# Wikipedia:Reference desk/Archives/Science/2007 May 21

Science desk
< May 20 << Apr | May | Jun >> May 22 >
Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

# May 21

## Resilience of light polarization

Does polarized light stay polarized when viewed through things? Specifically, I'm wondering about windows - both normal and frosted glass - and clouds. (I don't know if this question was answered in the polarization article or not -- it was too technical for me to understand.) CameoAppearance orate 00:08, 21 May 2007 (UTC)

Polarized light will rotate its polarization when passing through anything that qualifies as a wave plate due to birefringence. Also, when light passes through a transparent material, only some of the light will pass through and some will be reflected. How much is reflected/transmitted is dependent on the polarization of the light. At a certain angle (See Brewster's angle for more) this will actually result in only a single polarization passing through the material. So yes, if the sun and window are at the correct relative angle, only one polarization will actually reach your eyes (ignoring light that reflected off of other sources and then through the window). Someguy1221 00:14, 21 May 2007 (UTC)
Actually, this is also ignoring scattering in the atmosphere, the fact that the sun is not infinitely far away...You can see these effects very nicely if you're playing with a laser. Someguy1221 00:26, 21 May 2007 (UTC)
Or just a pair of polarized sunglasses. I've seen car windshields that look positively psychedelic, with rainbow colors and patterns from how the glass was prepared. —Keenan Pepper 05:18, 21 May 2007 (UTC)

I'm not sure if the above responses fully answer your question, CA. You wanted to know whether polarized light stays polarized after being reflected, scattered, transmitted, etc. The answer is, as usual, it depends. Firstly, some terminology. Polarized light includes the two extremes of linear polarization (where the electric field is fixed in direction but oscillates in magnitude) and circular polarization (where the field is fixed in magnitude but spins in direction), and all the cases of elliptical polarization in between. Any beam of light, polarized or unpolarized, can be considered as the combination of two components, linearly polarized in perpendicular directions (or equally validly, two circularly polarized components, left handed and right handed, but that may be a little harder to visualise). The distinction between polarized light and unpolarized light is that in polarized light, the two components are coherent with each other: their oscillations are correlated, and they have a constant phase difference. The actual value of the phase difference, in combination with the average magnitudes of the two components, determines the type of polarization and its orientation. In unpolarized light, the two components are independent, uncorrelated, randomly related.

If a beam of polarized light is reflected from or refracted through a smooth surface (smooth on the scale of a wavelength) then in general its components remain coherent, therefore it remains polarized, although its state of polarization may change, e.g. from linear to elliptical. If it is scattered from a rough surface or by a turbid medium, like clouds, then its components can be randomly altered in phase, making the beam unpolarized. So normal glass will not cause depolarization, but frosted glass may, depending on how dense or thick the frosting is. Clouds will depolarize the light if the droplets are large enough. Scattering from very small particles (Rayleigh scattering) will generally alter the state of polarization, but won't destroy it. Polarized light scattered from an ordinary cinema screen becomes depolarized, so to get 3-D movies to work, you have to use a silver screen. Sometimes the change in state of polarization depends on wavelength, which gives the startling colours and patterns seen in plastics between crossed polarizers (see photoelasticity), or car windscreens illuminated by partially polarized skylight and viewed through polarizing sunglasses. It's a fascinating and rich subject, which is why the polarization article may look a bit impenetrable. --Prophys 13:44, 21 May 2007 (UTC)

## Energy and mass - Binding energy

This question is in relation to the earlier question energy loss = mass loss.

The first reply from Cody Pope says "...bond energy makes compounds lighter than the sum of their parts. When the bonds are broken this mass is expressed as energy in the form of light/heat...". The binding energy article does appear to say the same thing. I can see how this may be the case where the compound is formed in an exothermic reaction, but it doesn't seem to make sense when formed in an endothermic reaction.

If there is energy stored in the bonds, according to E = mc2 shouldn't the mass be greater in the compound than the component parts, not less? If it was less, when the bonds are broken and energy is released (as stated earlier), further mass would be lost, and the component parts would have less mass than when they first formed the compound, ultimately meaning the elementary particles would have to lose rest mass.

Can someone explain this apparent anomaly please? --jjron 03:17, 21 May 2007 (UTC)

