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This is siad to be Venus true color is it close to white with faint yellow. If I was orbiting Jupiter will the planet look like a pearl, always playing with color. Seen from Saturn's low magnitude will Saturn still be around blueish silver if I was orbiting planet. If I was far away from the planet, it will look darker, if I was clser to planet, it looks brighter. if I was orbit one foot away from Uranus, it's methane gas will still give glows of violet color since it's magnitude is 5. For Neptune, it would be essentially dim, since the magnitude is 8, I beleive I will still see a dark indigo color when I am orbiting the planet since it have little light far away, for Pluto, and it's other dwarfs, the surface will look black, even if I stad on it--Freewayguy00:27, 30 October 2008 (UTC)[reply]
We discussed this stuff in some detail in your earlier question about Uranus's color. But what you are saying here is bogus. The 'magnitude' of an object depends on it's size as well as it's brightness. When you are up close to it - the size is irrelevant - so comparing the magnitude of these objects and using that to estimate surface brightness is pretty much meaningless. To do this right - compare the albedo divided square of the range of each of the objects from the sun. Do the comparison with the same numbers calculated for the Earth and you'll have a better idea of relative brightnesses from close to the planet. Both sets of data are available in the standardized info-boxes in the articles about each planet. SteveBaker (talk) 04:18, 30 October 2008 (UTC)[reply]
I don't think that you would see Pluto as black when you are on the surface, in the same way that the moon does not look black from earth, but a moon rock does when you have it on earth, and on the moon surface it looks bright grey. Pluto seems to be mostly an apricot colour. The light at Pluto may not be as dark as you think, instead it would be like an incandescent bulb, dim but still good enough to see colour. For Venus you are right it look white, you can see it with your own eyes. Graeme Bartlett (talk) 06:25, 30 October 2008 (UTC)[reply]
I've been to the moon-rock 'vault' at the NASA space center in Houston (it's open to visitors most days). You can see the actual moon rocks (LOTS of them!) in normal room lighting - and there is even one moon rock there that you can actually touch (although the surface of it is so covered with grease and dead human skin cells...I'm not sure you are actually getting in contact with it)! Those rocks are emphatically NOT black - they are mid-grey with very little color to them. So they match more or less what the moon looks like in the sky - with an albedo of 0.1 or so.
But here is the problem (and we talked about it before when the same OP asked about the color of Uranus): Human vision is a crappy way to measure brightness and color. We have eyes that adapt to the dark and to the very bright. So when you see Venus in a dark sky, it looks impossibly bright. But your dark-adapted eyes have shut down all of your color sensors and really wound up the "brightness" controls. So you can really only see in black and white and the fairly dim dot that is Venus is overloading the few 'pixels' in your retina because it's soaked in rhodopsin and the iris is wide open. This means that the light-sensitive cells that Venus impacts simply report "Ouch! Too bright!" and not much else. The atmosphere here on Earth is also scattering blue light due to the Raleigh and Mei scattering mechanisms...that leaches blue out of the image and further distorts any chance of getting good color information from distant objects. (Especially Venus which is only ever visible near the horizon where the atmospheric color distortions are greatest). Hence, Venus LOOKS white to the naked eye - an slightly yellowish in a telescope. When we fly a spacecraft up next to it or use the Hubble telescope - we don't have any of those annoying human visual wierdnesses - and the planet looks yellow.
So our OP asks "What color is it "really"?" - is it white or yellow?
Well, that's an utterly meaningless question. Humans can't even see yellow light properly - we don't have sensors for yellow light. We only notice it at all because it slightly stimulates our red and green sensors and we've decided to label that "red-and-green-together" sensation using the word "Yellow". However, small amounts of pure red and pure green light (such as you'd see in a photo of Venus on your computer screen) also appear "Yellow" - but for different reasons. If you stare for 30 seconds at a picture of the US (or British) flag - then look quickly over at a sheet of white paper, the paper "looks" like it has an orange and cyan flag hovering over it because our eyes have color-adapted to the strong reds and blues in the flag and take a few seconds to recover. So in orbit about Venus, our eyes would adapt to the colors and start to wash them out. So what we see and how the universe truly is are very different things - photos (either on photographic prints, or in CMYK magazine prints or RGB computer screens) have nothing like the range of colors that the planet actually has or that our eyes could potentially resolve (you simply cannot reproduce a really strong "cherry red" on a TV or computer screen for example). When NASA design a space probe - they need to capture specific data to determine chemical compositions, etc. So they may well decide to use a true yellow filter on their camera (so it really can tell the difference between a mixture or red and green light - which would appear dark - versus true yellow light - which would pass through the filter and look bright). So even the photo we took with a satellite is hopelessly "wrong" compared to what we'd see with our own eyes or with a more traditional RGB camera. Then it would depend on whether we needed to be dark-adapted to see it (in which case it would all be pretty much monochromatic) - or whether our eyes were overloaded by the brightness (in which case everything pretty much looks white). Hence it doesn't make a whole lot of sense to be asking about the "real" color of objects in space - where the lighting is so very different from here on Earth where our vision system evolved.
If you have two cars - an orange one and blue one. Park them out in the street. In daylight - they look orange and blue. At night (in the dark) they look light blue-ish grey and dark blue-ish grey. At night under sodium street lights - they look orange and black. At night under moonlight they look orange and blue again. If you have a weak-green sensor because you are mildly color-blind then the orange car looks red. If you have cataract surgery and lose the filtering ability of your corneas - then if the paint used on the orange car happens to be strongly reflective of UV for some reason and the blue car not so much so - and they both look to be different shades of blue in strong sunlight - but revert to orange and blue on cloudy days. If you stand 20 miles away from the cars and look at them through binoculars with the sun behind your back, the blue car looks darker than it did before and the orange car looks more pale - perhaps, almost yellow-ish. If you do the same thing with the sun in front of you then the blue car still looks darker but the orange car still looks even more intensely orange. What color are the cars "really"? Well, it's a meaningless question. By convention, we humans say that they are orange and blue because we have decided that normal daylight vision is what we're going to use to label the colors...but in space, there is no "normal daylight" - so that labelling scheme fails.
