Wikipedia:Reference desk/Archives/Science/2006 September 25

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Aeroplane advertising

Occasionally you see aeroplanes advertising by towing banners behind them. The banners always seem to stay vertical, rather than twisting and turning in the wind as you might expect - is this due to particular aerodynamic properties of the banner, or just the way it's attached to the plane?

Thanks, --Noodhoog 02:20, 25 September 2006 (UTC)

This link has a reasonable description of the stabilizing elements. ---Sluzzelin 05:48, 25 September 2006 (UTC)

Very high SPLs at bass frequencies

Can very high SPLs (>100 dBA)at bass frequencies induce heart palpitations?--Light current 03:53, 25 September 2006 (UTC)

That is the goal, and yes. Edison 04:40, 25 September 2006 (UTC)

No I mean fairly persistent palpitations after the exposure has ceased. BTW Im serious. Also do you have any refs? 8-|--Light current 13:02, 25 September 2006 (UTC)

Force Dynamics

Problem: 'A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of friction os .20 and the push imparts an initial speed of 4.0 m/s?'

...i don't see how i can solve this if i am not given the mass of the box.

Neither do I! 8-)

Go ahead and try the problem anyway, substituting "M" where you'd otherwise put the mass of the box. The M's should cancel out. --Allen 04:06, 25 September 2006 (UTC)

God, I hated freshman physics!Edison 04:41, 25 September 2006 (UTC)

Hint: deceleration is not dependent on mass. Clarityfiend 05:32, 25 September 2006 (UTC)

I think the answer does depend on the mass - I normally wouldn't solve the hw problem, but I think the question is incomplete, so I will. The velocity equation for this block is

$v(t)=v_{0}-F_{friction}t=v_{0}-\mu _{k}F_{normal}t=v_{0}-\mu _{k}mgt$

and

$v(t)=0$ at the stopping point, so $t_{stop}={\frac {v_{0}}{\mu _{k}mg}}$

The position equation (the integral over time of the velocity equation) is

$x(t)=v_{0}t-{\frac {\mu _{k}mgt^{2}}{2}}$, which at the stopping point is

$x(t_{stop})={\frac {v_{0}^{2}}{\mu _{k}mg}}-{\frac {v_{0}^{2}}{2\mu _{k}mg}}={\frac {v_{0}^{2}}{2\mu _{k}mg}}$

I could be doing something wrong, but I do believe the final answer depends on the mass m. --Bmk 06:20, 25 September 2006 (UTC)

PS: I should define my notation - v0 is the initial velocity, v(t) is the velocity at time t, mu_k is the coefficient of kinetic friction, x(t) is the position at time t, g is the acceleration due to gravity, etc, etc... --Bmk 06:25, 25 September 2006 (UTC)

The very first equation is wrong; it confuses acceleration with force. Melchoir 06:31, 25 September 2006 (UTC)

Aaaargh. Thanks. --Bmk 06:59, 25 September 2006 (UTC)

Look at it this way: what if you performed this experiment on two boxes, simultaneously, side-by-side? They would go the same distance. Now consider that having two boxes is a lot like doubling the mass of a box. So mass doesn't matter. Melchoir 06:33, 25 September 2006 (UTC)

Very nice thought experiment, Melchoir. Thanks. --Allen 06:39, 25 September 2006 (UTC)

But changing the surface area of the of box in contact would also alter the friction coefficient wouldnt it. Which means your thought experiment doesnt really work. Philc _T^E_C^I 17:41, 25 September 2006 (UTC)

In theory, the contact area does not effect the frictional force, so long as there is contact. This is rather counterintuitive, but true nonetheless. StuRat 18:17, 25 September 2006 (UTC)

In that case, surely there is a way to cancel m out of the equation, possibly by using some other known equations in terms of m. Philc _T^E_C^I 22:46, 25 September 2006 (UTC)

Yes, the mass cancels out. StuRat 22:52, 25 September 2006 (UTC)

Relativity theory

What is the relativity theory? With much explanations.