If you do a really accurate analysis of the energy levels of a molecule, there are no such things as bonds and it gets horribly, horribly difficult. Try to work with this premise: Every molecule (or atom, just any object) has a specific ground level energy. That energy includes everything, bounds, rest mass, internal stress, everything. It is a natural constant that can in principle be measured for any type of molecule by putting one molecule in the ground state on a scale and take the reading.
For any chemical reaction, take the sum of the energies for the old parts and compare it to the sum of the energies of the new parts. If the new parts are heavier, you have to add some form of energy to make the reaction happen. If the new parts are lighter, some form of energy needs to be removed.
Thanks for your answer, but it's not answering the question (or if it is, it's agreeing with what I said, which indicates a problem with the binding energy article and the answers to the earlier question). My quandary is that the binding energy article says:
• "A bound system has a lower potential energy than its constituent parts; this is what keeps the system together.", and
• "Because a bound system is at a lower energy level than its unbound constituents, its mass must be less than the total mass of its unbound constituents."
In other words, according to this, the mass of the bound system will always be less. That doesn't seem right to me (and also disagrees with what you're saying, which is that sometimes the products will be lighter and sometimes heavier). Can this be clarified? --jjron 04:55, 21 May 2007 (UTC)
The article seems correct to me. All bond breaking is endothermic. Give an example of a compound that you think has a positive-energy bond. —Keenan Pepper 05:16, 21 May 2007 (UTC)
Dihelium. Atoms bond to each other so they can sit in a nice potential well. You can smush two atoms together (not very technical, I know), and if they physically can't exist in a potential well (like two helium atoms) they'll never actually "bind." Molecules often break apart spontaneously and reform in a new form, with even less potential energy, but breaking the initial bonds always takes energy for any remotely stable compound. Someguy1221 06:32, 21 May 2007 (UTC)
Re Keenan. Take an exothermic reaction, such as regular combustion. Yes, you may need to add some energy to break the initial bonds, but overall the energy released is greater than that which you add to break the bonds. Now, energy (overall) has been released, which I would think according to E = mc2 would suggest mass has been lost, but according to the statements above, mass will have been gained by the now unbound constituents. Can you explain where this is wrong? Am I misinterpreting terms like bound systems/unbound constituents? --jjron 07:27, 21 May 2007 (UTC)
Okay, let's look at combustion. You start with oxygen molecules (O2) and a hydrocarbon. You break the bonds between the carbon and hydrogen atoms in whatever you're burning, and you break the bonds between the oxygen atoms in the oxygen molecules. This is what takes energy, this is the activation energy. But wait, combustion isn't finished yet! These unbound oxygen and carbon and hydrogen atoms shuffle about and new bonds are created, giving you water (H2O) and carbon dioxide (CO2). Creating these new bonds gives out energy. In this case, creating these new bonds gives off more energy than was needed to break the old bonds, making this reaction exothermic. So you go from oxygen and a hydrocarbon, to oxygen atoms, carbon atoms and hydrogen atoms, to water and carbon dioxide. Both the reactants and the products have bonds. For an exothermic reastion, the products will have less energy in their bonds than the reactants. Skittle 15:51, 21 May 2007 (UTC)
I like that answer, thanks. I was kind of thinking something like that may be the case. Thanks to all others that responded too.
(This may be too late to now get another response so I may need to reask...) however, to take it one step further and to clarify overall, to get back to the constituent components, say if we break the CO2 & water back up into their constituent atoms, would require an energy input, and thus a gain of mass? And this would always be the case, i.e., there's never any sort of compound that has more energy/mass than it's constituent atoms alone? And that explains the statements about the binding energy? --jjron 06:56, 22 May 2007 (UTC)
I haven't read all the posts as closely as I would like, but there seems to be a horrible misconception going around. It first appeared in the energy loss = mass loss question. The problem is that the mass-energy equivalence concept is not being applied correctly here. The mass of the measured gases over the city would indeed remain the same because this is a chemical reaction, with NO loss in mass (i.e. a mass defect). The released radiant energy and heat come from the difference in bond energies between the reactants and products. Reactants have a lot of energy stored in their bonds (energy not mass—bond energy has absolutely no effect on the mass of the compound; no conversion is taking place between mass and energy) and products have less energy. The "energy defect" is released in various form (light and heat). There's no such thing as "binding energy" for a molecule. Yes, there are chemical bonds, involving valence electrons, but in no way do atoms contribute some of their mass in the formation of chemical bonds. We speak of binding energy when talking about nuclei. In a nucleus, some of the mass of nucleons is lost in the form of binding energy that holds the nucleus together. This is why the constituents (the nucleons) have greater total mass when separated than when they are bound in the same system. This is why a nuclear reaction releases energy: the products are more stable, so they have more binding energy, which means more mass is last, which creates a mass defect—now you can apply the formula E=mc2 and calculate the energy liberated by the reaction from the lost mass. —LestatdeLioncourt 08:17, 22 May 2007 (UTC)
The energy/mass issue is completely a matter of interpretation. If you take an enclosed system of gas and let an exothermic reaction occur, yes, the system will weigh the same at the end. But if you then cool the system back down to its original temperature, it will weigh less (you'd better have a damn good scale, but it will weigh less). As far as saying there is energy in the bonds, these have mass...interpretation! Of course, the only truly accurate way to do this is using quantum wave equations, and mass/energy operators (which themselves are actually only approximations, but very very good ones). So I'll have to disagree with your assesment. If we are not permitted to offer interpretations that are not truly accurate, then the only remaining answer is quantum, which most people would fail to understand. Read any popular quantum physics text, there are incorrect interpretations abound! So anyway, these interpretations, while bad, are often useful, simply for lack of a simple and accurate explanation. Someguy1221 09:24, 22 May 2007 (UTC)

"A bound system has a lower potential energy than its constituent parts; this is what keeps the system together." - That may work most of the times, but it is not truein general. It works for two particles and electromagnetic force, maybe for more particles and electromagnetic force, but not in general.

Consider Alpha decay. The two participating forces - strong interation and electromagnetic interaction - form a resulting potential which has a "well". The system has higher energy when bound than when separated. It is kept together by the fact, that the intermediary position, an almost seperated system, has even higher energy than the bound or the unbound state.

Systems which are kept together in this way are called "metastable". When some activation happens, the system seperates. A fission bomb would be an example for the, ehm, "application" of this principle.

Alphy decay does not work by breaking some bond and creating another bond. As I said, bonds do not exist when you look really close. They are like "the surface of water". If you just look close enough, there is a fuzzy region of atoms jiggling around, hopping up and down, with no clear distinction whether they are above or below the surface. Bonds are a very useful concept, but like any other abstraction, they loose their meaning at some level of detail.

I am currently trying to think of a physical system that violates the "bound = lower energy" condition, if we do not brake it into arbitrary parts, but only in true elementary particles, but I can find none. Maybe the statement is true in that case.

As a last part, let me add, that "total mass" and "total energy" of a physical system are exactly the same thing. Any form of energy excerts gravitational pull and has inertia, not just the rest mass. The misunderstanding of this fact is often the root cause for problems to understand binding energy. I once had the same problem, but once I figured this crucial fact out, it started to make sense to me. I hope it does so to you too now. ^^

## Seeing imaginary colors

I just wrote Imaginary color, and I was going to add a section on the possibility of seeing them, but didn't because it would be nothing but OR. Is there some substance you can inject into your eye that chemically stimulates one kind of cone pigment much more than the other two? Or, for a less dangerous way of seeing imaginary colors, you could stare at a bright purple light to bleach out your red and blue pigments, and then immediately switch to a green light to see greener-than-green. Is there any literature on this subject? —Keenan Pepper 05:13, 21 May 2007 (UTC)

There is tetrachromacy. And you could always be missing a color receptor or two. Someguy1221 06:51, 21 May 2007 (UTC)
[ec] Yep, that's just what I was gonna say. --Cody.Pope 07:10, 21 May 2007 (UTC)
Some kinds of colorblindness is caused by the absence (or non-functioning) of red or blue detectors - those people would see these colors. SteveBaker 12:16, 21 May 2007 (UTC)
I have proposed merging the new article Imaginary color into Color vision. Please discuss the merger on the talk page of Color vision if you have interest in the matter. There does not appear to be a need to have a standalone article on imaginary colors. Imaginary supersaturated colors outside the CIE chromaticity diagram can be seen by staring at a saturated color until the receptors are fatigued, then staring at the complementary color. Do not inject anything into your eye. "Human Information Processing" by Peter Lindsay and Donald Norman, Academic Press, 1972 has a long section on this. They also suggest (in those days) alternating two slides, which have a square of one color and a surround of its complement, with a slide where the colors are reversed. Someone who does computer graphics should be able to create such a display for the computer monitor, and the result should be pretty dazzling. Edison 14:35, 21 May 2007 (UTC)
I agree with Edison's suggestions. This material would fit well at color vision as a section. We can create a redirect. Nimur 14:52, 21 May 2007 (UTC)
I suppose birds, with their superior color vision can see these colors Coolotter88 20:44, 21 May 2007 (UTC)
So stare fixedly at the center of the image below. The article Afterimage says that 30 seconds should be enough - so this image is a '.gif' that automatically alternates magenta and green every 30 seconds - just keep staring at it - and whatever magic happens should happen automatically every 30 seconds...personally, I didn't find it particularly exciting although there is clearly something happening:
SteveBaker 22:30, 21 May 2007 (UTC)
Nor I. [Mac Δαvιs] ❖ 22:47, 21 May 2007 (UTC)

