:I had toothache that was relieved by swirling water. It was due to a large cavity between my teeth - due to moving town I had not had regular check-ups for some time - my fault for not signing up with another dentist and not the NHS's. Coincidence - it was the last tooth on my upper left as well! Luckily it was just a wisdom tooth, so it was removed without any cosmetic or functional effects. Am I right in guessing from the OP saying that they could not afford treatment that OP is from the USA? (Why dont Americans revolt and demand free health care from their government?) [[Special:Contributions/78.146.180.118|78.146.180.118]] ([[User talk:78.146.180.118|talk]]) 20:17, 30 March 2010 (UTC)
:I had toothache that was relieved by swirling water. It was due to a large cavity between my teeth - due to moving town I had not had regular check-ups for some time - my fault for not signing up with another dentist and not the NHS's. Coincidence - it was the last tooth on my upper left as well! Luckily it was just a wisdom tooth, so it was removed without any cosmetic or functional effects. Am I right in guessing from the OP saying that they could not afford treatment that OP is from the USA? (Why dont Americans revolt and demand free health care from their government?) [[Special:Contributions/78.146.180.118|78.146.180.118]] ([[User talk:78.146.180.118|talk]]) 20:17, 30 March 2010 (UTC)
::Did you read the post? He does have an appointment to get it fixed, and presumably can afford treatment. But his regular dentist is backed up, so he has to wait a little while (or find another dentist, which he has not yet tried to do). —[[User:ShadowRangerRIT|ShadowRanger]] <sup>([[User talk:ShadowRangerRIT|talk]]|[[Special:Contributions/ShadowRangerRIT|stalk]])</sup> 20:22, 30 March 2010 (UTC)
::Did you read the post? He does have an appointment to get it fixed, and presumably can afford treatment. But his regular dentist is backed up, so he has to wait a little while (or find another dentist, which he has not yet tried to do). —[[User:ShadowRangerRIT|ShadowRanger]] <sup>([[User talk:ShadowRangerRIT|talk]]|[[Special:Contributions/ShadowRangerRIT|stalk]])</sup> 20:22, 30 March 2010 (UTC)
:::But I though this kind of delay was what Americans imagined happened with the UK NHS, and not what is supposed to happen in the US? I mean, '''May''' and he's in pain! Over a month away! This side of the Atlantic you'd probably get it done as an emergency appointment the same day, or the next day at most. [[Special:Contributions/78.146.180.118|78.146.180.118]] ([[User talk:78.146.180.118|talk]]) 20:26, 30 March 2010 (UTC)
== Cancer disease ==
== Cancer disease ==
Revision as of 20:26, 30 March 2010
Welcome to the science section of the Wikipedia reference desk.
The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
It's not when the mother and baby don't match, it's when the mother is Rh negative, has been previously sensitized to the Rh antigen, and the baby is Rh positive. The reverse situation (mother Rh positive, baby Rh negative) is not a problem. Or, simplifying and applying it to the original question: when the mother is Rh positive, no problems are anticipated, regardless of the Rh status of the father or the baby. - Nunh-huh01:41, 26 March 2010 (UTC)[reply]
Well, yes, that's my point, and why I felt compelled to clarify your answer. In the question asked, there are no "potential" problems. An answer that suggests that there are is potentially misinformative. - Nunh-huh02:32, 26 March 2010 (UTC)[reply]
There's no uncertainty at all that Rh incompatibility does not result when the mother is Rh positive - which is the question asked. - Nunh-huh14:31, 26 March 2010 (UTC)[reply]
Ok, I (and probably Cuddlyable3) didn't understood what you were trying to say, until just now. Thanks, for the clarification. StuRat (talk) 17:05, 26 March 2010 (UTC)[reply]
I agree that your title is better, but I also tried to keep the original title, in case it's used as a search term by the Original Poster to find their Q.StuRat (talk) 12:31, 26 March 2010 (UTC)[reply]
My wife is RH- and has to have the injection before giving birth and after, the reason given is as follows - the mothers immune system would atk the feotus - Thas what the doctors said to me and my wife when i asked why it is neededChromagnum (talk) 08:11, 30 March 2010 (UTC)[reply]
A cityscape is the profile of a city skyline, a skyline can be the view of any notable horizon, e.g. buildings, mountains or whatever. —Preceding unsigned comment added by 86.4.186.107 (talk) 07:02, 26 March 2010 (UTC)[reply]
My inner geekdom asks: hypothetically if "paxilon" here refers to a synonym for the herbicidemethazole (C9H6Cl2N2O3), is "PAX" a plausible chemical and what might be its formula based on the name "paxilon hydrochlorate"? How would "G-23" be a realistic part of its nomenclature?
