Wikipedia:Reference desk/Archives/Science/2009 May 13

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May 13

Blood in near vacuum

The article blood says that blood in a vacuum looses its oxygen and turns blue. How long would this take? Would it happen before or after drying, as a matter of fact, would blood dry at all? Thank you. 190.17.201.142 (talk) 01:54, 13 May 2009 (UTC)

The lack of a source for that statement in the article makes tracking down related information. The Vacuum article has some useful information and lots of references. I also found this article that has some more in-depth information along the same lines. One of the key points is that the body contains the blood, so it does not boil and you do not freeze (at least not right away). With no air in the lungs, the blood becomes oxygen depleted fairly quickly, causing loss of consciousness in about 15 seconds, but you still may have a couple of minutes where recovery is still possible. -- Tcncv (talk) 02:41, 13 May 2009 (UTC)

I think the pertinent article is probably oxygen dissociation curve. In a vacuum, the air pressure is 0, making the partial pressure of oxygen also 0, so when the vacuum occurs, you drop to the left end of the curve, oxygen dissociates from hemoglobin, dissolved oxygen in the blood proper will also be lost. All of this would occur before any drying would. Blood would turn "blue" only in the sense that venous blood is said to be blue: it's the color of hemoglobin without an associated oxygen molecule. - Nunh-huh 02:50, 13 May 2009 (UTC)

Our vein and cyanosis articles explain that oxygenated blood is red, and deoxygenated blood is darker-red. The blue color comes from Rayleigh scattering by tissues. Notice that the sentence from the blood article that you referenced has a "citation needed" after it. Ginogrz (talk) 04:06, 13 May 2009 (UTC)

"Citation needed" is there because I just added it. -- Tcncv (talk) 04:55, 13 May 2009 (UTC)

Really interesting. So what would happen if I shed blood on a surface in an atmosphere at, say, 10 percent pressure that of the earth surface, for example in protected conditions at 15km altitude ? 190.17.201.142 (talk) 05:23, 13 May 2009 (UTC)

Runway threshold markings

This is not really a science question, but I figure the aviation folks are more likely to see this here. While touring the taking a tour via Google Earth, I happened to notice differences in the threshold markings on runways at various major airports. Most have twelve bars (six on either side of the centerline, while other very similar runways have eight (four on either side). For example, of Atlanta's four major runways, only 8L/26R has the twelve bar markings. However, this diagram shows that the runway 9L/27R is the same width and longer, but it only has eight bars. Another intersting case is Seattle-Tacoma International Airport here, where both runways have eight bars on the south end, and twelve on the north. Can anyone clarify this? -- Tcncv (talk) 01:56, 13 May 2009 (UTC)

There are standards - but there are SEVERAL standards - and there doesn't always seem to be a lot of logic as to which standard each airport uses. Runway#Runway_markings explains this to some degree. SteveBaker (talk) 02:42, 13 May 2009 (UTC)

Thanks. I saw that section, but it does not go into enough detail. I am convinced there is a standard out there somewhere that precisely states when to use eight stripes and when to use twelve. Aviation is two heavily regulated for airports such as Atlanta and Seattle to have mixed markings without a well defined reason. -- Tcncv (talk) 02:56, 13 May 2009 (UTC)

Apparently there are two configurations for runway threshold markings mentioned by the FAA (See section 2F here). They consist of either 8 parallel bars, or various numbers of parallel bars to indicate runway width (e.g. 8 = 100 ft, 12 = 150 ft). Unfortunately, my quick read didn't reveal why there are two configurations, or when each can be used. And as for why the two systems are mixed at SeaTac, I have no clue. -- Flyguy649 ^talk 05:04, 13 May 2009 (UTC)

A bit more. According to this, the 8-bars-no-matter-what configuration was to become invalid as of Jan 1, 2008. So it's possible that the google images are from a time when the airports hadn't yet finished changing over the markings. Just a thought. -- Flyguy649 ^talk 05:09, 13 May 2009 (UTC)

Thanks for the excellent info. The transition from configuration A to configuration B would certainly explain things. -- Tcncv (talk) 05:47, 13 May 2009 (UTC)

As further evidence, looking at the third runway construction in Seattle, it is apparent that the image of the south end of the airport is older that the image at the north end. And in Atlanta, the twelve stripe thresholds appear to be newer pain jobs than the eight stripe thresholds, judging from the tire marks. -- Tcncv (talk) 06:14, 13 May 2009 (UTC)

