Wikipedia:Reference desk/Archives/Science/2006 December 7

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December 7

Runny nose in cold weather

Why does my nose run when I am outside in the cold. J. Finkelstein 00:21, 7 December 2006 (UTC)

Cold air tends to be very dry. Your nasal passages create moisture to keep from drying out. If you spend most of your time in the cold, your nasal passages will get used to it and your nose won't run as much. A similar thing happens in the desert. When I was in 29 Palms, we called it newbie snot because the new guys always had a "cold" - which was actually just a runny nose. --Kainaw ^(talk) 01:25, 7 December 2006 (UTC)

Heh. Happens to me everytime I change climates. X [Mac Davis] (DESK|How's my driving?) 04:43, 7 December 2006 (UTC)

Near and Far - The Automobile Rearview Mirror

I need glasses to read type at a close distance, and can see fairly clearly at a distance. However, my rearview mirror in the car is out of focus on its edges, yet the image of a vehicle several hundred yards behind me remains crystal clear. How can the image I see on the outer portion of the mirror be "out of focus" but the image on the mirror be so clear? 70.181.28.145 00:45, 7 December 2006 (UTC)--70.181.28.145 00:45, 7 December 2006 (UTC)Aloyshai

I would think it's because of imperfections in the mirror, and not because of your eyesight. --Bowlhover 01:53, 7 December 2006 (UTC)

When you look into a mirror at something's reflection, you're not looking at the surface of the mirror, you're looking at the virtual image behind the mirror. The virtual image is farther away than the mirror's surface, so you can see it more clearly. In other words, your farsightedness allows you to see objects as long as the light from them travels more than a certain minimum distance. The light from the mirror's surface only travels a short distance to your eye, but the light from a distant object travels a long way to the mirror first, so the total distance is enough for you to see it clearly. —Keenan Pepper 02:31, 7 December 2006 (UTC)

But the point is, the original poster doesn't see things in the mirror clearly, even though he sees them clearly without the mirror. --Bowlhover 03:00, 7 December 2006 (UTC)

I think you misread it and Keenan has it right. --Anon, 07:22 UTC, December 7. —The preceding unsigned comment was added by 66.96.28.244 (talk) 07:22, 7 December 2006 (UTC).

"and can see fairly clearly at a distance [...] However, my rearview mirror in the car is out of focus on its edges" Can you please point out how I'm misunderstanding this? --Bowlhover 03:21, 8 December 2006 (UTC)

I think the poster's reference to the edges of the mirror refers to the mirror itself, not reflections seen in it. --Anon, 03:27 UTC, Dec. 8. —The preceding unsigned comment was added by 66.96.28.244 (talk) 03:26, 8 December 2006 (UTC).

Oh, he was talking about the mirror itself. Now that I get it, I agree with Keenan's answer. --Bowlhover 04:40, 8 December 2006 (UTC)

Could it be one of those undesirable things one gets in 'less than high-quality' telescopes, namely COMA? Or just a 'cheap' mirror? Dave172.147.84.40 22:17, 8 December 2006 (UTC)

Moon

I have two questions: First Is it possible for balloon craft to still work on the moon or will it not work because there is no Air? Second is it possible to collect condensation from the moon from any atmosphere that is there by heating a plate and colling it? Thank You for any answers.

Balloon craft only work because they float on air. If there is no air, they'll drop as fast as anything else. Also, it's not possible to collect condensation on the Moon because there's no water vapour, and you need water vapour for condensation to occur. --Bowlhover 01:49, 7 December 2006 (UTC)

As far as has been ascertained there is minimal atmosphere around the moon. So the answer to both Qs is NO!--Light current 01:53, 7 December 2006 (UTC)

See: Moon#Presence of water. --JWSchmidt 02:39, 7 December 2006 (UTC)

NAD+ turning into NADH

I was trying to understand the concept of energy harvest via NADH and the electron transport chain, but I don't understand a particular step. Whenever the NAD+ accepts the two H+ ions and two electrons I can never figure out where the H+ ions came from. I've read that they come off of the substrate being broken down, like glucose. However, in all of the chemical equations, once the NAD+ has accepted its hydrogen ions and electrons, the original substrate molecule doesn't seem to have lost these. What am I missing? Thank you.

