk) 13:30, 24 October 2011 (UTC)[reply]

Remember, though, that various minerals on the Moon including anorthosite may have tints that are locally shades other than gray. ~AH1 ^(discuss!) 14:35, 24 October 2011 (UTC)[reply]

I am at a loss for a reference, but I want to say that it was during the Apollo 15 lunar excursion that the astronauts discovered one incredibly out-of-place bright orange rock. I seem to recall quite a scientific hullabaloo over this one lunar sample, debating whether it had formed in-situ, had ejected during an impact, or had deposited from space... but in any event, it was the only very non-grayish sample obtained on any of the missions. I will try to track down exactly which mission discovered this sample. Nimur (talk) 06:12, 25 October 2011 (UTC)[reply]

Ah, here we go: Orange Soil near Shorty Crater, discovered on Apollo 17. It is called a volcanic glass on this particular NASA page. Unfortunately, with only a few data-points of manned exploration, we don't really know how rare this stuff is on the lunar surface. Nimur (talk) 06:15, 25 October 2011 (UTC)[reply]

More information from astronaut Harrison Schmitt, the first and only scientist (a geologist) to walk on the moon: "The Missions of Understanding," ... "Finding orange soil near Station 4 on Apollo 17 at the time when oxygen was running low kept us on the jump. We dug a trench 8 inches deep and 35 inches long, took samples of the orange soil and nearby gray soil...." This link includes more high quality, better true-color photos (including some from the Hasselblad camera), and some micrographs of samples that were brought back to Earth laboratories. An entire chapter in this book is dedicated to Beads of Orange Glass. Nimur (talk) 06:21, 25 October 2011 (UTC)[reply]

Blue sky

Is it possible that the blue color of the sky has something to do with the water color reflection on the ozone layer? Some friends of mine seem not convinced with Rayleigh scattering and wanted me to further search about, reasoning that photos by satellites don't show enough blue color for land hemisphere.--Almuhammedi (talk) 13:01, 24 October 2011 (UTC)[reply]

No, it is not possible, why is the sky blue everywhere then? It is certain to be Reyleigh scattering, blue light is scattered more than red light as it has a shorter wavelength, and it has a shorter pathlength. At dusk and dawn, all blue light is scattered, and red light is seen, even though it hasa longer pathlenth. Plasmic Physics (talk) 13:50, 24 October 2011 (UTC)[reply]

Could the overland sky be less blue due to dust storms? ~AH1 ^(discuss!) 14:33, 24 October 2011 (UTC)[reply]

Yes, it can. Plasmic Physics (talk) 22:07, 24 October 2011 (UTC)[reply]

Ice

why is ice slippery? — Preceding unsigned comment added by 203.112.82.2 (talk) 15:54, 24 October 2011 (UTC)[reply]

Ice is only slippery when it has water on the surface. Water acts as a lubricant. When ice is very dry, it is not slippery. However, it takes very dry and very cold air to produce ice that doesn't have a layer of water on the surface. Even pressure from standing on ice will cause it to produce a thin layer of water. -- kainaw ™ 15:58, 24 October 2011 (UTC)[reply]

According to the academic genius of cracked.com this is disputed, here. Here is a NY times covering much of the same ground. There is no clear answer. Grandiose (me, talk, contribs) 16:24, 24 October 2011 (UTC)[reply]

Good articles. I'd like to invite them to northern Norway in February and ask them to try to slip on the ice. This is one of those things where scientists are stating one thing and people on the ground are walking around on non-wet ice (I'm avoiding the use of "dry ice") and noticing that it isn't slippery. -- kainaw ™ 16:38, 24 October 2011 (UTC)[reply]

Northern Norway doesn't get nearly cold enough in winter for this to happen -- even Tromso and Hammerfest hardly ever get colder than about -15 C. Now, if you invite your friends to some place like Vorkuta or Norilsk, then it's a different story altogether. Us Russians got antifreeze for blood... :-) 67.169.177.176 (talk) 04:39, 25 October 2011 (UTC)[reply]

There's not much snow in Tromso - kept too clean. Up the mountains - which is where I was stationed - it is all ice. Never slipped once. Then, took a ride to the airport to fly home. Slipped about 20 times walking from the airport to the plane. -- kainaw ™ 16:28, 25 October 2011 (UTC)[reply]

Please keep in mind that it's always colder in the mountains than in Tromso itself. 67.169.177.176 (talk) 01:03, 26 October 2011 (UTC)[reply]

Point Barrow would also be a good place to demonstrate the non-slipperiness of ice in extreme cold conditions. 67.169.177.176 (talk) 05:29, 25 October 2011 (UTC)[reply]

From personal experience, clean ice is definitely still slippery at -25 C. Dragons flight (talk) 16:02, 25 October 2011 (UTC)[reply]

This means that in the city of Tromso, ice is always slippery because it never gets below -18 C. In Vorkuta, on the other hand, -40 C is commonplace and -50 is not unheard of, and Norilsk is colder still, so in those places you could walk on ice and not slip. 67.169.177.176 (talk) 01:07, 26 October 2011 (UTC)[reply]

The article Ice on wikipedia also says that your explanation is in doubt. — Preceding unsigned comment added by 203.112.82.1 (talk) 17:18, 24 October 2011 (UTC)[reply]

The surface of a piece of ice is always liquid, even at absolute zero. This fact and not that ice melts under pressure, is now believed to be the cause of ice being slippery. Count Iblis (talk) 17:13, 25 October 2011 (UTC)[reply]

So then, could you explain why ice ceases to be slippery at extreme cold temperatures (as Kainaw testified from personal experience, and as every bush pilot knows)? 67.169.177.176 (talk) 01:09, 26 October 2011 (UTC)[reply]

It's to do with contact - as soon as contact is made, the thin layer of water is no longer a surface, but an interface between two solid objects. This means that the layer of water at the interface, solidifies, and bonds the two objects together. This is true for any temperature at which ice is stable. The time it takes for the water layer to freeze and weld the ice to the touching surface is a function of time. The colder it is, the less time it takes to weld. The reason why why ice is slippery at relatively higher temperatures, is that it simply hasn't had the time weld together. You'll find that if you have a metal disk or other wise good thermal conductor, it will easily slide along on a flat sheet of ice. However, take that same disk and rest it on the ice for 5 minutes, and you'll find it stuck to the ice. Plasmic Physics (talk) 02:14, 26 October 2011 (UTC)[reply]

This is really under the domain of nanoscience. Plasmic Physics (talk) 02:17, 26 October 2011 (UTC)[reply]

Related to that answer is just how cold it was when I was in the Norwegian mountains and didn't slip. I was trained to use duct tape to seal up my tent. I couldn't. The glue on the tape froze before I could get it to stick to anything. I don't know the actual temperature because I only had a mercury thermometer and it went "mushy" on me. I figure it was close to -35c if not colder. So, I believe that any water on the ice would nearly instantly freeze. -- kainaw ™ 02:24, 26 October 2011 (UTC)[reply]

Your report of the mercury getting "mushy" indicates that the temperature was close to -40 C (but not quite there). 67.169.177.176 (talk) 02:44, 26 October 2011 (UTC)[reply]

We just established that the thin film of water does not freeze, as long as it is on the surface and exposed, doesn't matter how cold it is. Regardless, it may have more to do with the glue being affected by the temperature than the ice. Plasmic Physics (talk) 02:51, 26 October 2011 (UTC)[reply]

How can a particle be its own antiparticle?