Hence questions about the "real" colors of planets are not ones we could (or should) answer - it falls into the realms of predictions and speculations. We should leave it up to the astronomers to create spectrograms of the light and print graphs in journals...or we should send a poet/artist along in a capsule to write movingly about the things he/she sees there and paint pictures that speak to us of the color impressions that person got.
Good answer, but a major mistake in there. We don't have red, green, and blue sensors in our eyes. The wavelength response of the three cone types is something like what's shown in this picture. The L (long wavelength) cone's peak sensitivity is actually in the yellow-green range, and it's as sensitive to blue-cyan frequencies as it is to red. But when reproducing colors we want to be able to stimulate the S, M, and L cones as independently as possible, and red (which stimulates L much more than M) is better for that than yellow (which stimulates L and M about equally). Blue stimulates S almost exclusively, and green preferentially stimulates M. Thus monochromatic red, green, and blue are useful for reproducing color. They're not good for capturing color, though. Monochromatic red, green, and blue filters throw away most of the spectral information that you need to determine the correct cone stimulus levels in the first place. To capture the perceived color reliably you need to use filters that match the cone response curves. Once you have the SML stimulus information you can derive from it the RGB intensities that will most accurately reproduce that stimulus. Of course, there are tons of details. The curves are somewhat different on different parts of the retina. Real cameras don't use the right curves. The eye compensates for ambient lighting so that white objects still look white even when illuminated by colored light, but this process doesn't work when you're viewing the photographs later in different lighting conditions. No set of three primaries can reproduce all SML stimulus triples, and on top of that the primaries used on typical computer monitors are highly suboptimal.
But the number one problem with making true-color astronomical images is not the difficulty, it's the pointlessness of it. There's nothing magical about human color vision. The particular cone types we evolved served us well in the environment we evolved in (I think the red-yellow-green distinction helps in finding ripe fruit). They're meaningless on Mars. True-color images aren't interesting to look at, and they aren't any more true to life than false-color images. Compare this false-color image and this "true-color" image. (I put "true-color" in quotes because the filters they used don't match the human cones, so the color calibration involves a certain amount of guesswork.) Our eyes don't do a good job of seeing what's interesting in these exotic environments, and it's ridiculous to copy their limitations in our technology when we can do better. That's what technology is for. -- BenRG (talk) 00:11, 31 October 2008 (UTC)[reply]
I would argue that "true color" images are "More true to life" than the false color ones. Imagine I sent my handy digital camera to Mars, snapped it a few pictures and brought it back to Earth. When I reviewed the images, the colors on the screen that I observed with my eyes would at least roughly match the colors I would have observed had I gone to Mars myself instead of sending my camera.
I completely grant that true color images are not nearly as scientifically useful. And perhaps even not worth the money required to capture accurately. But it's disingenuous to claim that there isn't an element of truth in true color images that a false color image lacks. There is a very real, common desire to know what faraway places would be like "if I were there". To satisfy this (rather unscientific) desire you need to know what things look like to human vision under natural lighting. (And perhaps what they would look like under what us earthlings would consider a white flashlight.) APL (talk) 18:41, 31 October 2008 (UTC)[reply]
Yeah, you're right. I suppose the images could even have scientific value as a preview of what future Mars colonists would see out their windows every day. I don't know whether your digital camera would reproduce the Martian colors accurately; I think they're calibrated for typical Terran lighting conditions. NASA's true-color images were made very carefully if this page is to be believed—they took all the data they had available (six frequencies), guessed the rest of the visible spectrum by cubic interpolation, then derived the XYZ coordinates from that and converted to sRGB or something closely resembling it. Reading that gives me more confidence in the accuracy of the colors. Some of the other descriptions I'd read gave the impression that they just took the blue, green, and infrared data and stuck it in the blue, green, and red channels and called it true color. -- BenRG (talk) 22:50, 31 October 2008 (UTC)[reply]
I've searched a lot of places but can't find anything that helps me answer...
Say that our planet was an Earth-sized Jupiter. (I.E. we're Jupiter, but not Jupiter-sized.) How big would The Great Red Spot be on here? -WarthogDemon01:16, 30 October 2008 (UTC)[reply]
Let's do a little math. According to our article on Jupiter, the great red spot varries in size, covering an area of 24–40,000 km × 12–14,000 km. Using the formula for the area of an ellipse (pi*a*b) we get an area of 9.04x109 - 1.7x1010 km2. The surface area of Jupiter, according to the article is 6.21796×1010 km2. Thus, at the smaller end of the scale, the red spot covers 14-27% of its surface. Given that Earth's land is about 29.2% of its surface, the great red spot would cover somewhere between 1/2 and all of the Earth's land area. As a further reference point, the Atlantic Ocean is 22% of Earth's surface area, which falls in the mid-point of the estimated sizes of the Great Red Spot. Thus, if transcribed onto Earth, it would be roughly the size of the Atlantic Ocean.