## Filter standard

To whom it may concerns I want to learn about Classification or Standard of dust filter media Would you please let me know Thanks in advance Quang TuanQuang Tuan 07:27, 21 May 2007 (UTC)

You might check the HEPA article... it discusses what qualifications are required to be labeled "high efficiency." [1] appears to be the official government regulatory and information website.Nimur 14:54, 21 May 2007 (UTC)

## AUTOMOBILES...................HELP

I am asking again bcz i am not satisfied with earlier answers.

why the tractors have their rear wheels lager than front ones? if the answers are in terms of torque plz clearly state driving or resisting torque.(the engine gives constant torque and rpm)125.63.107.131 11:49, 21 May 2007 (UTC)
Also I.C. Engines will have more efficiency in winter or in summers?125.63.107.131 11:49, 21 May 2007 (UTC)
So far the helpdesk said:
• Tractors have big rear wheels so that the tractor doesn't get bogged down in deep mud and to avoid over-compacting the soil - the front wheels can be smaller because they aren't powered and most of the weight is on the rears anyway.
• Car engines burn more gas but are able to produce more horsepower in the winter than in the summer.
Those answers were both good - what else do you need to know - or what didn't you understand? SteveBaker 12:12, 21 May 2007 (UTC)
i don't know exactly, but it is related anyhow to torque? but no body is telling abt that(formatting fixed)

I think the confusion is about the meaning of torque. The "traction" of a tractor is the amount of force it can apply on the ground, at the area where its tyres contact the ground. This force, multiplied by the radius of the tractor wheel, equals the torque. So even though the engine has a constant torque and RPM, the force that it exerts on the ground depends on the radius of the driving wheel: the bigger the radius, the less force for the same torque. Yes, it is related to mechanical advantage. --Prophys 14:02, 21 May 2007 (UTC)

Expanding on Prophys' point, I think the key to the traction of a tractor is the use of lugged tyres. Those large ribs on the tyres provide a concentration of the normal (downward) force of the tractor's weight and allow the transference of a very large traction force (front-to-back force) to the earth without heaving large clods of dirt out and back behind the tractor. Smaller lugs (on smaller tyre) wouldn't work as well because there'd be less volume of soil over which to distribute the tractive force.
By comparison, the front tyres are smaller because they don't have any need to be big; they only have to be so big as to adequately support the weight of the front of the tractor and, has already been pointed out, the weight of the tractor is mostly on the driving wheels, not the steering wheels. This is no different than for over-the-road tractors which commonly have 8 driven wheels and only two steering wheels.
Atlant 15:43, 21 May 2007 (UTC)
Let's be clear here: lugged tires are important, but not because they "concentrate force" or simply "increase friction". Frictional force is simply the product of the coefficient of friction, and the perpendicular or "normal" force -- which in this case is just the weight of the tractor. The surface area doesn't enter into it. Neither increasing the surface area (even though that reduces the normal force per unit area) nor reducing it (which, as noted, increases the normal force per unit area) ends up changing the overall frictional force at all.
What lugged tires do do, of course, is dig into a soft surface (such as soil) slightly, so that you have mechanical coupling between tire and surface (almost like rack and pinion gears), not mere friction. (But on hard pavement, the lugs don't accomplish a thing, other than making the ride rougher and maybe leaving marks on the pavement.)
ALso, there's nothing magical about putting all the weight over the rear wheels. Plenty of tractors (such as this one, on our tractor page, or this one, have front and back wheels the same size. Also, as tractors have gotten more powerful (and as the second picture clearly shows), on big tractors the tires are often doubled, for even more traction. --Steve Summit (talk) 23:30, 22 May 2007 (UTC)
Lugged tyres "dig in" (your words) precisely because the lugs concentrate the normal force on a small surface area (the outward-facing portions of the lugs). And you'll note that I did not mention friction at all. It is precisely the mechanical coupling (mentioned by you) that makes it work, but the large size of the lugs allows the distribution of the tangential force over a sufficiently large area of soil to absorb that tangential force (and couple it to the rest of the soil) without throwing clods of soil about.
Atlant 11:57, 23 May 2007 (UTC)
It's not about traction - it's about reducing the ground pressure so you don't compact the soil in the fields. The larger the wheels - and the more of them, the more square inches of tire is one the ground spreading the load. Indeed this will improve traction and help you to avoid getting bogged down - but doubled tires are complete overkill for that...no, it's all about reduced ground pressure. SteveBaker 01:30, 23 May 2007 (UTC)
Steve, I've personally seen a John Deere 5050 with doubled rear tires stuck to its axles in a muddy field. True, the driver was an idiot, but it can happen. This was in a pea field in Illinois in about 1969. For maximum freshness and sweetness, peas must be harvested within about a two-day window. Different fields are planted at slightly different times, and the harvest moves south-to-north, so the whole harvest takes place over about a six-week period, but any given field has a two-day window. This means that you cannot wait for the field to dry out, and you take risks with muddy fields. The big pea combines are owned by the packing companies and move from field to field. The tractors were leased from local farmers for the season. (I think modern combines are self-propelled.) Crews are were-sub-minimum-wage seasonal workers including poor college kids. We got that particular tractor and combine out without using a bulldozer (we used three other tractors instead) but is was a very near thing. -Arch dude 11:31, 23 May 2007 (UTC)

## Bucklling

what is the reason of buckling of structures?125.63.107.131 13:33, 21 May 2007 (UTC)

The article on Buckling explains it. -- Diletante 15:09, 21 May 2007 (UTC)
Others might say it was the illuminati, but I think such conspiracy theories are out of the mainstream (not to mention, riddled with factual inaccuracy and POV). Nimur 15:31, 21 May 2007 (UTC)

## Moments

Can anyone explaing the concept of moments in physics and mathematics. Like second moment of area, third moment of mass etc? Thanks. And why are they called moments?

Have you checked moment (physics) and moment (mathematics)? --Kainaw (talk) 15:07, 21 May 2007 (UTC)

"The n-th moment of a distribution is the expected value of the n-th power of the deviations from a fixed value."[2].
It's an obscure etymology; a lot of words in physics, such as Action (physics), have evolved from historical meanings and are now fairly different from the modern meaning of the word. "Moment" has more to do (linguistically) with "momentum," though you shouldn't let this word-quibbling be any issue. From the physical standpoint, each moment is just a mathematical operation to describe an object.
The various moments are merely a way of constructing a more complete description of a function. Mathematicians have come up with many ways to approximate an entire description of an object (such as its density, which is a continuous function) into one or more numbers (such as its total mass, and its moment of inertia). This saves the headache of using large integrals at later stages of the problem by creating descriptive "constants" (which are the results of simpler integrals). Put another way - a continuous function has an infinite number of values ("density is defined for every point in space", while most problems are easily solved with simplifications ("total mass is 5 kilograms"). Computing higher-order moments allows as much complexity as you are willing to compute, without resorting to the continuous-function approach.
The definition of a moment is given at Moment (mathematics); a specific case of first-order moments is available at Moment (physics). Each of these can be thought of as a specific case of the more general kernel integral.
Hopefully this math has not scared you off... the purpose of computing such integrals is to reduce the amount of work in harder problems by defining continuous functions in terms of their moments. Nimur 15:08, 21 May 2007 (UTC)
Hmm I will need a few moments to digest all that. THanks

