Wikipedia:Reference desk/Archives/Science/2007 March 21

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

March 21

Bullet velocities and wound types

This question has been bugging me for some time, and I could never find the answer, so I'll ask here:

When a human is shot in the torso with a 9mm bullet from a handgun, what causes more damage to the body; the laceration of the bullet passing through the tissue, or the blunt trauma which creates internal bleeding, bruising, etc.? And does this change with other types of ammo, eg. 5.56, 7.62, .50 Cal? Thanks guys, --71.197.149.164 00:17, 21 March 2007 (UTC)

See the article on stopping power. There is also have a stub at wound ballistics, but it's not very thorough yet. Nimur 01:28, 21 March 2007 (UTC)

The subject you are dealing with here is terminal ballistics, that and the above articles seem to have a lot of overlap. Bullets in the 9mm range generally don't cause a lot of 'blunt force' or mechanical type damage unless they are hollow point. People under the influence of drugs like pcp, and even just very strongly willed people, such as those with special forces training can function quite normally with several bullet wound, as long as you can 'block out the pain'. Unless the bullet actually hits your spine, brain or heart, it's generally the bleeding and 'shock' and stuff that eventually kills you. Apparently it isn't actually that uncommon for someone to faint from shock when they are being shot at, even if they aren't hit by a bullet. Vespine 02:40, 21 March 2007 (UTC)

As for the second part, yes of course, bullet type, speed and mass has a massive impact of what happens if a person is hit. Typically the more mass in the bullet the more damage it will cause. Continuing from my post above, a 9mm bullet is considered survivable if you get treatment and you are not hit in the spine, heart or brain, definitely that is not true for a .50 Cal, I imagine anything but a hit to the extremities would be fatal. Vespine 04:13, 21 March 2007 (UTC)

It depends on a lot of things. Mainly, damage is related to how much energy can be imparted from the moving bullet to the body in the quickest amount of time. Hollow points are designed to expand when they hit fluid. This tends to slow them down very rapidly and the energy is transferred very rapidly from the bullet to the body. This limits how far it penetrates, though, so there is a tradeoff. The ideal self protection round would be a very heavy bullet that expanded when it entered the body and stopped with 0 velocity in the spine. Military rounds such as the 5.56mm NATO ball and 9mm ball round are designed to wound consistent with the goals of warfare. Police in the U.S. usually use a .40 cal hollow point which is designed to rapidly stop (i.e. very lethal) someone without overpenetration. --Tbeatty 07:11, 21 March 2007 (UTC)

Thank you very much, I am quite satisfied. 71.197.149.164 16:40, 21 March 2007 (UTC)

lisps(speech impediments)

I would like to know if the lisp (speech impediment) is hereditary? —Preceding unsigned comment added by 69.112.118.126 (talk • contribs) 01:48, 21 March 2007 (UTC)

Physical lisps (as opposed to learned behaviour) are normally caused by physical issues. Thumb-sucking and missing teeth are usually attirbuted, but so can hereditary facial features. So, for the most part, it is not hereditary, but can be. --Kainaw ^(talk) 02:19, 21 March 2007 (UTC)

Travel to the Asteroid Belt

How long would it take to travel to the asteroid belt using minimal energy and current means? Say, the inner edge. I can't find a good distance for it anywhere.-- ×××jijin+machina | Chat Me!××× -- 06:56, 21 March 2007 (UTC)

• I think here is got it at about 4 months going 32,400mph http://www.spacetoday.org/SolSys/ThePioneers.html -- ×××jijin+machina | Chat Me!××× -- 07:02, 21 March 2007 (UTC)

That would be if you were just passing by. If you wanted to stop there, you'd have to decelerate, so it would take a good deal longer. Clarityfiend 08:19, 21 March 2007 (UTC)

You could decelerate once you got there, but that use much more fuel. Think outside the box 11:01, 21 March 2007 (UTC)