It may be that there is finally some effort to reconcile the multiple standards into a single one - but repainting those stripes is not a trivial matter. They look small from the air - but each one of those 'piano keys' is as big as a medium-sized parking lot - burning off and repainting that much paint isn't easy. Closing the runway in order to do the work is also difficult. Other aspects of the differing standards are even tougher to fix - lighting configurations, for example, can involve lights positioned out off the end of the runway on land that the airport authority may not even own...and a typical airport has many hundreds of lights. Some of the complications arise from updated requirements that 'grandfather in' the old standards for these kinds of reasons - others come about from military airfields being repurposed or shared for civilian needs - resulting in conflicting military versus civilian/FAA requirements, etc. It really is something of a mess. SteveBaker (talk) 14:00, 13 May 2009 (UTC)

Potential energy

We're given three charges, Q1, Q2, and Q3, which are arranged in an equilateral triangle and we're asked to find the total potential energy. Now normally I would have said that U = U1 + U2 + U3 = (k/d)(Q1Q2 + Q1Q3 + Q2Q1 + Q2Q3 + Q3Q1 + Q3Q2). But the answer says that U = (k/d)(Q1Q2 + Q1Q3 + Q2Q3). Why is this? —Preceding unsigned comment added by 65.92.6.148 (talk) 04:00, 13 May 2009 (UTC)

It appears that you are double-counting some energy. (Note that you're off by a factor of two). This is because the energy of two particles Q1 and Q2 is only counted once; if you also calculate it for Q2 and Q1, you are considering the same system twice. Nimur (talk) 04:34, 13 May 2009 (UTC)

(ec - partial duplication of prior answer) I'm no expert, but one way to check for consistency between the three change potential energy equation and the two charge potential energy equation is to set one of the charges, say Q3 to zero. When compared with the equation in Potential energy#Electrostatic potential energy, it appears that the textbook answer is correct. In short, don't double count Q1Q2 along with Q2Q1. That effectively is already built into the constant k. -- Tcncv (talk) 04:43, 13 May 2009 (UTC)

Put simply, the standard potential formula $U={\frac {k_{e}q_{1}q_{2}}{r}}$ already counts the potential for both of the charges involved; you do not need to calculate $U_{1}=U_{12}+U_{13}$ and $U_{2}=U_{12}+U_{13}$ and $U_{3}=U_{13}+U_{23}$ and then add those three. (Those three quantities are not nonsensical, but they apply only if you consider establishing two of the charges and then bringing in the third. They can't all three be that third charge!) --Tardis (talk) 04:51, 13 May 2009 (UTC)

Why is that so? Is there some sort of derevation for this? Because the formula kq1q2/r comes from assuming one charge stays put, so why would it also account for the work that's done on it as well? —Preceding unsigned comment added by 65.92.6.148 (talk) 00:35, 14 May 2009 (UTC)

I managed to derive that for a constant force (such as when we assume gravity to be constant), the equation for U doesn't change whether we assume one of the bodies (eg the earth) says at rest or not. I just said that ΔU=-W_T=-(W_1,2 + W_2,1) = -(F•s_1 + F•s_2)=FΔh_1 - FΔh_2=F(Δh_1-Δh_2)=mgΔd, where W_1,2 is the work done on body 1 by body 2, and • is the dot product. Can someone help me with the derivation for a non-constant force? —Preceding unsigned comment added by 65.92.6.148 (talk) 04:00, 13 May 2009 (UTC)

It's really just a matter of definition. We could go around calculating the potential energies of each charge separately, but the contribution to A's potential from B is always identical to B's from A, so we might as well call their sum "the potential of A and B" and be done with it. Clearly we cannot then also consider "the potential of B and A" because we've already counted both those interactions. As for deriving it, it should be sufficient to apply Newton's Third Law (and its corollary that the forces are radial) and note that $\int _{C_{1}}{\vec {F}}_{12}\cdot d{\vec {r}}_{1}+\int _{C_{2}}{\vec {F}}_{21}\cdot d{\vec {r}}_{2}=\int _{0}^{T}({\vec {F}}_{12}\cdot {\vec {v}}_{1}+{\vec {F}}_{21}\cdot {\vec {v}}_{2})\,dt=\int _{0}^{T}{\vec {F}}_{12}\cdot ({\vec {v}}_{1}-{\vec {v}}_{2})\,dt=\int _{0}^{T}F{\frac {dr}{dt}}\,dt=\int _{C_{1,2}}F\,dr$ (where we define the scalar F as positive when repulsive). --Tardis (talk) 18:03, 14 May 2009 (UTC)