Douglas Adams was asked many times during his career why he chose the number 42. Many theories were proposed,[7] but he rejected them all. On November 3, 1993, he gave an answer[8] on alt.fan.douglas-adams:
“The answer to this is very simple. It was a joke. It had to be a number, an ordinary, smallish number, and I chose that one. Binary representations, base thirteen, Tibetan monks are all complete nonsense. I sat at my desk, stared into the garden and thought '42 will do.' I typed it out. End of story. --Aspro (talk) 18:16, 26 March 2010 (UTC)[reply]
The Moon currently orbits the Earth and so it doesn't crash into the Earth (at least, in our lifetime). For this thought experiment, assume that the Flying Spaghetti Monster suddenly sticks out his noodly appendage and brakes the moon to a halt (we'll use the Earth as our reference frame just for convenience - from the point of view of an observer on Earth, the Moon is now at "absolute rest"). Now, instead of an orbiting moon, we have a moon that is in free fall and on a collision course with the nearest large body (aka Earth). My question is, how long will it take for the moon to collide with the surface of the earth? Assume that the only bodies that have significant gravity are the moon and the earth, the gravitational constant G is 6.67 x 10−11kg−1m3s−2, the mass of the earth is 5.9742 x 1024 kg, the mass of the moon (maybe not important, but here just for reference) is 7.349 x 1022 kg, initial distance between earth and moon (center to center) is 384403000 m, and the radii of earth and moon are 6378100 and 1737400 m, respectively.
It looks to me as though solving the problem involves a differential equation, which I don't know how to use. So instead I approximated a solution using Excel. According to my estimate, it would take 4.85 days for the two objects to collide. Is this close to the correct answer? 66.178.138.144 (talk) 18:02, 26 March 2010 (UTC)[reply]
Well, even without full diff-eq (which I'm not an expert at; I'm more of a number theory guy), an iterative solution should get you fairly close. Just use Newton's law of universal gravitation with your initial conditions to find the instantaneous acceleration involved. Then assume that acceleration applies for, say, an hour, and use basic calc to find the distance traveled.. Then recalculate the acceleration for the reduced distance. Repeat until the distance from the center of the Earth to the center of the Moon is less than their combined radii. It will actually happen a little more quickly than that (since it spends most of each hour accelerating more rapidly than it did at the very beginning), but it should be close enough. —ShadowRanger(talk|stalk)18:09, 26 March 2010 (UTC)[reply]
The Moon was in free fall even before the noodly FSM meddled with it. Hopefully he/she/it will allow the "textbook" equation for vertical motion of a falling object ignoring air resistance to give an approximate answer. Taking the OPs figures:
Distance to fall y = 384 403 000 - 6 378 100 - 1 737 400
= 375 287 500 m
Free fall acceleration g = 9.81 m/s²
0.5
Then time to fall t = (2 y / g )
= 8747 s
~ 2 hours 26 min
Also, while in normal gravitational calculations on Earth (where the object attracted to the Earth has trivial mass), the force exerted by the object on the Earth can be ignored. In this case, the mutual gravitational attraction between the Earth and the Moon would rapidly become non-trivial, so you'd need to keep track of the acceleration of the Earth as well, assuming you want a really precise answer. —ShadowRanger(talk|stalk)18:50, 26 March 2010 (UTC)[reply]
In other words, the gravitational attraction from the Earth is far less than 9.81 m/s^2 at the distance of the Moon (which would make it take longer) and the Earth falling toward the Moon must also be considered (which would make it shorter). I believe the former has more effect than the latter, so it ends up longer. StuRat (talk) 18:52, 26 March 2010 (UTC)[reply]
I'm aware it's a weird ethnic background. My dad was Eastern European Ashkenazi, my mother was a mix of Irish and Scots-Irish who converted. I was raised Jewish, but only the cultural aspects took. That said, I'm fairly sure neither the Invisible Pink Unicorn nor the Flying Spaghetti Monster nor the Abrahamic God is likely to post on this reference desk and give us the answer. :-) —ShadowRanger(talk|stalk)19:10, 26 March 2010 (UTC)[reply]
My kids will be even more confusing if someone tries to categorize them, as I'm engaged to an ethnically Chinese woman, born in Indonesia, but raised in the States. They'll get to check *at least* half the boxes on the Census form! —ShadowRanger(talk|stalk)19:16, 26 March 2010 (UTC)[reply]
If we make a boundary for 9.8>g>0.163 m/s2, then one can verify at least that the time must be between 2.5 and 19 hours. I don't think it will take days when solving the differential equations (neglect air friction).--Email4mobile (talk) 20:44, 26 March 2010 (UTC)[reply]
Hi, OP here. Are you sure that this is the correct equation, Dragons flight? 6.82 days would be an average acceleration of 0.00221 m/s2, which is smaller than the initial acceleration of 0.00270 m/s2! I would expect acceleration to increase in magnitude as the moon and earth get closer together.