For composite particles like mesons, it makes sense, but for elementary particles with no apparent internal structure, it doesn't seem right, yet it seems that this is the case for all bosons (except the W boson). How does annihilation work in these cases? --Goodbye Galaxy (talk) 16:21, 24 October 2011 (UTC)[reply]

The article Antiparticle indicates that for particles which are their own antiparticle will annihilate when they interact with another of their kind. Such particles obey a certain equation known as the Majorana equation which explains their behavior (well, it would explain it to someone who understood it, which I do not). --Jayron32 17:41, 24 October 2011 (UTC)[reply]

Jayron, the Majorana equation can only be used for fermions. Bosons will follow either a real Klein–Gordon equation if they have spin zero or a real vector boson equation if they have spin 1.By real I mean no imaginary numbers. If imaginary numbers are allowed than the particle and anti-particle equations are related to each other through complex conjugation. Dauto (talk) 18:20, 24 October 2011 (UTC)[reply]

Just adding a link to the Proca equation which is the spin-1 vector boson equation I was talking about. Dauto (talk) 18:35, 24 October 2011 (UTC)[reply]

That's good Dauto. You should fix the lead to the antiparticle article, because it implies that all particles which are their own antiparticle follow the Majorana equation. I was only reporting what I was told. --Jayron32 19:38, 24 October 2011 (UTC)[reply]

I removed the mistaken reference from the article. Dauto (talk) 17:04, 25 October 2011 (UTC)[reply]

So photons annihilate with other photons? Does this happen all the time? What do they annihilate into? More photons? --Goodbye Galaxy (talk) 18:05, 24 October 2011 (UTC)[reply]

Yes, photons can annihilate into more photons. It is called scattering of light by light - A purely quantum phenomenon absent from classical physics. Dauto (talk) 18:08, 24 October 2011 (UTC)[reply]

Although photons don't couple directly, they can interact through higher-order processes. See Two-photon physics. Red Act (talk) 19:40, 24 October 2011 (UTC)[reply]

(EC)Pick up a pair of gloves. They're mirror images of each other. One is the particle and the other is the anti-particle. Which one is the particle and which one is the anti-particle? It's just a matter of arbitrary convention. Now pick up a pair of socks. They still can be seen as mirror images of each other except that they are completely identical. The particle and anti-particle are one and the same. What that means is that two Z-bosons (for instance) can annihilate each other. Dauto (talk) 18:06, 24 October 2011 (UTC)[reply]

Now my feet are cold. Thanks for the explanation! --Goodbye Galaxy (talk) 18:35, 24 October 2011 (UTC)[reply]

In fact there is a reason for calling some particles "particles" and their CP-transformed brother particles "anti-particles" and not the other way around. In the universe there is more matter than anti-matter, which has something to do with the CP-violation. So one calls the normal type of matter "matter" and the other type, from which there is nearly nothing, "anti-matter".--Svebert (talk) 17:03, 26 October 2011 (UTC)[reply]

Yes, that's the criterion generally adopted. But that's still an arbitrary choice and occasionally it is convenient to use other criteria. Dauto (talk) 21:37, 26 October 2011 (UTC)[reply]

Its the same , matter adn anti matter , Its jast revers in time . thanks Water Nosfim

bird "song"

most of the birds I've seen make sounds that are too short (and repetitive) to be called "songs" are there any birds that really "sing"? like... making a melody or something? — Preceding unsigned comment added by 31.7.57.246 (talk) 18:58, 24 October 2011 (UTC)[reply]

How about this blackbird song, from our bird vocalization article: Blackbird song recorded at Lille, France Looie496 (talk) 19:10, 24 October 2011 (UTC)[reply]

Come to Australia and listen to our magpies. HiLo48 (talk) 19:44, 24 October 2011 (UTC)[reply]

I am not a musician. But Western music does seem to rely on the existence of a melody which build tension and then resolves it within a key. Resolution (music). (Think of the very characteristic "dut dut dut DAH" end of a piece of classical music.) So far as I am aware, certain birdsong may have harmonic, or, better, tonic characteristics. (The Blackbird song seems to.) But as far as I am aware, only human melody has tension and resolution. Perhaps some real musician can comment. μηδείς (talk) 02:33, 25 October 2011 (UTC)[reply]

Your example of Beethoven's Fifth there is extremely fortuitous; our article mentions that it may have been inspired by the song of the yellowhammer. Matt Deres (talk) 13:46, 25 October 2011 (UTC)[reply]

Good catch, but the funny thing is I wasn't actually consciously thinking of Beethoven's Fifth, but trying to summon to mind the dramatic flourish at the end of any symphonic piece. But yeah, I did know the attribution of the theme to a bird species. I am curious if parrots will sing entire songs. I know some do dance to them. http://www.youtube.com/watch?v=sV5bxaLDG-w μηδείς (talk) 18:27, 25 October 2011 (UTC)[reply]

Some of these might help. http://birdsinbackyards.net/feature/top-40-bird-songs.cfm(I find the grey butcher bird partcularly melodic, and this doesn't fully capture it's essence. when all the birds go crazy together in my neighbourhood at dawn, the butcher bird seems to cap it all off with the resolution Medeis mentions. HiLo48 already mentioned the magpie. There's always the Canary. I'm not a musician, so can only say what i like the sound of.. these guys might also help http://www.folkways.si.edu/albumdetails.aspx?itemid=1106 ("The Lyrebird: A Documentary Study of Its Song")WotherspoonSmith (talk) 07:34, 26 October 2011 (UTC)[reply]

kinetic theory and compression: explaining Boyle's law

It's long been intuitive -- by operating any double-action bicycle pump -- that compression of a gas beyond atmospheric pressure takes work, and that balloons undergo explosive decompression in space.... but on a kinetic level, how does the energy get transmitted? I know it's force x displacement and so forth, but suppose I am compressing a piston and I move walls ever so slightly and frictionlessly (let's say, the length of one tenth of the mean free path).

I guess I don't get the Newtonian argument that restricting the motion of a bunch of colliding particles would increase its kinetic energy -- wouldn't it have the same KE, but less freedom? I get the idea that as the walls become closer, collisions against the walls are more frequent and so the force against the walls increase.

But suppose I moved the piston so slowly that for every discrete move, I only feel the "bump" of one gas molecule at a time. In the time that I don't meet a gas molecule, force is zero, so F x d is zero. Why would moving the piston walls a little closer (cutting the mean free path of the next gas molecule short) speed up the next gas molecule? After all, if I do it slowly enough, the magnitude of momentum change should remain the same, right? elle _{vécut heureuse} ^{à jamais} (be free) 20:37, 24 October 2011 (UTC)[reply]

Kinetic theory laws and properties cannot be understood that well individually, but make much more sense in "bulk". Think about these ideas like "temperature". An individual particle cannot have a temperature, it also cannot exert a pressure, and we cannot think of the "volume" of a gas in terms of the volume of the individual particles. The basic properties of Kinetic Theory just don't make sense on a particle-by-particle basis. These properties only make sense in bulk, pressure and temperature and volume require mole-level numbers of particles to start to make sense. When you start to think of reducing the properties to the level of individual particle behaviors, the model breaks down. --Jayron32 21:22, 24 October 2011 (UTC)[reply]

Except, my professor used a particle-by-particle approach for a "proof", and then extrapolated to all particles. While this seems to prove having a certain pressure, given n, a volume and temperature, I was trying to think of a mechanism for transmitting K.E. elle _{vécut heureuse} ^{à jamais} (be free) 10:55, 25 October 2011 (UTC)[reply]

Actually, you can improve your though experiment by moving the wall by a small step only when there are no molecules striking the wall. So, no energy is transferred via work to the gas at all, yet we are compressing the gas. Clearly, in this case the temperature doesn't increase.

If you think about how you would move the piston to make sure you don't perform work, you see that you must actually move it very fast. At any moment you have molecules very close to the piston and there is always a clear space between the postion and the nearest molecule. You can then move the piston very fast to let that distance shrink a bit without hitting a single molecule. Then you wait until there is again a clear space and repreat this.

