Wikipedia:Reference desk/Archives/Science/2008 May 31

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May 31

Fermentation in eukaryotic cells

Where does fermentation or anaerobic respiration occur in eukaryotic cells. Is it the cytoplasm, the mitochondria, or what? I've been searching online, and I can't find any clear answer. Any help would be greatly appreciated. FlamingSilmaril (talk) 00:25, 31 May 2008 (UTC)

In the cytoplasm. Wisdom89 _{(T / ^C)} 00:30, 31 May 2008 (UTC)

Ok, thanks very much. FlamingSilmaril (talk) 00:32, 31 May 2008 (UTC)

EEG and epilepsy

Hi,

Does anyone knows if EEG data that are typical of epileptic seizures can be the sign of other illnesses than epilepsy?

Thanks! —Preceding unsigned comment added by 85.0.41.187 (talk) 02:37, 31 May 2008 (UTC)

Electroencephalogram#Comparison_table may be of some use to you. If you are seeking medical advise, however, it's advisable to see your physician. Regards, CycloneNimrod ^talk?_contribs? 17:12, 31 May 2008 (UTC)

Acceleration Due to Gravity

Hello. How is the following equation for acceleration due to gravity proved: $\vartriangle {\overrightarrow {d}}={\overrightarrow {v}}\vartriangle t+{\frac {1}{2}}{\overrightarrow {a}}(\vartriangle t)^{2}$? Thanks in advance. --Mayfare (talk) 02:49, 31 May 2008 (UTC)

In science there is no such thing as a proof, see Scientific method and Epistemology as well as the Science article. For example, for centuries Newton's law of gravitation was the widely accepted theory, but know we know that General relativity explains many things that Newton's theory did not. Jkasd 05:43, 31 May 2008 (UTC)

Although Jkasd is right, this doesn't mean there's anything wrong with your equation just because it's newtonian. On that note, newton did derive it from somewhere, he didn't just pull it out of hat. It's just straight up calculus. One of the major ideas underlying newtonian physics is that when a force is applied to an object, it ACCELERATES in the direction of that force. $\vartriangle {\overrightarrow {a}}={\overrightarrow {a}}$ for a constant force (such as gravity). Accerleration is given in units of m/s^2, so $\vartriangle {\overrightarrow {v}}={\overrightarrow {a}}\vartriangle t+{\overrightarrow {v}}_{i}$ . If you've done calculus, you may have noticed that as speed is the first derivative of accerlation, similarly $\vartriangle {\overrightarrow {v}}={\overrightarrow {a}}\vartriangle t+{\overrightarrow {v}}_{i}$ is the first derivative of $\vartriangle {\overrightarrow {a}}={\overrightarrow {a}}$ . The equation you gave is just the second derivative:

$\vartriangle {\overrightarrow {a}}={\overrightarrow {a}}$ 
$\vartriangle {\overrightarrow {v}}={\overrightarrow {a}}\vartriangle t+{\overrightarrow {v}}_{i}$
$\vartriangle {\overrightarrow {d}}={\frac {1}{2}}{\overrightarrow {a}}(\vartriangle t)^{2}+{\overrightarrow {v}}\vartriangle t+{\overrightarrow {d}}_{i}$

Ta Da! That's how it's derived. Your equation is just a fancy way of saying "Displacement is the second derivative of acceleration." As Jkasd pointed out, the most you can do is verify it to a high degree of certainty via experimentation. PS: this was totally the first time I played with making equations in wikipedia, and I found it to be a huge pain in the ass. For the record. --Shaggorama (talk) 08:09, 31 May 2008 (UTC)

Actually, I think you meant to say that the acceleration is the second derivative of the displacement function. What you did above was not differentiation, but integration. If you take the original definition of acceleration as the change in velocity over the change in time (valid only if acceleration is constant), and rearrange the equation you get the velocity equation above. If you integrate that with respect to time, you get the equation you originally asked about. PhySusie (talk) 11:32, 31 May 2008 (UTC)

Except (on closer examination) that the second equation in the box should have final velocity on the left side of the equation (not change in velocity) and the third equation should have final position on the left side (not change in position). PhySusie (talk) 11:35, 31 May 2008 (UTC)

Why is there ${\frac {1}{2}}$? --Mayfare (talk) 13:15, 31 May 2008 (UTC)

This is the standard result for integrating a polynomial of degree one; that is to say:

$\int (t)dt={\tfrac {1}{2}}t^{2}+C$

See integral. Very briefly, 1/2 is the result of calculating this integral from the definition of integration. Evidently you haven't had your first calculus class yet! All in good time; you will learn these tools if you continue your study of physics. For now, suffice to say that calculus is the mathematical tool developed by Isaac Newton while he was working on mechanics (physics). Calculus is essential to the study of physics, because it allows consideration of continuous quantities (smooth motion, for example). Integration and differentiation are not very difficult to learn if you already understand algebra; the twist is conceptual - applying limits, continuity, and functional analysis; and then practicing on a lot of sample problems. Nimur (talk) 15:12, 31 May 2008 (UTC)