Thanks everyone for the answers you've given so far; they've been helpful. I was assuming that the mass of the moon was negligible (as implied by my original question) but I tried my Excel "iterative solution" again with the mass of the moon factored in and it didn't make a significant difference. This is understandable - the moon is two orders of magnitude less massive than the earth. If someone knows how to do the diff-eq, I would appreciate them chiming in. :) Oh yes, I found this: [1] archived version of [2] which is currently down. They say it takes "about five days" so that tells me I'm in the right ballpark according to them. I'd still appreciate more confirmation. 66.178.138.206 (talk) 23:15, 26 March 2010 (UTC)[reply]
Dragons flight's formula is almost right. The correct one is
The extra square root of two that was missing changes the final answer to 4.82 days which is pretty close to the answer from the spread sheet. Dauto (talk) 02:05, 27 March 2010 (UTC)[reply]
The question is based on the moon suddenly accelerating towards the earth. To achieve this, the questioner suggested the Flying Spaghetti Monster has brought the moon to a sudden and complete stop. In fact, this is unnecessary because the moon is already accelerating towards the Earth. Whether the moon is travelling at its current speed, or has been brought to a stop by a Spaghetti Monster, is irrelevant because the acceleration towards the Earth will be the same in both cases.
The moon orbits the Earth about once every 28 days. Falling all the way from its current orbit to the level of the center of the Earth is nearly the same as completing one quarter of an orbit. One quarter of 28 days is 7 days, but of course that is based on the acceleration due to Earth's gravity at the radius of the moon's orbit. If the moon were falling directly towards the Earth it would progressively be subjected to increasing gravitational attraction and that would hasten the fall. Is this progressive increase in gravitational attraction sufficient to explain the difference between about 7 days and 4.82 days? Dolphin51 (talk) 05:17, 27 March 2010 (UTC)[reply]
The moon's orbital period is about 27.3 days and a quarter of that is about 6.82 which is what Dragons flight calculated above, but that misses a factor of square root of two. Dauto (talk) 14:52, 27 March 2010 (UTC)[reply]
I've read through the answers to the OP's question and I can't tell which estimate or range of estimates is most likely correct. Could someone sum up? 63.17.63.71 (talk) 00:13, 28 March 2010 (UTC)[reply]
Ok here are the highlights
The OP uses a numerical solution and finds T = 4.85 days which is essentially correct but he want an analytical solution
Cuddlyable3 uses a constant acceleration g = 9.8 but that will only give a lower bound since Earth's gravity decreases with distance: T > 2 hours and 26 min
Email4mobile uses the bounds 9.8> g > 0.0266 m/s2 and finds 6 > T > 0.1 days
Dragons flight points out that this problem can be solved using Kepler's laws and solutions but makes a couple of mistakes in the application of the formula. He uses the wrong value for the semi-major axis which actually is half of the moon-earth's distance and devides the result by four instead of two which is the correct way to do it.
Finally I point out those mistakes and find then result T = 4.83 days which is pretty close to the OP's original value.