To answer this question, we need to look into the fundamentals... what is potential energy of a system ? It is the energy required to assemble the system. Or the energy you would get if you blew up the system. This is basically the "potential" energy that the system has. So first coming for a one charge system. What is its potential energy ? Assuming there is nothing else in the surroundings, it is zero. You didn't have to do any work to put it there. Note that we don't take into account the energy to create a charge... charges are simply given to us and we just put them around. This is wise policy, as the energy of a point charge, is in fact, infinite. So for a system of two charges, we first bring in the first charge. This wouldn't take any work. Now, for the second charge. We bring it in from far away. How far away ? You could say from your house, or from a known landmark like the Eiffel tower or the sun, but the standard practice is to bring it in from infinity. But it is repelled (or attracted) by the first charge.So we integrate to get the work done, which is how we get your famous formula. Now for say three charges, we put the first charge in. Again no work done here. Next we put the second charge in. You know the formula. Now for the third charge, it is repelled (or attracted) by both the first and the second charges. So we plug in the formula for both of them, and add them all up, which is how you get your textbook formula. So once you understand why you get a certain formula, it is easy to build skyscrapers on you strong basement... Rkr1991 (talk) 08:00, 14 May 2009 (UTC)

Makes sense. But out of curiosity, how would you re-derive the equation for potential energy while taking into account both charges moving (I guess it's more of a math question than anything....). —Preceding unsigned comment added by 65.92.6.148 (talk) 08:40, 14 May 2009 (UTC)

Now that's interesting... The whole time we were talking about Electrostatics, where we assume everything to be in rest or in very slow motion. Now the question you are asking belongs to the domain of Electrodynamics. Now i need to talk a bit more in the basics, a point which i evidently forgot to mention. I had told that you integrate the differential work from infinity to the required position. The question is, along what path do i integrate ? That is, while getting from infinity to the point, what curve does to charge follow ? The answer is it doesn't mater. You can integrate along any curve want, it doesn't change the answer. This is because the electrostatic force, given by coulomb's law, is a term we call conservative [[1]]. Its line integral is independent of path, and we get the same answer always. Now in the realm of electrodynamics, coulomb's law doesn't hold good. Further, in addition to electric forces, there are also magnetic forces. Both of these are now non conservative. So, our answer would now depend on the path we choose. So, if we get different energies if the charges are all blown up in different directions, there isn't really much meaning in the term potential energy anymore... So in electrodynamics, there is no such term Potential energy. However, there exists a four vector potential term... but we don't really care about that do we ? Also, don't forget to sign you posts, by placing four '~' marks at the end... Rkr1991 (talk) 15:09, 14 May 2009 (UTC)

MICROPROCESSOR

Deleted. Poster has posted the same homework question on RD/C repeatedly. Tempshill (talk) 15:51, 13 May 2009 (UTC)

Hypnotic state of mind while watching t.v.

(kids' eyes glass-over, people walk by unnoticed, because of the fast 'flickering' of the tv)

I've heard this several times, but searching for it online, I can't find anything on it.

67.236.121.225 (talk) 06:32, 13 May 2009 (UTC)

Photosensitive_epilepsy#Symptoms has a subsection about Television, traditionally the most common source of seizures.Cuddlyable3 (talk) 07:49, 13 May 2009 (UTC)

Immersion (virtual reality)?--Lenticel ^(talk) 08:03, 13 May 2009 (UTC)

Hyperfocus is marked as needing work, but the term may help with googling studies. 71.236.24.129 (talk) 08:13, 13 May 2009 (UTC)

You can get the same effect reading a book, writing or doing other tasks. Any time you "zone out," you're so focused on the task at hand that other stimuli are ignored. — The Hand That Feeds You:^Bite 13:21, 13 May 2009 (UTC)

Also see tetris effect, highway hypnosis and automaticity. Lanfear's Bane | t 13:54, 13 May 2009 (UTC)

Sound Energy Converter

Is there and if there isn't, is it possible to create a device converting sound into kinetic, electrical, or electromagnetic energy?The Successor of Physics 12:56, 13 May 2009 (UTC)

See Microphone. -- Coneslayer (talk) 13:12, 13 May 2009 (UTC)

While we're at it, see geophone and hydrophone and piezoelectric crystal as well. Nimur (talk) 13:21, 13 May 2009 (UTC)

In the 1880's Thomas Edison invented a device which converted sound waves into vibration, and then into rotary motion, and it could power a drill. He noted that many a mother-in-law's voice had been thought to be capable of drilling holes. Edison (talk) 17:23, 13 May 2009 (UTC)

... really? That deserves a tag. Nimur (talk) 19:11, 13 May 2009 (UTC)