Our article on NAD+ may clarify things. The article also links to a Flash animation which may help. The original substrate molecule (which is oxidized) definitely does have to give up the hydrogen. TenOfAllTrades(talk) 04:07, 7 December 2006 (UTC)

I read the article and watched the flash animation, but I still am a little bit lost. Perhaps I've just read a typo, but the book I've been reading shows no change in the substrate's number of hydrogens after it has given them up. Additionally, it states that two hydrogens are given up, rather than the one that is explained in the flash animation. An example is converting pyruvate into Acetyl CoA. My book shows pyruvate as C3H3O3 - then shedding a CO2 to C2H3O - then shedding the hydrogen for NADH - and finally adding in Coenzyme A. However, the book still shows 3 hydrogen on the newly formed Acetyl CoA.

Yeah, the Flash animation oversimplifies a bit—it doesn't mention the hydrogen atom that is dumped into solution. Okay, to find the missing hydrogen, you can start with the structures for acetyl-CoA, CoA, and pyruvate, and try to work out where the missing atoms go.

Or, you can note that biologists are sloppy about their terminology. Pyruvic acid is C₃H₄O₃. Its dissociation into pyruvate ion therefore produces C₃H₃O₃^- and H⁺. If you look at the structural formulae in our article on glycolysis (and probably in many textbooks) you'll find that the structure labelled as pyruvate is actually the undissociated pyruvic acid. So the reason why you're missing a hydrogen atom is that your formulae for pyruvate has already accounted for the proton that wanders off into solution. Meanwhile, the other hydrogen atom comes from the thiol (-SH) tail of the CoA; that hydrogen is lost when the sulfer atom forms the bridge between the acetyl group and CoA. TenOfAllTrades(talk) 05:36, 7 December 2006 (UTC)

I always thought that the H+ is soluble in the particular solution (be it cytoplasm, mitochondrial matrix) and exists there due to the many metabolic processes occuring at a time (doesn't necessarily have to occur together incidently)- a result of the dehydrogenase reactions of pathways. --131.245.147.176 08:39, 8 December 2006 (UTC)

Since all these reactions take place in the aqueous environment, there are always hydrogen ions around to be pulled into the reaction. Water is readily ionizable, and is always in equilibrium with its ionized H3O+ and OH- forms, once you have the species that is ready to be reduced it just pulls a hydrogen 'as if from nowhere.' Since the water is in vast excess, this can happen without any other appreciable equilibrium effect.

Ionosphere page

Hello,

I was wondering where the page for the Ionosphere went. Did it get deleted for vandalism or something? I hope I spelt it right, but on word it doesn't show up a typo.

from Jinglebells ( I'm not a member here but didn't want to sign my real name) —The preceding unsigned comment was added by 58.107.41.170 (talk) 05:33, 7 December 2006 (UTC).

Seems ok: Ionosphere. Cheers, Antandrus (talk) 05:34, 7 December 2006 (UTC)

No one is expected to sign their real name. Just type four tildes. This can help us trace problems.--Shantavira 09:31, 7 December 2006 (UTC)

Driving behind a truck

It was mentioned to me that it is more fuel-efficient when driving to drive behind a truck, because the vacuum behind the truck helps to "suck" the car along. So I was wondering about a few things, because it sounds a little dubious to me.

• 1) How long would the cone of low air pressure be behind a large truck travelling at say, 100km/h? (For the effect to work, it would obviously have to be long enough to partly contain a car, plus a safe following distance)

• 2) Wouldn't there probably be sufficient air sucked under the truck and out the back to negate the effect?

• 3) Supposing that the cone was long enough and that there was negligible air sucked under the truck, would there be any noticeable difference - i.e., would the air pressure difference make a sufficient contribution to the work performed by the engine? BenC7 06:21, 7 December 2006 (UTC)

It doesn't have to be a truck. They do that a lot in the Tour de France. See Drafting (racing). Can't help you with the details though. Clarityfiend 06:42, 7 December 2006 (UTC)

Back when I rode a three-speed bicycle and they had trolleybuses in this city, I used to draft behind them at speeds up to about 25 mph. The effect was significant if I got close enough to the trolleybus -- I think I tried to stay within 5-10 feet for as long as I could, and the effect dropped off rapidly when it got above my top speed and pulled ahead. (There was no significant risk in following so close, as I could hear the trolleybus's motor and air brakes and respond quickly, and of course there were no exhaust fumes.) I have no idea of how to extrapolate this to larger vehicles and higher speeds, though. --Anonymous, 08:30 UTC, December 7. —The preceding unsigned comment was added by 66.96.28.244 (talk) 08:30, 7 December 2006 (UTC).