Someone can check my math on this, but it seems about right... --Jayron32.talk.contribs02:07, 30 October 2008 (UTC)[reply]
Something must have gone wrong with the maths there. The Red Spot can't cover anywhere near 27% of Jupiter's surface, the picture on the Jupiter article shows that it covers a much smaller proportion (less than 5% I would argue by looking at the picture). According to Jupiter, the radius of the planet is 11 times larger than the radius of earth. From Atmosphere of Jupiter, the Red Spot's "dimensions are 24–40,000 km west–to–east and 12–14,000 km south–to–north." Divide them by 11 and you have the approximate equivalent. - Akamad (talk) 02:13, 30 October 2008 (UTC)[reply]
I double checked the maths, and the calculations are correct. There must be a problem with the numbers I used then; either the dimmensions of the Great Spot are too large, or the surface area given is too small. And divide by 11 wont work. If jupiter is 11 times larger, than its surface area is 112 times larger, so we would need to divide by 121. Dividing by 121 gives an Earth-scaled area of the great red spot as: 7.47x107 - 1.40x108 km2. Since Earth's surface area is given as: 5.10x108 in our article on Earth, that gives us an overall ratio of 14.6% - 27.4%, or the same thing I got by calculating it the other way. --Jayron32.talk.contribs02:30, 30 October 2008 (UTC)[reply]
Sorry I meant divide the 24 - 40000 km by 11 and divide the 12 - 14000 km by 11 too. So you are dividing the area by 121. I think the problem in the maths was that you used 24000 km and 12000 km in the Pi*A*B equation. The numbers used should be half that (12000 and 6000) since the equation refers to the distance from the middle. See http://www.math.hmc.edu/funfacts/ffiles/10006.3.shtml. - Akamad (talk) 02:36, 30 October 2008 (UTC)[reply]
(I wrote this during and edit conflicted with above. We had the same thought at the same time) I assumed that a and b in the ellipse area equation were the major and minor DIAMETERS of the ellipse. Its not. Its the major and minor RADII of the ellipse. My answers were thus off by a factor of 4. Thus, the actual % area is 3.65% - 6.85%. Since the surface area of the earth is 5.10x108 km2, that gives us an "Earth-scaled" size of 18,615,000 - 34,935,000 km2. North America covers an area of 24,709,000 km2 , which falls dead in the middle of this range. Thus, the Great Red Spot is roughly the size of North America, scaled to Earth's dimmensions. --Jayron32.talk.contribs02:39, 30 October 2008 (UTC)[reply]
OK - move aside - let a professional through here. OMFG! This answer is a train wreck!
Great_Red_Spot#Great_Red_Spot says "Its dimensions are 24–40,000 km west–to–east and 12–14,000 km south–to–north." - so diameters - not radii. Jupiter has a radius of 71,500 km. Earth has a radius of 6,380 km - so to convert dimensions on Jupiter to "Earth scale" you've gotta multiply by 6380/71500 which is 0.09 - or for mental arithmetic: Divide Jupiter-linear-dimensions by 11 to get Earth-linear-dimensions. So an "Earth scale" spot is 2,200 to 3,600km by 1100km to 1300km. Since there are big error bars in the original data for the spot size - let's cheat and pretend it's a rectangle of the smaller size...that way we avoid all of the ikky ellipse math stuff. So the area is roughly 2200x1100 = 2.4 million square kilometers.
Now - for comparisons: According to List_of_continents#Area_and_population, North America covers 24 million square kilometers - so I don't know how you arrived at "the size of North America"...that's wrong by a factor of 10! (Which is bloody obvious if you compare a picture of Jupiter and Earth scaled to similar sizes!)....so a third the size of Australia...or Texas plus Alaska...or somewhere between Sudan and Algeria...the drainage area of the Black Sea...I dunno.
The image (at right) was made by taking two NASA photos - one of Earth, the other of Jupiter. I resized the two images to be identical in size - then overlaid one onto the other. Finally I erased all of Jupiter EXCEPT the red spot. So - here is an image of what the OP is imagining. Notice in the top of the image we have Africa. The red spot looks to be maybe four times the area of Madagascar - which is half a million square kilometers - so two million seems about right (although it's a pretty fuzzy "spot" - so there is plenty of room for error)...anyway - it's NOTHING LIKE the size of North America.
The original question described a counterfactual situation ("we're Jupiter, but not Jupiter-sized"). In answering, we have to figure out what exactly that means. To me the obvious interpretation is that the relevant linear dimensions are reduced in the same proportion, as if making a scale model. That's what Steve did. The initial derailment that started the train wreck earlier was to try working with areas, which of course vary as the square of the linear dimensions. --Anonymous, 06:00:00 UTC, October 30.00000 :-), 2008.