They had such a machine on exhibit at the reconstructed Edison Menlo Park lab at Greenfield Village when I last toured the place. I found a reference which (snippet view) refers to the rotary motion being used to run a sewing machine, voice powered. "STITCH! STITCH! KEEP STITCHING DAMMIT!" Treadle powered and later electric powered machines were More conducive to domestic tranquility. "Menlo Park Reminiscences," by Francis Jehl, (writen 1937, Dover reprint 1990) (who was present at Edison's lab in the 1870'2-80's) tells on pages 180-181 of the "phonomotor." On page 180 is an illustration of it from the Scientific American (apparently July 27, 1878). It was the Sci Am reporter who said "Mr. Edison says he will have no difficulty in making the machine bore a hole through a board, but we consider such an application of the machine of very little utility, as we are familiar with voices that can accomplish that feat without the mechanical appliance." The machine used a diaphragm connected via a rubber tube to a spring, which carried a pawl that acted on a ratchet on the flywheel. Certain voice sounds could cause the flywheel to rotate "with considerable velocity." See also [2], [3], [4], [5], [6], and [7]. The notability guideline says "If a topic has received significant coverage in reliable secondary sources that are independent of the subject, it is presumed to satisfy the inclusion criteria for a stand-alone article." Does it appear that this little "philosophical toy" or demonstration gadget qualifies for its own stand-alone article, since it caught the fancy of scientific journals and science writers in various countries? Edison (talk) 21:13, 13 May 2009 (UTC)

Sounds like a great idea for an article...and once you get it into a halfway reasonable shape - you should definitely send a quote from it to the "Did You Know?" section of the Wikipedia front page. SteveBaker (talk) 22:41, 13 May 2009 (UTC)

Certainly a microphone fits the bill. It produces alternating current electricity from sound. SteveBaker (talk) 22:41, 13 May 2009 (UTC)

Thanks, guys! I forgot even about simple stuff like microphones! I must have gone mad!The Successor of Physics 13:03, 14 May 2009 (UTC)

megawatt to kg, dm and minute

can anyone tell me the answer to these four questions: (i dont need hints)

1. convert a power of one megawatt on a system whose fundamental units are 10 kg, 1 decimeter and 1 minute. 2. find the value of 60 joule/min on a systemwhich has 100g.cm and 1 min as fundamental units. 3. write the dimensions of a/b in the relation v is velocity and t is time. 4. write the dimensions of a * b in the relation E= b - x²/at, where E is energy, x is distance and t is time. —Preceding unsigned comment added by 122.50.139.37 (talk) 13:17, 13 May 2009 (UTC)

If we do your homework for you (instead of helping you do it yourself), will we get your course credit? -- Coneslayer (talk) 13:23, 13 May 2009 (UTC)

i have already solved these questions. i just need to verify my answers (there's no one else left but you people, cos my friends are all out of town and i'm on vacation. i need to check my answers)

If you show us your work, we'll be happy to confirm the answers. This is a core tenet of our do your own homework policy. — Lomn 13:45, 13 May 2009 (UTC)

I think in that case, what might work better, would e for you to tell us the answers that you got, and then we could see if they are correct? 65.121.141.34 (talk) 13:46, 13 May 2009 (UTC)

(EC) You could send us your answers and ask us to check them for you! But we really do have strict rules about not answering homework questions - it's not just people being mean to you. SteveBaker (talk) 13:48, 13 May 2009 (UTC)

Please do your own homework.

Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know.

So I just measured my penis size, why am I not happy with it?

I am almost 8" inches long erect, definitely well over 7", but only 6" exactly in girth. Why am I not satisfied with my penis size?--Pipelinefine (talk) 14:28, 13 May 2009 (UTC)

The reference desk will not address medical or psychological issues. If you have a medical concern, see a physician. If you have a psychological concern, see a psychiatrist. Nimur (talk) 14:32, 13 May 2009 (UTC)

the reason probably has to do with your self-esteem (google self steem and impotence to see extreme examples), which is probably made worse by the stigma and discrimination you face for being a troll. 79.122.61.98 (talk) 16:58, 13 May 2009 (UTC)

Over 7" is above average, I believe, so your lack of satisfaction is presumably about you, not your penis, and since we only have information about your penis we cannot help you. If it is causing you significant distress, then see a psychiatrist (your GP/family doctor can refer you to one), otherwise just try and get over it! --Tango (talk) 17:15, 13 May 2009 (UTC)

Take it easy, no body else is measuring but you. Length graph. Girth graph. Mac Davis (talk) 03:42, 14 May 2009 (UTC)

Is this the time to say "It isn't the size of the ship, its the motion of the ocean"? A penis is to sex as a hockey puck is to hockey; it may be a needed component, but your enjoyment and understanding of the game will become far greater when you take a more holistic approach to it. Just as real hockey fans learn to appreciate the off-puck action, you may find that the enjoyment of all games improves when the focus is off of a specific "tool" in the game, but rather on the entirety of the game itself. --Jayron32.talk.contribs 04:07, 14 May 2009 (UTC)

Well said Jayron32, I hope your ocean is long in motion. Richard Avery (talk) 13:43, 14 May 2009 (UTC)