Believe it or not, there's actually very little experimental data about the size of truck wakes, so it's difficult to answer the question, but I'll answer the question about the generic effect. As to whether the effect would be negated, think of it this way, in a stationary frame of reference. Until the truck impacts with an amount of air, it is stationary, so upon impact it is given a net forward momentum. This means that the following car will have to do less work when it collides with the same air. Any air that comes from under the truck would be there anyway if the car was alone on the road, so the effect couldn't be negated. Close to the truck, it's wake will be dominant over any air flowing underneath. Of course the further back you get, the smaller the wake will become. In general the effect is significant and measurable, as long as the following car is close enough to be in the wake. Readro 08:49, 7 December 2006 (UTC)

Never heard it called drafting before - you should look at Slipstream - a similar term. The car infront pushes the air in front out of the way (a bit like a piston) - if you are close behind (a few metres or less) you can follow in it's wake as others have said. Have you never noticed a strong wind as a fast big truck goes by, or even better stand on a train platform as an express train goes through - they really 'suck' literally.83.100.174.147 11:38, 7 December 2006 (UTC)

1. Not a very long cone - probably not safe unless you are a racing driver - see however road train.

2 Depends on how high the truck rides - many european trucks have skirting so even if you're in a low slung car you should still get sucked along..

3 Yes - at high speeds you would be in a low pressure area - which does effect engine performance.

Sorry I can't give you any solid figures.83.100.174.147 11:44, 7 December 2006 (UTC)

While there may be a fuel efficiency gain, I must advise against following a truck so closely, for numerous safety reasons:

• The turbulence behind the truck could make driving difficult and dangerous.

• The truck may stop suddenly, leaving you insufficient time to react.

• Your view of the road ahead will be obscured by the truck, causing you to be unprepared for upcoming dangers.

• Water or road debris kicked up by the truck tires may strike your car, doing body damage, temporarily blinding you, smashing the windshield, etc. StuRat 12:26, 7 December 2006 (UTC)

Actually, if you follow close enough, this last one is unlikely, especially if the truck has mud flaps. The debris would pass through the gap between the mud flap and the road, giving it a flatter trajectory, so you would be more likely to be hit if you were further back.

Except that rocks may have quite a spin on them, so may not follow a straight-line trajectory. StuRat 12:12, 9 December 2006 (UTC)

In addition, such tailgating is rude, and the truck driver will be annoyed. StuRat 12:26, 7 December 2006 (UTC)

Yes, ever see the bumper sticker - 'If you can't see my mirrors, I can't see you'?

Yes and illegal (in some countries at least) for the reasons given above.83.100.174.147 12:36, 7 December 2006 (UTC)

If an 18 wheeler is driving along with the back open, is it windier inside the trailer, or on top?

--18.90.5.3 16:42, 7 December 2006 (UTC)

I'd say windier (higher average wind velocity), relative to the truck, on top, but perhaps more turbulent inside. StuRat 12:18, 9 December 2006 (UTC)

Data communication and Network

Hello sir I want to have some help from u. My questions are

1.How group leadres use to work in KaZaA, what are their responsibilities and how they used to lead their group in perfect manners?

2.what is bootstraping in KaZaA and how it works?

3.what is itrative model in DNS( Domain Name Server ) and what is the recursive model in DNS?

4.how Gnutella works and how bootstraping occures in Gnutella?

5.What is TLD( Top Level Domain Name Server ) in DNS and how it works?

6.What is DHCP( Dymanic Host configuration Protocol )? —The preceding unsigned comment was added by Honey65 (talk • contribs) 06:46, 7 December 2006 (UTC).

Definately you should ask on the computing page Wikipedia:Reference desk/Computing83.100.174.147 11:33, 7 December 2006 (UTC)

Also be aware that we don't do homework questions here (assuming this is what it is). They'll send you in the right direction over at the computing reference desk though. For now try: Gnutella P2P and DNS. Benbread 13:06, 7 December 2006 (UTC)

the homeostatic nature of skin repair.

when the skin is injured, how does homeostasis play a role in its repair? —The preceding unsigned comment was added by 81.154.244.237 (talk) 10:42, 7 December 2006 (UTC).