Then this process violates the assumption of slow quasistatic changes under with the formula for work done by the gas of p dV is valid. If you increase the volume then it is more straightforward to see that increasing the volume slowly will lead to the gas doing work. But if you suddenly increase the volume, the gas won't do any work at all.

In a quantum mechanics treatment you recover this result. A gas in some volume can be in states will well defined energies (so-called energy eigenstates). These are wavefunctions psi_r(x1, x2, ..., xN; V) of the coordinates x1, x2, etc. of the N molecules that depend on the volume and on a qantum mumber r that enumerates the energy levels. The energy of this state then has some value that depends on the quantum number r and on the volume, we can denote it as E_r(V). If you change the volume V very slowly, then the probability that the system will jump from one quantum number r to another becomes very small. This then means that the change in the energy if the system is given by the change in E_r(V) for fixed r.

If you increase the volume of the system very fast, then the wavefunction of the system will not change. This means that the the expectation value of the energy will stay the same. Count Iblis (talk) 21:54, 24 October 2011 (UTC)[reply]

Count Iblis's initial paragraph, describing how you wait to move the wall until there are no molecules to hit, is a varient of Maxwell's demon, which is essentially a thought experiment which shows the breakdown of the laws of thermodynamics when taken to the level of individual particles rather than bulk materials; it would provide an excellent primer on the OPs question as well. --Jayron32 22:18, 24 October 2011 (UTC)[reply]

Knowing when to move the wall would require measuring the molecules hitting it, which cannot be done without energy exchange. I guess in theory it would be possible <ith very few atoms. Given that the average free path at 1 atm is 60 nm, it's not that simple...

There's a 2D simulation of particles confined between a thermal reservoir and a piston at opensource physics: Lennard-Jones PVT System http://www.opensourcephysics.org/items/detail.cfm?ID=10857 download link at bottom, java .jar file. You can move the piston and in the top right you can enter the termal conductivity of the reservoir, 0 in this case.

DS Belgium (talk) 01:05, 25 October 2011 (UTC)[reply]

I'm thinking of the case of free expansion in vacuum. No work is done, but entropy increases (irreversibly), and yet the gas must heat up if attempt to compress it back into its original volume (adiabatically). It seems I could make the gas ridiculously hot by repeating this indefinitely. So we are dealing with a sort of path function. I guess my question is, why should moving the walls transmit K.E. on a molecular scale? I know it increases K.E. from a force * distance perspective. Kinetic theory says that smaller walls means more frequent collisions -- hence higher pressure, but it doesn't seem to imply a higher temperature. [in fact we can see that a smaller volume at the same pressure implies a colder gas]. elle _{vécut heureuse} ^{à jamais} (be free) 10:55, 25 October 2011 (UTC)[reply]

When molecules collide with moving walls they gain energy, increasing the gas temperature. Dauto (talk) 12:42, 25 October 2011 (UTC)[reply]

October 25

Speed of the average water molecule at room temperature?

Can anyone give the speed in miles per hour of an average H2O molecule in a body of pure water at 1 Atmosphere and 68° Fahrenheit? The proper equation would be nice. μηδείς (talk) 02:23, 25 October 2011 (UTC)[reply]

v_ave = √ [8k_BT ÷ π ÷ m]

Plasmic Physics (talk) 02:34, 25 October 2011 (UTC)[reply]

You need to apply the equipartition theorem, because a water molecule is not a point-particle. The equipartition theorem states that the energy will be equal in all modes (on the average); so, you need to estimate the vibrational and rotational modes of energy; then count all the degrees of freedom, and divide the average total energy per molecule amongst each mode. Then, compute the velocity that corresponds to the translational motion modes. Nimur (talk) 02:42, 25 October 2011 (UTC)[reply]

Here it is: Stowe's Introduction to Statistical Mechanics and Thermodynamics. I have a hardcover red copy that I refer to as my little red book of thermal physics (they changed the cover in the 2nd edition, apparently). Stowe's textbook is essentially the only physics books you will ever need (because everything else can be derived from these principles - at least in statistical ensemble - and what else matters, besides ensemble behavior?) Unfortunately, reading this text does require at least a little preparation and familiarity with some more elementary physics and mathematics concepts. Nimur (talk) 02:53, 25 October 2011 (UTC)[reply]

Well, I would actually be happy with an approximation for ideal point particles. I just want something accurate on an order of magnitude. And I can figure out the equations with the ideal gas laws, or at least follow the derivation. But I figure there are complications for liquids, given they are not gasses. And I was hoping there must be something more obvious for water than an entire book, based on something llike pv=nRt. μηδείς (talk) 03:39, 25 October 2011 (UTC)[reply]

Okay, well here is a summary from HyperPhysics, Molecular Kinetic Energy. Unfortunately, to compute this molecular translational velocity correctly, you have to dive deep a little bit - the physics of molecular motion is nontrivial, even treated in a purely classical non-quantized way. There's no practical utility in incorrectly approximating molecular motion. Be sure you understand the difference between average speed, most probable speed, expected-value of velocity in one dimension, and so on. Ideal gases follow the Maxwellian distribution, elaborated here. Water is not an ideal gas, because it has viscosity, its molecules are asymmetric, hydrogen bonding and van der Waals forces are nonnegligible, and so on. Nimur (talk) 05:56, 25 October 2011 (UTC)[reply]

The redlink above is probably in reference to the Maxwell–Boltzmann distribution (sometimes Maxwell distribution) Grandiose (me, talk, contribs) 21:20, 25 October 2011 (UTC)[reply]

Is k_B the Boltzmann constant? Dualus (talk) 04:36, 25 October 2011 (UTC)[reply]

Yes. Plasmic Physics (talk) 04:37, 25 October 2011 (UTC)[reply]

Is there such thing as an ideal liquid, or would that just be the same thing as an ideal gas at extreme conditions? (I am wondering what would happen to the p variable.) Yes, I understand the difference between mean, median and mode. Yes, I understand that water molecules are polar, and that the angle between the hydrogen atoms varies. I can even figure out that unless a body of water is moving, the net average velocity of its particles will have to be zero. But since the atoms are indeed moving it seems reasonable to me to ask for some average speed at a given temperature and pressure. All I can remember from the gas laws is that the answer is calculable and on the order of 1000mph (or was it kph?) for air molecules, with a mean free path of ~10cm, if I remember correctly. Surely there is some sort of analog for water molecules. μηδείς (talk) 11:27, 25 October 2011 (UTC)[reply]

The assumptions of an ideal gas are dilute concentration which, a liquid is not, and very low temperatures. Plasmic Physics (talk) 11:38, 25 October 2011 (UTC)[reply]

The Van der Waals equation gives a fairly good description for a liquid while still keeping the math reasonably simple. Dauto (talk) 12:34, 25 October 2011 (UTC)[reply]

The system not being an ideal gas is irrelevant here . The translational part of the partition function factors out, so barring quantum effects, you have the Maxwell distribution for the velocity. Also that H2O is not a point particle is also irrelevant, because, again, the partition function for the translational part factors out of the configurational part rotational, vibrational parts (note that these vibrational, rotational and other internal parts may not factorize as they are coupled, but the translational part will factor out due to physics being invariant under translations).

The only issue with H2O being a liquid as far as the velocity distribution is concerned, is that there will be strong correlations between nearby molecules. Count Iblis (talk) 15:23, 25 October 2011 (UTC)[reply]

In other words, the only issue is that the actual molecular velocity depends on the non-ideal parts? I agree whole-heartedly. Except for the non-ideal parts, water is an ideal gas. The ideal velocity can ideally be calculated by ignoring any non-ideal effects. PV = nRT and such, for n spherical non-interacting water-molecules and cows. Count Iblis, while you are correct in stating that translational motion is not coupled to the rest of the molecular motion, your assertion implies that these other energy modes do not contribute to the temperature of the system: I think that assertion is incorrect.