Note that the 1/2 here has a very simple geometric interpretation: it comes from the fact that the area of a right triangle is half that of a rectangle with the same sides. To see the connection, plot speed (v) as a function of time (t): by the definition of acceleration, it's a straight line with slope a. Assuming you start at v = 0 when t = 0, the distance (d) travelled at time t > 0 will be equal to the area under the line and to the left of t (i.e. the integral of v with respect to t), which forms a right triangle with base length t and height at. This has 1/2 times the area of a rectangle with sides t and at, which of course has the area t·at = at². If you start at some non-zero initial velocity v = v₀ > 0, this moves the line upwards so that the area under it (and left of t) now also includes a rectangle with sides t and v₀. Putting these together, you get d(t) = 1/2 at² + v₀t. —Ilmari Karonen (talk) 04:15, 1 June 2008 (UTC)

Another way of looking at it that may offer some insight is that the speed of the object at time t is at + v₀. Thus, d₁ = at² + v₀t = (at + v₀)t (without the 1/2) is the distance the object would've travelled if it had been moving at that speed all along. But since the object only gradually accelerated up to its full speed, d₁ is obviously an overestimate of the real distance. On the other hand, the distance the object would've travelled if it had been moving at its original speed of v₀ is simply d₀ = v₀t, which is clearly an underestimate. To better estimate the real distance travelled, we can try taking the average of these two distances, which is d_avg = avg(d₀, d₁) = 1/2 (d₀ + d₁) = 1/2 (at² + v₀t + v₀t) = 1/2 at² + v₀t, which we might expect to be closer to the real distance than either d₁ or d₀. As it happens, since, in this case, the change in speed is linear, this averaging "trick" actually does yield the exact distance the object actually travelled — but even if the acceleration had varied a bit over time, it still would've been a better approximation.

Also, in both of my comments above, I've implicitly assumed a and v₀ to be positive scalars, since that's the simplest case to visualize. Once you have a good handle on that, it's a good exercise to look at the case where either a and/or v₀ might be negative, or even non-collinear vectors, and show that the same rule still works in those cases as well. —Ilmari Karonen (talk) 20:36, 1 June 2008 (UTC)

Ice in a Hot Room?

How effective, if at all, would a jar of ice be at cooling down a hot room? Digger3000 (talk) 03:22, 31 May 2008 (UTC)

Evaporative cooler has a section on performance.--76.111.32.200 (talk) 04:12, 31 May 2008 (UTC)

Evaporative coolers start with liquid water. Ice can absorb more heat. Just a moment, let me do some arithmetic... --Anonymous, 05:00 UTC, May 31, 2008.

Okay, say you start with 1 kg of ice at -15 C from a freezer, and you want to finish with water vapor at a room temperature of 25°C. The specific heat of ice is 2.05 J/gnbsp;K (or equivalently kJ/kgnbsp;°C), so you'll need 15x2.05 kJ to reach the freezing point. Then the latent heat of fusion of water is 333.55 kJ/kg, so you'll need that many more kilojoules to get a liquid. Now liquid water has a higher specific heat, 4.18 kJ/kgnbsp;°C, so it's 25x4.18 kJ to reach room temperature. And finally the latent heat of vaporization is a further 2,270 kJ/kg. So the total heat energy needed adds up to over 2,700 kJ.

Now say that the room is 5 m square by 3 m high; that's 75 m³. The density of air is about 1.2 kg/m³, so that's 90 kg of air. It could give up 2,700/90 = 30 kJ/kg. The specific heat of air is 1 kJ/kg °C, so a 1 kg chunk of ice could lower the temperature of the air by a full 30°C.

But there's a big flaw in the above calculation, which is that I assumed that the same 90 kg of air would be present the whole time! In practice, air in a normal room is circulating in and out all the time, even when the doors and windows are closed. That's why you don't suffocate when inside a building. The melting and evaporation of ice is a slow process and there's not much you can do to speed it up.

In other words, it's not the heat capacity of the ice that's the issue, it's the transfer rate of heat from the air to the ice. You not only need to cool the air, you need to cool it before it's replaced with fresh warm air from outside. It is possible to cool air with ice -- it used to be done routinely on trains, before head-end power was available to run air-conditioners - but in practice it takes a lot of ice, way more than my calculation above would suggest. Forcing the air to blow over the surface of the ice would only help a little, because it doesn't have much surface area.

--Anonymous, 05:36 UTC, May 31, 2008.

On trains in India during the 19th century, this type of air-conditioner worked by misting ice water into the train car. The device was called a thermantidote. I had a heck of a time looking this word up when I read this novel (a broken thermantidote ended up being some kind of ... hiding place for a criminal, or something like that). This terminology is decisively not in current use in America (we've still got a redlink article)! I figured it sounded like some kind of medication. Nimur (talk) 15:23, 31 May 2008 (UTC)

Oh, mist coolers are used all the time in tropical places! Every open-air bar in Key West, FL has them! --BenBurch (talk) 01:08, 1 June 2008 (UTC)

Rather than cooling the whole room, try just cooling yourself by blowing air over the block of ice onto you. That should be noticeably cooler than just using a fan alone. StuRat (talk) 23:49, 31 May 2008 (UTC)

Or as a short cut, misting yourself near a fan for evaporation effects. Julia Rossi (talk) 00:41, 1 June 2008 (UTC)

That's an effective cooling method, but doing so inside will result in damp (and possibly moldy) furniture, drapes, carpet, etc. StuRat (talk) 15:58, 2 June 2008 (UTC)