Theory of relativity with much explanations.  freshofftheufoΓΛĿЌ  12:04, 25 September 2006 (UTC)

See also our articles Introduction to special relativity and Introduction to general relativity. --Lambiam^Talk 13:37, 25 September 2006 (UTC)

Your relatives always have theories about how to fix your problems/shortcomings - the special ones and the general ones - and are more than willing to provide long explanations. Clarityfiend 19:21, 25 September 2006 (UTC)

I'm probably gonna get angry reactions for this, but I'd like to point out that Lambiam, Clarityfiend, etc... are all humans, not bots. In fact, bots don't come here that often. And "please" costs nothing.Evilbu 20:03, 25 September 2006 (UTC)

Maybe the questioner is an AI. --Lambiam^Talk 20:55, 25 September 2006 (UTC)

If English isn't your native language, you may like this explanation in words of four letters or less. --ByeByeBaby 02:16, 26 September 2006 (UTC)

That is... so... strange odd....  freshofftheufoΓΛĿЌ  05:03, 27 September 2006 (UTC)

Engine behaviour (with film)

I have made a few observations concerning my one-cylinder, four-stroke lawn mover engine that I was hoping you could help me explain.

1. When increasing throttle rapidly, the rotational speed seems to oscillate before settling at a high value. I demonstrate this in a little film.
2. Immediately after having changed oil, it was very difficult to start. It took several tries, some resting (for the engine) and then a few tries again before it started coughing slowly, almost stalling, and then eventually picking up speed and running normally.
3. Also soon after changing the oil, it was spewing out a lot of smoke. Some of it is visible in the above film. Later, it got back into shape, not exhausting anything visible.

—Bromskloss 12:37, 25 September 2006 (UTC)

I only hear sound but see no image. I guess that 2 and 3 are due to excess oil in the combustion chamber. --Lambiam^Talk 13:47, 25 September 2006 (UTC)

That's funny, it works for me (VLC). Well, the sound is the important part. You can hear how the engine speed sways up and down a couple of times when I throttle up. This has nothing to do with oil change, btw. It has always done this. Ok, oil in the combustion chamber, you say. How does it get in there? Aren't the valves (fuel-air mixture in, exhaust gases out) the only entrances? And even if those were open, is the oil really supposed to be there? —Bromskloss 14:30, 25 September 2006 (UTC)

What's the vintage of the mower? My old lawnmower would just belch out tons of smoke once in a while. Finally replaced it, probably made a good go-kart engine. --Zeizmic 20:19, 25 September 2006 (UTC)

Hmm, what do you mean, "vintage"? —Bromskloss 20:08, 26 September 2006 (UTC)

Past experiences effecting future perception - psych reference wanted

I remember reading in an undergrad psychology textbook about an experiment that had been performed. A picture was shown to two groups of people, one very young (i.e., between toddler and pre-pubescent) and the other older (teenager and up). The picture was very similar to the goblet illusion, in that depending on which elements you took to be the foreground or background, two different images could emerge. One image was of two people intimately and amorously entwined, possibly even having sex, while the other image was of a pair of dolphins. The study showed that the younger group always/usually picked up on the dolphin image first, whereas the older group always/usually picked up on the sexy-couple image first. The idea was that young children, having no experience with most things sexual, had no context to recognize that the sexy-couple image was of anything meaningful, and so the dolphin image was predominant. And of course, vice-versa for the adults, for whom sex is much more salient than dolphins.

Now my questions are these: does this phenomenon have a specific name? Can anyone point me to the name of this study, the researchers, or any textbook in which the study appeared? Can anyone point me to a digital copy of the image in question? I've been Googling off and on every few months for a couple of years now, but since I can't remember any of the relevant terms relating to the study, I'm at a bit of a loss. Help is appreciated! --PeruvianLlama^(spit) 18:52, 25 September 2006 (UTC)

It took me more than 5 minutes to see the dolphins :) I don't know the name of it though --froth^{T C} 19:28, 25 September 2006 (UTC)

This isn't quite what you're looking for, but maybe it will help: Gestalt psychology. --Allen 19:30, 25 September 2006 (UTC)

Is this the picture you're looking for? It's not a couple of dolphins, but several. --Allen 20:11, 25 September 2006 (UTC)

Yes! That picture is it exactly! I guess the exact number of dolphins had faded from my memory over the years, but that's precisely what I was looking for. Thanks! --PeruvianLlama^(spit) 20:20, 25 September 2006 (UTC)

But where are the dolphins? --Lambiam^Talk 20:48, 25 September 2006 (UTC)

And what does it mean if you see both images, at the same time? That you're an adult-child? Skittle 20:51, 25 September 2006 (UTC)