If you want information on normal range of variation in sizes, we do have it here in Wikipedia. See the detailed article Penis size. We may not answer medical questions at the desk, but we do have information in the encyclopedia from which you can answer them yourself. 8 x 6 is in the highest few percent, as you can see from the graphs there.DGG (talk) 15:27, 14 May 2009 (UTC)

Maybe because your penis is too large? If your measurements are accurate your penis is exceptionally above average, probably I expect in the top 1-2%. If this is a problem, I recommend as with most above you seek help from a professional who can help you learn to deal with your exceptionally large penis and whatever complications that arise. Simple advice would be take it slow, learn to communicate and use sufficient lubrication Nil Einne (talk) 12:38, 18 May 2009 (UTC)

Why does sticking your thumb on a water hose make the water come out faster?

I'm not sure if this is the right place to ask this question, but our article on the List of common misconceptions states (depending on which edit you happen to look at):

It is not true that a nozzle (or a person's thumb) on the end of a garden hose makes the water squirt farther because the same amount of water gets forced through a smaller opening. The rate of flow of water through the hose is not a set constant; in fact, putting one's thumb over the end of the hose reduces the rate of flow. What is constant is the water pressure at the source. When water is flowing, the pressure decreases the farther from the source one gets due to friction between the water and the pipes it's flowing through. The faster the water moves through the pipe, the greater is the friction that cuts down pressure at the output end. A thumb over the end of the hose decreases the flow rate, causing the friction from the source to decrease, causing the remaining water to have more speed.

A discussion is here[8].

Would some editors who know more about friction and pressure help settle this issue? A Quest For Knowledge (talk) 14:56, 13 May 2009 (UTC)

Take a look at the above section, WP:RD/S#Submarine hull breach. SteveBaker's math is pretty applicable to the hose problem. Note the important distinction between mass flow rate and flow velocity. Reducing the aperture size (with your thumb) will increase the flow velocity but decrease the mass flow rate. Nimur (talk) 15:00, 13 May 2009 (UTC)

To put the above in layman's terms, covering the end of the hole with your thumb both reduces the amount of water coming out and increases the speed at which it comes out. These are not necessarily contradictory ideas. --Jayron32.talk.contribs 15:09, 13 May 2009 (UTC)

Jayron32, thank you for putting it in layman's terms but is the increased speed caused by pressure or friction? This seems to be the heart of the issue. A Quest For Knowledge (talk) 15:21, 13 May 2009 (UTC)

Both change and both are variables that change the velocity. David D. (Talk) 15:55, 13 May 2009 (UTC)

Having said that if you turn the tap off nothing happens so pressure is the primary issue. David D. (Talk) 15:59, 13 May 2009 (UTC)

I don't believe that pressure increases. You have the same amount of pressure pushing from the tap, which is forcing a similar amount of volume per second to flow through the hose (the rate will decrease somewhat due to increased flow resistance, analogous to a resistor in an electric circuit, but that is subtracting from the spraying effect, not adding). Because you have the same (or similar) amount of volume going through a smaller area, the water must speed up in a linear, meters/second way (if you think about it, volume flow rate, q, is equal to A*V, decreasing A while keeping a similar q will invariably lead to a higher V). The pressure drop should equalise (after a few seconds of flowing through your thumb) to approximately the same, it's going from whatever it is at the end of the pipe to atmospheric. To get a more mathematical idea about this phenomena, you might find Bernoulli's principle an interesting article. TastyCakes (talk) 16:14, 13 May 2009 (UTC)

The pressure at the nozzle does change. I agree the pressure at the valve does not. Also bear in mind there is not a constant flow rate in this example. David D. (Talk) 16:33, 13 May 2009 (UTC)

q is not approximately constant. In fact, for a fixed pressure difference the volumetric flow rate is strongly dependent on the size of the outlet. (Consider a basin full of water, the rate of water flowing out clearly depends on the number and size of the holes provided.) Also, water in a real hose is subject to viscosity, which affects the pressure reaching your thumb and is a factor not considered under Bernoulli. Dragons flight (talk) 16:35, 13 May 2009 (UTC)

For the purpose of Bernoulli, is water considered inviscid? I have only seen Bernoulli used in reference to gasses such as air, which is about 50-100x less viscous then water. 65.121.141.34 (talk) 16:23, 13 May 2009 (UTC)

Ya, Bernoulli's principle is true for all incompressible fluids as I understand it. In fact I think it should be simpler with an incompressible fluid than you've probably seen with air. TastyCakes (talk) 16:26, 13 May 2009 (UTC)