Yes - it's a very common error to assume that since "A is 1/11th the diameter of B" that "The area of A is 1/11th of B" (when in fact it's the SQUARE of the linear dimensions - so A is 1/121th the area of B) and that "The volume of A is 1/11th of B" (when it's really the CUBE of the linear dimensions - so A is 1/1331th the volume of B). If I had $1 for every time I'd seen that mistake made I'd have...well, actually, maybe just enough for lunch today. The trouble is that the word "size" is vague - so you hear things like "Jupiter is 11 times bigger than the Earth" and "Jupiter is 1300 times bigger than the Earth" - both of which are true in some fuzzy-thinking way. That's why I was so careful to say "Divide Jupiter-linear-dimensions by 11 to get Earth-linear-dimensions." in my answer. SteveBaker (talk) 13:12, 30 October 2008 (UTC)[reply]
The trouble is, physics isn't scale-invariant, so it's really hopeless to conceive of an "Earth-sized Jupiter". The physics that formed and sustains the Red Spot, whatever it is, wouldn't form a scale-model Red Spot on an Earth-sized planet. I don't think an Earth-sized gas "giant" can exist in any case since it wouldn't have enough self-gravity to hold it together. So the only way to understand this question is as a visualization aid: to get a feel for the size of the Red Spot, scale it down to something more familiar. Personally, though, I find these comparisons singularly unhelpful. (As the saying goes, there is so much sand in Northern Africa that if it were spread out it would completely cover the Sahara desert.) -- BenRG (talk) 12:58, 30 October 2008 (UTC)[reply]
Yeah, I was curious about a size comparison. I understand how a Great Red Spot couldn't exist on Earth. (Though hurricane speeds can get pretty close to the same wind speeds.) And plus, if I put my question further and said what if we had the Great Red Spot composed of the same materials, it wouldn't last long: hydrogen-rich storm in an oxygen-rich atmosphere? Nope, nope, nope. :) -WarthogDemon16:46, 30 October 2008 (UTC)[reply]
That's true - but sometimes it helps people to bring vast or tiny things to "human scales" in order to think about them. Saying that the spot is "big" isn't helpful - and saying that it's 40,000 km across when Jupiter itself is the largest planet in the solar system doesn't give you a feeling for it either. But knowing that (compared to the size of the planet) it's like a third the size of Australia really gives you the strong idea that this is a really big thing. On the other hand - compared to the size of the planet - it's not that much bigger than the "great white spots" we get a couple of dozen times a year here on Earth. (We call them hurricanes and cyclones). Those can be up to 1000km in diameter - which (proportionate to the size of the planet) is only about half the size of the Great Red Spot. What's different about our hurricanes is that they can't sustain themselves over land - so they inevitably fizzle out after a week or two. Jupiter has no land masses - so there is really nothing to stop the spot from staying pretty much together for (at least) hundreds of years - we don't know whether it's truly a permanent feature. The Great Dark Spot on Neptunes southern hemisphere was a similar kind of storm - and it fizzled out sometime between 1989 and 1994, to be replaced by another one in Neptune's northern hemisphere. Saturn has shown similar features (eg Dragon Storm (astronomy)). SteveBaker (talk) 13:12, 30 October 2008 (UTC)[reply]
Well, that probably depends on the size of the alien more than the size of the planet. People have occasionally speculated that the Giant Red Spot IS an alien! But since Jupiter has no solid surface - anything living there must live in a gaseous environment - so it has to complete it's entire life-cycle flying or floating or something. The most likely form of such creatures would be massive solar-powered gas-bags floating around - and they could easily need to be very large indeed. So perhaps the spot doesn't seem all that big to them after all. SteveBaker (talk) 00:39, 31 October 2008 (UTC)[reply]
LANDFILLS - why are garbage cells covered with a layer of soil?
Having read the landfill page, I see that individual cells are covered with a layer of soil or in some cases another material. Innumerable other google-found webpages explaining landfills mention the same process. However, I can find no direct explanation as to WHY this is done?
Thank you.
To trap all that rubbish in a contained way. It will reduce smell, reduce vermin and scavengers, stop plastic bags blowing in the wind, and speed decay. Graeme Bartlett (talk) 06:08, 30 October 2008 (UTC)[reply]
To close the top is sometimes done to prevent rain water going in. The water will be contaminated and has a chance to reach ground water. Other possibility is air thight closing to collect the natural gas from decomposition within the landfill. You also get rid of the smell and the animals lifing in an open landfill (rats seagulls).--Stone (talk) 11:15, 30 October 2008 (UTC)[reply]
Also, gas (methane) from the landfill is sometimes collected for use as a fuel...the soil helps to trap this in. The main reason is to keep smells and litter in and vermin outGaryReggae (talk) 12:59, 30 October 2008 (UTC)[reply]
I'm the original poster, and I wanted to say thank you. :-)
The decomposing organic matter produces methane, and if the landfill is not adequately vented by perforated pipes driven through it or other means, the methane can work its way through the ground and come out in people's basements. Sealing the top of the landfill might increase the likelihood of this. Edison (talk) 19:14, 30 October 2008 (UTC)[reply]
Careful Steve! When the landfill is ultimately covered and landscaped, then yes grass can grow (although you want to be careful with trees due to the roots). But just covering each cell with soil is done at the end of each day as the landfill develops, for all the reasons given above. You're likely to be putting another layer of rubbish on top of that soil tomorrow, so there's no time for grass yet! 79.66.32.150 (talk) 21:12, 30 October 2008 (UTC)[reply]
Ah! I think I got it. It's a trick question. There is a measuring device. It is called the water or the water level. Using the water level as a measuring device, measure out half the volume of water in the cylinder. 122.107.157.9 (talk) 11:07, 30 October 2008 (UTC)[reply]
Did you understand what Dmcq was saying? You still need to know a way to measure exactly half. You can't just eyeball it from the glass when it's level. If you don't get it, read his statement more carefully and think Nil Einne (talk) 12:31, 30 October 2008 (UTC)[reply]
As you very slowly pour the water out, there will come a point when the surface of the water just touches the bottom of the glass (and also just the lip of the glass on the opposite side) - at that moment - it's exactly half-full....like this:
That's one way to solve it. Another would be to completely fill the glass, and then get an identical glass and pour the water from the full one to the empty one until they both have equal amounts. --Russoc4 (talk) 19:59, 30 October 2008 (UTC)[reply]
Or you could get REALLY crazy. You could put the glass into a larger sealed container whose overall volume would allow the evaporation of the water to proceed until the vapor pressure of the water and partial pressure of the water vapor in the larger container were in equilibrium. Per the ideal gas law, you would only need to know the initial amount of water in the container you started with, and as long as you maintain a constant temperature, you make the volume larger container such a size as to allow exactly 1/2 of the original volume of water to evaporate. Its a whole bunch of algebra, but its certainly doable... --Jayron32.talk.contribs20:13, 30 October 2008 (UTC)[reply]