Well, let's do our own homework okay? Maybe you should take a look at the article on skin and see if it sheds any light on it. Laurən^whisper 18:42, 7 December 2006 (UTC)

Scooting

Why do dogs scoot? Dismas|^(talk) 11:19, 7 December 2006 (UTC)

Worms? or for fun - am I describing the same thing?83.100.174.147 11:32, 7 December 2006 (UTC)

I thought the answer was something along the lines of, "If you could, you'd never leave the house." --Kainaw ^(talk) 13:25, 7 December 2006 (UTC)

You might find something useful in anal glands, assuming the above repartee failed to answer your question. Apparently, one of the reasons for a dog to scoot lies in discomfort caused by overly full anal glands. -- Ec5618 16:16, 7 December 2006 (UTC)

how..................

i need more information about diabetes........n its types(type 1,type 2 and gestational...)culd u plzzz add sum more information.....which give in a nut shell......?'coz i need it for my board xam...... —The preceding unsigned comment was added by 220.226.140.51 (talk) 13:28, 7 December 2006 (UTC).

Have you considered reading the article on diabetes? --Kainaw ^(talk) 13:43, 7 December 2006 (UTC)

"Board exam"? Does anyone else find this a deeply disturbing inquiry? alteripse 17:53, 7 December 2006 (UTC)

Not me. I'm too jaded from past RD questions. --Kainaw ^(talk) 18:18, 7 December 2006 (UTC)

Atmospheric chemistry units

I would like to know what molecules per square centimeter/second means, when talking about chemical reactions in the atmosphere.

Liza (uni student)

originally posted by User:195.215.9.140 at Wikipedia:Requested articles/Natural Sciences --User:Ceyockey (talk to me) 14:32, 7 December 2006 (UTC)

It might be a flux; does that article help? Melchoir 16:11, 7 December 2006 (UTC)

Calculating a change in temperature during irreversible adiabatic expansion.

I know it should be relatively trivial to calculate the change in temperature of an adiabatically expanding gas, but I just can't seem to figure it out. If I know, for example, that 1 mol of gas expands from a pressure of 30 atm to 10 atm, irreversibly and adiabatically, do I then have enough to calculate the work done by the gas? What other parameters do I need to know? What formula can I use?

As I said, I have a feeling the answer is relatively simple, but it eludes me and the puzzle is beginning to bug me. Thanks. -- Ec5618 16:12, 7 December 2006 (UTC)

"Irreversible" could mean many things. If you're asking about a sudden expansion due to, say, a removed partition, then the gas does no work, since its bounding walls don't move in the direction it's pushing. Another way to see this is that its internal energy doesn't change, and since no heat is transferred by assumption, no work is done either. Melchoir 16:18, 7 December 2006 (UTC)

...so, to hit on your other questions, I guess the parameter you need to know is the work done by the gas! Or, at least, a description of the physical situation that allows you to figure it out on physical grounds, not necessarily through any particular equation. And once you know the work, you know the change in internal energy, which directly gives you the change in temperature. Melchoir 16:21, 7 December 2006 (UTC)

Ah no, work may be done on a partition that is suddenly free to move, or that is suddenly moved outward. This forces the gas to expand, and thus to do work on the surroundings. The only scenario on which no work is done, as I understand it, is one in which the expansion is into a vacuum, as the external pressure would then be 0. Work, after all, equals external pressure times change in volume, assuming thr external pressure is constant.

And yes, I had come to the conclusion that it was the work that must be found. I had thought the situation was clear. A chamber containing a single mol of gas is forcibly expanded so that the pressure drops. Work is done, and it is equal to the change in internal energy. The change of temperature then follows from the heat capacity of the gas, which I can probably find in a table somewhere.

What exactly is unclear? -- Ec5618 16:34, 7 December 2006 (UTC)

Well, I did say in the no-work situation that the bounding walls don't move in the direction it's pushing. You can accomplish this by sliding away a partition or deactivating a force field in a large box, allowing the gas to explore a greater volume. There will still be a vacuum present in the problem, but it's a different physical situation.