That's what happens when you've seen the picture before. You can't unsee either image, the damage is already done. Speaking of which, ever see the arrow in the Fedex logo? Find it, and it will be burned onto the surface of your brain forever and it'll stick out every time you look at a Fedex truck. Gary 04:47, 26 September 2006 (UTC)

[hushed voice] I see dead people. Clarityfiend 06:34, 26 September 2006 (UTC)

Check out Priming. --Lambiam^Talk 20:50, 25 September 2006 (UTC)

Creation week (moved from misc)

If the sun, planets, and other stars all disappeared and the earth was just zooming along through deep space, could the atmosphere stay stable for a day and the oceans stay liquid for two days? In other words how fast would heat energy bleed into space if we received none from any star?

I ask this because the creation account in Genesis describes the atmosphere and oceans as being created before the sun. Incidentally, plants were as well, but I suppose they can last a day! --froth^{T C} 16:50, 25 September 2006 (UTC)

It's an interesting question. For now, I would like to say that some people who believe every word in the bible argue that without the Sun etc... "one day" meant something different, something longer, thus allowing more things to happen before mankind appeared. Evilbu 16:57, 25 September 2006 (UTC)

Hmm...God created plants and then created sunlight, hoping the plants won't die in the process (due both to the lack of sunshine and the freezing cold)? And he created an liquid-water ocean just to watch it freeze? What a crazy God! --Bowlhover 04:26, 26 September 2006 (UTC)

Right. God, being omnipotent, is capable of miracles: striking interpositions of divine intervention in the universe by which She overrules, suspends, or modifies the ordinary course and operation of Nature. This doesn't answer the original question, of course, where it is assumed that the laws of physics as we know them are back into presumably unmodified form after some Cosmic Jester has whisked away the inspirational source of astrology. --Lambiam^Talk 18:25, 25 September 2006 (UTC)

Being a mathie, I'm actually interested in this question, but I don't know that much about thermodynamics etc... Froth, I think more people will be able to help you on the "Science Reference Desk" [1].

I think it has to do with black body temperature. IF we can assume there is no greenhouse effect (and then I'm not just talking about the manmade greenhouse effect), AND that all of earth has the same temperature, then maybe I could turn it into a differential equation (separation of variables should make that a relatively easy equation). However I do not know the specific heat of Earth. I welcome any help!Evilbu 18:52, 25 September 2006 (UTC)

It would be simpler just to figure out how much energy we get from the Sun in two days. Since the Earth is in thermal equilibrium, that would be the amount it loses during the same period. As for the atmosphere, we'd still have the Earth's gravity to hold onto it. Clarityfiend 19:02, 25 September 2006 (UTC)

That will give a different result. If you "shut off" the Sun now, Earth will not emit power at a constant rate, the colder it gets, the less Joules(or calories if you like) it will send out into outer space per second. However, it is possible that for just a few days both calculations give a nearly identical result. Anyway, Clarityfriend's method and mine both need the specific heat of the earth :( Shouldn't I be bold and move this to the Science Reference Desk? Evilbu 19:12, 25 September 2006 (UTC)

Done. --froth^{T C} 19:22, 25 September 2006 (UTC)

According to Solar radiation, we receive 1.74 x 10¹⁷W/s from the Sun. So, we'd lose it at roughly (as Evilbu pointed out) the same rate. Who's this Clarityfriend? Clarityfiend 19:40, 25 September 2006 (UTC)

Hmm, that's you, sorry for the typo. But in a long forgotten dark past I took an astronomy class and I was thought that eventually every atmosphere escapes. (it's just a matter of particles attaining the escape velocity (Boltzmann distribution) ). However, I had ANOTHER professor that Earth's atmosphere is a growing system due to stuff flying into it. Just guessing now... (surely those meteorites aren't neatly packed bottles of nitrogen and oxygen???)Evilbu 20:00, 25 September 2006 (UTC)

Yeah, the atmosphere leaks, but not a significant amount in two days. Plus, gases dissolved in the oceans would be released due to the dropping partial pressures. Clarityfiend 20:11, 25 September 2006 (UTC)

Throwing around gazillio-watts lost from a 10 thousand mile wide highly complex system is tough to visualize! What kind of heat loss could be expected for your average rock laying around assuming God created the world at average modern temperatures? --froth^{T C} 20:19, 25 September 2006 (UTC)

Thinking completely unscientifically, the earth cools dramatically during nighttime hours, so much that in most regions without other circumstances it can drop 30F between day and night. In places near warm water (Key West, FL) temp drops average 10F overnight. In areas away from water (Delaware, OH) temp drops average 25F overnight. This cooling takes place in less than 12 hours (much less at times but we will use this conservative estimate). Places without good temperature buffers would stand to lose around 50F per day, and those with them would only lose 20F per day. After two full days with no sun, I would say the less fortunate areas would be completely uninhabitable (overall loss approaching 100F) and the more fortunate would only be days away from demise. Even with warm water around, the cooled air from land masses will quickly create harsh winter storms over the entire planet.