## Carbonic acid & stomach acid

Does drinking carbonated drinks lower the pH of stomach acid/bile?--69.118.235.97 14:52, 21 May 2007 (UTC)

Probably not much; if anything, it most likely dilutes the stomach acid. Of course, the system is not static - as soon as you ingest anything, the digestive system may begin to deliver additional digestive chemicals as a result of the stimulus. This would change the results dramatically as external reactants are added. I think you were probably asking about the restricted case, where no new stomach-acid is created. If you have access to a chemistry lab, you could easily perform a controlled experiment - simulate the gastric acid with a small amount of hydrochloric acid and add a small amount of carbonated beverage, and measure the pH change. Be sure to follow safe lab procedures and obtain permission if necessary. Nimur 15:13, 21 May 2007 (UTC)
This page gives the pH of several soft drinks. Coke, Pepsi, and a generic store brand cola are all reported at about 2.5; the remaining non-cola beverages range upward from there, though none have a pH higher than about 4. Our article on gastric acid ('stomach acid') puts the normal pH in the stomach at between 2 and 3. (Acid secreted by parietal cells in the stomach can be as concentrated as pH 0.8, but this is quickly diluted by liquid present even in the 'empty' stomach.)
So, cola beverages are at about the same pH as gastric juice. Adding cola to the stomach will leave the pH relatively unchanged, as it has the same hydrogen ion concentration. Other soft drinks may temporarily raise the pH, as they are less acidic than the colas. The stomach will compensate for this dilution by secreting more acid to restore the 'normal' low pH. TenOfAllTrades(talk) 15:35, 21 May 2007 (UTC)

Another question. Why are diet colas and non-colas less acidic than colas? bibliomaniac15 23:58, 21 May 2007 (UTC)

I wasn't aware that they were, but I could speculate as to why this might be the case for diet colas. Sugar is used in regular colas to counter the sour taste from the acid. If you want to avoid using sugar, then it makes sense to use less acid, as well. This is especially true since you need to restrict the quantities of most artificial sweeteners, or they will give off a distinct "chemical taste". StuRat 07:21, 22 May 2007 (UTC)

## Tension

Every once in a while I get a pulling feeling on the top of my head? ???Nick 15:55, 21 May 2007 (UTC)

First off, in all seriousness, are you really coming to wikipedia help desk for medical advice? If you think you have a serious problem with headaches, see a doctor. Before that, ask yourself: Are you sure there is not a hook on top of your head attached to a pulley?Gaff ταλκ 17:21, 21 May 2007 (UTC)

Go to a doctor and say "I think I have headache number 14, can I have some asprin?" :] Actually you probably should tell him your symptoms, not just take asprin all the time, but... :) HS7 19:50, 21 May 2007 (UTC)

## planetary conjunction

Hi. I heard about the near-conjunction of Saturn and Venus coming up this canada day, and I wanted to know if I should be able to see both planets in the fov of my telescope. If you need some specifications, here they are:

• aperture: 50mm
• focal length: ~600mm
• magnification: ~<50x
• field of view: uncertain, about 30 arcmins
• eyepiece fov: uncertain, about 30°
• magnitude capability: uncertain, about mag. +10
• type: non-acromatic department store ( :( ) refractor

So, will I be able to see both planets through the ~50x eyepiece? What about the ~35x eyepiece? Thanks. -- AstroHurricane001(T+C+U) 16:12, 21 May 2007 (UTC)

If you can successfully locate Saturn, you will probably be able to view it with a weaker telescope than even that. Provided that light pollution is low, you can even see Saturn with the naked eye. Our article mentions that it is the farthest planet visible to the unaided eye. So, your only task will be locating their position in the sky. You might consider a free astronomy software; Stellarium is often recommended, but I find CyberSky to be a bit more "chart-like." Stellarium is free and open source; CyberSky has a free trial version available. Nimur 17:35, 21 May 2007 (UTC)
Hi. Thanks, but I don't need planetary software. Usually, I can locate planets just fine. I've seen Saturn's rings through my telescope, so I can locate it just fine. I've seen Venus for a long time now in the west, so I should know where the 2 planets are during near-conjunction. If I don't, I could just check an online star chart (I know where to find one). Oh, and here's an example of a free online sky chart (no purchase or download required): [3], among many more. My place has low enough light pollution so that I can see to about mag.+5 with the naked eye. What I'm asking is, will I be able to see both planets in my FOV of the telescope, so I wouldn't have to move the telescope to locate only one planet at once, I'm asking if I will see both planets at once, in the eyepiece, during near-conjunction. Is it possible to calculate whether or not I will see them in my FOV at once given those parameters, or do you need more? Brightness or ability to locate shouldn't be a problem. Thanks. -- AstroHurricane001(T+C+U) 18:26, 21 May 2007 (UTC)
Why don't you look at the sky tonight, find two stars that barely fit in the FOV and then look the stars up, and calculate your actual FOV from that. Then you can predict whether the near conjunction will fit in that fov. -- Diletante 18:59, 21 May 2007 (UTC)
Do you have the data about how close they will appear? My advice would be just try, Saturn and Venus should be spectacular this close even to the naked eye. Vespine 22:30, 21 May 2007 (UTC)
Conjunction says 46' will be the closest they come, which is close but sounds like no luck. Still, good luck. Eldereft 12:27, 22 May 2007 (UTC)

A hypothetical question that has been floating around the Internet for some time goes as follows:

Imagine a regular passenger jet, a 747 if you will. This plane is in normal condition.
This plane is placed on a giant treadmill (approximately the size of a normal runway).
The treadmill is connected to a computer which will measure the forward speed of the plane
and make the treadmill move at that exact speed in the opposite direction.
The net effect is that the plane will (supposedly) stay still in relation to the air.
Now when the plane starts to take off, will it rise into the air or will it stay still
due to the treadmill moving it backwards?
Assume that the treadmill can go at any speed, and that its movement is not limited by
normal mechanics.
Note that the engines (not the wheels) are what power the plane.