In short, I think saying it's a misconception about the water flow is incorrect. While it is true that the volumetric flow rate is not constant because of changes in friction etc is completely accurate, the concept is true (in my opinion) in that the area of the end of the hose is decreased much more than the volumetric flow rate is decreased, and as such the q = V*A equation dictates an increase in velocity, even though the effect is somewhat lessened by q's own reduction. The pressure will increase at the end of the pipe because the flow rate has reduced the friction loss along the pipe. But I think in your average garden hose the difference will be quite small, since the change in volumetric flow rate would also probably be quite small. It is not the main cause of the effect, in my opinion; the area reduction is. TastyCakes (talk) 16:36, 13 May 2009 (UTC)

In the laminar, viscous limit, the volumetric flow rate is proportional to A² and so changes more rapidly than A (see: Hagen–Poiseuille equation). Obviously your thumb is not conducive to laminar flow, but I think your assumption that the change in q is small is still incorrect. Dragons flight (talk) 16:51, 13 May 2009 (UTC)

mm isn't it P that's proportional to A*r² rather than proportional to the area squared? And isn't that only for circular cross sections, which wouldn't be true for the thumb over the hose analogy? TastyCakes (talk) 17:00, 13 May 2009 (UTC)

So HP would be correct for the most part but your average velocity leaving the pipe still follows the simpler formula:

$Q={\frac {\Delta P\pi r^{4}}{8\mu L}}=Vavg*A$

Perhaps you are right that Q is greatly reduced, but it seems quite clear from the equation that in the hose example it is reduced much much less than A, else the water wouldn't come out of the pipe faster. TastyCakes (talk) 17:11, 13 May 2009 (UTC)

I don't think you're factoring in the resistance along the hose. As the flow rate decreases the resistance along the hose decreases too. The effect of this is an increase of pressure at the nozzle. David D. (Talk) 20:44, 13 May 2009 (UTC)

That problem cannot be correctly solved by either the Hagen-Poiseuille's equation or the Bernoulli's equation alone. The first one assumes an uninterrupted pipe of circular cross-section which is not true here because the thumb interrupts the flow at the end. The second one neglects friction all together and cannot explain the phenomenon. But a combination of both can solve the problem. Lets stablish three points of interest, $A\,$, $B\,$, and $C\,$. $A\,$ will be a point somewhere in the water reservoir being kept at a constant pressure $P_{A}\,$, $B\,$ will be a point at the end of the hose right before the nuzzle (or thumb) whose pressure is yet to be determined, and $C\,$ will be a point outside the nuzzle kept at atmospheric pressure $P_{C}=P_{0}\,$. Lets define the pressure drops along the hose before and after (that is without the thumb and with the thumb).

$\Delta P=P_{A}-P_{C}\,$, and

$\Delta P'=P_{A}-P_{B}\,$.

We can use Hagen-Poiseuille to compute the water flow before and after.

$Q={\frac {\pi R^{4}}{8\mu L}}\Delta P$, and

$Q'={\frac {\pi R^{4}}{8\mu L}}\Delta P'$.

To find $P_{B}\,$ we use Bernoulli

$P_{B}-P_{C}={\frac {\rho }{2}}(v_{C}^{2}-v_{B}^{2})={\frac {\rho }{2}}v_{C}^{2}(1-{\frac {A_{C}^{2}}{A_{B}^{2}}})={\frac {\rho }{2}}Q'^{2}({\frac {1}{A_{C}^{2}}}-{\frac {1}{A_{B}^{2}}})$

Where we used the fact that the flow through points $B\,$ and $C\,$ must be the same,

$Q'=v_{B}A_{B}=v_{C}A_{C}\,$.

$A_{B}\,$ and $A_{C}\,$ are the cross-section areas at point $B\,$ in the hose and point $C\,$ at the thumb.

Now we place the expression we got for $Q'\,$ into the expression obtained from Bernoulli's equation.

$P_{B}-P_{C}={\frac {\rho }{2}}\left[{\frac {\pi R^{4}}{8\mu L}}\Delta P'\right]^{2}({\frac {1}{A_{C}^{2}}}-{\frac {1}{A_{B}^{2}}})=X(P_{A}-P_{B})^{2}$

Where we introduced the quantity $X={\frac {\rho }{2}}\left[{\frac {\pi R^{4}}{8\mu L}}\right]^{2}({\frac {1}{A_{C}^{2}}}-{\frac {1}{A_{B}^{2}}})$. Solving the quadratic equation for $P_{B}\,$ we get

$P_{B}=P_{A}+{\frac {1-{\sqrt {1+4X\Delta P}}}{2X}}$, and

$\Delta P'={\frac {{\sqrt {1+4X\Delta P}}-1}{2X}}$.

