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May 26

Alka-Seltzer Neutralization

Hello. The chemical equation of Alka-Seltzer's effervescence is C₆H₈O₇ (citric acid) + 3NaHCO₃ → 3H₂O + 3CO₂ + Na₃C₆H₅O₇ (trisodium citrate). If I add hydrochloric acid to a solution of water and Alka-Seltzer, then is the neutralization chemical equation Na₃C₆H₅O₇ + 3HCl_(aq) → C₆H₈O₇ + 3NaCl? If so, then how does Alka-Seltzer neutralize the hydrochloric acid? Isn't C₆H₈O₇ an acid? Thanks in advance. --Mayfare (talk) 01:24, 26 May 2008 (UTC)[reply]

If dissolved Alka-Seltzer is trisodium citrate as you have written (I had no idea), then it can't neutralize HCl. I expect Alka Seltzer works because it contains less citric acid per sodium bicarbonate than you describe in your equation; perhaps it's more like 6 moles of bicarb per mole of citric? Woodlore (talk) 03:41, 26 May 2008 (UTC)[reply]

About 4 to one. The result is excess bicarbonate, and that buffers both excess acid or excess base:

HCO₃^-_(aq) + H⁺_(aq) → CO_2(g) + H₂O_(l)

HCO₃^-_(aq) + OH^-_(aq) → CO₃^2-_(aq) + H₂O_(l)

Sound right? So the effervescence is just a nice and flashy delivery system for bicarb. --jpgordon^∇∆∇∆ 04:12, 26 May 2008 (UTC)[reply]

Yup, as well the carbonic and citric acids makes taking a sodium bicarbonate solution far more palatable. Sjschen (talk) 23:52, 26 May 2008 (UTC)[reply]

Partial UV protection in sunglasses

When a sunglass or any other lens such as contacts offers partial, say 90%, UV protection, is it referring to 90% of the UV spectrum of the less energetic portion of the spectrum? In other words, if I wear a contact lens with 90% protection behind sunglass with 50% protection, am I exposed to .1x.5=5% of UV or closer to 10%? Thanks. —Preceding unsigned comment added by Imagine Reason (talk • contribs) 03:59, 26 May 2008 (UTC)[reply]

Most UV absorbing substances have strong absorption responses with relatively sharp cutoffs as a function of frequency, i.e. everything above wavelength X passes and nearing nothing below it does. As a result, my expectation is that the effects are not additive and that the cumulative effect will be closer to 10%. However, I have no specific experience looking at the frequency response of either sunglasses or contacts, so I could be mistaken. Dragons flight (talk) 04:14, 26 May 2008 (UTC)[reply]

Most sunglasses polarise light as well, reducing the overall intensity of the light, see polaroid--Shniken1 (talk) 12:31, 26 May 2008 (UTC)[reply]

Indeed - you take two sets of polarising sunglasses and put them one in front of the other and rotate one of them you'll find at one angle they block virtually all the light, since one only lets through light with one polarisation and the other the opposite polarisation, so no light can get through both. --Tango (talk) 14:13, 26 May 2008 (UTC)[reply]

The notion that each polarizer independently removes part of the light is flawed: by Malus' Law, some light does pass through two crossed polarizers so long as their transmission axes are not orthogonal. This leads to the counterintuitive result that, given two crossed polarizers passing little or no light, a third may be inserted between them at an intermediate angle and increase the transmissivity. It reduces the intensity, but the light that passes it is not so poorly aligned with the last polarizer and so survives with a larger final intensity. --Tardis (talk) 21:43, 28 May 2008 (UTC)[reply]

When I put two of my sunglasses one in front of another, I can still see, no matter how I rotate them. What gives? Imagine Reason (talk) 13:13, 29 May 2008 (UTC)[reply]

If you see no changes in brightness with rotation at all, then at least one of your two pairs is unpolarized. If you are rotating only one of the pairs and see only weak changes, then the one you are rotating is polarized and the other is not (and the light passing through them is already mildly polarized, as it often is). If you see strong changes, but never complete darkness, you have discovered that polarizers are not perfect: they always block some light in the "correct" direction and pass some light in the "wrong" direction. --Tardis (talk) 15:27, 29 May 2008 (UTC)[reply]

Oilfield Volcano

If say a volcano erupts under a oilfield, wouldn't it be possible that phenols be created and keytones be created, and then bisphenol a be created?