I don't think a 100F temp drop would necessarily make a place uninhabitable, depending on the starting temp. Storms are also caused by uneven temps due to uneven solar heating. I suppose there might be storms as the world temps even out, but, eventually, we would be left with a very cold, still, dead planet, at least on the surface. Life living off sulfur vents at the bottom of the oceans would likely survive. StuRat 09:43, 26 September 2006 (UTC)

Yeah, but we'd have time to update Wikipedia! Get your priorities straight. Clarityfiend 18:07, 26 September 2006 (UTC)

We need to teach the tube worms living near the undersea sulfur vents, to update Wikipedia, just in case. :-) StuRat 10:34, 27 September 2006 (UTC)

Electrical resistance caused by gravity

I was reading Electric_power_transmission#HVDC and it got me wondering... if you had two HVDC lines running down the side of a mountain (one each way), would it take more voltage to push the electrons (who do have some mass, at least enough to visibly weigh down carrier lines) up the mountain than push them down the mountain?

Also, I'm trying to learn more about the properties of electricity- what if you used the same voltage for each line? Would there then be a slight difference in amperage? Or would the voltage be reduced on the upward line?

Thanks --froth^{T C} 19:53, 25 September 2006 (UTC)

The electrons in the cable contribute less than 0.03% of the weight. What makes you think the contribution to the "weigh-down" is visible? I have no data for the current through these cables, but for every Ampere per vertical metre the potential-energy difference is about 56 pJ. That must be negligible compared to other losses. Since you don't store the electrons at the top of the mountain but just "pump" them around, there is no actual energy difference involved. --Lambiam^Talk 20:44, 25 September 2006 (UTC)

First: the electron mass is really so low, that they cannot pull down the carrier line. Only about 1/1800 of the mass of ordinary matter is due to the electrons, nearly all weight is due to the nucleii of the atoms. You got misled by the statement in the HVDC article that power lines sag in case of high load. But the reason for this is not the minuscule weight of the electrons but the fact that the wire gets heated by the current (due to the friction of the moving electrons). If the current is large, the wire gets heated even more, and as metal expands when heated, the wire gets longer -- and hence sags down. (The effect is most prominent if you compare power lines in winter and sommer.)

Second: Even if electrons had a significant mass, a wire carrying current still would not be heavier. That is because in a current-carrying wire there are no more electrons than in a a wire without current. Only, the electrons are moving instead of staying where they are. And, by the way, they are actually moving quite slowly, much slower than the speed of the electrical signal. But if an electron starts moving at the beginning of the wire, it pushes ahead (by means of the electrical forece due to its negative charge) the electrons in front of it. So, just like the tail of a train starts moving immediatly, when the locomotive starts, the current starts immediately (or with only speed-of-light delay, to be precise) all over the wire. Simon A. 20:58, 25 September 2006 (UTC)

Agreed. Think of it like sounds waves, which move much faster than the air which carries them. Similarly, electrical current moves much faster than the electrons which carry it. In A/C current, I don't believe that there is any net movement of electrons, on average. StuRat 22:49, 25 September 2006 (UTC)

I'm not familiar with the wire sizes used in HVDC transmission lines, so consider a residential situation. A 14 American wire gauge cable goes straight up from a basement to an attic, a distance of 6 meters. To calculate the energy needed to lift an electron this distance, use the formula energy = (mass)(change in height)(accelleration due to gravity)

= 9.1•10⁻³¹•6•9.8 joules

= 5.4•10⁻²⁹ joules

The potential difference (voltage) between two points is defined as the amount of work required to move a charge between two points divided by the charge, so the voltage between the bottom and top of the wire is

5.4•10⁻²⁹/−1.6•10⁻¹⁹

=−3.8•10⁻¹⁰ volts

By comparison, the maximum voltage change due to electrical resistance in the upward wire, if it is carrying 15 amperes of current, would be about −0.9 volt. The difference between the effect of resistance and the effect of gravity is so extreme that it would not be practical to measure the effect of gravity.