- 2-16 17:04, 21 May 2007 (UTC)

How can it take off? It's just standing there. There's no air flowing over and under the wings. Clarityfiend 17:11, 21 May 2007 (UTC)
Check out the discussion in an earlier reference desk question: [4], search for "conveyor". Weregerbil 17:12, 21 May 2007 (UTC)
The treadmill is irrelevant. As you said, the jet engines push the plane forward- it doesn't have drive wheels. Friday (talk) 17:16, 21 May 2007 (UTC)
The difference in air pressure above and below the wing is what creates lift. If the plane is not in motion relative to the air, then there is no lift generated and the plane stands still, with engines blaring. Read up on basic Aeronautics and the Science of Wings. Note that both of these articles would benefit from a drawing explaining wing dynamics.Gaff ταλκ 17:19, 21 May 2007 (UTC)
So long as there is sufficiently high friction between the plane and the treadmill, it won't go anywhere. If friction is low enough, on the other hand, the plan will slip along the surface and still take off. For amounts of friction inbetween these critical values, the plane will slip along the surface but not reach take-off velocity. Assume the plane must achieve an acceleration "a" to reach take off velocity by the end of the runway, and its mass is "m," then the force provided by its engines is "F." In this case, the static coefficient of friction must be less than than F/(mg), where "g" is theacceleration of gravity for the plane to move. The kinetic coefficient of friction must be less than (F-ma)/(mg) for the plane to take off. Note here that I've ignored thee fact that any motion of the plane will cause some lift, and lower the friction between it and the treadmill. I also ignored rolling friction, would would tend to raise it. I'm not sure off the top of my head how to account for these in making the calculations. Someguy1221 17:24, 21 May 2007 (UTC)
Oh, I also assumed the treadmill itself is massless and internally frictionless, but who wouldn't? ;-) Someguy1221 17:25, 21 May 2007 (UTC)
When discussed at the Straight Dope a few years back, the core contention lies with the wording of the question, not the question itself. Friday summarizes the issue above quite well with his answer: so long as you strip out any cleverly-worded paradoxes, the plane takes off. — Lomn 17:45, 21 May 2007 (UTC)
To summarize that, if the treadmill is free to move, the plane will still take off. If the treadmill is moving with constant velocity, well, the plane is already on a giant spinning sphere, will no ill effect on its ability to take off. In my answer above, I essentially assumed the treadmill to be accelerating, always moving as fast as the airplane is trying to. Someguy1221 17:59, 21 May 2007 (UTC)
What does "moving as fast as the airplane" mean? The airplane (meaningfully) moves with respect to the air, not the ground. Either the treadmill "moves as fast as the plane", in which case the wheels spin at double speed and the plane takes off, or the treadmill "moves fast enough to keep the plane stationary", in which the plane doesn't move. They are not, however, the same question (though the question as worded above attempts to posit that they are) -- this is the core of the paradox that the Straight Dope points out. — Lomn 18:36, 21 May 2007 (UTC)
My original answer was meant to determine whether the plane could take off in some fixed distance, as I assumed the treadmill is only as long as the original runway, and we assume something nasty happens if the plane leaves the treadmill and is still on the ground. But my answer (don't ask me why I imagined it this way) assumes the treadmill to begin stationary and then accelerates backwards. For any speed the plane's wheels might be turning, it can't move forward if the treadmill is spinning fast enough unless it slips. I realize this is a bizarre way of thinking about it, and I still don't know why I figured it would be this way. Someguy1221 19:19, 21 May 2007 (UTC)
Seems to me that no matter what else is involved the lift of the plane is determined by the velocity of the wings relative to the air around them. So regardless of how much force is output by the engines, if the wings remain stationary relative to the air then the plane will not take off. Someguy above mentions that the forward velocity of the plane relative to the air, and hence the velocity of its wings relative to the air, correspond to the various coeficiencts of friction in and between the wheels and the treadmill. But either way the only way for the plane to achieve lift is for the wings themselves to move relative to the air, so unless the engines are putting out enough force to make the plane move horizontally despite it being on a treadmill it won't take off. (In other words, at least initially the engines will expend their energy just making the plane's wheels rotate really, really quickly. However, if it can put out enough force that the wheels start accelerating faster than the treadmill can accelerate to keep pace, then the plane will start moving horizontally. If the horizontal velocity in that case passes the critical velocity for take off, then the plane can take off.) Dugwiki 18:03, 21 May 2007 (UTC)
Of equal significance to the question itself, is the question 'Why does this draw anything more than a 30 second analysis and dismissal from so many people?'. It is the human phenomenon of 'armchair engineering' taken to new levels by the ubiquity of the Internet, that allows ideas like this to live in flame-war infamy. Anyone with the tiniest shred of critical thinking, given time *on their own* to ponder this, can easily come to the correct conclusion. What is it about the internet that turns people, for lack of a better word, 'dumb', and keeps topics like this quite absurd plane and treadmill theory afloat? --66.195.232.121 21:05, 21 May 2007 (UTC)
Why not concoct bizarre interpretations if you have the time? Obvious answers are boring :-p Someguy1221 21:30, 21 May 2007 (UTC)
Yeah! Maybe we should be thinking of a treadmill with respect to however the entity normally moves...planes fly, so its "treadmill" is really a wind tunnel. A plane with engines running same speed as the air is circulating (or the computer controls the air to move to match the engine speed), so the plane stays motionless horizontally, but rises vertically. Assuming its flaps and other control surfaces are set appropriately! So maybe the "trick" of the question is that everyone is assuming the plane is horizontal and not pointed nose-down and starting in the air: treadmill runs, plane buys the farm perpendicularly to the treadmill motion. DMacks 21:37, 21 May 2007 (UTC)
What I don't understand about this question is how the plane would be able to accelerate. A plane does not have the engine connected to the wheels like a car, instead, it moves on the ground by "grabbing the air". That means that (in the ideal case) the wheels of the plane can rotate freely independently of the movement of the plane itself, so whatever the conveyor belt does, it won't have any effect on the movement of the plane. If the plane starts to move using its engines, the conveyor belt cannot keep it stationary; all it can do is rotate the wheels of the plane faster and faster. – b_jonas 12:14, 22 May 2007 (UTC)
The conveyor belt moving at constant speed only serves to raise the minimum speed the wheels need to turn, yes. Imagine yourself in the plane, staring at the conveyor belt, and ignoring the rest of the airport. From your perspective, you're standing still, and there is a light breeze moving in the same direction the plane is pointing. From your perspective, all you have to do is move the plane a little faster to take off. For anyone outside, the plane is actually moving the normal speed at take off, and the wheels are simply rotating faster. Someguy1221 19:11, 22 May 2007 (UTC)

## stick towers

when challenged to build the highest tower that can support itself with things like paper, spaghetti etc and some something to hold the supports together tape, blue tack , even seen marshmallows once. what is the best shape to build it? assuming you've only got a limited amount of materials, and if the challenged was changed say to build it to a certain height but then be able to hold the most weight at the top, how would the shape have to change?--137.205.79.218 17:19, 21 May 2007 (UTC)

If there was one "best design", then all buildings would look the same. The available materials, building environment, budget, and goals will all need to be taken into account. Without any details into those, I'd suggest looking at the Pyramid shape, it's naturally stable and able to hold large loads, but it really depends on a bunch of factors you haven't provided. - CHAIRBOY () 17:28, 21 May 2007 (UTC)
• I think that you should build it out of brownies. Because then, even if its not the best one in your physics class, at least you get to eat the brownies!Gaff ταλκ 17:36, 21 May 2007 (UTC)
I say, why bother coming up with best design, when someone's already done it! My bet would be to take a look at nearby electrical transmission tower and build a similar support scheme. Someguy1221 17:37, 21 May 2007 (UTC)