Temperature is comprised by all the energetic degrees of freedom in a system - not just the translational kinetic energy; this is the definition of "temperature." You can't ignore the other energetic modes, especially in a molecule as complicated as the water molecule. You must define all the degrees of freedom and apply equipartition theorem. Unlike, say, Helium, these other degrees of freedom are non-negligible for water - which is why we call it a non-ideal gas (or fluid). Nimur (talk) 21:08, 25 October 2011 (UTC)[reply]

Count Iblis is right (as usual). Note that the question is not about the total energy of the molecules at a given temperature in which case you would be right and the equipartition theorem would have to be invoked. The question is about the speed of the molecules and that will be given by the Maxwell-Boltzmann distribution. Note that Count Iblis never said that the ideal gas is a good description for the liquid water. That would have been a ridiculous statement. What he said was that the Maxwell-Boltzmann distribution is a good description for the liquid water, and in that he is right. Dauto (talk) 03:38, 26 October 2011 (UTC)[reply]

Correct me if I'm wrong, but my impression from Temperature is that E_k = 1/2 kT for each degree of freedom; thus, there should never be any need to divide by the number of degrees of freedom to figure out the speed of any given particle at a given temperature. If you have a liquid made up of a mix of molecules that can stretch and rotate in a hundred different ways, and individual atoms floating free, the individual atoms don't have any way to know they're supposed to go faster than the others when they hit something (unless they have less mass, that is). Of course, those hundred different extra rotations do carry energy and are in equilibrium, so if you cool such a liquid I suppose it should release quite a bit more energy over time / larger heat capacity. Wnt (talk) 23:19, 26 October 2011 (UTC)[reply]

Lightsabers

Would plasma or lasers be better suited for a lightsaber, and which would be more practical? →Σ τ c. 05:59, 25 October 2011 (UTC)[reply]

Neither is practical. Light sabers are impractical; they make for exciting fiction, but they defy any attempt to explain their behavior according to known physics. The real behaviors of both plasmas and lasers are very dissimilar from the special-effects that were used in Star Wars. Nimur (talk) 06:05, 25 October 2011 (UTC)[reply]

Could you explain further? →Σ τ c. 06:14, 25 October 2011 (UTC)[reply]

Sure. Let's start with plasma. Think of a plasma as a very hot gas. It has other special properties, but your starting point for intuition should be "very hot gas." In what way do you think very hot gases behave like a sword?

Now, let's move on to a laser. Think of a laser as a very bright light. It has other special properties too, but your starting point for intuition about lasers should be "very bright light." In what way do you think very bright lights behave like a sword?

Ultimately, I'm sorry that you were exposed to fictional ideas about lasers and plasmas before you were ever exposed to nonfictional facts about them. Rest assured - both lasers and plasmas are fascinating for all sorts of reasons - but they have nothing to do with light sabers.

In fact, there are cold plasmas and dim lasers, but this doesn't change the fact that you can't build swords from either.

If you have some prior scientific knowledge, I recommend a better intuitive starting point for both concepts - albeit this requires more than laymans' terminology. A plasma is an ionized gas whose ensemble behaviors are determined by electromagnetic effects, rather than just by kinetic effects. A laser is a machine designed to amplify light, typically resulting in a very coherent, monochromatic, high power output signal. These more technical descriptions are much more accurate than the ones I gave above, but they are harder to "really" understand for a lot of people - which is why we have sci-fi stories about laser/plasma/swords. Nimur (talk) 06:43, 25 October 2011 (UTC)[reply]

The clearest way I know to explain the problems with a lightsaber in a layman's terms is to ask the following: if it is a laser, a beam of light, what caps it so it doesn't shoot off into infinity like any other beam of light, and if there is some physical cap, what holds it in place and why wouldn't a cap at the tip get caught up on things? If, on the other hand it is a plasma, what, like the glass of a neon light, holds the excited gas in place? I always figured it has to be some field-bound mono-molecular filament in some excited state. Surely there is a book on the physics of Star Wars? μηδείς (talk) 11:41, 25 October 2011 (UTC)[reply]

Better than that - there's a Wikipedia article: Physics and Star Wars. Alansplodge (talk) 12:52, 25 October 2011 (UTC)[reply]

The force bends the light. That's why you have to have the power of the force to use a light saber. Otherwise, it is just a clunky flashlight. -- kainaw ™ 12:51, 25 October 2011 (UTC)[reply]

This blog has some conjecture about light sabers. Alansplodge (talk) 12:58, 25 October 2011 (UTC)[reply]

You can purchase a lightsaber in most toy shops, or you could. They are basically a translucent plastic tube with a flashlight at one end. Have fun.--Shantavira|^{feed me} 14:02, 25 October 2011 (UTC)[reply]

I wouldn't utterly rule out a kind of deliberately shaped ball lightning, but ... it's a really big stretch. Also I remember reading a news report recently (<1 month) about transmission of light in packets which somehow shape themselves, even in vacuum, to keep a bit of light in a tight bunch, but I can't find it now; you might be able to devise some handwavy explanation about how you put "english" on the original composition of such clumps of light so they boomerang back. Still, until someone turns up with a bona fide light saber it's all mist and vapors. Wnt (talk) 14:27, 25 October 2011 (UTC)[reply]

The most practical light sabers are made from computer graphics. ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:35, 25 October 2011 (UTC)[reply]

How about magnets and mirrors? →Σ τ c. 22:14, 25 October 2011 (UTC)[reply]

It is well-established that a lightsaber consists of plasma encased within a force-field. See the Wookieepedia article on lightsabers for more information. --130.216.172.199 (talk) 22:18, 25 October 2011 (UTC)[reply]

Ok, but since "force-fields" are complete make-believe that just brings us back to where we started. APL (talk) 03:02, 26 October 2011 (UTC)[reply]

Oh, well that's reassuring. Since the plasma is kept safely contained within a force field, you can't accidentally cut your fingers on it ... ? Wnt (talk) 23:06, 26 October 2011 (UTC)[reply]

Watch the following video series, http://www.youtube.com/watch?v=xSNubaa7n9o This is about as close as you will get to a working lightsaber, but obviously there are more practical uses of the technology used in Michio Kaku's design. ScienceApe (talk) 15:43, 26 October 2011 (UTC)[reply]

Hello chaps - digging way back into the Lightsaber article history gives us this choice info :-

"Lightsabers, made of immaterial beams of light, collide when they should pass right through one another without a sound. Moreover, laser beams propagate in a straight line as long as they do not meet an obstacle, therefore scientifically correct lightsaber blades would cut through a starship hull. This would mean that dueling with lightsabers would be like dueling with flashlights because the blades would pass through each other and continue to travel until they hit an object, not reflecting back onto themselves.

Instead of a laser-based device, the most believable design for a lightsaber-like device would use plasma confined by a magnetic field. Plasma is a super-heated gas and is also the fourth state of matter, the color and luminosity of which varies depending on the temperature and composition. Plasma could be ionized by a particle beam from a compact particle accelerator; at relativistic energies, the beam would produce its own blue glow along its axis from Cherenkov radiation.

Keeping a gas in the plasma-state requires considerable energy: 40 kW are necessary for a 10 centimeter (4 in) saber at 10,000°C (18,000°F). It would be difficult to fit the required generator into the saber's hilt. To control and increase the length of the blade, the plasma would need to be confined within a magnetic field. Although this design would behave like a lightsaber from the Star Wars movies, it is considered foolhardy to confine plasma magnetically. A handful of magnets would disrupt the confinement field, and plasma would spill onto the saber's wielder. Furthermore, the magnetic field would prevent the plasma from performing any cutting action because it would always be shielded from whatever the blade struck by the magnetic field.