If say a volcano erupts under a oilfield, wouldn't it be possible that, phthalate be created?68.148.164.166 (talk) 02:26, 26 May 2008 (UTC)[reply]

The temperature our article gives for magma is 700°C to 1300°C the autoignition point given for bisphenol A is 600°C. You'd need just the right coincidental conditions to create it and then there's a question of whether it would be around long enough for anyone to detect it. (And they'd preferably have to stay alive long enough to tell anyone, with a volcano erupting under their pants and all.) That sounds like an awful lot of "what ifs" happening simultaneously. Lisa4edit (talk) 05:10, 26 May 2008 (UTC)[reply]

Thanks. What about phthalate?68.148.164.166 (talk) 06:51, 26 May 2008 (UTC)[reply]

Given enough heat and pressure, you'll get one of everything. As Lisa notes though, it will be largely decomposed by the time it gets to the surface - you'll only have to worry about the deadly carbon monoxide gas. And the sulfur dioxide. And the pyroclastic flows. There will likely be a few phthalates in there, but nothing to worry about. Franamax (talk) 11:31, 26 May 2008 (UTC)[reply]

as you may have gathered from the above responses, the short version is: if a volcano erupts anywhere, you have much bigger health concerns than BPA and phthalate, such as lava, fire, earthquakes, flying rocks and toxic fumes. --Shaggorama (talk) 19:12, 27 May 2008 (UTC)[reply]

Solar panels on the Moon

Every plan I've seen for long-term human activity on the Moon calls for using solar panels for power. But those solar panels would be exposed to solar flares, cosmic radiation, and other phenomena not present on Earth or in near-Earth orbit. What effect would those have on the solar panels? --Serie (talk) 07:35, 26 May 2008 (UTC)[reply]

If I'm reading the articles right, the geostationary orbit is above the Van Allen belts, so we already have many solar panels exposed to full intensity solar and cosmic radiation. Also the SOHO spacecraft is way out there, looking right at the sun, and it has solar panels [1], so they must work. Presumably though, the high-energy particle flux will progressively degrade the panels and eventually, like most other things, they would stop working and need replacement. Franamax (talk) 10:35, 26 May 2008 (UTC)[reply]

It's a feasible idea, theoretically. However, the main issue is down to the economics of it all. Solar panels in space aren't as efficient as those used on the ground and theres a huge cost involved in getting them into space and installed. Regards, CycloneNimrod^Talk? 13:53, 26 May 2008 (UTC)[reply]

Huh? Space-grade solar cells are more efficient than typical earth use ones (and much more expensive). Where would the lower efficiency come from? You would have a higher solar influx on a more efficient cell, does temperature variance make them that much more inefficient? Rmhermen (talk) 20:12, 26 May 2008 (UTC)[reply]

Remember efficiency is the amout of energy input that you convert to electricity (your output). It is easily possible I would suspect that current solar panels are significantly less efficient in space because they are generally designed for earth-like conditions (yes you would modify them somewhat for space but given the fact the field is of minute interest it's likely it still lags far behind earth-based solar panel design) including temperatures and light levels (amount and spectrum of light). However it does seem extremely likely that given the far greater amount of light reaching solar panels in space, they will generate a significantly higher amount of electricity per cell even though they are a lot less efficient at converting the solar energy that reaches them to electricity so I'm indeed not 100% sure if CycloneNimrod is correct. Nil Einne (talk) 20:22, 26 May 2008 (UTC)[reply]

In fact I think it is almost the opposite - that solar cells for space use led the design which trickles down to earth use. At least that is always the impression I was given.[2] Our article on Solar cells seems to support my memory as well. Rmhermen (talk) 20:47, 26 May 2008 (UTC)[reply]

This quote is the closest I have come to understanding the efficiencies: "The space solar spectrum or air mass zero (AMO) spectrum is richer in ultraviolet light than the typical terrestrial solar spectrum (air mass 1.5 or AM 1.5). The ultraviolet light is typically converted into electrical power less efficiently than other parts of the spectrum, resulting in lower AMO efficiencies. For example, production muInjunction solar cells are currently -28% AMO, but these cells are over 32% efficient in the AM 1.5 spectrum. Since cells are typically measured under the spectrum for their intended use and efficiencies are not easily converted..."[3] So I still don't know if a panel actually produces less power in space than on Earth. Rmhermen (talk) 21:09, 26 May 2008 (UTC)[reply]