If the comparison had been closer, we could go on and consider the interaction between the upward cable and the downward cable, but I don't think it is necessary to carry the calculation that far. --Gerry Ashton 00:11, 26 September 2006 (UTC)

---

Hmm, no one's really answered the question yet. Yes, the effect would be miniscule, if it exists at all; it's also true that the effect would not need to be considered with AC, which just doesn't have that way of getting your electrons moving.

So, if you have a (non-superconductor, though I recommend the article) wire, and you make a current flow through it uphill, will:

1. the resistance be the same
2. the average speed at which the electrons are moving be the same
3. anything change if static electricity is the source of the current rather than, say, a galvanic cell

compared to the case of a level conductor?

If it helps you think about it, imagine the whole setup to be in an extremely strong gravitational field, such as near a black hole's event horizon.

Without having fully thought this through yet, my guesses are:

1. yes
2. no
3. yes

I'll write more on my reasons (and details about 3.) later.

RandomP 00:57, 26 September 2006 (UTC)

Hmm, I might be contradicting myself here:

I'm beginning to believe that where electrons are considered, their gravitational potential must be added to their electric potential to make things accurate. In other words, the voltage in a galvanic cell will depend on its orientation relative to a gravitational field.

Thus, it's not the resistance that changes, it's the voltage that's reduced in the wire-running-uphill case, which (at the same resistance) means a lower current.

Note that this also means that in a very strong gravitational field, you should be able to build a galvanic cell where the same material is used for anode and cathode, by placing one of them higher than the other.

What's different with static electricity? Well, quite simple: with static electricity, only the gravitational potential energy of the electrons needs to be taken into account: that's fairly low. However, in a galvanic cell, a transfer of matter to the cathode happens, so there is a whole lot more work to do (or energy to gain, if the cathode is lower than the anode).

All of these effects will probably be masked by brownian motion, however.

RandomP 16:55, 26 September 2006 (UTC)

I think the question is answered fully above with the answer "No". --Lambiam^Talk 18:32, 26 September 2006 (UTC)

There's a difference between an effect being negligible (and this one is, on earth) and that effect not existing. There's no argument above that the effect doesn't exist.

When electrons move uphill, they gain gravitational potential energy. It's got to come from somewhere.

RandomP 18:42, 26 September 2006 (UTC)

Above there is a sentence: "Since you don't store the electrons at the top of the mountain but just "pump" them around, there is no actual energy difference involved." It was meant to convey the idea that there is no actual energy difference involved. In other words: the effect doesn't exist. I also gave a quantification of the self-cancelling part X in the equation X − X = 0, namely 56 pJ/Am. --Lambiam^Talk 00:37, 27 September 2006 (UTC)

permethrin as it relates to sulfa

I have been doing research regarding the many contradictory statements and articles regarding permethrin. Through my research I found an article stating that " individuals with defects in sulfate-related enzymes may be unable to easily break down permethrin leading to increased suseptbility to motor neuron disease." I myself am highly allergic to sulfa containing drugs and sulfonimites (found in ie. wine). Would this also make myself more likely to have an allergic reaction to this medication? They are mandating that everyone who works and lives in the nursing home I work at be treated and I have serious concerns related to the safety. This and any other info. would be of great value considering there are not any contraindications listed in my drug reference. (Which I find hard to believe after reading numerous other articles) Thank you. Trudi Heiselman LPNTrudiLPN 20:59, 25 September 2006 (UTC)

Why would you expect to find information about an insecticide in a drug reference? Try looking in Medlilne for articles on "permethrin" and "toxicity" and "human". One such is here. - Nunh-huh 06:54, 26 September 2006 (UTC)

Fresh frozen plasma (FFP)

Does fresh frozen plasma have to be grouped against a person's ABO bloodtype? If so, can one give FFP from a Group O patient to a Group A patient? (Getting different answers from different places.) Thank you very much for your help!Mmoneypenny 21:04, 25 September 2006 (UTC)