TLDNR, but have you taken into account which side of the Earth the moon is stopped on? Ie, is the moon in front of Earth, or behind it? Aaadddaaammm (talk) 19:19, 28 March 2010 (UTC)[reply]
That makes no difference since we assume the moon's stopped with respect to the earth. It makes some difference if the moon stops at apogee or perigee but we've been neglecting that. Dauto (talk) 19:53, 28 March 2010 (UTC)[reply]
Has anyone produced a black and white only digital slr? The sensors aren't wavelength-sensitive, so in order for them to differentiate wavelength, they need a filter in front of them of the right colour. However, if you remove the filters, you could squeeze in more pixels per square inch of the sensor, but they'd only sense light and dark. I would imageine it would have a highly specialist use in high iso, high shutter speed, low light photography. But has anyone built one? —Preceding unsigned comment added by 81.147.56.162 (talk) 18:23, 26 March 2010 (UTC)[reply]
The digital output of the camera doesn't "know" about the colour filters so the full pixel resolution is there already (before jpeg compression). Removing the filters would gain some sensitivity. Since they are added last to the CCD chip it might be possible to bleach them out. Cuddlyable3 (talk) 18:49, 26 March 2010 (UTC)[reply]
If nobody has done this, it sounds like a good idea to me. It would be good for pics of color subjects where black and white is used for artistic reasons, and also for subjects which lack color. StuRat (talk) 19:01, 26 March 2010 (UTC)[reply]
For subjects that lack colour the individual pixels are immediately available in the BMP image format and only need relative weighting of Red/Green/Blue. Cuddlyable3 (talk) 19:39, 26 March 2010 (UTC)[reply]
It's not a new idea (not necessarily in a DSLR; possibly a digital back for medium or large format cameras, or a rangefinder). I don't think any such camera has gone into production, though, presumably because of the relatively small market. -- Coneslayer (talk) 19:52, 26 March 2010 (UTC)[reply]
I know that some microscope cameras are monochrome only. Over the past couple of months I have been using a Leica fluorescence stereomicroscope as part of a research project (if you are particularly interested I was using this one [3]) in conjunction with a digital camera (this one [4]). It doesn't need to be full colour since it is being used for fluorescence microscopy - so the only colour it will see is the fluorescence emission wavelength coming through the emission filter of the selected filter set. If I needed multiple fluoresence channels I took individual images with different filter sets, false coloured them and overlayed them in photoshop. Sensitivity is very important here, which is another reason to use monochrome. It's also interesting to note that the camera was only 2 megapixels, since minimising noise is far more important in this application than maximising resoultion (If you need to enlarge something you use a higher magnification lens, whereas if there is too much noise to precisely interpret/measure fine details in the image you're stuck). Equisetum (talk) 19:53, 26 March 2010 (UTC)[reply]
Yes. Several models in the Kodak/Nikon Pro DCS line were available in three chip styles, Color, Infra-Red, and Black And White. I see that it's not mentioned in our article, but see here.
The deal is that a color digital camera has groups of (typically) four sensors grouped into a single "pixel" - one red, two green and one blue. But a slightly different part of the image lands on the red, green and blue sensors. Hence, if you can get the "RAW" image (not JPEG or whatever) and if you know the physical layout of those four sensors, then with the right software you can get a good monochrome image at somewhat higher resolution than you'd get a color image from the exact same camera. Since monochrome digital cameras are not in high demand, they'd be costly specialist items. So you'd be better off using a color camera that saves "RAW" format images - and use some really good image processing software to make your final monochrome image. SteveBaker (talk) 23:06, 26 March 2010 (UTC)[reply]
From some more searching, it looks like the Kodak DCS cameras mentioned above may have been some of the last SLR cameras to offer monochrome chips as a standard option. (In certain forums they seem to be fondly remembered for this reason.) It looks like a monochrome conversion may be possible. (ie: They remove the colored filters from the chips.) Check it out [5].
Within the context of talking about food safety, would it be correct to call ethyl alcohol a "toxin"? Our article does not say so, and neither do the otherarticles I've looked at it, but it would seem to (loosely) meet the definition. Created by a living thing, harmful if swallowed, etc. Our articles treat it within the context of it being a "drug", which it certainly also is, but I wonder if the term "toxin" would also make sense. Matt Deres (talk) 18:37, 26 March 2010 (UTC)[reply]
More importantly, it has an LD50 that can be achieved without exceeding the ability of the human in question to consume it. There are toxins with theoretical LD50s that, practically speaking, are never hit, because they must be consumed and the amount required to kill you is so great that it can't all fit into a normal human before the excretion processes end up removing it. —ShadowRanger(talk|stalk)18:55, 26 March 2010 (UTC)[reply]