If $1>>4X\Delta P\,$ we then get $\Delta P'\approx \Delta P\,$ and the flow does not change very much. but if $1<<4X\Delta P\,$ we then get $\Delta P'={\sqrt {\frac {\Delta P}{X}}}\,$ which gives a flow proportional to the area and a constant exit velocity. Dauto (talk) 21:03, 13 May 2009 (UTC)

why do women love chocolate

or is it a myth —Preceding unsigned comment added by 79.122.61.98 (talk) 16:43, 13 May 2009 (UTC)

Probabily the same reason why men love chocolate. Dauto (talk) 17:10, 13 May 2009 (UTC)

I'm talking about the stereotype. Either its a myth or women love chocolate more and differently than men. 79.122.61.98 (talk) 17:24, 13 May 2009 (UTC)

A few studies seem to suggest women do have a higher preference for chocolate than men do, but the reasons for this are not known (PMID 1799282, PMID 12954417). Rockpocket 17:34, 13 May 2009 (UTC)

Yeah, I prefer chocolate to men any day!--80.3.133.116 (talk) 17:38, 13 May 2009 (UTC)

Touché, (I've edited to clarify my meaning). Rockpocket 17:54, 13 May 2009 (UTC)

What are your views on men buying you chocolate? --Tango (talk) 17:59, 13 May 2009 (UTC)

Our survey says... I specifically linked to the account at confectionerynews to highlight that this survey was commissioned by a chocolate manufacturer. 80.41.120.247 (talk) 19:31, 13 May 2009 (UTC)

I'd always thought it was just that women tend made a bigger deal about going against whatever fad diet they were currently on and 'being naughty' by eating the oh-so-fattening chocolate that <whoever writes those womens' mags> says is bad. Guys seem to be much less concerned about their weights and figures. --Kurt Shaped Box (talk) 17:49, 13 May 2009 (UTC)

Well, yeah, but I think there's a little more to it. Since we're dealing in stereotypes, I would think of a guy chowing down on a massive steak or similar "manly" meal rather than a box of assorted chocolate. Matt Deres (talk) 18:11, 13 May 2009 (UTC)

I seem to recall women reacting more strongly to the serotonin-related effects than men, but maybe I'm misremembering this. I did get the impression that I have read numerous scientific discussions of why women do seem to have a much stronger psychological reaction to chocolate than most men do. --140.247.252.198 (talk) 18:57, 13 May 2009 (UTC)

Yeah, I remembered reading the same thing (not sure where) but was too afraid to post for fear of being flamed. Google scholar seems to remember it too. SpinningSpark 19:35, 13 May 2009 (UTC)

Can you link to a specific study showing evidence that chocolate per se affects serotonin (not "carbohydrates" or whatever, as lots of other foods contain this)? The "Chocolate: Food or Drug?" (J Am Diet Assoc.99:1249-1256.(1999)) study is pretty unscientific when it comes to the neurobiology part - acknowledging that PEA doesn't reach the brain (and higher quantities are in cheese) but despite that still speculating that because "7 MDMA abusers [displayed] an intense craving for chocolate [...] it is worth noting that MDMA is structurally related to PEA and may produce similar effects". Also I only see higher cravings for chocolate during pre-menstrual stage, along with a mention of cravings for carbohydrate being affected by serotonin. --Mark PEA (talk) 10:17, 14 May 2009 (UTC)

what is this "I did get the impression that I have read numerous" sentence?? This phrasing sounds schizophrenic to say the least...79.122.61.98 (talk)

What, you say? — The Hand That Feeds You:^Bite 21:29, 13 May 2009 (UTC)

When I write sentences like that it usually means I changed by mind about what I was going to say half way through writing it and didn't go back and change everything so it was consistent. --Tango (talk) 21:44, 13 May 2009 (UTC)

I fucking hate chocolate, and ice cream, and anything like that

A distinction should be made here between milk chocolate and dark chocolate. Dark chocolate has less sugar than milk chocolate, and perhaps is less preferred by women. In fact, chocolate alone is quite bitter. The Wikepedia article on chocolate says, "Aztecs made it into a beverage known as xocolatl, a Nahuatl word meaning 'bitter water' ". The Wikipedia article is well written, information dense, and interesting, especially if you are interested in making and manufacturing things. It includes a lead to "mouthfeel" which has some good definitions. - GlowWorm.

Perspective of an electron

I was thinking the other day what it would be like to be a person on an electron while it orbits the nucleus of an atom. I tried to calculate it out but it got confusing. What I tried to do was make the proportion of a height of a person (about 2 meters) over the diameter of Earth (12,756,200 meters) equal to X (the height of a person on an electron) over the diameter of an electron (about 5.636x10^-15 M). I got ~8.836x10^-22 meters. After I got here though I couldn't figure out where to go, in other words how I would be able to use this information to help me relate my size relative to Earth to what it would like on an electron. I thought of trying to compare the size of the Sun in Earth's sky to the nucleus of a hydrogen atom in an electrons "sky". If this makes any sense to anyone, I would appreciate some help! Thanks!