The overall efficiency is a little less because there is more UV in the spectrum, but this is outbalanced by the far greater light flux (what with there being no clouds and all). This GaAs 3-layer solar cell for use in space rates 135 mW/sq.cm. at 28% efficiency, whereas this writeup for a terrestrial silicon cell works out to ~9 mW/sq.cm. Check my math on this, but there is a lot more sunlight in space. Franamax (talk) 22:26, 26 May 2008 (UTC)[reply]

Solar power is considered more likely at least because launching Radioisotope thermoelectric generators large enough to power a moon base is probably politically infeasible and perhaps environmentally questionable. Rmhermen (talk) 20:16, 26 May 2008 (UTC)[reply]

The 28 days of night on the moon make solar cells good for production of electricity, but the storage of the power is more important. The moon station at the north pole with sun at some peaks with no night are the place to go!--Stone (talk) 07:26, 27 May 2008 (UTC)[reply]

Just a nit, but would you believe ~14 days of daylight and ~14 days of night?

Atlant (talk) 14:33, 27 May 2008 (UTC)[reply]

I can't see why a solar flare would damage the panel. Solar flares can temporarily or permanently scramble digital systems; they can permanently affect analog system performance by introducing a total radiation dosage significant to affect the dopant concentrations; but solar panels are basically the simplest kind of semiconductor - just two layers of material which create current when light hits them. I can't think how a large blast of solar wind (even way beyond the normal levels) would permanently damage the panel. Nimur (talk) 15:09, 27 May 2008 (UTC)[reply]

In any event, solar panels on the moon are bound to work far better than would lunar panels on the sun. Edison (talk) 15:43, 27 May 2008 (UTC)[reply]

The data sheet I link above for the space-use solar cell has a specific section called "Radiation Degradation". High-energy particles will eventually disrupt the crystal structure of the material. How long it would take for 10E15 1MeV electrons to hit 1 sq.cm. of the panel and reduce its performance by 14%, I have no idea. Franamax (talk) 17:04, 27 May 2008 (UTC)[reply]

To what other uses can a water pump/ motor be put to?

I am a resident of New Delhi, India. Due to frequent low pressure of water supply in our pipeline, I bought a brand new self priming mono block pump/ motor a few days back. After I bought it, to my astonishment, water started coming at good pressure. Now it is lying idle.

(i) Can anyone please tell me as to what other uses can the pump/ motor be put to? (ii) If I want to use it as a shower to take bath, what (modifications, if any) should be done? Bye. Signed; Kvees. —Preceding unsigned comment added by 164.100.5.4 (talk) 08:16, 26 May 2008 (UTC)[reply]

One other use would be to set up a storage tank and attach a fire hose so you are prepared for emergencies. If your local water pressure is not enough to get a decent shower, you would need to add a pressure tank to control the pump. I'm not sure that connecting a pump to the waterline will help much when the line pressure is low, you might just cause cavitation in the pipeline. In any case, it sounds like installing a storage tank would be a good idea, waterline -> storage tank -> pump -> pressure tank -> house. Franamax (talk) 10:05, 26 May 2008 (UTC)[reply]

You could drop it from the top of a building and measure the time taken for the fall to calculate the height of the building. You'd need a stopwatch 125.21.243.66 (talk) 14:35, 26 May 2008 (UTC)[reply]

In Turkey they heat their water in tanks on top of the roof. They're just barrels painted black. (See Franamax's storage tank for details). You might try that. A garden water feature would also be possible, but those are rather a bit wasteful water wise (and waste power, too, now that I think of it.) Your low pressure might have been caused by an air-bubble lodged somewhere. Once it became dislodged the pressure went back up. I haven't encountered that with municipal water, but know of it with well water. Dropping a perfectly working object from a roof just to see it shatter should be punishable under some law. You could always see if there's a charity that would like to have it, if you don't want to keep it.71.236.23.111 (talk) 03:54, 27 May 2008 (UTC)[reply]

You might rent it out to other users. Maybe that would give you some extra money.--Lenticel ^(talk) 00:39, 28 May 2008 (UTC)[reply]

Vesna Vulović's fall duration (pt. 2)