To make a semi-solid beam of energy which could interact with both matter and energy would require containing a quasar and quantum singularity inside the hilt. The gravitational field would pull all the quasar's expelled plasma back moments after the quasar releases them. The speed of the returning plasma would form a chainsaw effect allowing it to cut through the matter with ease, while still being stopped by an opposing beam. A modulated gravity field would bounce, allowing for the reflection of energy beams. So far, no known substance exists which is able to contain a quantum singularity or to contain and/or focus a gravitational field.

A lightsaber in reality would be difficult to wield, since its center of gravity would be in the hilt next to the hand. A real sword has its center of gravity further from the hand, which allows the user more control and power in the swings and thrusts. With a lightsaber, a user would need to have much more control and wrist strength in order to accurately use a lightsaber with strength and power. However, because the blade is supposed to be made of light, it would have no weight at all. In that instance, the blade would move easily while the hilt moved slowly. According to commentaries on the prequel trilogy DVDs, the lightsaber props were indeed built so that their center of gravity was centered in the hilt, as if there was no blade at all."

Quintessential British Gentleman (talk) 19:56, 26 October 2011 (UTC)[reply]

" Furthermore, the magnetic field would prevent the plasma from performing any cutting action because it would always be shielded from whatever the blade struck by the magnetic field." -- Only if the "whatever the blade struck" was magnetic in its own right. 67.169.177.176 (talk) 06:01, 27 October 2011 (UTC)[reply]

" dissolving " water in other solvents

How can I calculate the amount of water I can " dissolve " into another solvent. For instance how can I calculate the amount of water that can dissolve in octanol before one can percieve two different phases? — Preceding unsigned comment added by 137.224.252.10 (talk) 09:48, 25 October 2011 (UTC)[reply]

You can't, you have to find it through experimentation. Plasmic Physics (talk) 12:33, 25 October 2011 (UTC)[reply]

Are there tables where you can look up this kind of thing? — Preceding unsigned comment added by 137.224.252.10 (talk) 12:48, 25 October 2011 (UTC)[reply]

See solubility chart and solubility table. μηδείς (talk) 17:42, 25 October 2011 (UTC)[reply]

Octanol doesn't mix well with water, so you'll only be able to dissolve a small amount of water in it before you get phase separation. (Best way to find out how small is to actually try it.) 67.169.177.176 (talk) 01:29, 26 October 2011 (UTC)[reply]

Just checked the solubility chart: the solubility product is negligible, so you might not be able to dissolve any water in octanol at all. 67.169.177.176 (talk) 01:32, 26 October 2011 (UTC)[reply]

Are you sure you looked under octanol, and not octane? "At 25 °C, the solubility of water in octanol is quite large, approximately 0.275 mole fraction, but the solubility of octanol in water is just 7.5 × 10 −5 mole fraction." [[9]]. See the Ropel reference. Dominus Vobisdu (talk) 02:07, 26 October 2011 (UTC)[reply]

Yes, I checked under octanol, C₈H₁₇OH. However, it does appear to me that I was looking at the octanol-in-water number rather than water-in-octanol. I stand corrected. 67.169.177.176 (talk) 02:47, 26 October 2011 (UTC)[reply]

Everything disolves, a concentration measureable in nanomoles or picomoles per cubic decimentre is still a concentration. Plasmic Physics (talk) 01:42, 26 October 2011 (UTC)[reply]

Thanks to everyone for their replies. They have been very helpful! — Preceding unsigned comment added by 137.224.252.10 (talk) 11:12, 26 October 2011 (UTC)[reply]

Price of airplane ticket: weight vs. volume

First or business class tickets are much more expensive than tourist class. However, you weigh the same, no matter what class you fly, but occupy more space, depending of class. Does that means that for airplanes, the volume you occupy if more important for determining the cost than your weight? Or is it simply a market dynamic: they charge more because they can. Strangely, when you send a package they normally charge you by weight. Quest09 (talk) 13:20, 25 October 2011 (UTC)[reply]

Well, skimming the article, a Boeing 747 has a maximum takeoff weight of 735,000 to 970,000 pounds, and seats up to 550 passengers. (There are all kinds of variations painstakingly documented there, which I basically ignored; a more careful reading may get you more accurate figures) At anywhere from 1330 to 1760 pounds per passenger, the passenger weight is a pretty small factor. Wnt (talk) 14:17, 25 October 2011 (UTC)[reply]

Don't forget fuel weight, it makes a huge difference. Our 747 article doesn't contain enough info for a proper calculation, but for the Boeing 727 one can work out that fully loaded and fully fueled, there is capacity for somewhere in the neighborhood of 250 pounds per passenger, including luggage. Looie496 (talk) 20:17, 25 October 2011 (UTC)[reply]

Well, I suppose that's how they define "fully" fueled. But jetliners don't usually fly with full fuel. Wnt (talk) 16:49, 26 October 2011 (UTC)[reply]

Also, the US postal service has recently been ggressively advertising a Flat Rate Priority Mail program with presized boxes and a slogan "if it fits it ships", so I think this applies to some extent to air shipping also. Wnt (talk) 14:19, 25 October 2011 (UTC)[reply]

The smallest USPS Flat Rate envelope comes with a 70lb domestic and 4lb international weight limit.μηδείς (talk) 16:00, 25 October 2011 (UTC)[reply]

Drat, I thought I had found a cheaper way to mail plutonium to people I dislike. :-) StuRat (talk) 03:04, 26 October 2011 (UTC)[reply]

The cost of flying a plane from A to B is basically fixed. You then want to maximise the amount of money you make from the flight. That's determined by business concerns, not engineering ones. First class doesn't really exist to make money by being sold. I don't have any figures, but I don't believe many first class tickets are sold. The seats are often either empty or filled with people that got upgrades (either free or with airmiles). They exist to serve as a reward for their frequent flier programs (which exist to get people to buy more economy and business tickets), and also to make business class look cheap by comparison. The improvements from business to first are minimal, at least in the air (you do get some special treatment on the ground with a lot of airlines), so very few people would pay the enormous increase in price just to get them. You travel first class either because it's a free/airmiles upgrade, someone else is paying, or you want to do so for the sake of it (eg. to show off how rich you are, or because it makes your honeymoon more special). Once you are rich enough for the extra price not to be significant to you, you're probably chartering private jets rather than travelling on commercial flights. Business class is a significant improvement on economy, so is worth the money to a lot of people - I'm not sure airlines actually make more from a business class seat than they would by filling the same space with economy seats, but they would struggle to get businesses to name them are their preferred airline if they didn't offer it, even if the business usually flies people economy. (Apologies for the complete lack of references - I'll try and find some later and will come back!) --Tango (talk) 18:17, 25 October 2011 (UTC)[reply]

Minor comment: the cost of operating an aircraft does vary as a result of fuel consumption; and fuel efficiency does vary as a result of gross takeoff weight; so it's not a totally fixed cost to fly a plane from A to B. Otherwise, Tango's pretty spot-on; most of the short-term decisions in commercial aviation are made on the basis of marketing and business choices, not on engineering limitations of the aircraft. Over the much longer term, major airlines choose how to strategically invest in their fleets in order to keep their operations in line with their business objectives: for example, purchasing large aircraft versus small aircraft; regional- vs. transcontinental- aircraft; and so on. Southwest Airlines is often cited for operating an all-737 fleet - excellent, versatile, large aircraft. (No first-class on SouthWest, either). This consolidates planning, maintenance, training, and other costs, but also consolidates risk. Two business analyses, both from Forbes: Southwest Gives Fleet A Facelift, And Saves Some Dough Too; and following a 737-related aircraft safety incident, Southwest shares drop.... Corporate strategy is a finely-tuned game of highly-quantitative guess-work. Nimur (talk) 19:20, 25 October 2011 (UTC)[reply]