But there must be all sorts of things that we could use to kill ourselves by eating too much, does that make them all toxins ? StuRat (talk) 18:57, 26 March 2010 (UTC)[reply]
Yeah, I'm not real sure on that definition. Water can kill you by drinking too much of it (no need to drown), though usually only if you are drinking truly absurd amounts and not peeing or sweating out an appreciable amount, and/or you keep it up long enough that you manage to disrupt the electrolyte balance in your body. That said, alcohol is clearly a toxin; it can kill you by the direct metabolic effects of the alcohol, not just by displacing or disrupting the balance of various chemicals in your body. —ShadowRanger(talk|stalk)19:02, 26 March 2010 (UTC)[reply]
Perhaps the word 'acute' should have gone before 'dose' and 'low' before 'median'. And of course, semantics can make these things confusing. This may account for the habit of clinicians to say 'poisoning' for when it is life threatening.--Aspro (talk) 19:31, 26 March 2010 (UTC)[reply]
No, ethanol is not a toxin. Toxins and poisons are two different things. Ethanol can be a poison in some doses but is never a toxin. alteripse (talk) 19:43, 26 March 2010 (UTC)[reply]
Can you explain why? Our article says that "a toxin is a poisonous substance produced by living cells or organisms". Ethanol can be produced by fermentation with yeast. This would seem to fit that definition - why doesn't it? 94.168.184.16 (talk) 20:06, 26 March 2010 (UTC)[reply]
I just wrote you a long answer that was completely lost when Stu answered the next question. Here are 6 more definitions of toxin: [6]. If you define toxin so broadly that it refers to any substance produced by an organism that can cause harm, then you must consider water, oxygen, carbon dioxide, chalk, and glucose to be toxins because they are all substances produced by organisms that can cause harm in sufficient quantities, rendering that usage pretty meaningless. alteripse (talk) 20:26, 26 March 2010 (UTC)[reply]
Yes, you've discovered my master plan to train everyone to be concise by destroying any responses which fail the test. :-) StuRat (talk) 03:29, 27 March 2010 (UTC)[reply]
Yea, with any modern web-browser you should be able to simply hit the [BACK] button and retrieve what you have written. Admittedly, this isn't immediately obvious. APL (talk) 21:27, 26 March 2010 (UTC)[reply]
“rendering that usage pretty meaningless”
To whom? Without identifying a cohort that statement may apply to, don't you think that statement is, err . . . pretty meaningless. Are not clinicians in The Royal Navy, oil recovery industries etc. familiar with Oxygen toxicity?. Maybe I should have said "And of course, some people use semantics to make these things confusing"--Aspro (talk) 23:48, 26 March 2010 (UTC)[reply]
If you wish to use terms imprecisely, go ahead, but I cannot think of any scientific fact that would justify putting ethanol in different "toxin" category than other simple substances produced as part of normal metabolism by many organisms. I think you are confusing a cultural classification with a scientific one (i.e. "clean" and "unclean" vs healthy or harmful)--- and yes I acknowledge that all semantic distinctions can be considered "cultural". alteripse (talk) 11:23, 27 March 2010 (UTC)[reply]
Ethanol is considered toxic because even consuming it in small amounts kills some brain cells, starts some cirrhosis, etc. Other substances, like water or glucose, are necessary for life and are not toxic in normal amounts.--Cheminterest (talk) 20:59, 26 March 2010 (UTC)[reply]
Alteripse has a good point when he linked to those other definitions of toxin; the term seems to be commonly applied only to proteins, in which case ethanol is definitely not a toxin. But again, it's largely a matter of semantics. The ill effects of alcohol are widely studied and well known, so there's no real need to group it into a vague category like "toxin" that can have multiple meanings and be applied to substances with widely varying mechanisms and magnitudes of harm. Buddy431 (talk) 03:52, 27 March 2010 (UTC)[reply]
Abstract for Ethanol Toxicity Paper Ethanol exposure during development is teratogenic. The central nervous system (CNS) is particularly susceptible to ethanol toxicity. In fact, heavy gestational ethanol consumption is one of the leading known causes of mental retardation in the Western world. Ethanol exposure disrupts the proliferation of glia and neuronal precursors in the developing CNS. Depending upon cell population and blood ethanol concentration, ethanol can either inhibit or stimulate cell proliferation. Two features of cell proliferation that are affected by ethanol exposure are the growth fraction (the proportion of cells that is actively cycling) and the cell cycle kinetics, particularly in the length of the G1 phase of the cell cycle. Cell proliferation in the developing CNS reflects the action of positive (mitogenic growth factors) and negative (anti-proliferative factors) regulators. Increasing evidence shows that ethanol interferes with the action of growth factors. In vitro systems are a good model to investigate ethanol neurotoxicity, since the effects of ethanol on cultured cells parallel the effects of ethanol in the developing CNS. The inhibitory effects of ethanol on cell proliferation may result from interference with mitogenic growth factors (e.g., bFGF, EGF, PDGF, IGF-I). Conversely, the stimulatory effects of ethanol may result from the interference with growth inhibiting factors (e.g., TGFβ1). Interestingly, both in vivo and in vitro studies show that proliferating neural cells display differential sensitivity to ethanol. This differential sensitivity correlates with their response to mitogenic growth factors; that is, cells that are actively regulated by mitogenic growth factors are much more susceptible to ethanol than cells that are less or unresponsive to such factors. Ethanol interference with growth factor action could occur at three levels: ligand production, receptor expression, and/or signal transduction. Thus, ethanol-induced alterations in the developing CNS that characterize fetal alcohol syndrome apparently result from alterations in the regulatory action of growth factors.