Brought across from the misc helpdesk - How long did it take Vesna Vulović to fall the 10,000 meters? So far, the best guess is about 3 minutes. TIA. Lugnuts (talk) 09:20, 26 May 2008 (UTC)[reply]

Using a terminal velocity of 193 Km/h I got 2.77 minutes.61.68.255.246 (talk) 10:12, 26 May 2008 (UTC)[reply]

Vesna didn't "fall 10,000 m ". She was in a plane that traveled at an altitude of 10,0000 m when it disintegrated. As Algebraist already stated back there, there is no way of telling how much residual forward motion the part of the plane she was in had, nor what it's aerodynamic properties were (see Drag (physics). All we can say is that it probably took quite a bit longer than 3 min. One might get a certain upper range by applying average airspeed for that type of aircraft and treating the plane fragment as if it were a parachute, but it'd be mostly guesswork. --Lisa4edit (talk) 10:12, 26 May 2008 (UTC)[reply]

(That's a typo - it should be ~10,000 meters, not 10,0000! Minor typo, major implications!) Nimur (talk) 15:11, 27 May 2008 (UTC)[reply]

Why is the airspeed/forward motion important? It would make no difference to the time it takes to fall down, would it? (Unless the airspeed is so large as to make the earth curve under before the aircraft hits the ground)? -- ReluctantPhilosopher (too lazy to login) —Preceding unsigned comment added by 125.21.243.66 (talk) 14:34, 26 May 2008 (UTC)[reply]

As lisa explained, forward motion can generate lift. Almost any shape has at least some aerodynamic lift. This will increase the time to impact because the debris is still partially "flying". SpinningSpark 21:40, 26 May 2008 (UTC)[reply]

Oh yeah, I never thought about the lift due to forward motion. Silly me. 125.21.243.66 (talk) 14:12, 27 May 2008 (UTC)[reply]

stomach

why do our stomach move like a heart beat when we hold our breathe? —Preceding unsigned comment added by Miley 7 (talk • contribs) 09:28, 26 May 2008 (UTC)[reply]

What happens when you hold your breath is that you expand your lungs and tighten your diaphragm. This stretches the area around your stomach and now you can feel your pulse more easily. I don't think anything actually changes with how your stomach works, you can just more easily feel the arteries in that area. Franamax (talk) 10:14, 26 May 2008 (UTC)[reply]

What is SUTWELL Nursing Device?

Search on google and other search engines present SUTWELL as a nursing device. The question is: What really is SUTWELL device? Does it have a generic name? If it does not have a generic name can it be given one? What is the purpose of this invention? Who invented it? How far is it in the development process? Can it pass as another artefact to our stock becoming a household appliance like spoon, cup, knife, chair etc? —Preceding unsigned comment added by Sutwell (talk • contribs) 10:04, 26 May 2008 (UTC)[reply]

There's lots of useful information on the Sutwell device at your user page, I'd start there ;) Franamax (talk) 10:07, 26 May 2008 (UTC)[reply]

Only Black and White pictures from Mars

Why are the pictures take from new spacecraft such as Phoenix still in black and white (like [[4]]? I have serious doubt that it's due to them not being able to afford a color camera, so is there any other reason why a color camera wouldn't be able to operate on Mars? Is the data transfer rate to slow to support this, or are the technology involved not space-proof (doubtful)? 90.230.54.138 (talk) 11:48, 26 May 2008 (UTC)[reply]

The Mars rovers can see in 11 colours from the cameras made for colour vision, but they do this by using filters over the camera. I think the basic reason for B&W is that you can transfer 3 (RGB) to 4 (CMYK) times the information in the same time = higher resolution. Franamax (talk) 12:06, 26 May 2008 (UTC)[reply]

How long does it take to send an image through? Zain Ebrahim (talk) 19:59, 26 May 2008 (UTC)[reply]

A lot longer than you would on a dialup connection, according to this, except when an orbiter is overhead. Franamax (talk) 21:46, 26 May 2008 (UTC)[reply]

The reason is to get a higher resolution and a greater spectrum range. There is no such thing as a full-color sensor -- your color camera has red, green, and blue sensors in a pattern known as a Bayer array. This lets you take a full-color picture with just a single shot (essential for moving objects), but the interpolation used to give the appearance of full-color sensors makes things slightly blurry and sometimes causes colored fringes to appear that don't really exist.