Boas vs. pythons

After reading this summary of the similarities and differences between boas and pythons, I had the following insight: without the anatomic and physiologic distinctions between the two families, the matter of inhabiting different locations would be irrelevant. Is difference in worldly habitat really, then, a difference between the two families? As a contrast, I offer the distinction between polar bears and giant pandas, which not only possess anatomic and physiologic differences, but a uniqueness in their areas of distribution that is, I think, somewhat necessary to their survival -- pandas could not exist in the arctic circle, although polar bears could probably live in China, as they do in zoos all over the world. DRosenbach ^{(Talk | Contribs)} 13:36, 25 October 2011 (UTC)[reply]

Well, I think you're missing a key concept here: There is a relationship, an interrelatedness between the environment an organism lives in and the particular physiological and genetic traits that produce them. Two important things: 1) None of the three factors is absolutely responsible: organisms change their environment, environmental pressures cause genetic changes, genetic changes cause changes in form and behavior, which cause further environmental changes. It is a series of complex feedbacks all affecting each other. 2) Nothing is static. The system of relationships changes continuously and never settles into any static set of relationships. There are periods of precarious equilibrium where some locations and some times show periods of stability, but nothing is ever permanent. --Jayron32 15:25, 25 October 2011 (UTC)[reply]

Your points are well taken, Jayron32, but my question revolves around the notion that, despite the loosely-fitting old-world/new-world classification of pythons and boas, respectively, introducing boas into Australia or Africa or pythons into Central America (or South Florida) doesn't seem to affect the species, unlike moving a panda into the arctic. If a difference is that boas don't naturally inhabit Australia and pythons don't naturally inhabit, but that difference is really only consequential as a side point (as can be demonstrated by invasive species of feral pythons in S. FL) -- is it really accurate to refer to it as a difference? DRosenbach ^{(Talk | Contribs)} 16:40, 25 October 2011 (UTC)[reply]

The fact that a species can survive in a place doesn't mean it isn't different from the species native to a place; invasive species can thrive and actually greatly alter an environment; it doesn't mean that they are somehow indistinct from native species. C.f. rabits in Austrialia. --Jayron32 20:38, 25 October 2011 (UTC)[reply]

Modern taxonomy uses cladistics which groups species by their evolutionary history. One cause of populations drifting apart evolutionarily into separate species is that two groups become geographically separated so they no longer interbreed. The differences in the geographic ranges of two families can say a lot about evolutionary distance. Rckrone (talk) 03:54, 26 October 2011 (UTC)[reply]

EMP effects on artificial pacemakers

What would be the effects of an exposure to EMP from a short distance on artificial pacemakers (will they stop working? for a limited period of time? permanently? and why?), and the person who owns it? (please correct my English)

It is difficult to find precise information, but the general conclusion seems to be that artificial pacemakers are susceptible to damage from a sufficiently strong EMP. Looie496 (talk) 18:04, 25 October 2011 (UTC)[reply]

I'd think it would have to be a very strong and/or close EMP, since the main susceptibility to EMP is due to long wires which build up charge over miles (such as electrical transmission wires on high tension towers). The wires on a pacemaker are relatively short. StuRat (talk) 03:09, 26 October 2011 (UTC)[reply]

...and the voltage required to screw it up is relatively small. Spinning Spark 21:46, 26 October 2011 (UTC)[reply]

Decapitation

Could any of you give me a brief description of what ligaments, muscles and bones would be cut through by a killer using an axe to sever a human head please? — Preceding unsigned comment added by IsonomicJedi (talk • contribs) 16:37, 25 October 2011 (UTC)[reply]

You can check google images for an "axial section of the neck," but this diagram looks like it answers your question. If you include the term 'Netter' (he's a very famous and accomplished anatomy illustrator) you'll likely get better results. DRosenbach ^{(Talk | Contribs)} 16:43, 25 October 2011 (UTC)[reply]

Note though that using an ordinary axe to sever a human head is essentially impossible. Even professional executioners, tremendously strong and well-trained and using a much broader and much sharper axe on a victim lying stretched out motionless in front of them, frequently failed on the first try. With an ordinary axe all the killer would do it make a gory mess. Looie496 (talk) 17:55, 25 October 2011 (UTC)[reply]

Hence the guillotine, leading edge technology of its day. ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:28, 25 October 2011 (UTC)[reply]

Cutting edge technology, surely? Greglocock (talk) 00:28, 26 October 2011 (UTC)[reply]

Or bleeding edge technology, perhaps ? StuRat (talk) 03:17, 26 October 2011 (UTC) [reply]

D'oh! I missed that obvious punning situation. And don't call me... you know. :) ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:02, 26 October 2011 (UTC)[reply]

...sharp? Pfly (talk) 03:26, 27 October 2011 (UTC) [reply]

Earthquakes and Brick Buildings

In terms of safety, would brick buildings be a good choice in an earthquakeprone area? Compared to, for example, timber construction or steel frame buildings. Thanks, Wanderer57 (talk) 18:20, 25 October 2011 (UTC)[reply]

Brick buildings are the absolute worst. In earthquake-prone parts of California they are forbidden. Steel is best, timber next. Concrete is bad, brick just disintegrates. Looie496 (talk) 18:34, 25 October 2011 (UTC)[reply]

Around these parts of California, you hear the terminology "un-reinforced masonry construction" - it's not merely that the bricks are present, its that structural integrity of the building is entirely dependent on the "glue" (the mason's mortar) that holds each brick together. An unreinforced brick building will "disintegrate," if the shaking gets bad enough. If the building had rebar steel inside it, and brick on the outside, the problem is less severe: the building may remain standing in an earthquake even if the mortar cracks and disintegrates, and bricks fall off. However, you still have the serious problem of falling bricks coming off the side walls: this can be very hazardous. Even if the building doesn't collapse, it only takes one loosened brick falling from just a few feet up on a wall to land on your head, causing a serious casualty or fatality. (I used to think - how unlikely is that? What's the probability of one brick falling and landing on your head? ... But after surviving a few earthquakes, I realized ... the entire region is shaking - somewhere in a fifty mile radius, somebody is standing near a brick wall that ~~could~~ will collapse). It's safer to avoid such construction altogether. Here's more information from the State of California: Typical Unreinforced Masonry Building Damage from our multi-county regional planning commission in the San Francisco Bay area; and here's Putting Down Roots in Earthquake Country from the Federal Government (the USGS). Lots of good photos and statistical information in that one; you can order a free paper-copy from the US Geological Survey if you want one. Nimur (talk) 19:05, 25 October 2011 (UTC)[reply]

Generally there are requirements to retrofit such buildings in earthquake-prone areas by putting in cross-braces, so that there is something that actually holds the building together should a quake occur. Mikenorton (talk) 19:20, 25 October 2011 (UTC)[reply]

Instead of bricks, if you use cinder blocks, and run rebar through the hole in each, you can come up with a reasonably safe construction. StuRat (talk) 03:15, 26 October 2011 (UTC)[reply]

Wood frame construction is surprisingly durable in an earthquake, provided it's tied to its foundation properly. As noted above, unreinforced masonry performs poorly: it can just come apart. While it's possible to engineer such structures to withstand earthquakes, that work involves inserting what amounts to a parallel structure into the masonry building to deal with the forces that masonry can't withstand. Examples of unreinforced masonry include the Washington Monument and Washington National Cathedral, which suffered significant damage in a rather moderate earthquake. Acroterion (talk) 17:16, 26 October 2011 (UTC)[reply]