Make origami boxes. They are very strong, made of paper, and very quick, especially if you are my sister (who can make one in a few seconds)

Bonus point: If you use more than one type of material, you could make a case for describing it as 'composite construction'. The school project you're doing this for will probably award points for being able to authoritatively and accurately use a term like that. See Composite construction for more info. - CHAIRBOY () 20:20, 21 May 2007 (UTC)

The sphere is naturally a very sturdy shape, but it won't do you much for height. Maybe try a bullet/egg shaped structure, like that building in dubai? -- Phoeba WrightOBJECTION! 12:06, 22 May 2007 (UTC)

## Weird unusual bays and lakes and islands etc

Hi. I've been pondering this for a long time, and can not seem to figure this out. There are many bays with unusual shapes, for which I can't find an explanation to other than they are volcanic calderas, meteorite craters, or collapsed mountains, etc. Is there a possible explanation for these? Here is the list:

• Lake of the Woods, canada
• Death Valley, usa
• Great Salt Lake, usa
• Yerba Buena island, usa
• Lake Okeechobee, usa
• Bahía de Campeche, mexico
• Isla de la Juventud, cuba
• Grand Cayman, uk
• Turks and Caicos Islands, uk
• Lago de Nicaragua, nicaragua
• Laguna Caratasca, nicaragua
• Golfo de Panamá, panama
• Lago de Maracaibo, venezuela
• Lago Titicaca, peru
• Lago de Poopó, bolivia
• Guanabara bay, brazil
• Gulf of Cádiz, spain
• Golfe de St-Malo, france
• Lough Neagh, uk
• Isle of Man, uk
• Tyrrhenian Sea, italy
• Krim, ukraine
• Cyprus, cyprus
• Sea of Maramara, turkey
• Van Gölü, turkey
• Ozero Balkhash, kazakhstan
• Ozero Baykal, russia
• Ozero Issyk-Kul', kirgyzstan
• Lake Tiberias, israel
• Daryācheh-ye Orūmīyeh, iran
• Dasht-e Lut, iran
• Dasht-i-Margo, iran
• Kathiawar, india
• Sri Lanka, sri lanka
• Ebinur hu, china
• Turfan Depression, china
• Qaidam Pendi, china
• Qinghai hu, china
• Bo hai, china
• Hainan dao, china
• Hong Kong, china
• Borneo, indonesia
• Sulawesi, indonesia
• Biwa-ko, japan
• Tōkyō-wan, japan
• Ō-shima, japan
• Erg Chech, mali
• Tanezrouft, algeria
• Lac Mai-Ndombe, zaïre
• Lake Victoria, tazania
• Lake Ngami, botswana
• Gulf of Carpentia, australia
• Lake Erye, australia
• Tasmainia, australia
• Mount Waesch, antarctica
• Berkner island, antarctica
• Alexander island, antarctica
• Mount Terror-Erebus island, antarctica

So, what are some poissible causes? Sorry about the length. Thanks. -- AstroHurricane001(T+C+U) 17:24, 21 May 2007 (UTC)

Can you possibly expand on what you mean by "unusual"? A cursory glance doesn't really show me an obvious connection between the items in the list, and many should have articles noting their origins (Grand Cayman is an atoll and Mount ErebusMount Terror is a volcano). Also, please consider reducing the size of your sig; it runs across 6 lines (390 characters) in the edit window on my system. — Lomn 17:37, 21 May 2007 (UTC)
You might want to read about geology, geography, oceanography, the littoral zone, coastlines, bays, erosion, cartography, ice sheets, glaciation, sedimentary rocks, island formation, vulcanology, calderas, and meteor craters. Nimur 17:39, 21 May 2007 (UTC)
Hi. Thanks for your help, what I meant by "unusual" was that many of the places seem to be created by the causes above, is this sig better? Thanks. -- AstroHurricane001(T+C+U) 18:51, 21 May 2007 (UTC)
I can assure you that of your list, at least Lake Nipigon, canada , Lake of the Woods, canada ,Death Valley, usa ,Great Salt Lake, Yerba Buena island, usa, Lake Okeechobee, usa, Grand Cayman, uk , Turks and Caicos Islands, uk, Golfo de Panamá, panama , Lago de Maracaibo, venezuela , Lago Titicaca, peru, Guanabara bay, brazil , Gulf of Cádiz, spain , Golfe de St-Malo, france , Lough Neagh, uk , Isle of Man, uk , Tyrrhenian Sea, italy , Krim, ukraine , Cyprus, cyprus , Sea of Maramara, turkey, Ozero Balkhash, kazakhstan , Ozero Baykal, russia , ,Ozero Issyk-Kul', kirgyzstan , Dead sea, israel , Lake Tiberias, Israel, Dasht-e Lut, iran, Sri Lanka, sri lanka, Turfan Depression, china, Bo hai, china , Hainan dao, china , Hong Kong, china, Borneo, indonesia , Sulawesi, Indonesia, Lake Chad, chad, Lake Victoria, tazania , Lake Ngami, botswana , Gulf of Carpentia, australia , Lake Erye, australia , Tasmania, australia

are not significantly affected in their geographic shapes by volcanic calderas or meteorite craters. Likely many others as well. Hope this helps - cheers Geologyguy 21:58, 21 May 2007 (UTC)

Hi. Thanks for such a long list, I've read through most of it, and I appreciate it. Thanks. -- AstroHurricane001(T+C+U) 00:36, 22 May 2007 (UTC)

I'm not sure if any of those qualify, but lakes near rivers in roughly semi-circular arcs are often oxbow lakes. StuRat 07:07, 22 May 2007 (UTC)

The original poster wrote "There are many bays with unusual shapes". While oxbow lakes are interesting I don't think this addresses the question. Try endorheic basin. Certainly some you mention are in that category. David D. (Talk) 07:22, 22 May 2007 (UTC)

## Sirius and Procyon

Hi. I observed this phenomenon over a year ago, but I want to ask now. When I pointed my telescope towards Sirius and Procyon (see specifications above), I noticed a dim star beside each. There weren't any stars as close or as bright near these stars. I highly doubt they're Sirius and Procyon B, because they are too close and dim for my 50mm 'scope to observe, and they appeared like ~mag.7 anyway. Also, I looked it up on this website, and it seems that I found what those stars actually were. However, it seems too much of a coincidence, so I'm not really sure. So, can you tell what it was? Do you think it was the stars I saw at that website? It also seems too much of a coincidence that the star near Procyon was dimmer than the one near Sirius (if I remember correctly). What do you think? Thanks. -- AstroHurricane001(T+C+U) 17:33, 21 May 2007 (UTC)