The Mars landers and rovers use a single wide-spectrum sensor (sensitive to all light from mid-infrared to near-ultraviolet) and a collection of color filters that can be put in front of the sensor. If you want a color picture, you combine a red-filter picture, a green-filter picture, and a blue-filter picture. Since there's not much movement on Mars, this works well, and because there's no interpolation being used, there's no blurring or fringes.

Further, since the camera has a wide-spectrum sensor, it can take pictures using frequencies outside of the relatively narrow visible spectrum. Those "black and white" pictures you're talking about are probably actually near-infrared photos, because near-IR shows better detail in shadowed areas. By using combinations of infrared and ultraviolet filters, scientists can identify ice (ice has a distinctive reflection in certain IR bands), make estimates of mineral content of rocks, measure atmospheric dust, and do other things you can't do with a mere color camera. --Carnildo (talk) 23:42, 26 May 2008 (UTC)[reply]

Another note is that these images just arrived. Surely the first batch was a test to make sure the camera was operational after the descent and landing... the faster it could be transmitted to eager controllers, the sooner they could relax and get to serious work. Sending back a black/white image requires lower bandwidth ("smaller file") and thus can be transmitted in less time. Over the next few days, the ground team will have time for post-processing of the data that comes back, and we will certainly see re-colored images (combining the multi-band spectral photos into standard color or false color images). Nimur (talk) 15:15, 27 May 2008 (UTC)[reply]

polymers

what is the identification of hdpe/lldpe —Preceding unsigned comment added by ASIMDASH (talk • contribs) 12:03, 26 May 2008 (UTC)[reply]

If you type LDPE and HDPE into the Wikipedia search box, you should get lots of information, including the recycling symbols. Franamax (talk) 12:09, 26 May 2008 (UTC)[reply]

Article with explanation

I was looking for an article to explain what the difference is beween specific gravity, density and specific weight to somebody who knows even less than I do. I read the definitions in these three articles, but what, in layman's language they mean and how they differ, I am at a loss. (I am not really totally ignorant, but I am not very good in explaining.) Can anyone help? It is almost worth an article, I think. --VanBurenen (talk) 12:07, 26 May 2008 (UTC)[reply]

Density is mass per unit volume – the quantity of something in a certain space.

Specific gravity is this density relative to that of water – how much more (or less, of course) of something you get in a certain space than you would get if you filled that space with water.

Weight is the force that a mass experiences in a gravitational field – specific weight is how strongly something is being pulled by a gravitational field, given a certain volume of the substance.

HTH? Angus Lepper^{(T, C, D)} 12:22, 26 May 2008 (UTC)[reply]

(since I hate edit conflicts so much, I'll put my answer in too, but I think Angus did it better :) Density is the fundamental measurement, the mass in a given volume - in other words, if I pick up a 1-foot cube of this stuff, how hard it that to do? Specific gravity is a comparison to water - will it float or sink? (If it has less density than water, it floats, the less dense it is, the higher it floats) Specific weight is how heavy the 1-foot cube is on the Earth's surface, I lied a bit about density, which is true everywhere in the universe. Mass is about how hard it is to accelerate the object any place in the universe, weight is about how hard it is to do it here on Earth. Maybe an article, or a mathematical disambiguation page, 'cause they're all the same thing. Does that help? Franamax (talk) 12:25, 26 May 2008 (UTC)[reply]

And how does relative density fit the picture? 125.21.243.66 (talk) 12:47, 26 May 2008 (UTC)[reply]

It fits relatively. Relative density == specific gravity. Franamax (talk) 12:56, 26 May 2008 (UTC)[reply]

Are you sure? Because my high school physics teacher scared the bejesus out of every student by asking the difference between relative density and specific gravity. 125.21.243.66 (talk) 14:27, 26 May 2008 (UTC)[reply]

Specific gravity is a particular instance of relative density (which is, generally, the ratio,

\rho _{A}/\rho _{B}

, of the densities of two substances A and B) where 'substance B' is water. i.e. specific gravity is a special case of relative density, or relative density a generalization of specific gravity, depending on which way you want to look at it. Angus Lepper^{(T, C, D)} 18:41, 26 May 2008 (UTC)[reply]