--Cheminterest (talk) 20:32, 29 March 2010 (UTC)[reply]
Rusted hammer
Someone borrowed my hammer and now the head is all rusted. It's got a fiberglass handle (no wood) so I was wondering if there's anything I can do (soak it in something, perhaps) that will de-rust it? DRosenbach(Talk | Contribs)20:11, 26 March 2010 (UTC)[reply]
If it's a Craftsman (Sears) hammer, you can return it for a replacement, they are quite generous with their terms. Otherwise, I'd just sand off the rusted spots. But beware if the head is now loose in the jacket, as it could fly out during use. And you could always tell whoever borrowed it that they owe you a new hammer; I would. StuRat (talk) 20:21, 26 March 2010 (UTC)[reply]
There are chemicals used to remove rust, but they might cost as much as a new hammer, so that's why just sanding it off might be best. And be sure to keep the metal portions of this hammer oiled from now on, as obviously it has a tendency to rust. StuRat (talk) 20:23, 26 March 2010 (UTC)[reply]
I would second the suggestion for WD40 and steel wool. You can also paint it with a rust converter, which will turn it black, but protect it from future rust (it converts iron oxide to iron tannate). Ariel. (talk) 20:33, 26 March 2010 (UTC)[reply]
That will teach you to lend your tools to somebody else! Three rules for tools:
I'd have to disagree with 1 and 3. Number 1 because there are times when you only need a tool for a single use, and buying the premium model would be a waste, and number 3 because that will result in your friends not lending you any tools either. If you and a buddy each lend the other tools and treat them with respect, you can double the number of tools each of you have at your disposal. StuRat (talk) 21:00, 26 March 2010 (UTC)[reply]
Wow! OK, I am envious. I'm left handed and thus find their sharpened on the wrong side, can they by ground the other way?--Aspro (talk) 21:38, 26 March 2010 (UTC)[reply]
I'm left handed too, and my santoku is actually sharpened on both sides, and isn't Japanese, which I guess means it isn't a proper santoku. If you're interested the knife is [7] and it only costs $20, I have several of that series of knives, recommended by my uncle who takes his cooking quite seriously. Neither of us has any idea how Kuhn Rikon makes such good knives so cheaply - my only concern is that at that price I don't know how long they'll last, but for $20 it's worth taking a chance. The reason I won't lend it is mainly that it's a very thin blade and if you tried to cut inappropriate things with it (like bones) it would get damaged. Equisetum (talk) 22:39, 26 March 2010 (UTC)[reply]
A dip in hydrochloric (muriatic) acid to dissolve the iron III oxide followed by a dip in phosphoric acid may remove and prevent rust. --Cheminterest (talk) 20:55, 26 March 2010 (UTC)[reply]
If you are exceptionally proficient with a tool, even if that proficiency is largely a self-held opinion, you would not want to loan such tools to anyone. The other person will always be using that tool in a less desirable way. And most tools are not indestructible. Bus stop (talk) 21:48, 26 March 2010 (UTC)[reply]
It's worth periodically squirting a little WD40 onto a shop cloth and wiping down all of the metal parts of your tools once in a while - especially those used to cut wood such as saws and chisels. That thin film of water-displacing oil keeps them shiney for months. I usually use a Dremel with a wire-brush head for removing superficial rust from old car parts and such like...but once something has gotten rusty, any surface coating it may once have had is gone and it'll rust again, soon after you've cleaned it - so using some kind of water-repelling oil on the surface is a good idea. (Bonus Fun fact: WD-40 is so called because it was Norm Larsens 40th attempt at creating a Water Displacing oil! You've gotta admire Norm's determination - even if you're less than impressed by his product naming skills!) SteveBaker (talk) 22:53, 26 March 2010 (UTC)[reply]
Are you sure you weren't thinking of Coca Cola? Its phosphoric acid content makes it a good mild rust remover. (And, in fact, a demonstration of its rust-removal properties is often used as an object lesson in how bad the stuff is for your teeth...) —Steve Summit (talk) 17:25, 27 March 2010 (UTC)[reply]
There's the electrolytic technique [8] (I don't remember where I found this link but I bet it was from the ref desk). The molasses fact is excellent, I wonder if there's a way to speed the reaction up though. 213.122.27.130 (talk) 19:14, 28 March 2010 (UTC)[reply]