FFP can contain anti-A or anti-B antibodies, depending on the blood type of the donor. So type specific or type compatible FFP is required for transfusions. However, unlike transfusions of RBCs, specific donor/recipient compatibility tests are generally not performed. Group A patients should receive FFP from a donor who is type AB or type A. Group A patients should not receive plasma from a donor who is type O or type B, because a type O or type B donor's plasma contains anti-A antibodies. - Nunh-huh 00:49, 26 September 2006 (UTC)

For more information, see Blood type#Plasma compatibility InvictaHOG 01:16, 26 September 2006 (UTC)

Thanks to both of you! (and apologies for not reading that article better the first time round)Mmoneypenny 06:17, 26 September 2006 (UTC)

Pressure lost from air flowing through a pipe

If you have a length of pipe that air can flow through, is there a formula that you can use to calculate the pressure that is lost from one end to another? I seem to have forgotten everything I ever learned about physics, but I remember being able to do this for liquids at least. The fluid dynamics pages aren't making too much sense.

The air is at atmospheric pressure at one end, with a slightly lower pressure at the other end. And how would I calculate the total drop in pressure over a series of pipes? Is there a simple way of relating the pressure difference at the two ends of the pipe to the airflow speed and a constant that depends on the pipe? --129.110.195.26 21:07, 25 September 2006 (UTC)

I have completely forgotten the formula, but there are formulae out there. I think knowing the velocity, the pipe diameter, the start pressure, air density and viscosity and having access to tables or graphs, you would be able to do it. I'll have a look for notes... Skittle 21:11, 25 September 2006 (UTC)

See fluid dynamics, Reynolds number, Nernst coefficient, Bernoulli's equation.--Light current 21:14, 25 September 2006 (UTC)

Bernoulli's equation was the one I remembered using. I think I can find out what I want using that and the darcy friction factor formula, but I'm not entirely sure how. I can solve that equation and get that the change in pressure per meter of pipe equals velocity times a constant. But according to bernoulli's equation, velocity depends on the pressure. So I'm not really sure how to continue. I imagine I would have to integrate across the entire pipe. So I have dp/dx = vC and p = d(c - (v^2)/2). And I need to solve this for p at one end given the constant in the second equation. I don't know how to do that, though. Should I got to the math desk? --129.110.195.26 22:03, 25 September 2006 (UTC)

Hmm, I can solve that to dp/dx = C squrt(2c-2p/d) where C = 2fd/D. Then I can plug in all the values, but I get a diferential equation in the form dp/dx = Asqrt(B + Cp) Where A, B, and C are constants. And I don't know how to solve something like that. --129.110.195.26 22:23, 25 September 2006 (UTC)

No dont go to the math desk: theyre all asleep over there!8-)--Light current 22:13, 25 September 2006 (UTC)

$p(x)={\frac {-4B+A^{2}C^{2}x^{2}+2AC^{2}xk+C^{2}k^{2}}{4C}}$ where k is the constant of integration needed to get the boundary conditions you expect. If this happens to be p(0) = 0, then k = sign(dp(0)/dx)2sqrt(B)/C. If your boundary conditions were different, then you'll need to solve for k to match your conditions. -- Fuzzyeric 04:13, 28 September 2006 (UTC)

From the Bernoulli page you have

${v^{2} \over 2}+gh+{p \over \rho }=\mathrm {constant}$

This means ${v^{2} \over 2}+gh+{p \over \rho }={v^{2} \over 2}+gh+{p \over \rho }$

where the symbols on the left would have little subscript 1s and those on the right, 2s, if I were any good at script. If you assume air is incompressible this is, of course. If you know v1, v2, p1, rho, g (9.81), h1 and h2, all you have to do is rearrange the thing. Skittle 22:10, 25 September 2006 (UTC)

Or are you saying you don't know v2? Skittle 22:15, 25 September 2006 (UTC)

Bernoulli, I think, is only applicable in laminar flow situations 8-|--Light current 22:16, 25 September 2006 (UTC)

Am I going about this the wrong way? My basic question is based on the fact that sucking on a short pipe is easier than sucking on a longer pipe. I assume that this difficulty is related to loss in pressure as you go through the pipe. I know how to use bernoulli's equation to find the difference in velocities and pressures between two areas of the pipe with different surface areas and whatnot, but not with something like this. --129.110.195.26 22:32, 25 September 2006 (UTC)

Am I actually looking for the amount of energy required to exert the pressure on the end of the pipe? --129.110.195.26 22:36, 25 September 2006 (UTC)