Our article on Hohmann transfer orbit says that it is "almost always the most economical way to get from one circular orbit to another". It also gives a formula for the time taken for the transfer. Plugging in the numbers, and assuming that the inner edge of the astreoid belt is at about 2 AU, I get a transfer time from earth orbit to the asteroid belt of about 340 days - say 11 months. Gandalf61 11:29, 21 March 2007 (UTC)

Enzymes

Does anyone know who discovered enzymes? Mattyatty 13:45, 21 March 2007 (UTC)

From Enzyme: As early as the late 1700s and early 1800s, the digestion of meat by stomach secretions[2] and the conversion of starch to sugars by plant extracts and saliva were known. However, the mechanism by which this occurred had not been identified.[3] [...] In the 19th century, when studying the fermentation of sugar to alcohol by yeast, Louis Pasteur came to the conclusion that this fermentation was catalyzed by a vital force contained within the yeast cells called "ferments" [...] In 1878 German physiologist Wilhelm Kühne (1837–1900) coined the term enzyme, which comes from Greek ενζυμον "in leaven", to describe this process. The word enzyme was used later to refer to nonliving substances such as pepsin, and the word ferment used to refer to chemical activity produced by living organisms. Does that answer your question? grendel|khan 15:15, 21 March 2007 (UTC)

Permanent magnets drain

is it true that if a permanent magnet held against another permanent magnet with the same poles facing for a long period of time they will lose their charge ? is there any exception to this ? and would this ocur in a system where both poles face each other very often but then move away from each other? clockwork fromage —The preceding unsigned comment was added by 216.113.96.124 (talk) 15:00, 21 March 2007 (UTC).

Permanent magnets vary in how "hard" they are. "Soft" magnetic materials can be easily depolarized/repolarized by an external magnetic field whereas "hard" materials are harder to depolarize/repolarize. Essentially, there's a minimum magnetic field needed to change the polarization of any given permanent magnet. Some things help cause repolarization: Heat (see Curie point) and ordinary mechanical shock and vibration (which jostel the individual atoms and allows them to flip to conform to the new magnetic field).

It's unlikely that two identical magnets unvibrated and at ambient temperature would depolarize each other.

Atlant 16:04, 21 March 2007 (UTC)

so theoricaly, if a device relied on magnetic as its source of power it would never run out right? clockwork fromage

The problem with that, assuming you want to use electricity, is that an electric generator has to rotate, whereas a magnet will only move things one way by itself :( HS7 19:18, 21 March 2007 (UTC)

Energy must be conserved. If you tried to extract a meaningful amount of power from the apparatus as you described, the magnets would rapidly demagnetize. They are not an infinite source of energy. In simple problems, they can be treated as constant sources of magnetic flux but in reality they will run out of energy if you extract it. Nimur 19:50, 21 March 2007 (UTC)

Normal permanent magnets could not be "drained" in such a fashion; they are in a lower energy state when they are magnetized than otherwise. I think you would just never find a way to extract anything from them. --Tardis 21:34, 21 March 2007 (UTC)

Magnets are not like batteries - they don't store or produce energy - so they don't "run down" in the sense that a battery might. However, they can get demagnitised under various complicated situations. The idea that you could build a motor that used the 'power' of magnets is an old one - but this won't work - magnets simply aren't like that. However, that doesn't stop various 'nut jobs' from imagining that this is possible. SteveBaker 20:22, 21 March 2007 (UTC)

In spite of everything written above, magnets really do get weaker if you store them with an open magnetic circuit (or, worse, with a closed-the-wrong-way magnetic circuit, like the original poster was describing). This effect can be practically significant. Many types of permanent magnet motor cannot be disassembled without ruining the magnet. (When the motor is assembled, other parts of the motor short out the magnetic field.) This is also the reason why permanent magnets are sometimes stored with an iron "keeper" across the poles.

And yes, conservation of energy holds. No perpetual motion. —The preceding unsigned comment was added by 24.91.135.162 (talk) 20:42, 21 March 2007 (UTC).