Colour photos of Mars

Inspired by the question above, I'd love to see some true-colour photos of Mars; I know that there are colourized versions of the greyscale photos, but does anyone know if there are any 'real' colour photos produced by taking photos with red, green and blue filters separately? Angus Lepper^{(T, C, D)} 14:00, 26 May 2008 (UTC)[reply]

As I understand it, most of the colour images are done like that. It's a little difficult to work out what "true colour" really is, though - they have to determine how to balance the different images (one in each colour) to try and get it as close as possible to what a human eye would see, and since there have never been any human eyes on Mars, that's a little tricky. I think the modern images are pretty good approximations, though. --Tango (talk) 14:08, 26 May 2008 (UTC)[reply]

Ah, I thought the colour images were colourized! Thanks, Angus Lepper^{(T, C, D)} 14:09, 26 May 2008 (UTC)[reply]

I think some are. Some are certainly as close to true colour as you can get. --Tango (talk) 15:32, 26 May 2008 (UTC)[reply]

See [5] for one example of an "(Approximate True Color)" image and a short description of how it was produced. Additionally the calibration target on the rovers includes color samples to provide known comparison values.[6] Rmhermen (talk) 19:57, 26 May 2008 (UTC)[reply]

For accurate color you need filters that match the response of the three cone types, which is not red, green and blue but more like yellow, green and blue. Red is used for reproduction because it stimulates the L cones while minimizing crosstalk with M and S; it's not what the L cones are primarily sensitive to. The first page linked by Rmhermen says they used filters centered at 753 nm, 535 nm and 432 nm, which doesn't make sense since 753 nm is way outside the visual range. I assume 753 is a typo for 573. -- BenRG (talk) 23:07, 26 May 2008 (UTC)[reply]

753 nm isn't a typo. Most of the "true-color" photos use near-infrared rather than red, because it shows more detail in shadowed areas. --Carnildo (talk) 23:51, 26 May 2008 (UTC)[reply]

Disagreeing a bit with the claim that accurate color reproduction requires matching yellow, green and blue cone response curves. Edison (talk) 01:43, 27 May 2008 (UTC)[reply]

There's been a continuous storm of comments from people who are rather unhappy with NASA's ideas and rendering of color. Look at these e.g. [The Colors of Mars: Reality and Illusion Sky and Telescope, volume 97, number 4, page 116], [7], [8], [9] 71.236.23.111 (talk) 08:22, 27 May 2008 (UTC)[reply]

Carnildo, having now read a lot more about the rovers I see that you're right. But then the whole thing makes no sense. You say they use near infrared for the R channel to show more detail, but if the goal is to make a better looking or more useful image then they might as well go all the way. And they do—the Pancam images also come in false-color versions which are prettier and more interesting than the "true color" versions. I can't believe that the boring brown "true color" images exist for any reason but to satisfy a public demand for pictures showing the colors that an astronaut on Mars would see. And they don't show that. It seems like NASA's best course of action, faced with a demand for true color, would be to educate the public about color vision and why true color images aren't "truer" in any meaningful scientific sense. Then they could pick all their filters based on scientific merit and use the prettiest possible false colors for their pictures of the day. The next best course of action would be to ship the correct color filters and give the public what it wants. The worst course of action would be to try to pass off false color images as true color, which is what they seem to be doing.

Edison: Okay, but the farther you get from the cone response curves the more assumptions you have to make about the shape of the spectrum to get accurate colors. I'm beginning to wonder if I'm missing something fundamental about color photography, though. I can understand a layperson thinking that the best way to get the red/green/blue channels of a digital photograph would be to use red/green/blue sensitive detectors. That's not true, but it's a reasonable mistake to make. What I don't get is that real cameras designed by experts do seem to use RGB sensitive detectors. At least that's what I gather from articles like Bayer filter. Do they really use a red filter? If so, why? Is it just cheaper, or is there something else going on?

Atlant (below): I assume the calibration target is for adjusting the white balance; I don't see how it would help in computing the right tristimulus values if you started off with the wrong color filters. -- BenRG (talk) 23:33, 27 May 2008 (UTC)[reply]

Well, the camera isn't a tri-stimulus device. Instead, it has many different filters (spectral channels) that can be switched-in. If the color swatch chart contained either filters that are accurate to the human trim-stimulus response or are fairly narrow-band filters, you could calibrate the camera's many spectral channels to come up with a transform function that will produce the same color images a human eye would see from the many-channel images produced by the camera.