There's not much to measure really. Just maintain a constant temperature, and the ratio of the size of the larger, air-tight container should be some ratio of the size of the smaller, water filled cylander. This ratio is dependent ONLY on temperature and nothing else, so as long as you maintain a constant temperature, there is some ratio of large box/smaller cylander which will cause exactly 1/2 of the water to evaporate. You don't even need to know the volume of the smaller cylander; you just need to build a box that has a volume that is some number of times larger than that cylander. Here, let me work it out for you. At, say, 25oC, the vapor pressure of water, using this site: [2] to calculate, is 0.0312 atm. Now, once the partial pressure water vapor in the air is equal to this pressure, the water stops evaporating. So, lets assume, to make the calculations easier, that our cylander is 1 liter. So, we want 500 mL of water to evaporate. The density of water being 1 g/mL that makes this water have a mass of 500 grams. Divide by the molar mass of water (18 grams/mole) to give us 500/18 = 27.8 moles of water vapor. Now, by the ideal gas law, PV=nRT or for our purposes V=nRT/P, where V is the volume of the box we need, n=27.8 (moles of water vapor), R = .08206 L*atm/mol*K (the Gas constant), T = 298 K (25oC), and P = .0312 atm. Solve and you get V = 21789 liters. Now, the ratio of the amount of water to the size of the box should be constant, since less water would need a proprotionally smaller box, while all the other numbers remain constant. So as long as the box is exactly 21789 times the size of the cylinder of water, and assuming of course the air in the box was fully dry (leave a dessicant in there for a few days, then remove it right before putting the cyilnder in it), once the system equilibrates, the cylinder will be EXACTLY 1/2 empty. Oh, and for our 1-liter example, it IS a pretty big box. There are 1000 liters in a cubic meter, so the box is 21.789 cubic meters, or if a perfect cube would be 2.79 meters to a side, or about the size of the average bathroom. To answer your question, the ONLY measurement you need to make is the temperature. The ratio of the volumes is ONLY depedndent on this number. Once you know that ratio of volumes, you can simply create a box that is that much bigger than the cylinder. I never said it was an EASY solution, from a practical standpoint, but who cares about being practical. Plus, unlike the "tip the glass to the diagonal" solution, this one doesn't have the pesky meniscus problem. --Jayron32.talk.contribs02:43, 31 October 2008 (UTC)[reply]
All of these ideas (apart from my 'tip the glass' suggestion) fail miserably because the problem states that you don't have a measuring device of any kind. If you are going to argue that getting the temperature just right in a particularly-dimensioned box doesn't constitute using "a measuring device" then I might as well say "I'll just take a second glass of exactly half the height of the original but of the same diameter, fill it to the brim and carefully pour the contents into the first glass" - which is vastly easier and gets you to the correct answer much more realistically. But I'd argue that the terms of the question say you have a glass - and water - and absolutely nothing else of known dimensions or other definite properties - since anything like that confers the ability to measure. If you're allowed to invent your own devices to help - then there are easier and more direct ways. SteveBaker (talk) 18:17, 31 October 2008 (UTC)[reply]
Oh, you are entirely right. I was just having fun coming up with the most rediculous and esoteric way to remove half of the water from the glass. Your method of using the diagonal of the glass is, of course, the only scrupulously correct way. My solution had an air of Rube Goldberg to it, and I was just goofing. --Jayron32.talk.contribs19:02, 31 October 2008 (UTC)[reply]
There is a tradition at some US college or other that every year the following question is always asked on their 1st year general science term paper: "Given an accurate barometer, how do you determine the height of a tall building?" - the "correct" answer is obvious - but there has come a tradition of awarding extra marks for "other" answers. So, for example, one may climb to the top of the building, drop the accurate barometer over the edge and count the number of seconds for it to hit the ground - and thus determine the height. Or one could tie the barometer onto a length of string and use it as a pendulum - and by accurately timing a large number of swings, determine the variation in the force of gravity at the top and bottom of the building - and thereby determine it's height. However, the neatest solution is that you take the barometer to the custodian and say "I'll give you this beautiful (and highly accurate) barometer, if you'll tell me the height of the building". SteveBaker (talk) 02:07, 1 November 2008 (UTC)[reply]
On a pedantic level, visual perception when tilting the cylindrical container certainly is a conceptual measuring device. Clearly, you are - visually - measuring the property of the surface of the liquid in relation to the side walls / bottom of the cylinder.
I added a subject header to your post. You may want to take time to read the guidelines if you want to receive a better response. You also should sign your post Nil Einne (talk) 12:31, 30 October 2008 (UTC)[reply]
There are three basic types of rock:
"Sedimentary" rocks (like sandstone and chalk) form as layers of mud and other 'stuff' accumulate at the bottoms of lakes and oceans ("sediments"). Over millions of years, enough layers build up that the weight of all of that stuff pressing down on the lower layers compresses them into rocks. Since the sands and other silts that make up sedimentary rocks came from older rocks that had been eroded by wind, rain, rivers, etc - it's often the case that pebbles and other small bits of other rocks get mixed up into the sedimentary rocks.
"Igneous" (volcanic) rocks form when lava or ash comes out of a volcano. As it cools, it turns into various kinds of rock. Sometimes, small bits of other rocks that got crushed up by the volcano end up mixed into the igneous rock.
"Metamorphic" rocks are formed when igneous or sedimentary rocks are crushed under yet more pressure from rocks in layers above them changes their form or composition (they change...or "metamorphose"). Some of the chemicals in a rock may get washed away by water flowing through them - and some rocks may have small voids and bubbles filled in by stuff deposited by flowing water. Some rocks get heated up by lava or by being so far underground - those undergo all sorts of chemical changes.
See also Rock cycle for more info. Rocks are mostly various forms of silica and other silicon-containing compounds, though there are some other components that make for some sharp differences. The relative amounts of silica to other components will determine the differences between the rocks. --Jayron32.talk.contribs13:04, 30 October 2008 (UTC)[reply]
Some rocks contain metals, either in a pure form like gold or in combination with oxygen or other substances like hematite(which contains iron), and these are called Ore. Edison (talk) 19:12, 30 October 2008 (UTC)[reply]
A delivery boy collected 5.35 pesos part in 5 centavo coins, 8 part in centavo coins. If the number of 5 centavo coins were 7 more than one-half the no. of 10 centavo coins, how many 5 centavo coins?