It would be possible to make something "move" using magnets for a seemingly perpetual period of time, but a condition of perpetual motion machine is that energy must be extracted from it, trying to extract energy from the system would cause it to slow and eventually stop. Similarly someone asked a while ago why the moon, if in perfect balance, couldn't be considered a perpetual motion machine (forgetting the sun exploding etc.) the answer was that if you tried to extract any energy from the moon's orbit, it would slow down and eventually fall to earth. So I think the rule is similar here, I saw one of those desk toys that would almost convince you perpetual motion was possible with magnets, but if you tried to extract any energy from it, it would slow down very quickly. Vespine 01:19, 22 March 2007 (UTC)

Refrigerators

How do the doors on refrigerators seal? Most modern fridges have that rubber gasket and when you close the door there is that signature "thoop" sound. But there doesn't seem to be a mechanism of any sort for a physical lock. Do the gaskets have magnets in them to hold the door shut? Dismas|^(talk) 16:26, 21 March 2007 (UTC)

Try it...take a piece of metal that sticks to a magnet and see if it sticks to the seal. DMacks 17:56, 21 March 2007 (UTC)

I'm fairly sure that most of these have a magnetic strip that runs the length of the seal. I've seen some high-end refrigerators that maintain negative pressure in the cabinet, but your home fridge probably isn't doing this. Note that in days past, mechanical latches were used to hold the door shut. -- mattb @ 2007-03-21T17:58Z

It's ferromagnetic [Mαc Δαvιs] (How's my driving?) ❖ 18:10, 21 March 2007 (UTC)

There should be a rubber strip around the edge of the inside of the door (as seen in this photo). If you cut it open (not a good idea unless you're throwing out the fridge anyway!), you can the magnet; it looks and feels like a long piece of smooth, flexible black rubber. The magnet serves two purposes: not only does it keep the fridge door shut, but it keeps the rubber strip pressed right up against the frame of the fridge, ensuring an airtight seal which increases efficiency and hygiene and keeps smells inside the fridge. Laïka 18:18, 21 March 2007 (UTC)

It works all by itself. Warm air from the room gets inside your refrigerator while it is open. When you close the door the air is rapidly cooled down by the surrounding refrigerator walls, which causes its pressure to decrease and seals the door. The effect increases for a few seconds, while the air is cooled down, then it diminishes over a longer time, due to leackage. Try pulling at a refrigerator door, when it has been closed about 5-10 seconds ago. You will notice a significantly higher resistance than usual. The effect also causes a notable sound. I have no idea, whether it is your "thoop".

—The preceding unsigned comment was added by 84.187.7.58 (talk) 06:00, 23 March 2007 (UTC).

Cooking wine & spoilage

How long can a (large) bottle of red wine with screw-top last (with the top on) once opened? Would there be any valued added in putting some salt in it, as is done with commercial cooking wine? Thanks --64.56.107.126 16:49, 21 March 2007 (UTC)

Depends how much air is in the bottle, and how much you shook it up and what the wine was like originally. Assuming it is a robust young red I'd stick it in the fridge, and if only a glass or so is missing it'll probably last a week quite happily, might even improve. Worst case, if it is half empty, and was an old wine, it would probably start to taste a bit 'thin' on the third day, and the temperature shock from the fridge might not do it any good either. Salt won't have much effect at a guess. I have had white wines that went from good-ish to undrinkable in 1/2 an hour, but they were very old and on their last legs. I keep a bottle in the fridge that started as a bottle of sherry, it gets topped up with anything that is too good to pour away, it is fine after a year (I'm not bothered about colour when I use wine in cooking). I would be a bit careful with red wine in food - I'm not convinced that it is always a good thing, certainly cheap red wine can create some pretty odd tastes and smells in stews Greglocock 07:18, 22 March 2007 (UTC)

I have read that the salt is put in cooking wine not so much to preserve it as to make it undrinkable. The salt makes it taste bad and might cause vomiting, but usual;ly tastes ok in a cooked dish. Edison 14:28, 22 March 2007 (UTC)

genetic epidemiology questions

Hello, I was asked a question in the exams regarding genetic epidemiology please verify if my attempted solution is correct. please pardon if the question itself is incomplete , im quoting them from memory

• question1:

how many affected individuals with gene DD? given P(DD)= 1 P(Dd)=P(dd)= 0,0001.

couldn't solve this. perhaps this question is incomplete and it would be nice if you could insert the missing parameters and give a sample solution.