I was thinking face, neck, chest, and a desk fan for a short time dear Sturat, not the room! At that point it evaporates, I hope. You had me worried for a sec. : )) Julia Rossi (talk) 23:41, 2 June 2008 (UTC)

Water vapor is difficult to contain, and will get on everything around the spray (and in the direction the fan is blowing). If the humidity level is low, it will evaporate readily. If not, it may remain until a nice little mold forest grows. I once stayed at a motel with a window A/C unit that was spraying water on the floor. I then thought "isn't the carpet quite a bit thicker over here ?". They were reluctant to give me my money back until I pointed out this little biology experiment and mentioned how photogenic it might appear on the evening news. :-) StuRat (talk) 03:00, 3 June 2008 (UTC)

Point taken. I guess where there's heat and water, there's humidity. Nice action, our man in Havana,  : ) Julia Rossi (talk) 02:04, 4 June 2008 (UTC)

Human Biology

The subject is Sweat Glands. The question is: The body has many sweat glands, but what parts of the body or where on the body sweat glands CANNOT be found?? 75.89.250.154 (talk) 04:09, 31 May 2008 (UTC)

Next question, "When's your homework due?" Here's a hot tip on how to find answers for such questions. Your teacher has probably assigned a textbook for this class. I's a great idea to actually get that. (Check for used books if your budget's strapped or check your local library.) Then go to the table of contents in the back. Look for "sweat" or related subjects. That should either get you a page, or at least the section. skim that section for things that say "sweat" or "glands", and start reading around it. Pay attention to things saying "except", "no" or "not". That should get you the answer and the skills you're building that way can let you survive not just to the next test, but throughout all your studies.76.111.32.200 (talk) 04:58, 31 May 2008 (UTC)

WP:SARCASM. Careful with the tone. This doesn't appear to be a homework question. Regardless, to the user who asked the question, simply scour the internet. Seriously, there's loads of info out there about this topic. Wisdom89 _{(T / ^C)} 05:03, 31 May 2008 (UTC)

To me it does sounds either like a homework or quiz question Nil Einne (talk) 05:27, 31 May 2008 (UTC)

You need to qualify the answer you're looking for, or the question is trivial: I don't think you can find sweat glands in any of the internal organs. If you're restricting the answer to places on the surface of the body, you need to decide whether, say, finger nails count as an acceptable answer. --72.78.237.206 (talk) 12:56, 31 May 2008 (UTC)

Eyeballs don't have sweat glands either, do they? And have about scar tissue, does that grow back sweat glands? Graeme Bartlett (talk) 21:44, 1 June 2008 (UTC)

You don't have sweat glands in lips.

Joe

Night Luminous Pearl (Ye Ming Zhu in Chinese PinYin)

How is the Night Luminous Pearl (frequently given as gifts to Emperors in Ancient China) formed in nature and how are they harvested ? What is the difference between this Luminous Pearl and normal Pearl from Oysters e.g its mineral contents ? Is it found only in China, as no other countries seem to refer to it in history. Currently, where can one view / purchase a real Night Luminous Pearl ?

218.186.72.68 (talk) 05:20, 31 May 2008 (UTC) KKC

I think you'll find this interesting[1] in that the term is also used poetically and about applying phosphorescence as well as the luminosity that made pearls special in times when there was not so much that shone in its own right. Julia Rossi (talk) 23:37, 2 June 2008 (UTC)

The Night Luminous Pearl is nothing but a Chlorophane, a variety of flourite with some REE (Rare Earth Elements). These elements act as activators and deactivators. Due to these rare earth elements the flourite stores heat energy from natural sun light and release same in darkness. Some Fluorites would need UV short wave(SW) or long wave(LW) light to glow. This fact was published in many chinese journals as early as 16th century and was known to west only in 19th century, one such article was published in Harper's Monthly magazine in 1890 (http://digital.library.cornell.edu/cgi/t/text/pageviewer-idx?c=harp;cc=harp;rgn= full%20text;idno=harp0080-4;didno=harp0080-4;view=image;seq=0546;node=harp0080-4 %3A6). Copy and paste the above link in new browser window.

The biggest piece produced from Tungsten mine in Guangdong China was 6.1 ton valued at 3.1 Billion USD. You can read the artilce at following link.

http://www.articlesbase.com/nature-articles/yemengzhu-the-3-billion-dollar-rock-298407.html

The similar Night Luminous pearl or chlorophane is found in siberian mines of Russia and few other places. The cobra also makes use of small luminous pearls to hunt insects and rodents and carry this kind of pearl, it is also called cobra pearl/serpant stone. But it is not formed inside the body of snake and it is picked from dry river beds. A small pea size cobra pearl is valued at few million USD, which is actually fluorite with night glow.

logic programming

(1)why is logic programming regarded as the corner stone of knowledge based programming41.209.23.34 (talk) 06:01, 31 May 2008 (UTC) (2)can an informal model theoretic argument be established for statements like this,(has niece(x)if(3x)daughter(x,y)sibling(x,y)) {x/brother(y)}.

>::cough::HOMEWORK::cough::< --Shaggorama (talk) 07:26, 31 May 2008 (UTC)

Got Lost In One Of The Questions (I'll Post This On Entertainment, As I'm Not Sure Which Desk Is Most Appropriate)

3 media of amplitude can be expressed in keyboards, per Keyboard_expression. I am assuming different instruments, such as Brass instruments, or instruments that use water, have different ways of changing dynamics, or the amplitude of the sound that can be normally produced by them, or not, say banging a saxophone against a wall. Are there any others, and if so, could you list them all? Also, please list all theoretical, possible and/or probable ways of changing dynamics (or, in other words, the amplitude of a sound).