If you're dealing with a forcibly expanded chamber, it depends on what external agent is applying the force, and how much work it does or absorbs itself. Assuming that the compound system of chamber + agent + outside atmosphere itself evolves reversibly, then the work done by the irreversibly expanding gas plus the work done by the agent should equal the work that a reversibly expanding gas would have done. Melchoir 16:57, 7 December 2006 (UTC)

Hmm, thank you, but I'm still puzzled. Suppose, if you will, that my setup is a bit more complicated. I have a box of gas at 30 atm. Given that the outside pressure is just 10 atm, the gas excerts force on the walls of the box. If I now allow my box to expand, the gas will do work on the surroundings, without the need for additional work. The box will expand until the pressure is normalised; that is, the pressure in the box will drop to 10 atm. How much work will the gas have done, assuming that the gas expands adiabatically and irreversibly, and that the box contains just one mol of gas? The answer will require some calculations, but without the right formulas, I'm stuck. -- Ec5618 17:36, 7 December 2006 (UTC)

I don't know... this is so much more complicated than the reversible case. At this point I'm going out on a limb. If the box's walls are too massive then you have to worry about them absorbing kinetic energy as they move through the pressure differential, and what happens to that energy? If the box's walls are too flimsy then they'll expand so quickly that in an effort to balance pressures, they build up an overpressure zone outside and an underpressure zone inside. Now you're really in trouble because the gas in question is out of global equilibrium, so forget about defining a single temperature and pressure, let alone applying the ideal gas law. You probably also have to worry about the bang carrying away sound energy, and the efficiency of that process probably depends on more variables than just the outside pressure.

If I'm missing some simplifying principles or assumptions, someone please tell me. But I suspect the problem is so hard that it requires experiment. Melchoir 05:04, 8 December 2006 (UTC)

How could you forget pV=nRT?

See ideal gas law (use this approximation please until you feel more confident) If a gas expands adiabatically then no energy is exchanged , and it can be shown that for an ideal gas pV is proportional to energy. So your gas will expand to three times its volume as the pressure drops to one third of its original pressure. (note this is a response to your original question)

For work done take a look at http://www.grc.nasa.gov/WWW/K-12/airplane/work2.html

If you let the gas do work as it expands it may loose some energy. You need the initial volume or temperature of the gas to proceed further here. Using pV=nRT (n=number of moles, R is a constant, p the pressure) with one more of the two the system can then be fully described.

The work done is Integral (Vbegin to Vend) of P(at volume V) x dV.83.100.174.147 17:50, 7 December 2006 (UTC) Please comment as to whether this helps/makes sense..83.100.174.147 17:51, 7 December 2006 (UTC)

Thank you for your help, but pV = constant only when T is constant. When the gas expands adiabatically, T changes, because the system does work, as per $\Delta T=\Delta U\times C_{v}$ (and in an adiabatic change, the internal energy is equal to the work done, as per $\Delta U=q+w$).

Now, you're right in saying that I could easily calculate the work if I knew the initial and the final volume of the gas, but for the purpose of this calcation I am assuming that the change in pressure is given, not the change in volume. I have tried to use $w=p_{e}\Delta V=p_{e}(V_{f}-V_{i})=p_{e}({\frac {nRT_{f}}{p_{f}}}-{\frac {nRT_{i}}{p_{i}}})=p_{e}\times nR({\frac {T_{f}}{p_{f}}}-{\frac {T_{i}}{p_{i}}})$ but since the change of temperature is unknown, I am stumped. -- Ec5618 19:17, 7 December 2006 (UTC)

Isn't it delta U = delta T x C (specific heat is Joules per kelvin,yes? deliberate mistakes - I make them..). With two unknowns (temperature and volume) you are surely stumped.83.100.174.147 20:29, 7 December 2006 (UTC)

Hmm, then what about a leaking gastank, for example? If a gastank containing gas at 30 atm springs a leak into an infinite chamber at 10 atm, is it impossible to calculate the temperature of the expanding gas? In this case, volume is irrelevant, and, assuming the change happens quickly enough, the change is adiabatic. Are you saying it would be impossible to calculate the temperature the gas would reach?

You're right, about the equation being backwards, of course.