Can you find us the address of the relevant frame? Fourmilab is a big site and all you've given us is a link to the container page. (Er, what's the usual word for a webpage that has frames?) —Tamfang 20:22, 21 May 2007 (UTC)
...sucky... SteveBaker 21:45, 21 May 2007 (UTC)
In a web development class we recited a phrase: "tables good, frames bad." [Mac Δαvιs] ❖ 22:44, 21 May 2007 (UTC)
They're often called framesets, just like the HTML tag. --Tardis 23:00, 21 May 2007 (UTC)
Could your extra stars be caused by internal reflections in the glass in the telescope? THis effect occurs when you look at a bight star through a window. GB 03:57, 23 May 2007 (UTC)
Not just that, some really bright objects (Sirius and Procyon are both pretty bright) could easily be reflected from your eye back onto the eyepiece, then reflected again back into your eye. I've certainly had this through a larger telescope, but I'm sure that a 50mm one could give the same effect. Richard B 09:14, 23 May 2007 (UTC)

difference between deadlock,livelock and indefinite postpondment? —Preceding unsigned comment added by 59.93.195.191 (talkcontribs) 19:16, 21 May 2007

Try starting with deadlock. Please remember that the ref desk will not do your homework for you. — Lomn 19:22, 21 May 2007 (UTC)
For extra credit, also read and understand priority inversion.
Atlant 21:41, 21 May 2007 (UTC)

## Eggs

138.89.123.95 20:57, 21 May 2007 (UTC) I found a grayish colored egg in a shallow hole in my garden mulch. I live in NJ. It is the size of a chicken egg. Does anyone know what kind of egg this could be?

                         Robin


i have a pet turtle and i don't know what kind of turtle it is

Could it possibly be a Red eared slider? Edison 00:47, 24 May 2007 (UTC)
Fascinating! Nimur 22:26, 21 May 2007 (UTC)

Is the egg hard or leathery ? Typically birds have hard, brittle eggs while reptiles have leathery eggs. StuRat 07:01, 22 May 2007 (UTC)

## "Hydrogenated" vs. "partially hydrogenated" fats

If an ingredient of a commercial food is labeled as "hydrogenated fat" (or "hydrogenated oil"), what does that mean exactly? What fraction of the fatty acids is still allowed to be unsaturated, and at what fraction of unsaturated fatty acids are the manufacturers required to call it "partially hydrogenated"? Icek 22:25, 21 May 2007 (UTC)

There's a slightly technical explanation at hydrogenation, and a bit more at fatty acid and saturated fat. I've heard more of polyunsaturated, monounsaturated and saturated fats, which I believe roughly correspond to unhydrogenated, partially hydrogenated and hydrogenated, and it looks (according to the saturated fat article) like fat is hydrogenated (saturated) to increase shelf life and alter texture, but the unsaturated fats are the ones that are meant to be better for you. Confusing Manifestation 22:39, 21 May 2007 (UTC)
Mostly correct. Executive summary: hydrogenation refers to the chemical addition of hydrogen atoms to carbon-carbon double bonds (i.e. alkenes) in the fat molecule. Hydrogenation is a molecular property rather than a bulk property - in other words "partially hydrogenated" means that the individual fat molecules in a sample have had hydrogens added to some (but not all) of their carbon double bonds. "Poly-"unsaturated means that a molecule has multiple carbon-carbon double bonds; "mono-" and "un-"saturated are exactly what they sound like. In theory, one could have a partially-hydrogenated, poly-unsaturated fat (hence the minor quibble), but I'm not sure I've ever heard of one.... -- MarcoTolo 23:00, 21 May 2007 (UTC)
Thank you for trying to answer. The terms "partially hydrogenated fat" and "hydrogenated fat" do not describe a certain chemical makeup, but they describe more or less the process by which the fats where produced. Butterfat has got a large fraction of saturated fatty acids (SFA) although it's not a "hydrogenated fat". It's true that hydrogenation is used to increase shelf life, and, for most purposes, unsaturated cis fatty acids seem to be healthier than SFA. However, some trans fatty acids (TFA) seem to be even unhealthier (i.e. stronger causative to cardiovascular diseases) than SFA. TFAs occur in fats derived from ruminants like butterfat (about 4% TFA); but a significant fraction of the ruminant TFAs are conjugated fatty acids, and conjugated TFAs seem to be not unhealthy. The other major source of TFAs in human diet are "hydrogenated" and "partially hydrogenated" vegetable fats, the cis-trans-isomerization is a side effect of the hydrogenation process. If a fat really were fully hydrogenated, i. e. all unsaturated fatty acids were transformed to SFAs, there couldn't be any TFAs in this kind of fat; however, looking into the USDA food database examples of "hydrogenated" fats still contain large amounts of unsaturated fatty acids, and therefore likely contain relatively large amounts of TFAs.
That's the background of my question; my question is: what are the legal regulations for calling an ingredient "partially hydrogenated" or "hydrogenated" in the USA and in the EU? Icek 23:15, 21 May 2007 (UTC)
In the US, the FDA requirement is listed here; I wasn't able to dig-up the EU definitions. -- MarcoTolo 00:08, 23 May 2007 (UTC)
Thank you. But the text says that it has to be labeled as "partially hydrogenated" if it is partially hydrogenated and as "hydrogenated" if it is completely hydrogenated. The USDA nutrition data says that a "hydrogenated" fat contains plenty of unsaturated fatty acids. Icek 09:05, 23 May 2007 (UTC)
Maybe the issue is that you expect government agencies to agree with one another.... <grin>. Can you give me a direct USDA link? I've tried searching via the USDA form but haven't come up with the same page you're seeing. -- MarcoTolo 20:45, 23 May 2007 (UTC)
Unfortunately a direct link is not possible, but detailed instructions are: Go to this page, enter "fat hydrogenated" as keywords, search in all groups. Then you'll find several entries, some "partially hydrogenated" and some just "hydrogenated". For example, select "Oil, vegetable, industrial, canola (partially hydrogenated) oil for deep fat frying" and click on "Submit". Select 1 x 100 g and click on "Submit". Then you'll see a table ... scroll down to the lipids, where you'll see information about total SFA content, content of some specific fatty acids or specific fatty acid groups etc. Here the total SFA content is 10.1 g, as opposed to 7.1 g which you would get when searching for "vegetable oil canola" - something you would expect from a partially hydrogenated oil.
Now go back to the search results page and select a fat or oil which is just "hydrogenated", e. g. "Oil, industrial, palm kernel (hydrogenated), filling fat". This one contains 88.209% SFA, 5.705% monounsaturated fatty acids (MUFA) and zero polyunsaturated fatty acids (PUFA). The MUFA fraction contains 4.658% (of the total) of TFA.
5.705% is maybe not that much, but it gets more extreme if you search for "oil hydrogenated" instead of "fat hydrogenated": Select "Oil, soybean, salad or cooking, (hydrogenated)" - that one has actually only 14.9% SFA (here there is no data for TFA however). I think that the "hydrogenated" oils with little SFA are mislabeled, while the "hydrogenated" fats with only 5 or 6 % of non-SFA are not, as any cheap industrial hydrogenation probably leaves about that amount of non-SFA. But how much unsaturated fatty acids are allowed for "hydrogenated" oils or fats? Icek 02:50, 24 May 2007 (UTC)