So what I'm saying is that for different Keyboard instruments, as per the article Keyboard_expression, there are 3 ways to express dynamics.68.148.164.166 (talk) 07:30, 31 May 2008 (UTC)

I think you're asking, what are the different ways of varying the volume on different musical instruments? I suggest looking at Musical instrument classification, and following the links from there. I would point out that asking people on reference desks to list all of something is not usually a very productive request: the people who reply on these desks do so voluntarily out of interest or a desire to help, but unless somebody happens to be a completist about a topic, they may not want to spend the time compiling lists. --ColinFine (talk) 00:23, 1 June 2008 (UTC)

Humidity and electronic equipment

Why do specifications for electronic equipment (TV, computer, etc.) specify a range for humidity, eg. 10-70%? If the air is too humid or not humid enough, how does that affect the equipment? I've heard from friends that it would catch on fire, but I doubt it. Rilak (talk) 07:42, 31 May 2008 (UTC)

I'm not entirely sure, but if it's too humid, I'd say condensation might occur in the device which might short circuit the whole thing. A lack of humidity doesn't sound like it would do much, although dry air does increase the chance of static electricity building up and whatnot.84.198.96.249 (talk) 08:37, 31 May 2008 (UTC)

I agree with the above for high humidity, now let me try to explain the low humidity issue, as well. It might seem odd, but quite a few electronic devices include paper (say a circuit diagram glued to the inside of the case) or paper-based tape. When either loses all of it's humidity it can fall apart. Those pieces can then fall onto some hot component and catch fire. StuRat (talk) 23:40, 31 May 2008 (UTC)

Birds and hot chips

moved from misc desk
When I eat a very hot chip (or fries, depending on your preference) I can burn my mouth. But when I drop a hot chip on the floor a seagull will swallow it down whole. Do they not feel the heat of the chip in their mouth?Iiidonkeyiii (talk) 08:48, 30 May 2008 (UTC)

Good to see another seagull question here at last. First, I would think that for a bird to swallow it whole, it would be a relatively small chip so it is already cooler than the fat ones. Secondly, birds don't chew their food and gulls usually swallow their food whole, so they probably don't notice its temperature until it's too late.--Shantavira|^{feed me} 09:24, 30 May 2008 (UTC)

Given that they do not chew, can we conclude that they have additional mucous or other sturdy membrane linings in their esophagus? I imagine they would need to be made of tougher stuff if they regularly swallow food whole. Nimur (talk) 15:30, 31 May 2008 (UTC)

Gulls also seem to be capable of quickly horking up anything that doesn't feel right in their crops. Sometimes they'll make several attempts at eating that large/hard/spiky thing to see if they can get it to sit comfortably inside. The other week, I saw a gull making three attempts at keeping a whole tomato down before finally deciding just to bite into it. --Kurt Shaped Box (talk) 22:23, 31 May 2008 (UTC)

Power Quality Managment

Is it possible to improve power by improving power factor? If yes upto what percent we can improve —Preceding unsigned comment added by Syedshahid (talk • contribs) 08:31, 31 May 2008 (UTC)

See power factor and power factor correction. Both of your questions are answered there. --Heron (talk) 10:11, 31 May 2008 (UTC)

Can you shoot yourself in the head twice?

Assume the following: a person has two pistols, one in each hand. Would it be possible for that person to fire both guns at their own head? Or would one lose muscle control faster than you can squeeze the second trigger? Pardon my morbid curiousity :) 84.198.96.249 (talk) 08:35, 31 May 2008 (UTC)

Rigor mortis—contraction of skeletal muscles due to depletion of ATP or cadaveric spasm might trigger the contraction of fingers of the other hand making them press the trigger. So, this might result in the situation you are speculating. I personally, have no problems with your curiosity. Regards. —KetanPanchal^taLK 09:05, 31 May 2008 (UTC)

You would be lucky to still be aiming right by the time rigor mortis sets in. It would be easier to fire both guns at the same time - that ought to work fine (even if you're off by a fraction of a second there is enough delay in the firing of the gun and the loss of muscle control to get away with it - or at least, I expect there would be, I've never actually tried it!). --Tango (talk) 12:04, 31 May 2008 (UTC)

If you're very, very lucky, you could get the 'classic' sort of suicide story where both your bullets hit each other and so don't kill you (since it's likely both your bullets will be at an angle, I presume they will still go forward and may still hurt you but I guess could potentially lose enough momentum so that they don't kill you although I can't be bothered doing the maths) Nil Einne (talk) 16:10, 31 May 2008 (UTC)

I was imagining one gun on each side of your head, so if the bullets hit each other they would do so in the middle of your brain. Where are you imagining the guns? --Tango (talk) 16:46, 31 May 2008 (UTC)

I'm also curious to know what's going on inside Nil Einne's head. I imagine Nil misunderstood the question since I can't imagine any configuration that conciliates two shots aimed at the head with both bullets hitting each other to save the experimenter. ----Seans Potato Business 13:03, 1 June 2008 (UTC)

He (she?) was considering the scenario where each gun's bullet is aimed towards the side of the head farthest from the gun. The bullets would then have a possibility of colliding in mid-air prior to entering the head. here's a diagram.