There is a mathematics reference desk located at Wikipedia:Reference desk/Mathematics, but if you post the question there you are going to get the same response I am going to tell you. Wikipedia is not here to do your homework for you. If you need help finding information, we can point you towards articles (like Currency of Mexico, which contains info on pesos and centavos. But seriously, you can push the buttons on your calculator by yourself... --Jayron32.talk.contribs15:37, 30 October 2008 (UTC)[reply]
There is a standard method for this kind of question. Assign letters to all your unknowns (number of each type of coin) and then rewrite the problem as equations. You can then solve those equations to get the answer. --Tango (talk) 16:29, 30 October 2008 (UTC)[reply]
There is probably something wrong in the way you have rewritten/translated your problem, or else there is something wrong in the problem itself. As written, it's not solvable. The first sentence mentions five-centavo coins and centavo coins (i.e. one-centavo coins). The second sentence mentions ten-centavo coins and five-centavo coins. So you say you have three unknowns, but as far as I can see, you have only provided information for writing two equations. You need as many equations as you have unknowns. Also, the meaning of the first sentence is not clear. I don't understand what you mean by "8 part in centavo coins". However, if the first sentence should be "A delivery boy collected 5.35 pesos, partly in five-centavo coins, partly in ten-centavo coins", the problem can be solved as Tango suggested, by letting x be the number of five centavo coins, and y be the number of ten-centavo coins. I'll help you along by showing how you make the second sentence into an equation:
If the number of five-centavo coins were 7 more than one-half the number of ten-centavo coins
Just substitute x and y for the number of five- and ten-centavo coins:
If x were 7 more than one-half of y
Then replace the remaining words with numbers and the mathematical operators (= + - * / etc) that have the same meaning as the words:
x = 7 + 0.5 * y
Do the same thing with the first piece of information. Now you have two equations with two unknowns. Solve these, then you know the value of x and y. The answer to the question, "how many 5 centavo coins?", is the value of x. --NorwegianBluetalk22:44, 30 October 2008 (UTC)[reply]
My question is why women who are menstruating easily get or feel electrocuted once they simply touch a door knob or any thing that is made of metal?
That sounds unlikely to me. I can't see any reason for it. The chance of getting static shocks depends on what you're wearing (particularly your shoes), what kind of floor you walk on and also the weather (dry air doesn't conduct as well as humid air, so you can build up a charge easier). I can't see why it would depend on whether or not your menstruating. I guess it's possible that some women like to wear big fluffy jumpers when menstruating and that's what causes it, but that's about it. --Tango (talk) 18:22, 30 October 2008 (UTC)[reply]
It may be possible menstruating women are more sensitive to the sort of thing for some reason. But I agree with you it seems unlikely and there's no evidence so far to suggest otherwise. Nil Einne (talk) 12:46, 31 October 2008 (UTC)[reply]
Pay attention to the pi-bonding and the location of the hydrogen atoms in 3D space, and what effect that has on the strength of the C-H bond. Remember that an acid is any substance that easily loses an H+ ion, and that will be directly related to the strength of that C-H bond (i.e. the weaker the bond, the better acid it is). Another factor is the relative stability of the resulting carboanion (i.e. if you lose an H+ as an acid, you will leave behind a C- ion; and the pi-molecular orbitals in the triple bond system has an effect on the stability of that carboanion. --Jayron32.talk.contribs18:19, 30 October 2008 (UTC)[reply]
This page here: [3] also contains lots of good info on the reactivity of alkynes (the class of compounds that acetylene is the smallest member). Read closely the section titled "Acidity of Terminal Alkynes". --Jayron32.talk.contribs18:23, 30 October 2008 (UTC)[reply]
An sp hybridized carbon also has a higher electronegativity than sp2 and sp3, making the resulting carbanion relatively stable. --Russoc4 (talk) 20:01, 30 October 2008 (UTC)[reply]
Hello. I'm playing Merlie Ryan in an adaptation of Carson McCuller's Ballad of the Sad Cafe. Merlie is known as "Crazy Merlie", and is generally the village idiot. In the book, it says he has the "three-day malaria", which means for two days he's dull and cross, then on the third day he livens up and has an idea or two, mostly foolish.
The entry for malaria says it can cause brain damage if you get it when you're young, so I'm assuming that's why he's foolish. But the play lasts for about 8 years. Could he have malaria the whole time? I thought you'd get better or get dead. And is it right that it might have caused some sort of cognitive problem? In the play he seems to be uninhibited, as if his sense of social appropriateness is missing. Might that come from the disease?
Slightly rambling question there, but I'd be grateful for any light anyone can shed on the effects of the disease.
(ec) Well, "three-day malaria" is "tertian malaria", which is that caused by Plasmodium vivax. The three-day cycle corresponds to the release of the merozoites from the liver and their invasion of red blood cells. Vivax malaria is not as bad as other forms of malaria, and is seldom fatal. The usual treatment is with chloroquine, which was developed in the 1930s -and was acknowledged as first line therapy since 1946 - I'm not cure if The Ballad of the Sad Cafe is explicitly set after this. But chloroquine alone doesn't cure the disease; to wipe out the liver infection primaquine is necessary, and this became available in the late 1940s, and wouldn't be widely available until after that, and so probably wouldn't figure in "Sad Cafe" published in 1951. So Merlie definitely would be expected to suffer from a chronic form of malaria that would last years and years. I don't think Merlie's foolishness is necessarily attributable to his malaria, though. But in answer to your specific question, absolutely he could have malaria the whole time. But no, people with malaria are not noted for being socially inappropriate in the way that, say, drunks, are. The brain damage in those who have it seems to result in cognitive difficulties: trouble in memory, in concentration, and learning. You could play Merlie as not caring much about social niceties because of his preoccupation with his own suffering; you could play him as unable to concentrate, with a short attention span; you could play him as mentally "slow". BTW, today we think of malaria as a disease of travellers, because it's been eradicated in the U.S. now, but in Merlie's time could certainly have been acquired here. - Nunh-huh21:28, 30 October 2008 (UTC)[reply]
Thanks for the prompt replies! We have set it finishing in 1945, so none of the treatments mentioned would be available to him. And it's useful to know that he could have suffered for the whole time without actually dying. "Unable to concentrate" seems to match most with my impressions of the character, but it's the way she links the three-day cycle of the disease with his activity and behaviour that confuses me - two days cross and dull, then livens up on the third to have a few foolish ideas. I thought at first this was a poetic way of saying he had some form of bipolar condition, but given that three-day malaria is a genuine thing, this seems less likely. Maybe having a fever involves less time shivering in bed than it does in my imagination, or maybe it's feverish in the sense of a feverish imagination.