I still find it hard to believe that it would be easy to calculate work when only the external pressure and the change of volume are known, but impossible when the external pressure and the change of pressure are known. Surely this sort of problem should be routine, in some way. -- Ec5618 20:50, 7 December 2006 (UTC)

your gas tank - yes I think it's impossible to calculate the temperature of the released gas without knowing it's original temperature. Maybe it would be possible to calculate the temperature change (ie the temperature drop).

Your third point - in the first example P(external) and V(change) are known (two variabels), in the second P (external) and P (change) are known that's only one type of variable - I think that's where the difference lies. Maybe you should ask think point again - I think it's explainable but I'm not very good at it.87.102.36.136 22:23, 7 December 2006 (UTC)

Magnets

hello, if I have lactate with an iron atom on it, or maybe EDTA with iron captured inside, with concentrations on the order of parts per billion, can I use a magnet to control these molecules? How big/small would thi s magnet need to eb? Thanks. -Steve —The preceding unsigned comment was added by 74.134.95.124 (talk) 17:24, 7 December 2006 (UTC).

I'm afraid not. The reason metal is affected by a magnetic field is that several atoms of the metal have the same alignment. A substance containing single atoms of metal will not be affected, as it will not be magnetic. See magnet. -- Ec5618 17:42, 7 December 2006 (UTC)

Suggest tentatively ignore the above.

The iron complex of edta or lactate should/will be paramagnetic - it will be attracted to a magnet. The concentration is relatively irrelevant here I think. But you would need a very big magnet to control the molecules - to control the molecules the force would have to be sufficient to overcome diffusion in the liquid. I can't give you a figure but expect 10tesla or more.

See Paramagnetism I think there should be enough data there to start calculating approximate figures.

There is a force an paramagnets resulting from magnetic fields - but don't expect it to be very great. Hope this helps.83.100.174.147 17:58, 7 December 2006 (UTC)

effect of oil spills on frogs

What is the primary effect that an oil spill would have on frogs? —The preceding unsigned comment was added by 86.42.77.243 (talk) 16:39, 7 December 2006 (UTC).

I'm afraid your question is rather vague, and it seems like a homework question. Oddly, we don't seem to have a lot of information on his topci in our oil spill article, but an obvious result of an oil spill on local animal life is death. -- Ec5618 16:48, 7 December 2006 (UTC)

But since oil spills tend to happen at sea, and frogs do not form part of the local animal life at sea, it is difficult to see how it would have any direct effect on them at all.--Shantavira 09:00, 8 December 2006 (UTC)

Concievably, oil spills could occur on fresh-water lakes such as the great lakes, or occur over land in a refining or pipline accident. This reference seems to indicate oil fields having an impact on frogs. There are some papers in google scholar as well that seem to indicate it has been studied. --TeaDrinker 16:35, 8 December 2006 (UTC)

Frogs require their skin to breath (they use their lungs, but it aids, especially when in water). It is also used to absord water into the body. A frog which is covered in oil will not be able to exchange water with its environment, and will die of dehydration. --liquidGhoul 12:37, 11 December 2006 (UTC)

High tension transmission

We don't appear to have an article on this and I'm looking into it for some school stuff. I found EHT but it's a stub and I'm not totally sure that's it's the same thing. I've searched on Google and can't find anything useful that I can understand so would appreciate advice on where to look, or at least what this concept is. —Xyrael / 17:03, 7 December 2006 (UTC)

Is this possibly related to HVDC? —Bromskloss 17:10, 7 December 2006 (UTC)

Power engineering contains some info.--Light current 17:17, 7 December 2006 (UTC)

I seem to be unable to locate any information on this specifically, as it doesn't appear to be HVDC - I don't think that power lines use AC all the way through. As for the latter page, I've looked through and can't spot it - what is it under? Thanks. —Xyrael / 18:00, 7 December 2006 (UTC)

Could you more specific about what you need to know if Electric power transmission doesnt cover it??--Light current 18:09, 7 December 2006 (UTC)

See also Overhead powerline. If you have further questions after reading these article, then we will attempt to assist you to find answers. Edison 18:50, 7 December 2006 (UTC)

Xyrael wrote "I don't think that power lines use AC all the way through." Many, but not all, high voltage transmission lines are AC. One way to find out is to walk under them while holding a flourescent tube at night. If it lights up, it's AC. One reason for using DC transmission lines is if the grids on the two ends of the line are not in phase. --Gerry Ashton 17:44, 8 December 2006 (UTC)