Atlant (talk) 17:37, 28 May 2008 (UTC)[reply]

The landers have all contained a "colo[u]r test chart" within view of their cameras. This allows the camera (and its "filters of many colors") to be calibrated so we can come up with a pretty-accurate representation of the colors in the scenes as a human would perceive them.

Atlant (talk) 14:29, 27 May 2008 (UTC)[reply]

Carrier concentration in doped semiconductors

For Intrinsic semiconductors $np=n_{i}^{2}$ and this makes sense as mentioned in semiconductors. However my textbook states that this is true for extrinsic semiconductors as well. I would say in this case $np>n_{i}^{2}$ , but I can't see why that's wrong? Thanks in advance, Gnorkel (talk) 15:35, 26 May 2008 (UTC)[reply]

Regardless of whether it is pure (intrinsic) or doped (extrinsic) the number of electrons (n) and holes (p) must be the same. You cannot have an electron hole without an electron on average in the complete semiconductor. There will be more electrons than holes (and vice versa) in some parts of the semiconductor and this is what give it the properties employed in the semiconductor devices. The doping changes the number not the ratio of n and p. -- Alan Liefting (talk) - 07:23, 27 May 2008 (UTC)[reply]

That's not true. Changing the ratio of electrons in the conduction band (n) to empty/available states in the valence band (p) is exactly what doping does. Besides, the fomula is showing that the product of n and p is the same, not the ratio. I believe the relationship is understandable through quantum statistical mechanics (see Semiconductors#Carrier generation and recombination). --Prestidigitator (talk) 01:50, 29 May 2008 (UTC)[reply]

Gravity powered electricity generator

Why isn't there such a thing? Why can't we turn gravity into electricity? I can only think of turbines in adam as a type of generator that uses gravity. Are there others? Thank you. 200.127.59.151 (talk) 18:49, 26 May 2008 (UTC)[reply]

Tidal generators can be said to use gravity as a power source. --Milkbreath (talk) 18:55, 26 May 2008 (UTC)[reply]

Pumped-storage hydroelectricity is a way of storing energy, then using gravity to recovery most of it. Rmhermen (talk) 19:54, 26 May 2008 (UTC)[reply]

Going on the "gravity into electricity means 'dropping things'" idea, the problem is "where do you get the things to drop?" Either you need a steady supply (hydroelectricity, for instance) or you lift things -- which uses the energy you're about to gain, and then some. All practical energy sources have to be built on the former, so unless you've got a ready stream of something falling, gravity won't do you much good for a power plant. — Lomn 20:08, 26 May 2008 (UTC)[reply]

You need some prime mover to put energy into raising the thing that generates electricity when it drops. For hydroelectric, I suppose solar power fuels the evaporation of water which falls as rain or snow and runs down the river to create the head which allows the hydro turbine to operate. A tidal generator is more interesting: it seems to extract energy from the tidal sealevel changes which result from movement of the moon around the earth, which implies that if you harness the tidal power, it would have a tiny effect of bringing the moon into a lower orbit. Edison (talk) 01:40, 27 May 2008 (UTC)[reply]

Hmm, not sure here. My understanding is that tidal effects are moving the moon into a higher orbit by energy exchange as the Earth's orbit slows in reciprocal action. Physicists needed here: does obstructing tidal changes increase or decrease the energy transfer that extends the moon's orbital distance? Franamax (talk) 07:03, 27 May 2008 (UTC)[reply]

It boils down to more drag, which is more energy lost in the transfer, which accelerates the Moon's progress to a tidally-locked orbit. — Lomn 13:19, 27 May 2008 (UTC)[reply]

flu

HOW LONG AFTER YOU ARE IN CONTACT WITH THE FLU VIRUS WILL YOU BECOME SICK. Does it happen right away or in a week or so? Thanks, \ —Preceding unsigned comment added by 70.73.149.192 (talk) 21:26, 26 May 2008 (UTC)[reply]

Our flu article does not say, but accoding to this site it is one to two days. -Arch dude (talk) 21:47, 26 May 2008 (UTC)[reply]

Pathogens in body? —Preceding unsigned comment added by 220.235.59.49 (talk) 23:55, 26 May 2008 (UTC)[reply]