## Microwave Auditory Effect - Quack Science?

This article was published in the Journal of Applied Physiology in 1961. The author, a scientist from Cornell and General Electric, claims to have documented and quantified human auditory system response to modulated electromagnetic energy. Is this quack science? Have there been follow-up experiments? Were the energy levels involved unsafe, and is this the reason why further work does not seem to be very common? Any other comments or suggestions? We have this article on microwave auditory effect, which doesn't seem to settle the issue. Nimur 22:44, 21 May 2007 (UTC)

"...produced experiments that demonstrate that materials as commonplace as aluminum foil, thin wires, pine needles, and wire-framed glasses can act as suitable transducers." This sounds like a tin-foil-hat-wearing conspiracy theory. Nimur 22:48, 21 May 2007 (UTC)
If that is true, then the article is likely a joke article. I've seen joke articles before in journals. It's not meant to be serious. [Mac Δαvιs] ❖ 22:58, 21 May 2007 (UTC)
Google Scholar has a few articles that cite this lead. DMacks 22:51, 21 May 2007 (UTC)
Clearly we can't directly hear microwaves - but it's maybe possible that the microwaves are having some kind of effect on other materials - which then vibrate producing sounds that we might be able to hear. We know that a microwave oven can heat things up - perhaps a modulated microwave signal might heat something up a teeny-tiny bit - then let it cool off a tiny bit - over and over. The object might then expand and contract in response - which could cause a sound to be emitted. Seems like a long-shot though - it would need to be a really powerful microwave signal and a really feint sound. I wouldn't dismiss the idea out of hand - but then if this theory is true then it's not a particularly exciting phenomenon anyway. SteveBaker 23:40, 21 May 2007 (UTC)
See rectifier or demodulator
Before ascribing it to quackery, I'll note a similarly little-known effect: if your eyes have adjusted to the dark, you can see x-rays. It's not known how this works, and because it's unsafe to try in humans (and presumably not particularly important), there hasn't been much of an inquiry. This case could be a parallel, I suppose. Link. --TotoBaggins 02:51, 22 May 2007 (UTC)
I see no reason that the human auditory system could not act as a detector of electromagnetic waves. In the early days of radio there were a number of wet (chemical) detectors. Note Galvani's finding that frog muscle tissue responded to weak DC voltages. Also any imperfect metallic contact (such as a broken filling in a tooth) should function as a detector of modulated AM radio waves, rectifying the radio wave. Human ears and even skin have reportedly been able to sense audio frequency signals as well, as was reported in scientific publications in the late 19th and early 20th century. Edison 15:27, 22 May 2007 (UTC)
This is a real effect. There's at least one standard relating to pulsed RF systems that sets the maximum RF level at below that which generates an auditory response. The reason for this based on the principle that any physiological response is a bad thing so being able to hear RF should be avoided.
I can't actually remember which standard though, I'll try to look it up tomorrow. JMiall 22:59, 22 May 2007 (UTC)
Here's a highly used document that mentions the auditory response several times ICNIRP GUIDELINES FOR LIMITING EXPOSURE TO TIME-VARYING ELECTRIC, MAGNETIC, AND ELECTROMAGNETIC FIELDS (UP TO 300 GHz) using it to set exposure levels. JMiall 16:59, 23 May 2007 (UTC)

## paracetamol synthesis

paracetamol synthesis? The chemical formula is fairly short. [Mac Δαvιs] ❖ 22:59, 21 May 2007 (UTC)

According to this document from a chemistry lab, paracetamol is made from: p-aminophenol + H3PO4 + acetic anhydride. Rockpocket 02:08, 22 May 2007 (UTC)

## Ideal gas laws and molar weights

Okay, this is sort of homework even though I finished school a decade ago. This is a course I'm doing for fun (and it's getting distinctly less fun when I look at questions I know are easy but can't remember how to do).

The question is regarding Arsenic Sulfide as a gas it gives a mass (0.345g), temperature (650C), pressure (760mmHg why not an SI unit I don't know) and volume (60.5cm3) and asks you to calculate the molecular formula at that temperature 650C. My solution to this is to calculate the number of moles and hence molar mass. I do this: (I think this may be over complicating things and I should just be doing something with relative densities but I can't think what).

${\displaystyle pV=nRT}$
${\displaystyle n={\frac {pV}{RT}}={\frac {(760\times 133.3)\times (60.5\times 10^{-6})}{8.314\times (650+273)}}}$
${\displaystyle n=8\times 10^{-4}mol}$

Now going the number of moles, I calculate molar mass: ${\displaystyle M={\frac {m}{n}}={\frac {0.345}{8\times 10^{-4}}}=432gmol^{-1}}$.

This seems sensible to me, giving As4S4 (which would be 428).

Now what makes me think this method of calculation is wrong is that the next part gives a different Temperature/Volume (1573C/242cm3), and asks you to to state and explain what the molecular formula is (it comes out as As2S2 according to the method above, but if it's state and explain it should mean I don't need to calculate it and that there is some "obvious" reason why it is.

What other method could I have used? Caffm8 23:07, 21 May 2007 (UTC)

You may be over-working! I think your numbers come out right, and you appear to have correctly followed procedure. I'm searching for a physical/chemistry explanation about why Arsenic Sulfide molecules might be found in different combinations... our arsenic sulfide article does not mention a gaseous phase. Perhaps the "quick method" for "part 2" is just to compare the Volume/Temperature ratios, without recomputing the entire answer. Note that the V/T ratio is twice as large; so you expect a molar mass twice as high. Nimur 23:32, 21 May 2007 (UTC)
[5] indicates that different forms are found (at least in the solid phase); As4S4 and As2S3 seem prevalent, but other trace species are found as well. Nimur 23:37, 21 May 2007 (UTC)
I'm doubtful that you have to explain with regards to Arsenic Sulfide, these types of questions in my experience tend to just use random elements (something they can change slightly next year). They should all be gaseous at these sorts of temperatures. The point is that the molar mass changes, from AsS->As4S4->As2S2 during heating. But I'm not sure there is any general explanation (which seems to be what it's looking for) why it should go to As2S2. I'll probably just put that about the V/T ratios being different but it's not much different than just reworking the original equation. Basically the question is:
i) Calculate the molecular formula at Temp-1, ans: As4S4.
ii) State what it would be at T2 (As2S2), and explain how you arrived at this conclusion. (note the lack of calculating).
Seems to suggest there is some obvious explanation. Caffm8 23:49, 21 May 2007 (UTC)

Write the temperatures in K instead of C, and you can see the relationship to the first case you already solved. --Reuben 03:37, 22 May 2007 (UTC)