It's surprisingly hard to find descriptions of this standard problem online, but I found an excerpt from a textbook on the subject that you might find useful. In particular, you want the area around page 115 (page 5 or so of the PDF). The energy lost (and thus reduced velocity for a given effort) with a longer pipe is due to viscosity in the flowing fluid, which must be dragged along the stationary wall for a greater distance. Does this help? --Tardis 16:10, 26 September 2006 (UTC)

Your short description helps, but I don't have access to that textbook excerpt, and can't even see which textbook it is. --Crazy Wolf 02:44, 27 September 2006 (UTC)

Ah, sorry -- the site seems to be IP-restricted. The book is ISBN 0-340-67649-3: Introduction to Fluid Mechanics by Y. Nakayama and R. F. Boucher (Elsevier, 2000). Perhaps more useful: I found it by searching for "fluid mechanics" pipe introduction, so perhaps more can be found with a similar search. --Tardis 16:14, 27 September 2006 (UTC)

Perhaps you're looking for pressure drop calculator. It uses empirical fits for the nonlaminar flow (which may or may not be fine depending on how rough your pipe's interior surface and inter-pipe joints are). -- Fuzzyeric 23:42, 28 September 2006 (UTC)

I think you're looking for Poiseulle's equation (I think that's how its spelt). I don't have it handy to give you but any reasonable physics text should have it. Concerning the rate of flow of fluids in tubes, the rate of flow is proportional to the force (pressure) applied, and inversely proportional to the length of the tube, the viscosity of the fluid and the 4th power of the radius of the tube. cheerio! Mattopaedia 03:55, 30 September 2006 (UTC)

why does me knee go crack

when I bend it?

My right one does but my left one doesn't. my fingers do it as well though not as loud.

According to the German Wikipedia, de:Gelenk#Gelenkknacken, the most common cause for creaking cracking sounds of joints is air bubbles in the synovial fluid. Another reason might be that the surface of the bone ends are not fully smooth. The article goes on to state that the creaking cracking is much less harmful than commonly thought and that it is untrue (though often claimed) that producing the sound intentionally causes arthritis (though it may reduce grip strength or overstretch adjacent joint). Simon A. 21:48, 25 September 2006 (UTC)

No not creak like a door crack like a (very quiet) gun.

That's what I meant, of course. I actually even checked the dictionary, while writing, but it was inconclusive. Simon A. 22:38, 25 September 2006 (UTC)

I'm a bit dubious about the air in the joint thing. I've heard that one before, but how does the air get there? I've never seen it. I think the real reason is sudden synovial fluid shifts as the joint travels through its range of motion. Also sometimes the patella and quads tendon may not track smoothly and give rise to cracking. Damage to the menisci or articular cartilage can slao produce cracking, but these latter are usually painful. If its painful, see your doctor. Mattopaedia 11:48, 28 September 2006 (UTC)

Could you elaborate on this business with the air? I don't understand how an air bubble can make a noise like that.

While we're on the topic, I'd like explanations for two similar phenomena:

1. Fingers cracking when you pull on them
2. My right elbow cracks sometimes when I fully straighten it. The interesting thing is that when it's ready to crack, I feel a bit of resistance when it approaches the cracking point.

--Smack (talk) 04:48, 29 September 2006 (UTC)

IANAD, but I'd always thought it was a ligament uncatching itself from a bone protrusion. Wouldn't air bubbles be very dangerous anywhere in the body? Anchoress 05:23, 29 September 2006 (UTC)

I dont think it is air either. We dont see air in the sinovial fluid in Magnetic Resonance Imaging (M.R.I.) or on Computerized Tomography (C.T.) either. --Ehsan 14:49, 29 September 2006 (UTC)

Ligaments and tendons do sometimes snap, creak, crack etc with joint movement. The amount of danger associated with air bubbles depends both on the size and location of the bubble. Air bubbles in the blood = generally bad, but they have to be of sufficient size to act as a foreign body and block an important piece of microvasculature (eg brain) for a long enough time without being absorbed long enough to cause irreversible ischaemia. Air in joints = ho hum. Joints don't really care if they get a bit of air in - otherwise most joint surgery would be impossible , or performed underwater. Mattopaedia 04:23, 30 September 2006 (UTC)