Thank you all. My question was prompted by the situation of Bangladesh. I'm led to believe that Bangladesh is in an active seismic area, is substantially a delta susceptible to shaking, and uses bricks as a major construction material because the soil can easily be used to make bricks. Comments? Wanderer57 (talk) 05:05, 27 October 2011 (UTC)[reply]

It lies close to the plate boundary between the Indian Plate and the Eurasian Plate and the earthquake hazard will increase in the northern and eastern parts of the country where the boundary lies, although it is significant everywhere. This suggests that the country is relatively unprepared for a major earthquake, such as the 1897 Assam earthquake, which severely affected the area that is now Bangladesh, with most of the brick buildings in Dhaka being destroyed (according to this). Mikenorton (talk) 07:22, 27 October 2011 (UTC)[reply]

October 26

Phase space

How to calculate number of quantumstaes or unit cells within energy range E and E+dE in the phase space to prove the eqn: g(E)dE=[(8π√2V)/h^3]*m^(3/2)√EdE — Preceding unsigned comment added by Intr199 (talk • contribs) 13:41, 26 October 2011 (UTC)[reply]

The number of states in a volume d^3p of momentum space is

V/h^3 d^3p

You can write this in terms of dE by integrating over all angles. This leaves you with:

V/h^3 4 pi p^2 d|p| = V/h^3 4/3 pi d|p|^3 =

(using E = p^2/(2m))

V/h^3 4/3 pi d (2 m E)^(3/2) =

V/h^3 4/3 pi (2m)^(3/2) 3/2 E^(1/2) dE =

V/h^3 4 sqrt(2) pi m^(3/2) E^(1/2) dE

If the particles have spin 1/2, then you need to multiply this by 2. Count Iblis (talk) 15:34, 26 October 2011 (UTC)[reply]

"using E = p^2/(2m)"

why kinetic energy is being considered only?Intr199 (talk) 16:40, 26 October 2011 (UTC)[reply]

This is for free particles. There is actually a potential, the particle is confined to a volume, so you can consider the case of a particle in a box, solve the Schrödinger equation and then derive this result. Count Iblis (talk) 23:22, 26 October 2011 (UTC)[reply]

Can just anybody get fat?

I'm not sure how to search for this, so I'll just throw out as a question and see what I get: can a great majority of average men of normal height reach BMI of 45/body weight of 150 kg or so (330 pounds) simply by eating a lot, if they tried? Or would many of them develop type II diabetes and just pee out the sugar or otherwise run into obstacles preventing the attempt? (I'm not interested in psychological/social factors here, only the metabolism) Wnt (talk) 17:28, 26 October 2011 (UTC)[reply]

Sure, when energy input exceeds energy used you gain weight. It's a simple equation. You can also get type II diabetes. It isn't like an either/or proposition. --Jayron32 17:43, 26 October 2011 (UTC)[reply]

There are a few people whose metabolism doesn't allow them to get fat, whatever they eat. This seems particularly true during growth spurts in the teenage years, and for those who are athletic. However, the majority of people seem to pack on pounds quite easily. The only question, then, is if they could reach that weight without dying first. Diabetes doesn't keep you from gaining further weight, but it could kill you. However, due to modern medicine, insulin should keep them alive until they reach the weight you want. (Planning a cannibal feast ?) StuRat (talk) 17:55, 26 October 2011 (UTC)[reply]

^{[citation needed]} I know some people can get a lot of junk and don't seem to get fat. This doesn't mean they have a metabolism that doesn't allow them to get fat. It's possible if they really tried, e.g. if someone forced them to eat 10 Big Mac combos and a 10 pack of KFC a day, they would get fat. They may vomit or have other problems at first but it wouldn't surprise me if they would eventually learn to adapt. Nil Einne (talk) 18:22, 26 October 2011 (UTC)[reply]

StuRat: I'm wondering if there is any real difference between a highly obese person with type II diabetes, and a normal person with no propensity to become overweight, except for the obesity. It seems like what would be called insufficient insulin production from the pancreas for an obese person would still be enough to provide for all the cells of a person of normal weight; likewise the insulin resistance would likely be much reduced or eliminated for such a person at normal weight. If that is true, then in overweight people, is type II diabetes a disease at all, or merely a symptom of obesity? To take this reasoning to the furthest extreme, what if hyperinsulinism is the real disease, leading to the excessive weight gain (as suggested by places like [10]) - is it possible that the subsequent onset of type II diabetes in obese people actually represents the remission of the condition of excess insulin production, so that it is then possible to lose weight more readily and avoid all adverse health effects? Wnt (talk) 19:50, 26 October 2011 (UTC)[reply]

You might be interested in a documentary called "Why Thin People are not Fat. I believe it is available online for free. From memory they said it was largely genetic and it's not easy to get some people to put on weight, even when they're trying to force it. Vespine (talk) 21:16, 26 October 2011 (UTC)[reply]

I saw that documentary, but I think that most people would fall in that category if they only made sure they were physically very fit well before they reached adulthood and stayed that way. The real problem is that the Western way of living has been made the standard for healthy living, while it is actually the cause of problems. Things are slowly changing, it has recently been realized that walking alone is not good exercise (unless you are past 75 years of age), you really need to exercise intensely, (at a heart rate of 140 bpm or higher). Count Iblis (talk) 23:37, 26 October 2011 (UTC)[reply]

To pick up on one point above, I read over and over again that a modest amount of walking IS good exercise. Is giving this advice an act of desperation to try to encourage ANY level of activity? Wanderer57 (talk) 04:28, 27 October 2011 (UTC)[reply]

Photon to electron

Can a photon be converted into an electron, neutron or proton?--195.94.11.17 (talk) 18:30, 26 October 2011 (UTC)[reply]

Sort of. There are certain conditions where a photon may spontaneously produce an positron-electron pair, see Pair production. Pair production is used in the explanation of Hawking radiation and Pair-instability supernova. As far as my limited understanding of particle physics tells me, it wouldn't work with neutrons or protons because those are composite particles, not fundemental particles. That is, protons and neutrons are themselves composed of smaller particles, called quarks. Electrons, however, are fundemental particles, and may be produced as via pair production, as may other fundemental particles. --Jayron32 19:05, 26 October 2011 (UTC)[reply]

Composite particles can be pair-produced. There's very little difference between composite and fundamental particles in particle physics. It has been proposed many times that the Standard Model fundamental particles are actually composite, and it was even proposed that there's no such thing as a fundamental particle (bootstrap model). -- BenRG (talk) 01:52, 27 October 2011 (UTC)[reply]

So it's turtles all the way down then? --Jayron32 01:56, 27 October 2011 (UTC)[reply]

Apparently yes. Turtles on top of turtles, even unto the Nth generation. I wonder if that is occasion for encouragement or despair. Wanderer57 (talk) 05:11, 27 October 2011 (UTC)[reply]

There are certain rules that must be followed when considering whether a group of particles can transform in a different group. Those rules include conservation of energy, conservation of momentum, angular momentum, conservation of electric charge, and so forth and so on (There are too many for me to list them all here). Because of all that, a single photon by itself cannot transform into anything else. There just isn't any way to avoid breaking all the rules, so it doesn't happen. But if the photon collides with something else, allowing it to shed some excess energy, then pair production as described by Jayron in the previous post can happen. Dauto (talk) 20:03, 26 October 2011 (UTC)[reply]

Dry ice snow

According to our article on Vostok Station, the coldest naturally occuring temperature recorded on earth is -89.2°C. This is colder than the freezing point of carbon dioxide. Has dry ice snow ever actually been observed? Spinning Spark 21:32, 26 October 2011 (UTC)[reply]