Suppose both guns are pointed towards the head so that their barrels, if extended, touch the head. The bullets cannot be travelling directly towards the head's centre since their collision point would then be inside the brain. Each bullet can be considered to be concurrently travelling towards the centre of the head and in the perpendicular direction. Upon collision, the bullets' velocities perpendicular to the path to the head's centre partially cancel out, while their velocities toward the head add. Whether the resultant amalgamation of the projectiles would have a greater speed than its components would have had without a collision depends on the angle between the head and the guns.

The probability of two bullets intercepting each other in mid-air, though, is likely less than the probability of winning the lottery. --Bowlhover (talk) 15:26, 1 June 2008 (UTC)

A simple fail-deadly system would be to put a strong rubber band around each trigger finger that requires you to actively resist pulling the trigger. Point the guns at your head such that one bullet's exit won't knock the other gun out of position, and then just ... relax. --Sean 17:12, 31 May 2008 (UTC)

Fail-safe, you mean? Since the goal is suicide and contingencies are planned to meet it, an initial failure is "safe" because it would not render the goal impossible. --Bowlhover (talk) 15:26, 1 June 2008 (UTC)

I'm disturbed at how practical these answers are getting. Doesn't this fall under the aegis of no medical advice? ;-)--Fangz (talk) 18:59, 31 May 2008 (UTC)

It's no medical advice, not no immoral advice. Discussion of war and other violent means is allowed here. SpinningSpark 19:55, 31 May 2008 (UTC)

Is it true that rigour mortis doesn't set in if you're shot in the eye? Bastard Soap (talk) 20:26, 31 May 2008 (UTC)

While being shot in the head is usually fatal, a surprisingly large number of people survive. Here are some examples [2][3][4][5]. However, my favourite is Michael Moylin who woke up with such a severe headache he thought he had suffered an anuerism. The hospital soon put him straight though, and pointed out that his wife had shot him in the head while he had been sleeping [6][7]. So yes, it would be quite possible to shoot yourself in the head multiple times because you might survive the first bullet. This is not a reliable method however - don't try this at home. SpinningSpark 20:47, 31 May 2008 (UTC)

I was trying to find a newspaper story on a woman who shot herself in the head in (possibly) the 1980s and survived. She had shot herself in the right temple but the surgeon removed the bullet from her left temple. The bullet had not, as one might imagine, passed through the brain. It had, in fact, skidded across the top of her skull underneath the skin. There was a great x-ray pic in the papers at the time - if anyone has a link please post it, I couldn't find it. Some women are just lousy shots I guess[8]. SpinningSpark 20:47, 31 May 2008 (UTC)

Well, Bastard Soap (am not calling you names! That's your user name), considering the mechanism by which rigor mortis develops (depletion of ATP) in the individual skeletal muscles, I don't think what happens to the brain (shooting in the eye) should have much influence upon the setting in of rigor mortis. Though there are factors like temperature, emotional state, poisoning by arsenic, etc. that could alter the rapidity and depth of rigor mortis, I don't know if the factor you are talking of could also be one of these. But, remember all these factors I enumerated simply determine the intensity and timing of rigor mortis not whether it develops or not, per se.

With all this talk going about medical advice and some one stating "Discussion of war and other violent means is allowed here.", this point from WP:Common sense could be instructive here:

"Similarly, just because something disruptive is not forbidden in a written rule doesn't mean it's a good idea (e.g., don't disrupt Wikipedia to illustrate a point). The spirit of the rules is more important than the letter."

But since, I believe the user who originally started the topic must be well aware that it can be fairly assumed that firing of both the pistols doesn't really greatly alter the probability of death, he'she is not asking for medical advice. Regards. —KetanPanchal^taLK 20:52, 31 May 2008 (UTC)

With rigour mortis I mean the sudden contraction at the moment of death (don't think it's the right term but don't know the right one). I remember reading somewhere that snipers aim for the eyes in hostage situations so even if the terrosit has a gun pointed he won't fire in death. Bastard Soap (talk) 15:49, 1 June 2008 (UTC)

Yes, I'm indeed quite aware that one shot to the head is usually quite fatal enough without the need for a second one ;). I just get the occasional bout of morbid curiousity. I think the rubber band guy might just be on to something though, very clever :D84.198.96.249 (talk) 22:22, 31 May 2008 (UTC)

Going back to the original scenario, let's say that the guns involved are Walther P99 pistols loaded with american .40 calibre ammunition with a muzzle velocity of 320m/s. Let's further assume that the bullet is decelerated to a complete stop just before exiting the head (this assumption gives you the maximum possible reaction time, a "through and through" shot will give you less time as the bullet travels faster). With those assumptions, the travel time through my 16cm head would be exactly 1ms. So you have, at most, a 1ms reaction time to fire the second gun before your brains are scrambled and unable to respond. According to the article reaction time for an audio stimulus is about 150ms. So the answer is no unless you contrive to fire both guns simultaneoulsy. Simultaneous in this context being $\pm 500\mu s$. This is humanly next to impossible, but probably unverified by experiment. Don't try this at home. SpinningSpark 22:15, 31 May 2008 (UTC)