• question2:

given p(A)=.7 ,and there are 100 individuals what is fAa? (fAa is the conditional probability that an individual is affected given he has the genotype Aa)

attempting to solve: p(a) =1-.7= .3 assuming hardy weinberg equilibirium: p(Aa)= 2*.3*.7=.42

p(affected|Aa)= P(affected and Aa)/p(Aa) = .001/.42 =.0024

• question3:

given p(AA and affected)=.08, p(A)=.2 calculate p(affected and AA)?

attempting to solve: perhaps the question is wrong, maybe one needs to calculate p(affected|AA).

p(AA)= .2*.2=.04

p(affected|AA)= P(affected and AA)/p(AA)= .08/.04 = 2

probably a wrong answer, how can conditional probability greater than 1? -

Thank you very much for patiently reading through this list of questions. Iam perplexed, lecture notes are of little use, any help would be wonderful.

Thanks in advance.

212.201.73.208 17:05, 21 March 2007 (UTC)

i think the answer to question 1 is that only 1 individual is affected and that its puff daddy =)

Vasopressin and caffeine

i ave just read articles on Vasopressin and i was wondering if there might be a relation between long term caffeine intake and the brain chemical fonction of Vasopressin like agressivity to other males and bonding to sexual partners clockwork fromage —The preceding unsigned comment was added by 216.113.99.149 (talk) 17:48, 21 March 2007 (UTC).

I think you're a little ahead of the game. I don't believe such a link has been established, which doesn't mean it doesn't exist. However, the caffeine article states the over use and intoxication effects of caffeine, which has been around for a long time, and none of it seems to mention aggression or relationship issues. Vespine 21:58, 21 March 2007 (UTC)

There is known to be cross-talk between caffiene induced and vasopressin induced release of intracellular Ca2+ stores, somewhere at, or upstream of, an endoplasmic reticular Ca2+-ATPase. However, knowing there is a molecular interaction between the two signalling pathways is very different from knowing their behavioural consequences. Rockpocket 22:05, 21 March 2007 (UTC)

Nuclear bombs

Hello,

On the British Vanguard am i right in thinking that there are 48 warheads on each sub and that each warhead in 100x more powerful than the bomb dropped on Hiroshima?

If so what would happen if they were all targeted in the middle of, say, London. What would be the scale of damage?

Thanks, --12345 wiki 20:29, 21 March 2007 (UTC)

Submarines of the Vanguard class can each carry 16 missiles, each missile can have up to 8 warheads, each warhead is either 100 kilotons or 475 kilotons - depending on which type of warhead is selected. The Hiroshima bomb was between 13 and 16 kilotons. So the sub carries 128 wardheads - each of which could be around 30 times the power of the Hiroshima bomb. Just one of those warheads would be more than enough to reduce the whole of central London to rubble. SteveBaker 21:34, 21 March 2007 (UTC)

I always wonder about threats to drop a zillion nukes on some specific site...like here, would 127 more do much more damage other than a deeper hole and more fallout? DMacks 21:40, 21 March 2007 (UTC)

(afer edit conflict)According to the article, it had 12 missile tubes capable of firing the trident missile which has a maximum payload of 3.8 megatons, the bomb dropped on Hiroshima was 13 kilotons.. It's hard to speculate the extent of the damage caused by a weapon like this and a description is probably impossible for a human to even fathom. Imagine this, a circle in the middle of London maybe 3 kilometres in radius is a crater in the ground, a circle 8 km in radius most buildings are levelled and even 20km windows are blown out, there is moderate damage to buildings and there are still fatalities. The nuclear explosion article is quite good. Now I am become death, the destroyer of worlds. As for the multiple nuke scenario, I don't think the "blasts" would add up, but there would definitely be an effect on fall out. The radioactive material comes from the blast it self, so more blasts, more radioactive material. I'm sure the big glass crater in the ground would reduce the amount of material gicked up in each subsequent blast but it would still add. Vespine 21:47, 21 March 2007 (UTC)