My understanding, without being able to give references, is that bubbles of gas can form within the synovial fluid in a joint, and is composed of dissolved gases in the fluid. The bubbles are more likely to form when certain pressure and volume conditions are present, eg, when there is a differential between external air pressure and internal joint pressure. This is why people who pull on the fingers to crack the knuckles are more likely to have them crack in cold and damp conditions, because it is associated with lower barometric pressure. The crack is the collapse of the gas bubble when the differential is suddenly removed, ie, altering the volume within the joint. It also may explain why my grandmother kept saying 'It's going to rain... I can feel it in my bones'. People with arthritic conditions often find they are bothered less by their symptoms when they live in a climate with steady air pressures. I'd welcome comments on this theory, as I've always wondered how it fits physical theories.81.174.167.6 11:22, 30 September 2006 (UTC)Steph

Adventurist Theory

Is there such a thing as a religion based on video games? I am referring to RPGs in specific and am also referring to stat points and basing them on real life. Plus, when dead, being able to decide where you want to go rather than having an established course and having your stats increase wherever you decide to go until you reach the top person(boss), and then you are born again with your base stats of your previous life evened out with your new stats. The base belief of this religion is: Life and Death are games. Play them. Would anyone believe in this religion?

Of course. People will believe anything. You just have to figure out a way to sell it. Melchoir 00:24, 26 September 2006 (UTC)

Most radioactive element

What is the most radioactive element found on the Periodic Table? Jamesino 23:33, 25 September 2006 (UTC)

Well, it's not completely clear what you mean by "most radioactive", but by at least one measure, the most radioactive nuclide would be the one with the shortest half-life (because it's the one that gets used up quickest). There are quite a few whose half-lives are so short that only their decay products can be observed. --Trovatore 23:37, 25 September 2006 (UTC)

By radioactive, I mean one that is the most harmful to organic matter and unleash the most Alpha, Beta and Gamma particles. Jamesino 00:59, 26 September 2006 (UTC)

Yeah, the most radioactive isotopes/elements are those that do least damage to anything (including us)--Light current 01:13, 26 September 2006 (UTC)

Also depends what you mean by radioactve. Do you mean alpha, beta , gamma or neutron radiation? All have different penetrations and dangers.--Light current 02:22, 26 September 2006 (UTC)

Radium is the best candidate I could find. I thought it would be plutonium, but that article says that "Naturally-occurring radium is about 200 times more radiotoxic". It's also over "a million times more radioactive than the same mass of uranium." You can't really use half-life as a criteria; some forms of radioactivity are more harmful than others. Clarityfiend 01:33, 26 September 2006 (UTC)

Well, first of all, there's no such thing as "a criteria". --Trovatore 01:57, 26 September 2006 (UTC)

Cornered by the Grammar Inquisition! Safe me! Saves I! Clarityfiend 04:58, 26 September 2006 (UTC)

Whenever I catch a loose criterion, I step on it immediately. :-) StuRat 09:34, 26 September 2006 (UTC)

That said, you're right, half-life by itself is not a completely reliable criterion for how dangerous a radioactive substance is. But it does give you a good first cut at it. Plutonium is less dangerous than radium precisely because its half-life is so much longer. Depleted uranium, to a first approximation, can be considered non-radioactive; with a half-life of four billion years there just aren't many decompositions going on in a sample of a given size. --Trovatore 01:57, 26 September 2006 (UTC)

Mmmmm! Depleted uranium. Heavy! (and filling)--Light current 02:10, 26 September 2006 (UTC)

• On a slightly related question, which radioactive element would leave an area uninhabitable for the longest period of time? Titoxd^(?!?) 03:37, 26 September 2006 (UTC)

Radium:

The SI unit of radioactivity is the becquerel (Bq), equal to one disintegration per second. The curie is a non-SI unit defined as that amount of radioactivity which has the same disintegration rate as 1 gram of Ra-226 (3.7 x 1010 disintegrations per second, or 37 GBq).

What do you mean by uninhabitable?--Light current 03:39, 26 September 2006 (UTC)

A place where nuns dare not go. Clarityfiend 04:58, 26 September 2006 (UTC)

OK I ve been saving this. You have now triggered the device by mentioning the second secret word 'nun'.:

• You can kiss a nun once, you can kiss a nun twice.
• But you must never get into the habit

--Light current 05:54, 26 September 2006 (UTC)

Where are the admins?! Light current is booby trapping this reference desk! And he's already caught one. Wait a sec...that looks like a certain unmentionable seabird. Clarityfiend 17:45, 26 September 2006 (UTC)