Molasses may prevent corrosion by blocking off the supply of free oxygen needed to corrode the iron. Also, if you put the tool as the cathode in an electrolytic cell containing copper sulfate or silver nitrate and use a copper or silver anode, respectively, you may copper or silver-plate your tool. Copper and silver will get tarnished eventually, though; then just dip them in hydrochloric acid. --Cheminterest (talk) 20:24, 29 March 2010 (UTC)[reply]
What substances would spontaneously ignite in a nickel-metal hydride cell? I took one apart, and some particles from one electrode contacted particles from another electrode. After a couple of minutes, both the electrolyte paper and the cathode material were up in flames. All of the black corrosion on the cathode was burnt away, and the paper was charred. I quenched the fire by putting a heavy piece of metal on top of the materials. It immediately extinguished it. There also seems to be a strong oxidizer in the anode, because it produced large quantities of chlorine gas when reacted with hydrochloric acid. --Cheminterest (talk) 20:53, 26 March 2010 (UTC)[reply]
Diet defined by required and forbidden foods
I have found a table of foods which distinguish carnivore from vegan diets and Kosher from Hindu diets. Is there a list of required or forbidden foods that can be added to the table to distinguish Muslim (or Islamic) diets and Christian diets from the rest? 71.100.5.192 (talk) 21:46, 26 March 2010 (UTC)[reply]
I don't think there is a "Christian diet" for mainstream Christian churches. At some time in the early first millennium what was to become the orthodox church decided to interpret Matthew 15:11 as abolishing Jewish food laws and allow Christians to eat everything, even Big Macs. But see Pentabarf. --Stephan Schulz (talk) 22:18, 26 March 2010 (UTC)[reply]
Taking that last point seriously, I don't think Christianity explicitly forbids cannibalism in all circumstances (though obviously it forbids most actions that would enable cannnibalism, such as murder). I recall reading that after the events of the Andes flight disaster, in which some survivors ate flesh from the frozen bodies of those killed in the crash in order to last out until rescue, the then Pope stated that they had acted correctly. If this was true and not misreporting, it might be added to the article. 87.81.230.195 (talk) 07:22, 27 March 2010 (UTC)[reply]
That's pretty funny! Matthew 15:11 says: "Not that which goeth into the mouth defileth a man; but that which cometh out of the mouth, this defileth a man.". So an off-hand comment that's clearly intended to mean that you shouldn't talk nonsense is taken to overturn an entire raft of detailed rules about diet?! It's just as well nobody actually reads the Bible isn't it? Anyway - there is at least one dietary law for many christians - where does the Catholic "Fish on Friday" rule come from? SteveBaker (talk) 22:42, 26 March 2010 (UTC)[reply]
That's a very interesting point, Steve (if I may call you by your first name). Hospital dietitians of today normally recommend a 30% fat, 60% carbohydrate and 10% protein diet with exceptions made for particular aliments on doctors orders. I found a table more or less for meals based on various combinations of foods but I do not know where in the Wikipedia to find or simulate this. 71.100.14.194 (talk) 23:16, 26 March 2010 (UTC)[reply]
Re Matthew: Well, I suspect that the conflict between Christianity as a Jewish sect, with all the inconveniences of food restrictions and Sabbath and Circumcision, and Paul's more radical "it's a new religion. Eat pork and keep your bits" had more to do it than the actual (pre-)text. If you are looking for a justification for anything in the bible, you're likely to find a suitable passage. And Paul won... --Stephan Schulz (talk) 23:31, 26 March 2010 (UTC)[reply]
I thought the no-meat-on-Friday thing was obsolete? I thought it was now acceptable for a catholic to substitute some other penance? Even a penance that was not food-related. APL (talk) 02:51, 27 March 2010 (UTC)[reply]
It was also circumvented by theologically defining beavers and other aquatic animals as fish, I seem to remember. I wonder if Christians thought their God would be hoodwinked or favourably impressed by such manouevring? 87.81.230.195 (talk) 07:22, 27 March 2010 (UTC)[reply]
In southern Germany there is a kind of ravioli called Maultaschen. The local rumor is that Catholics can eat them on Fridays, since god won't see the meat inside the pasta ;-). More seriously, this kind of dissimulation makes sense if the purpose of many religious rules is understood to not be in the commandment itself, but rather in the function of setting followers of a religion apart from others. --Stephan Schulz (talk) 14:43, 30 March 2010 (UTC)[reply]
I suppose it all depends upon whether the diet plans are based on religious belief or have a health or scientific basis. For instance, not eating pork may have a religious basis but prior to that the basis appears to be health. 71.100.14.194 (talk) 23:34, 26 March 2010 (UTC)[reply]
That table probably shouldn't be there - it's redundant to the one directly above, with the only difference being that it is much less attractive. Intelligentsium23:36, 26 March 2010 (UTC)[reply]
The new Wikipedia is about being attractive? What ever happened to the idea of presenting content in a usable and meaningful way. You can always add color and take care of the attractive end but I find it completely atypical of a lack of anything better to do than for administrator to remove content for such a stupid rational. 71.100.5.167 (talk) 00:52, 27 March 2010 (UTC)[reply]