You're right about the 48 warheads (that's what Nuclear weapons and the United Kingdom says), but not about their yield. That article indicates the British bombs are similar to (or perhaps derivatives of) the US W76 warhead, which has a yield of 100 kilotons. The Fat Man (Nagasaki) bomb was around 21 kilotons, and Little Boy (Hiroshima) 16 kilotons (max). So Vanguard's warheads are maybe six times as powerful as Hiroshima. There's really no military need to make ones larger than that: you'd just end up digging a deeper hole. Atomic bombings of Hiroshima and Nagasaki says the area of total destruction at Hiroshima was about 2 miles across - lets guess that a 100kt bomb maybe destroys 3 miles across (like anything, more power gives diminishing returns) - that's around 9 square miles. That's per warhead. Now you wouldn't just launch all your warheads to one location, or all at one time - they'd all blast one another (what the nice nuclear men call "fratricide"). So you'd launch them over a space of a couple of hours, and you'd spread out their targetting so their effects would be maximised. 48x9 is about 400 square miles, or around 2/3s of Greater London would (in some evil-optimal scenario) be entirely destroyed. But really this is total overkill - five or six explosions, spread across the central section of the city, would be enough to horribly devastate it and kill maybe a million people. That's why Britain feels it needs only one sub at sea at a time - even accounting for losses due to (largely non-existent) missile defenses, those 48 warheads around enough to kill at least ten million people in any enemy country. I really cant for the life of me figure out why the US thinks it need 14 Ohio class subs (each with more tridents and far more warheads than the Vanguard boats) and why the Russians think they need 10 Borei class submarines. Darryl Revok 21:51, 21 March 2007 (UTC)

It is also worth remembering as well that blast effects were not the cause of the majority of deaths at Hiroshima — fire was. Many nuclear planners during the Cold War vastly underestimated the effects of nuclear weapons by ignoring fire effects (which are harder to calculate in any case) and as such vastly overestimated how many weapons would be needed to take out a city. (Lynn Eden's Whole World On Fire is my source for this, btw). (As for why the US and Russia employ so much overkill, it is because they aren't really committed to exclusively second-strike forces. Which is scary as hell. China and the UK, by comparison, have purely second-strike forces — they could never expect to launch a nuclear attack that could take out the nuclear weapons of an enemy country, but they could counter any attack launched against them.) --24.147.86.187 23:12, 21 March 2007 (UTC)

Keep in mind that there are maybe three countries in the world that could survive a nuclear strike to the largest city: the US, the Soviet Union, and China. That's a large part of why the US and the USSR had so many nuclear weapons of all sorts. --Carnildo 20:03, 22 March 2007 (UTC)

One of the other not-so-nice parts of the madness that goes into planning nuclear attacks is that you really can't use nuclear weapons on tiny small towns and villages away from the big cities - but you can more effectively crush resistance by deliberately wounding and otherwise incapacitating people in the big cities without conveniently vaporizing them. Thaking care of these casualties then completely overwhelms whatever rural resources remain - spreading despair, uncontrolled infections and consuming valuable food, water and medical care facilities. So it's not a great idea to turn major cities into smoking craters - you want a small smoking crater and then a vast area of lesser damage. Using lots of tiny warheads achieves that far more effectively than one large one. It's a nasty business. SteveBaker 04:02, 23 March 2007 (UTC)

As a child, and before I learned about the sorts of things you mention, and how most strategic weapons were intended to knock out the other guy's strategic weapons, I was mystified about why the superpowers would have 10,000 nuclear warheads. I would try to figure out what the 10,000th-largest town in the US was. It's probably not very big. Madness. --TotoBaggins 05:29, 24 March 2007 (UTC)