There's also going to be some time between squeezing the trigger and it actually firing. Also, you don't need to respond to a stimulus to fire the second gun, you can have already made the decision to fire it before firing the first. If you fire them simultaneously your margin of error is the 500 microseconds you mention, plus the time the gun takes to fire and the time it takes for the signal to get from your brain to your finger (you'll still pull the trigger even after your brain is gone if the signal has already been sent). I don't know what that works out to, but I expect it's significantly more than 500 microseconds. --Tango (talk) 23:18, 31 May 2008 (UTC)

One could argue philosophically that if it is a reflex or rigor mortis that causes you to squeeze the trigger rather than a conscious choice, then you, personally, in fact did not shoot yourself twice. Wrad (talk) 23:24, 31 May 2008 (UTC)

You could, but I don't think anyone mentioned reflexes... --Tango (talk) 00:22, 1 June 2008 (UTC)

They did further up. Wrad (talk) 01:29, 1 June 2008 (UTC)

The breakdown of reaction time is (very approximately) 10ms ear to brain transmission, 100ms decision time, 40ms brain to hand transmission. SpinningSpark 13:13, 1 June 2008 (UTC)

Trapezius muscle—A doubt!

If all the fibers in the trapezius muscle are supplied by the same nerve, will the simultaneous contraction of the superior and inferior fibers not antagonize each others' actions? Hope someone answers it.

PS: I have posted the same question on the talk page of the article trapezius muscle.

Regards.

—KetanPanchal^taLK 09:05, 31 May 2008 (UTC)

Not all neurons that form the part of the accessory nerve that supplies the trapezius need fire at the same time. Indeed, upper and lower fibres of trapezius have opposite actions on the scapula. -- Flyguy649 ^talk 14:29, 31 May 2008 (UTC)

Hi flyguy! You mean these neurons come from different spinal segments? Yes, that is a distinct possibility considering that the spinal accessory nucleus has the root values of C1-C5. Thanks. Regards. —KetanPanchal^taLK 17:43, 31 May 2008 (UTC)

Pauli Equation

I I wanted to use the Pauli equation to describe a spin-4 particle, would I just have to replace the Pauli matrices with the spin-4 equivalents, or would I have to change something else, such as changing the part at the beginning fron 1/2m to 4m? Thanks, *Max* (talk) 16:01, 31 May 2008 (UTC).

The 1/(2m) part doesn't change. You just increase the dimension of the spinor part of the wavefunction (for spin 4 it has dimension 9), and replace the Pauli matrices. —Keenan Pepper 02:50, 1 June 2008 (UTC)

Good, that's what I'm doing. Looking at the equation again, it's obvious that the 1/2 doesn't chamge. Will changing the matrices get rid of the Pauli exclusion principle? *Max* (talk) 22:16, 1 June 2008 (UTC) It was never there. *Max* (talk) 19:52, 2 June 2008 (UTC)

NUCLEOPHILICITY VS BASICITY

Hello,can some1 please help me to undstand what is the difference between NUCLEOPHILICITY and BASICITY.After going through books,i have learnt that they are somehow directly proportional in terms of strenght.But I'm bit confused as I find sometimes stronger nucleophiles are weaker bases.also i found that type of solvent affects nucleophilic nature.My text books dont explain clearly why.Can Some1 please help? Abhiroopron (talk) 17:06, 31 May 2008 (UTC) Abhiroop.

The strongest bases typically bear negative charge and lone pairs on the same atom. This property makes them both excellent nucleophiles and lewis bases (electron pair doners). This actually creates a problem in protic solvents, as they are typically very good at accepting hydrogen bonds if not outright reacting with the solvent. As a result, stronger bases (especially when comparing halide ions) can often be very poor nucleophiles in such a solvent. This can be thought of in either of two ways; in the kinetic sense, having the nucleophile lone pairs bound up by the solvent reduces their availability to attack the electrophile, lowering the rate of reaction; in the thermodynamic sense, the interaction with the solvent lowers a nucleophile's (and thus, the total reactant's) energy, reducing the favorability of a reaction. As an example of such a thing, iodide ion (an extremely weak base) will complete its reaction with methyl iodide (that's an interesting reaction, isn't it? But you can measure it via iodide isotopes) in 17 minutes in methanol, and 8.7 seconds in Dimethylformamide (a polar aprotic solvent). Faster in the aprotic sovlent, as expected from what I just told you. Cyanide, a much stronger base than iodide, reacts in 0.011s in DMF (the stronger base reacts faster than the weaker base, as you yourself probably wanted to say). In methanol, however, cyanide ion takes 1.5 hours to completion, about six times slower than iodide. Thus, as you can see, stronger bases are often worse nucleophiles in a polar protic solvent. Someguy1221 (talk) 23:36, 31 May 2008 (UTC)

Thanks a lot,that made some things much clearer to me.But smthng still confuses me.I now undstand about this nucleophilic strenght being reduced in protic solvents due to H-bonding(i know now after i looked up the terms here after your reference about the action of such solvents),but why is the basic strenght not reduced too.i mean isn't the strenghth of a base also measured by its ability to co-ordinate with a proton.if its lone pair is hindered(thereby making it a weaker nucleophile evidently),then why doesn't its(or does it)proton abstraction ability decrease hence making it a weaker base according to the definition of a lewis base if i'm not wrong.I'll be really grateful if you can clear my doubts.Thanks mr...umm Someguy —Preceding unsigned comment added by Abhiroopron (talk • contribs) 11:40, 1 June 2008 (UTC)