This has been asked before -- the answer is no; here is a thorough explanation of why (basically, because the freezing point is pressure-dependent). Looie496 (talk) 21:44, 26 October 2011 (UTC)[reply]

I asked myself the same question a few years back. I came to the conclusion that CO2 snow was unlikely -as it requires colder temperatures to form. I thought it possible, that liquid CO2 might form on the floors of basement/cellars were it would natural collect (colder temperatures would make it sink to these places). However, I never found any accounts that this had been observed, only that people reported 'coughing' due to the high levels of CO2 when descending down into them.--Aspro (talk) 21:55, 26 October 2011 (UTC)[reply]

CO2 doesn't become liquid at sea-level pressure. You need at least five times more pressure to produce liquid CO2. That's why dry ice is dry. Dauto (talk) 22:05, 26 October 2011 (UTC)[reply]

(EC)Just because the temperature is below the freezing point doesn't mean the CO2 will freeze. A more familiar situation might help you understand. Right now where you are the air temperature is below the 100 Celsius boiling point of water but that doesn't mean that the water vapor in the air will start condensing. In fact, the opposite is more likely to happen and if you spill some water on the table it will evaporate over time. That happens because the air is not saturate with water vapor - that is the water vapor partial pressure is below the saturation partial pressure. The CO2 partial pressure in our atmosphere is just to small to get even close to saturation (even at Vostock). Dauto (talk) 22:02, 26 October 2011 (UTC)[reply]

October 27

At the time of when Islam was first arose...

Did we know of any other planets in the solar system? ScienceApe (talk) 02:55, 27 October 2011 (UTC)[reply]

That would depend on who you talked to. Nevard (talk) 03:17, 27 October 2011 (UTC)[reply]

The existence of five other planets has been known since at least ancient times, if not prehistoric, by pretty much everyone everywhere--at the very least Venus, Mars, Jupiter, and, I would think, Saturn. Mercury might have been harder to notice by prehistoric peoples everywhere, I'm not sure. I have never knowingly seen Mercury, but then I've always lived with urban light pollution. Anyway, see History of astronomy, Venus#Historic understanding, Mars#Historical observations, Jupiter#Ancient mythology, Saturn#Ancient observations, and Mercury (planet)#Ancient astronomers. Islam first arose in the early 600s AD, of course. What prehistoric and ancient people thought these planets were and why they moved through the sky the way they did is a different question. Pfly (talk) 04:11, 27 October 2011 (UTC)[reply]

It should probably be mentioned here that the dawn of Islamic astronomy marked a real advance on earlier knowledge - an emphasis on empirical observation had long-lasting effects on the development of astronomy as a science. One should note that even today, many of the visible stars have names of Arabic origin: e.g. Aldebran, Deneb, Rigel, and all the other stars found in our List of Arabic star names. AndyTheGrump (talk) 04:25, 27 October 2011 (UTC)[reply]

If a proton were the size of our sun...

How big would an electron be approximately? And if it were just one electron, about how far out would its probability cloud be if the proton were where the sun is? Just to be clear I'm aware that the bohr model isn't accurate. ScienceApe (talk) 02:58, 27 October 2011 (UTC)[reply]

An electron is a point particle, so it's unclear what is meant by the phrase "how big would an electron be". However, I can give an idea of some of the relative sizes involved. Expressing everything in the same units (picometers) for easier comparison, in increasing size, the charge radius of a proton is about 0.000877 pm, the classical electron radius (which isn't really the radius of an electron) is about 0.00282 pm, the Compton wavelength of an electron is 2.4 pm, and the Bohr radius is 53 pm. Multiplying those four numbers by the same constant gives 1 solar radius, 3.2 solar radii, 2,700 solar radii, and 60,000 solar radii, respectively. In comparison, 1 astronomical unit is about 215 solar radii, so divide those four numbers by 215 if you'd prefer to measure everything in AU instead of solar radii. Red Act (talk) 07:12, 27 October 2011 (UTC)[reply]

Rods and cones in the male/female eye

I've noticed that my brother and I can usually see quite well in the dark, whereas many women I know have to strain their eyes to see things at night (like mountains) that I can see easily. This set me wondering, do females have less cones than males? If so, do they have more rods? And are the effects of this noticeable? Can females differentiate between colours better than males? Thanks, --T H F S W (T · C · E) 03:00, 27 October 2011 (UTC)[reply]

I'm not sure all of the specifics, but one key difference is that males are 16 times more likely to be colorblind than females, see Color_blindness#Genetics. My understanding is that a lot of color perception genetic coding is on the X-chromosome and consists mostly of dominant genes. In females, with two X-chromosomes, the presense of the recessive "color blindness" gene can be masked by the dominant gene on the second X chromosome. Males, with only one X-chromosome, don't have that second gene, so if they have the color blindness gene, it always manifests itself. I'm not sure what this means for general (non-defective) vision differences between males and females, but there may be something there. --Jayron32 03:42, 27 October 2011 (UTC)[reply]

You appear to be stating your question backwards of how you're intending it. Rod cells are the ones responsible for night vision, and cone cells are responsible for color vision.

Number of rod cells would be only one potential difference that might affect night vision.

This study, at least, would seem to suggest that women might if anything tend to have better low-luminance visual acuity than men. Red Act (talk) 04:57, 27 October 2011 (UTC)[reply]

I think this would be a hard comparison to make, because it takes a long time for night vision to reach maximum sensitivity. You would have to be sure that you were comparing males and females who had both been in the dark for equally long. Looie496 (talk) 05:00, 27 October 2011 (UTC)[reply]

Laws of Physics

Is there any law of advanced physics that says that a body may travel from point A to B without actually having to cross the actual distance between these two points ( or any other possible path )? Jon Ascton (talk) 04:30, 27 October 2011 (UTC)[reply]

There are several features of quantum mechanics that pertain to your question. For one thing, at extremely small scales, objects don't really take any specific path from A to B, so much as take all possible paths from A to B; see Double-slit experiment. Also, objects can go from A to B by going through a region where the object would classically be prohibited from going due to energy considerations; see Quantum tunneling. Also, a qubit can be transmitted from A to B, without being transmitted through the intervening space; see Quantum teleportation. Red Act (talk) 05:24, 27 October 2011 (UTC)[reply]

Night vision

Do people from the Far North tend to have better night vision than average? Seems to me that this would be highly advantageous, considering that it's dark for six months up there... 67.169.177.176 (talk) 06:08, 27 October 2011 (UTC)[reply]

How much heat can Asbestos withstand?

Pure or mixed with concrete or other materials. By withstand, I mean both how hot till it burns and how hot till whatever it's protecting gets significantly hotter. Thank you. Cliko (talk) 07:09, 27 October 2011 (UTC)[reply]

Revision as of 07:22, 27 October 2011 edit Mikenorton (talk \| contribs) Autopatrolled, Extended confirmed users, Pending changes reviewers, Rollbackers 44,009 edits →Earthquakes and Brick Buildings: re ← Previous edit		Revision as of 07:34, 27 October 2011 edit undo Cliko (talk \| contribs) 64 edits No edit summary Next edit →
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	== How much heat can Asbestos withstand? ==		== How much heat can Asbestos withstand? ==

	Pure or mixed with concrete or other materials. By withstand, I mean both how hot till it burns and how hot till whatever ~~its~~ protecting gets significantly hotter. Thank you. [[User:Cliko\|Cliko]] ([[User talk:Cliko\|talk]]) 07:09, 27 October 2011 (UTC)		Pure or mixed with concrete or other materials. By withstand, I mean both how hot till it burns and how hot till whatever it's protecting gets significantly hotter. Thank you. [[User:Cliko\|Cliko]] ([[User talk:Cliko\|talk]]) 07:09, 27 October 2011 (UTC)