78.86.213.238 (talk) 22:45, 30 October 2008 (UTC)[reply]
Well, it would certainly be fever in the sense of increased body temperature, but not all fevers involve shivering in bed or cold sweats. In 1945, he would probably have been taking quinine, which is not terribly effective (especially as it had to be made from cinchona bark rather than synthesized in 1945) but would have given some relief. - Nunh-huh01:14, 31 October 2008 (UTC)[reply]
There were some experiments with using malaria as a way to create fever and control syphilis, specifically neurosyphilis. The idea was that since malaria was a known quantity and would produce a fever that could be controlled by use of quinine, the patient would avoid the worst symptoms of the syphilis in exchange for the discomforts of malaria. Before you get too excited about the character's possible lurid past, the disease can be congenital (passed from mother to child). SDY (talk) 09:35, 2 November 2008 (UTC)[reply]
Well, its either a weasel or a ferrit (I thought it was a rat at first when I found it rummaging in my upturned kitchen bin and I was going to brain it with a claw hammer). The RSPCA man can't come to pick it up until tomorrow. I've put it in a cat carrier with some straw and given it water but what do I feed it? Sunflower seeds? Nuts? Hay? Carrots? They're similar to rabbits and other small rodents aren't they? --84.64.99.103 (talk) 21:54, 30 October 2008 (UTC)[reply]
It could also be a Stoat. You might not want to keep a ferret in the house - they REALLY, REALLY STINK! Anyway - it won't die if it goes without food for 24 hours - and you know it just ate when you caught it. You aren't doing it any favors by getting it used to getting free handouts from humans - so you should probably avoid giving it anything at all. So long as it has water and fresh air - it'll be fine. SteveBaker (talk) 00:28, 31 October 2008 (UTC)[reply]
Depending on where you live, it could also be a fisher cat or any other of a number of mustelids. Still, they are pretty much requisite carnivors, like a cat, so if you threw some hambuger or something in there, it should be fine. Cat food would probably work too... --Jayron32.talk.contribs02:23, 31 October 2008 (UTC)[reply]
Why not just let it go in the local woods where it can find its own food? I mean, unless it's injured I'd think the reason to call the RSPCA would be to catch it, which you took care of on your own, so why not release it on your own as well? --Shaggorama (talk) 05:58, 31 October 2008 (UTC)[reply]
Sorry, I should've made it clear. I'm quite sure that it's someone's pet. It's very tame. Either that or it's ill and weak (looks okay though). I fed it some beef last night and I gave it some cat food this morning. The guy is coming by to pick it up later. —Preceding unsigned comment added by 90.241.161.187 (talk) 11:54, 31 October 2008 (UTC)[reply]
I realize you boys are fisher-ing for laughs here, but if your dad told you were funny, he must'a lied. Frankly, you otter be ashamed at your stoatally stinky puns, but I'll stop badgering you now that I've skunked you. Matt Deres (talk) 14:05, 1 November 2008 (UTC)[reply]
Just a quick update. The mustelid I found turned out to be a ferret and was definitely an escaped pet. She was only a youngster too. She had a few ticks and was a bit skinny but otherwise seemed to be in reasonable health on first inspection. She's been taken to the animal shelter and all being well, will be rehomed if they can't find her original owner (apparently, they don't have much of a problem rehoming ferrets). --84.64.31.41 (talk) 17:19, 1 November 2008 (UTC)[reply]
I don't know exactly, but my initial guess is that ibuprofen might induce higher concentrations of the enzyme that metabolizes atenolol, therefore reducing the effectiveness of atenolol, assuming the parent compound is the active one. --Russoc4 (talk) 00:07, 31 October 2008 (UTC)[reply]
I'm pretty sure that atenolol is renally cleared, so that's not it. Also, ibuprofen reduces GFR (at moderate concentrations), which would be expected to reduce renal clearance (and increase atenolol concentration, all other things being equal). My guess is that the renal effects of ibuprofen would generally raise blood pressure, and that would counterbalance atenolol's effect. If there's a more direct mechanism, that would be interesting. --Scray (talk) 02:24, 31 October 2008 (UTC)[reply]
I had a further look around and found somewhere that it's an antagonistic effect due to NSAID-induced inhibition of renal prostaglandins, sodium and fluid retention. Thanks. —Cyclonenim (talk · contribs · email) 10:31, 31 October 2008 (UTC)[reply]
Yes, ibuprofen and other NSAIDs block prostanglandin synthesis, including PGE2. By the way, you know that beta-blockers are no longer recommended as treatment for hypertension? Axl¤[Talk]11:13, 31 October 2008 (UTC)[reply]
The problem was initially flagged by this classic paper in the Lancet. Beta-blockers will reduce blood pressure equally well, but they do not prevent the undesirable outcome: stroke, as effectively. There have been several other papers showing similar findings. Here is the Cochrane review. The NICE guideline states "the cost-effect analysis... [supports] the clinical data in that beta-blockers are the class of drug least favoured, and CCBs and thiazide-type diuretics appear the most cost-effective choices in most scenarios". Beta-blockers are now fourth-line treatment. Axl¤[Talk]11:37, 31 October 2008 (UTC)[reply]
That's 4th line as a a sole agent, I think most UK doctors would demote slightly further given that as multiple drugs are introduced, so patient's would typically already be on bendroflumethiazide... and thiazides in conjunction with atenolol is diabetogenic which I understood was final issue (on background of above cited findings of less good outcomes) - so likely alternative would be of an alpha blocker... but that's to digress away from avoiding NSAIDs in hypertensive patients, but also relevant to mention is that long term NSAIDs may cause deterioration in kidney function... David RubenTalk02:21, 2 November 2008 (UTC)[reply]
i'm 23 and my hands look fine, but I wouldn't like to have liver spots all over them when I'm 76. Do you think I can do anything (daily) or avoid doing something so I won't get them eventually?