Well, keep in mind that nuclephilic/electrophilic attacks are Lewis acid-base reactions. The coordination with a proton is Brønsted basicity. Now, something that is usually not discussed until advanced chemistry courses, the strength of an acid/base is dependent on the solvent. But for Brønsted reactions, I doubt you are going to see the complete reversal of order of rates that you do for Lewis reactions; and the reason would be that a horrible base (like iodide) still won't be able to hold on to a proton. Someguy1221 (talk) 22:58, 1 June 2008 (UTC)

Ok,i see.Thanks.Though i doubt i'll do chem hons. i wish they'd make our plus II text bks better.Its an interesting subject really!Abhiroopron (talk) 15:50, 2 June 2008 (UTC)Abhiroop

Properties of silk dope

Hi - our articles are a bit vague on the actual properties of the protein solution that passes through spinnerets to produce silk. Is it called 'silk dope' or is there a more technical general term? What does it look like en masse? What is its colour, viscosity etc?

Thanks Adambrowne666 (talk) 17:58, 31 May 2008 (UTC)

Which articles? The spider silk article has some good information. The properties of the silk ingredients are rather complex (the protein sequences are different depending on the species, and only a few have been decoded), and describes various attempts to synthesize it, with references for further reading. ~Amatulić (talk) 19:45, 31 May 2008 (UTC)

Yes, it's a good article, and thanks for the answer, but it doesn't address my questions above - what are the properties en masse of silk dope - if you had a cupful, what would it look/smell like? - and is there a more technical generic term for the stuff? Adambrowne666 (talk) 00:18, 1 June 2008 (UTC)

Otherwise known as gossamer. Seriously, where did you find the idea of "silk dope"? Julia Rossi (talk) 00:29, 1 June 2008 (UTC)

I get it, here at Softpedia[9] seems to refer to the fibre "stock" before being spun into thread. Julia Rossi (talk) 00:33, 1 June 2008 (UTC)

It's mentioned too in the Synthesis section of the Spider Silk article, but dope seems to often be contained in inverted commas - just thought there might be a more official generic term - and a description - I realise it's a difficult question... Adambrowne666 (talk) 04:48, 1 June 2008 (UTC)

Not difficult if the scientists were more pictorial, but here[10] and searching "synthetic spider silk stock", gives terminology of "stock" and "stock solution". Only found chemical descriptions (polypeptide etc), but no sensate accounts (smell, texture, colour) among these. Synthetic spider silk "stock" etc gets more hits than "dope" so I'd go with the stock/solution term which as you know, is the bulk before drawing out fibres as such. Interestingly there are some mixtures of both natural and synthetic stock. It's freezable, too. Cheers, Julia Rossi (talk) 06:37, 1 June 2008 (UTC)

Thanks for that, Julia - and thanks, all, for your efforts - I'm starting to realise it's a ridiculous question, like asking what a gallon of pure DNA looks like. Adambrowne666 (talk) 00:52, 2 June 2008 (UTC)

You're welcome. It raised an interesting aspect – the deficit of presenting something as its chemistry without a description of its substance. At least spider web solution is a quantity of something not so in-tangible as DNA – it could be described as fluid or lightweight, sticky, grey or odorless or whatever, but we can't know!  : ) Julia Rossi (talk) 23:30, 2 June 2008 (UTC)

btw, I have a vague memory of a newspaper article about a science project where they derived several grammes of pure DNA from onions - wasn't an extraordinary stuff, as I recall, pale grey goop - if it was up to me, it'd be a prismatic soup visibly fizzing and zinging with genetic potentialities - probably best if I don't know what silk stock looks like - more fun to make it up Adambrowne666 (talk) 10:47, 3 June 2008 (UTC)

I hope I'm not too late. I think it is feasible (but impractical) to isolate the unspun spider silk proteins. First isolate it from a producer that can provide it en masse like a Spider goat. Second, find a way to separate the milk from the silk using methods such as High performance liquid chromatography. I'm not sure how much protein per liter of milk though. Maybe around 1 gram.--Lenticel ^(talk) 09:06, 4 June 2008 (UTC)

Asthma + Other disease

When I was three, I had an athsma attack with another disease at the same time. I do not remember the name of the disease, but I could not breath in or out. The last I remember of that day was falling asleep after a mask was put on me, leaving me to believe I had surgery. Ever since then, I have only had acute bronchitis, not general asthma. Does anyone in the medical field know what treatment could possibly lead to this(As well as what the disease could be)? I have a highly curious mind, so semi-complex awnsers are okay, just try not to get all E=MC² over me. 67.171.165.4 (talk) 20:34, 31 May 2008 (UTC)

Do you mean chronic bronchitis? --Tango (talk) 21:22, 31 May 2008 (UTC)

We're not really supposed to give medical advice here but i'll give you some background, I suppose. Have you been confirmed to have bronchitis, or was it a disease similar to it? I ask this because both acute bronchitis and chronic bronchitis are pretty much always caused by an infection, be it by viruses or bacteria. The proportion of cases that haven't been caused by an organism are almost always due to breathing in industrial fumes, or even more commonly by smoking. I'm not sure it bronchitis can be caused by any physical treatment but i'll leave that to the more medically trained.