I think a high DC voltage gradient can also light a fluorescent. And please do not suggest holding up objects under power lines. Worst case: the line (might be 12kv distribution) has sagged or has contacted a tree limb such that it arcs to or makes contact with the fluorescent tube, causing electrocution. Edison 18:09, 8 December 2006 (UTC)

I'm sorry, I misphrased that - I meant that I thought all major power lines were AC because this is very easy to produce. Those articles appear to be in the right area, but I'm having difficulty finding mention of tension at all. Unfortunately, I don't actually know what 'high tension transmission' means on its own so it's difficult for me. So, I'd like to ask if there is perhaps a more commonly used phrasing of it, that I could then research myself. Thank you for your help so far. —Xyrael / 18:34, 8 December 2006 (UTC)

I've thought about this some more and think I've done all that I can. Thank you for all help given :) —Xyrael / 20:33, 10 December 2006 (UTC)

herbs

Could you please tell me and edible herb that is best used when the moon if full? Thank you for your help! —The preceding unsigned comment was added by 172.141.41.43 (talk) 17:57, 7 December 2006 (UTC).

Best used for what?--Light current 18:10, 7 December 2006 (UTC)

If you believe the myths, garlic is potent werewolf repellent, as well as a vampire repellent, which would be very useful during the full moon. Wolfsbane would be even better, although it is not edible but deadly poisonous... Laïka 18:16, 7 December 2006 (UTC)

It's edible, it'd just kill you. Something inedible would be like a lorry or something. Although if you were patient, you could probably cut it into bits and swallow the bits. Vitriol 23:29, 7 December 2006 (UTC)

The definition of "inedible" is "not fit to be eaten", not "impossible to eat". Neither wolfsbane nor a lorry are "edible". - Nunh-huh 20:11, 8 December 2006 (UTC)

I knew that, I was just pointing out how silly it is. Vitriol 20:40, 8 December 2006 (UTC)

This question is getting out of hand. --Russoc4 20:42, 8 December 2006 (UTC)

Hypothermia

In light of James Kim's death, I was just thinking about hypothermia. It says that he may have removed his clothing in a futile attempt to cool down because one of the stages of the condition involves overheating. I was wondering whether it's wiser to keep the clothes on, despite the heat, or if it doesn't matter either way? - Pyro19 18:54, 7 December 2006 (UTC)

See Hypothermia. I don't see that the article addresses the issue of victims feeling warm, but the article makes it clear that the victims need to be warmed up, so I'd assume this applies even if they feel hot. Friday (talk) 19:00, 7 December 2006 (UTC)

Yes, it has been observed (in some cases) that the victims of Hypothermia have undressed. This paper quotes another which suggested an explanation: "when core body temperature falls to a critical level, peripheral vasoconstriction fails. The resulting sudden vasodilation could lead to an exaggerated sensation of heat and a consequent attempt by the victim to undress". This is only an educated guess, however, and the reported undressing may simply be an artifact of the general delirium brought about by the hypothermia. Note that if this is true, it's a sensation of overheating, not the real thing. Clearly it's wiser to keep the clothes on, but by that point the person is likely to be quite seriously altered, and prone to doing unwise things. -- Finlay McWalter | Talk 19:12, 7 December 2006 (UTC)

My understanding, however, is that rescuers felt that Mr Kim's discarded clothing had been arranged in a deliberate manner, such that it seemed intended as a sign. That wouldn't be consistent with a delirious person haphazardly discarding clothing in the throes of such false fever. -- Finlay McWalter | Talk 19:19, 7 December 2006 (UTC)

I kind of thought it was possible that he was avoiding getting wet from his own sweat, which is what I think kills people in extreme cold a lot of times. You wear warm clothes and you exert yourself, and you panic, and you sweat, then you get wet and cold! That's what Les Stroud said! X [Mac Davis] (DESK|How's my driving?) 06:18, 9 December 2006 (UTC)

CAPITAL PUNISHMENT METHODS

HOW MANY DIFFERENT WAYS HAVE PRISONERS BEEN EXECUTED BY GOVERNMENT AGENCIES AND WHAT IS WRONG WITH PUTTING THE PERSON IN A HYPERBARIC CHAMBER AND PUMPING THE AIR OUT THUS DEPRIVING THE BRAIN OF OXYGEN AND CAUSING A HUMANE AND PEACEFUL DEATH? —The preceding unsigned comment was added by 70.226.82.194 (talk) 19:38, 7 December 2006 (UTC).