My textbook says concentration of pure solids and liquids is taken as unity.But no explanation is given as to why.Can some1 please why is it taken so?
"Concentration" in what sort of units? If you are asking about %, it should make some sense that a pure sample of some chemical X is "100% X" based on the standard meaning of percent ("X is what fraction of the whole?"). But for concentrations that aren't just "fractional of the whole", it's clearly a false statement: 1 g of water is 1/18 mole and has a volume of 1 mL, so it most certainly does not have a molarity of 1 mol/mL. DMacks (talk) 23:58, 30 October 2008 (UTC)[reply]
It is because 1 part in 1 is that pure substance. And 0 parts in 1 of anything else. In reality what you may call pure sugar for example will have air between the crystals, and the density of the bulk material may be below that of a pure uniform crystal. They would be saying the concentration of H2O in water is 1 and the concentration of NaCl (or anything else) in pure water is 0. Graeme Bartlett (talk) 00:03, 31 October 2008 (UTC)[reply]
Actually, the above is not really the reason. It sounds good, but it doesn't get to the heart of the problem. The one could certainly calculate the "concentration" of a pure solid. It would merely be the density (grams/liter) divided by the molar mass (grams/mol). You would get mol/liter, which is of course concentration. After all, we can and do use concentrations of pure gases, and their calculation can be done essentially the same way. The same problems where a pure gas concentration is used as calculated, the pure solid concentration is treated as unity. This is not the same thing as being equal to one, but we'll get to that in a minute. The question is "Why do you need to know the concentration?". For the problems I am guessing you are doing, you are probably working in kinetics and equilibrium, that is the problems are largely dependent on the rate of reaction. The rate of a chemical reaction is dependent on the number of collisions occuring between the reactants. In gases and aqueous solutions, the reactants are spread out over a large space, so how often they colide is a direct function of how close together they are packed. If you double the number of molecules of a gas or an aqueous solution in a given space, the number of collisions doubles as well, because the molecules are forced closer together. Thus, changes in amounts effect changes in rate. Now, here's the thing with pure solids and pure liquids. They are condensed phases, which means that the molecules are already touching essentially. If you, say, double the number of molecules of a pure liquid, you don't effect the rate at which those molecules collide since those molecules can't be forced closer together. Thus, for questions that involve rates of reaction, the amounts of condensed phases don't affect the rate of the reaction. In our calculations, we ignore these amounts because they don't change the outcome. Ignoring them is the same as multiplying by 1 (since multiplying by 1 doesn't do anything at all). So we say that we treat them as "unity", but we could just as easily say we ignore them all together, because they don't affect the outcome. When you do an equilibrium calculation, well, the equilibrium is just a situation where the rates of the forward and reverse reactions are equal. Since equilibrium is determined by rate, and solids and liquids don't effect rate anyways, we can ignore these here too... --Jayron32.talk.contribs02:18, 31 October 2008 (UTC)[reply]
Having been a chemistry teacher for 10 years, I think I am fairly qualified to make these assumptions. The only type of situations where one finds the statement "treat the concentrations of solids and liquids as unity" is in the kinteics and equilibrium chapters of your standard general chemistry textbook. Its the particular phrasing he used, and that he mentioned that it came from his textbook that cued me in on the kinds of problems he was working out. Trust me on this, he's working out rate law and equilibrium problems. --Jayron32.talk.contribs03:26, 31 October 2008 (UTC)[reply]
Yup. The real problem is when they think that context doesn't matter, which is why I led with that caveat. Otherwise they get into my chemistry class and tell me "the concentration of water is 1" or "solvents don't have a concentration themselves" because that's what they were told, not knowing what the heck it actually means:( DMacks (talk) 04:04, 31 October 2008 (UTC)[reply]
When working with reaction rate and equilibrium-type problems, you're not actually using the concentration of the substance, you are using the Activity (chemistry). This isn't an absolute measure, it's a dimensionless relative measure. Usually, the reference concentration is taken to be 1 mol/L, so that the activities match exactly with the concentration. However, this is not always the case, and very concentrated solutions can have activities which differ numerically from their molar concentration. In the case of solids, the activity doesn't change when increasing the amount of undissolved material. Thus the activities are constant, and are effectively rolled into the rate constant or equilibrium constant. The same thing happens with rates and equilibria in aqueous solution: the amount of water in most solutions is near constant (~55 M), and varies very little from one aqueous reaction to another, and is defined as unity. If you happened to be working in a solution where the concentration of water was significantly different from a normal aqueous solution (e.g. in a highly concentrated alcohol in water mix) you would have to explicitly account for the water concentration, and would have to use rate and equilibrium constants which account for this. -- 128.104.112.72 (talk) 15:51, 31 October 2008 (UTC)[reply]