As for the insight into what the disease is, it's essentially an inflammation of the bronchi. It's normally treated by either antibiotics (in cases where a pathological aetiology exists) and/or by bronchodilators. Regards, CycloneNimrod ^talk?_contribs? 12:41, 1 June 2008 (UTC)

With no medical background, but just based on "grandma says": child + trouble breathing + surgery brings tonsillitis to mind. That would however be an upper respiratory infection. Respiratory diseases has long lists of what it could be/have been. But that might just confuse matters further rather than clear things up. (No medical advice whatsoever.) Have you asked your physician? --76.111.32.200 (talk) 14:04, 1 June 2008 (UTC)

The user mentioned that they've had it ever since they were three, I highly doubt it was tonsillitis. Tonsillitis doesn't really lead to bronchitis. Regards, CycloneNimrod ^talk?_contribs? 14:13, 1 June 2008 (UTC)

Help with my daughter

My daughter is going through puberty and is starting to develop acne. I'm trying to find research papers and information regarding the effect of physical exercise on acne. Thank you.--Goon Noot (talk) 20:36, 31 May 2008 (UTC)

Our article acne mentions many things that may cause or help counter acne, but physical exercise isn't one of them. I suspect therefore that exercise has little impact on it. -- SGBailey (talk) 21:08, 31 May 2008 (UTC)

We can't give medical advice here, so there's not a lot of help we can offer. You would be best off consulting your GP. --Tango (talk) 21:21, 31 May 2008 (UTC)

I did not ask for medical advice. I specifically asked for research papers and information. Not advice.--Goon Noot (talk) 22:10, 31 May 2008 (UTC)

Try this search string [11] which seems to get a lot of relevant hits. however, most of them require a subscription to view the full article. SpinningSpark 22:46, 31 May 2008 (UTC)

Here's one you can get for free: [12] - if you come up against a journal requesting a subscription or fee, you can always check to see if the research institution has archived the paper for free on the internet. Just yesterday, Nature tried to fleece me for $32 but I found that the research institution in Germany puts their papers online for free. You also ought to lobby your political establishment and research charities to mandate open access since its the people's donations and taxes that support so much research. ----Seans Potato Business 12:59, 1 June 2008 (UTC)

Muscle Growth

I know this isn't probably the best place to ask for exercise advice, but I had some questions about the notion of between-lifting rest. I know you're not supposed to lift weights every day, because you need to give your muscles time to rest and regrow. But I've also heard that there are some muscles that it's OK to work every day (like, it's OK to run daily, I think), and I've also heard that abs work like this. I've also heard something, which doesn't make much sense to me, about how body-weight exercises (push ups, pull ups, etc) can be done daily, for some reason possibly having to do with lightness of weight. Do any of you know, from personal or scientific knowledge, how much of this is accurate, or what the real story is?

And, regardless of what particular muscles and exercise types call for what types of rest, what are the deleterious effects of not getting enough rest time? Will my muscles grow less if I do pushups daily than if I did them every other day? Will the muscle be more dense? Bigger? More wirey? I don't know.

And which is it that causes denser muscles, heavy weightlifting or many reps of lighter (bodyweight etc) weights? Or is the different-muscle-type-by-different-exercise stuff I've heard about pretty much made up?

Thanks,

70.108.222.173 (talk) 21:51, 31 May 2008 (UTC)

Running every day is mostly to build endurance and reduce fat. It may actually reduce muscle growth though if done excessively. I don't know about muscle density, but more reps means more tone, while more weight means bulkier muscles. Hope that helps. 24.46.50.159 (talk) 23:11, 31 May 2008 (UTC)

Genome and DNA

So a Gene is part of a DNA molecule. And a Genome is a complete set of Genes in a given organism. So does a single DNA molecule contain the entire genome of an animal? 24.46.50.159 (talk) 23:08, 31 May 2008 (UTC)

The pairs of chromosomes (which are the same in all cells of the body) contain the entire genome of an animal. Wisdom89 _{(T / ^C)} 23:11, 31 May 2008 (UTC)

A chromosome just contains one DNA molecule right? So just one cell will have all the chromosomes and DNA? How many chromosomes does a human have? 24.46.50.159 (talk) 23:28, 31 May 2008 (UTC)

A normal human has 23 pairs of chromosomes. Algebraist 23:32, 31 May 2008 (UTC)

I see. When you say pairs of chromosomes, do you mean 46 DNA molecules? 24.46.50.159 (talk) 23:55, 31 May 2008 (UTC)

Chromosomes are not the only DNA molecules in cells. According to chromosome, a normal human cell also contains hundreds of copies of the mitochondrial genome. Algebraist 23:59, 31 May 2008 (UTC)

And the genome isn't just the genes, it's all genetic information, coding and uncoding. - Nunh-huh 00:01, 1 June 2008 (UTC)

Nunh and Alge are absolutely correct. To answer more directly the last question, however, what we mean by "23 pairs" is 46 total chromosomes (see diploid), one copy from each parent. There are 22 pairs of autosomes and two sex chromosomes. Essentially, yes, 46 DNA molecules I suppose. But, that's misleading since "DNA molecules" can mean different things depending on the species we're talking about, the cell type (gametes are haploid for instance), or whether we're talking about in vivo or in vitro analysis. Wisdom89 _{(T / ^C)} 00:35, 1 June 2008 (UTC)