Wikipedia has a list of methods of capital punishment, which includes twenty or thirty different methods. The Capital punishment in the United States article links to this page with a list of methods by U.S. state. It looks like lethal injection is far and away most common. As for asphyxia, perhaps it's not as humane and peaceful as you think. Have you tried it? -- Plutor ^talk 20:20, 7 December 2006 (UTC)

HUMANE according to whom? 202.168.50.40 21:15, 7 December 2006 (UTC)

Should we have a "CAPITAL" punishment here? –mysid☎ 21:36, 7 December 2006 (UTC)

Looking for the name of a disease

I'm looking for the name of a "disease": newborn babies born with hard brittle skin, bloodred eyes. They're almost intolerable to look at. Anyone know the name? Jack Daw 20:11, 7 December 2006 (UTC)

Harlequin fetus  ?Namlemez 21:39, 7 December 2006 (UTC)

That's it, thanks. Jack Daw 22:24, 7 December 2006 (UTC)

Do they look scary? I just feel sorry they won't live. X [Mac Davis] (DESK|How's my driving?) 06:12, 9 December 2006 (UTC)

Scary depends on the person i suppose, but definitely not pleasant to look at if you're squeamish, the face especially. From the wiki: "The lips, pulled by the dry skin, are fixed into the semblance of a clown's smile (Eclabium), which many find extremely disconcerting." Cyraan 21:24, 9 December 2006 (UTC)

Weight/force

I have a 25-ton object that I want to lift and move. I have a crane that can handle a 75-ton load. Can the crane, from 65 feet away, at a 90 degree angle, lift the object up 45 degrees and move it to clear a building? How would one determine this? ST47Talk 21:36, 7 December 2006 (UTC)

I'm unsure of what you mean by 'at a 90 degree angle' and 'lift the object up 45 degrees'. Can you clarify? -anonymous6494 22:24, 7 December 2006 (UTC)

I guess the arm of the crane is horizontal..

and lift the object so that the arm of the crane is at 45 degrees

I assume the crane arm is 65ft.

Need some more info. How big is the house and where in relation to the crane is the house.87.102.36.136 22:28, 7 December 2006 (UTC)

Don't even worry about the house, I'm just wondering if the crane can lift that much weight with that much distance. ST47Talk 23:47, 7 December 2006 (UTC)

Under what conditions can the crane lift a 75-ton load? (Apparently not from 65 feet away at a 90 degree angle; otherwise you'd know it could handle your 25-ton object as well.)  --Lambiam^Talk 10:26, 8 December 2006 (UTC)

75 tons, straight up. ST47Talk 18:56, 8 December 2006 (UTC)

Retarded

How would it be possible to make a healthy intelligent individual retarded? Hit them in the head? Certain drugs? Lobotomy? Please advise ASAP thx. —The preceding unsigned comment was added by 128.61.113.155 (talk) 21:48, 7 December 2006 (UTC).

Feed them unlimited amounts of fine brandy, Cuban cigars, and the strongest cannabis, and they might go psychotic. You could probably line up some volunteers right here... --Zeizmic 22:09, 7 December 2006 (UTC)

See Mental_retardation#Causes for information. Friday (talk) 22:14, 7 December 2006 (UTC)

Mental retardation refers to slower achievement of developmental milestones by a child, culminating in a lower level of cognitive function than normal as an adult. This combination cannot be induced in an adult (because you cannot rewrite their childhood development). Loss of cognitive ability in a previously normal adult is not referred to as retardation but as dementia. There are many causes of dementia in adults, including mad cow disease, strokes, poisoning, radiation to the brain, encephalitis, head trauma, neurosyphilis, hypothyroidism, etc. Take your pick. alteripse 02:20, 8 December 2006 (UTC)

D

During which phases of cell separation could you see centromeres? Brittani

In mitosis: all the phases before anaphase. In meiosis: all the phases before anaphase II. Is this homework? --Bowlhover 00:12, 8 December 2006 (UTC)