And if this is still kinda confusing, I'll try to simplify what I'm trying to do:

As we stand here on earth, we can look out the Sun, the Moon, other planets, and other stars and galaxies. If you notice, electrons, although they don't orbit like Earth does around the Sun, still "orbit" around the nucleus of an atom. I'm curious to see what the "sky" would look like if an electron was earth and we were people on it. How far away would the Nucleus (Sun) be? How large would it look in the "sky"?

Hopefully you can get my drift, and help me out. Thanks again! —Preceding unsigned comment added by 129.21.109.153 (talk) 21:45, 13 May 2009 (UTC)

I don't know where you got the diameter number for your electron - but it's wrong. Electrons are 'point particles' - they have no size. The nearest thing to a 'size' you could use would be some function of the probability cloud that describes it's position...but that would give you a really odd view of what you're trying to understand. SteveBaker (talk) 22:33, 13 May 2009 (UTC)

The OP is referrring to the classical electron radius. Although Steve is right; the electron has no actual physical size in the way that macroscopic objects do. Someguy1221 (talk) 22:40, 13 May 2009 (UTC)

Yes, but the classical radius of the electron isn't a very useful conceplt. Compton wavelength of the electron might be a better choice? Dauto (talk) 23:39, 13 May 2009 (UTC)

How about, what would be its radius if it were a Kerr–Newman black hole of the same mass, charge, and angular momentum? That would seem to be a limiting value closer than which you couldn't really model it as a point. --Trovatore (talk) 08:11, 14 May 2009 (UTC)

Oh ok I see what you are saying. I was actually thinking that too, but I got a number for it so I figured it must have a diameter. I suppose this means that this couldn't really happen then, huh? —Preceding unsigned comment added by 129.21.109.153 (talk) 00:33, 14 May 2009 (UTC)

Some of our models of the atom treat an electron as a discrete little "planet" orbiting the nucleus of the atom. This can be a useful model for explaining some simple aspects of atomic structure, but is fantastically far from reality. Depending on what you are attempting to study about an electron, you can model it as either a) a discrete particle b) a "cloud" surrounding the nucleus of the atom c) a standing wave d) a quantum mechanical wave function. Like any models, these have varying degrees of proximity to reality, as well as varying degrees of "ease of use" and "usefulness" in particular situations. If, however, you stop trying to force an electron to behave like something you have experience with, like a little ball, and instead just accept it for what it is, it makes it easier to understand the more accurate models of electron behavior. --Jayron32.talk.contribs 04:02, 14 May 2009 (UTC)

Yep. There are some things (especially at the quantum level) that you just can't come to understand by analogy with macroscopic things. You have to resort to these rather unsatisfying ways of coming to terms with one property by itself - but never really getting a 'feel' for what the entire object is like. SteveBaker (talk) 04:52, 14 May 2009 (UTC)

This is especially true because you wanted to know what things would "look like." Your visual perspective is based on electromagnetic waves of a certain wavelength, visible light. Those wavelengths are about 300 nanometers to 700 nanometers. If you tried to "see" anything smaller than those length scales, you would not be able to form an image. At the very best, you could try to do some kind of interferometry or amplitude detection (LIDAR ?) And every time one photon of light arrived, it would blow your "planet" (electron) into a totally different quantum state (with new position and momentum). I don't think you could "see" anything. Nimur (talk) 13:22, 14 May 2009 (UTC)

Also worth noting is that while applying classical physics understanding to a quantum mechanical problem produces problems like you described, the reverse is not true. All macroscopic particles obey quantum mechanical equations perfectly, its just that within the realm of real measurement devices, the "classical approximation" is functionally identical to the quantum mechanical reality. In other words, all objects have a corresponding wave function; even YOU do, its just that your wave function is not terribly useful in getting information about your expected behavior. --Jayron32.talk.contribs 18:25, 14 May 2009 (UTC)

Grumbling Tummy

Childish question, this may seem, but why does your tummy grumble and gurgle? What's it doing? Eating itself? (Well, digesting, anyway).--KageTora (영호 (影虎)) (talk) 22:28, 13 May 2009 (UTC)

See Borborygmus. (I'll admit, I just learnt that word, and isn't it fantastic?!) --Tango (talk) 23:03, 13 May 2009 (UTC)

Excellent! Interesting to note that it even has a plural 'borborygmi'. Thanks Tango!--KageTora (영호 (影虎)) (talk) 23:12, 13 May 2009 (UTC)