The content was largely already there, you were simply duplicating it and in a non encyclopaedic manner to boot (wikipedia is not an instruction manual nor do we refer to readers in the first person). This is neither useful, nor meaningful to most readers who are likely to simply wonder why wikipedia has two tables that basically say the same thing except with the axis reversed perhaps wondering if wikipedia editors think they are so stupid they are incapable of using a table if the axis aren't the way they want them (fortunately most editors think more highly of our readers). And given your history, I strongly suggest you do not edit war further over this Nil Einne (talk) 10:06, 27 March 2010 (UTC)[reply]
I don't know whether any of you were raised Christian but these are some pathetic explanations of Christian doctrine related to diet. The new covenant between God and man in Christ frees people from old Jewish law, including circumcision and dietary restrictions. The Reformation was a "reboot" in a sense, shedding a thousand years of accreted non-scriptural Catholic customs and restrictions that most people could not distinguish from Christ's primary teachings. The Protestants explicitly disavowed this type of "rule-based salvation" since a person cannot "earn" salvation by his own power and actions and is wholly dependent on God's grace. In other words, there are no Christian dietary laws. The Catholic custom of avoiding meat on Fridays was a custom that arose to honor Christ's carnal sacrifice on a Friday but has never had the strength of law. I have no idea what Mormon teachings are because Mormonism appears to be about as closely related to Christianity as Christianity was to Judaism around 200 AD. Now if there is anyone left unoffended, let me know and I will try to include you too. alteripse (talk) 11:18, 27 March 2010 (UTC)[reply]
There's also a specific repudiation of dietary law in Acts 10. Now, it's been pointed out to me that, in this case, food is almost certainly a metaphor for humanity (i.e. Peter is being informed that Cornelius is not unclean). But Christians are, as a whole, happy to also accept the surface meaning, i.e., they get to eat bacon now. PaulStansifer13:19, 27 March 2010 (UTC)[reply]
Unless I missed it, no one seems to have mentioned blood, and food offered to idols, which are forbidden to Christians in Acts 15, verses 19-20. The only major grouping I know of that pays attention to this is the Jehovah's Witnesses. --Trovatore (talk) 03:53, 28 March 2010 (UTC)[reply]
your extra comma changes your meaning. The significance was not the food, it was participating, or even appearing to condone, idol worship that was the prohibition. This is not really a dietary prohibition but a co-worshipping prohibition. And I think the blood prohibition for the Jehovah's witnesses is from the Pentateuch, not Acts, but i am not familiar enough with that sect to know the scriptural justification. alteripse (talk) 04:32, 28 March 2010 (UTC)[reply]
Well, in the NIV, it clearly seem to be independent commandments[9]. The same seems to be true (if in archaic grammar) for Luther's translation[10] and in modern German translations[11]. Even the KJV seems to list these as independent[12]. --Stephan Schulz (talk) 11:56, 28 March 2010 (UTC)[reply]
Looks like you are corrrect about consuming strangled animals and blood being a separate prohibition. Never seemed to catch on, though. alteripse (talk) 14:00, 28 March 2010 (UTC)[reply]
Alteripse, I'm pretty sure you're right. See 1 Corinthians 8, where Paul argues that idols being powerless, the only reason to be concerned about eating food sacrificed to them is whether people will take that as an endorsement of the idols. PaulStansifer18:38, 28 March 2010 (UTC)[reply]
Chile
I heard that in plate tectonics, Chile is being slowly pulled away and separated from South America and will one day move away as an island (can't find this on wikipedia)- is this true? Thanks, --AlexSuricata (talk) 23:33, 26 March 2010 (UTC)[reply]
(ec)You're probably thinking of California. From what I can see on the article on the nearby Nazca Plate, the Nazca Plate is perpetually moving towards and underneath the South America Plate. That wouldn't cause Chile to split, just to become ever more mountainous. —ShadowRanger(talk|stalk)23:39, 26 March 2010 (UTC)[reply]
Yup. As the plate moving in from the west gets pushed underneath South America, it melts and rises upwards to form the Andes. That's what I learned in high school geography anyway. And yes of course it's not quite that simple, but I think it's a fair approximate. Vranak (talk) 16:12, 27 March 2010 (UTC)[reply]
You're right Vranak, it is a bit more complex than that. The subducting plate itself doesn't melt, water from the sediments carried down with it are driven off, hydrating the asthenosphere of the over-riding South American Plate, thus lowering its melting point, causing large-scale partial melting. The magma rises up causing chains of volcanoes. This doesn't explain the origin of the Andes on its own, as much of the uplift is orogenic even though there is no continent-continent collision involved. There is, as yet, no generally accepted mechanism for the shortening along the andean margin, although relatively high convergence rates or high levels of coupling across the plate interface are amongst those suggested.[13]Mikenorton (talk) 16:45, 27 March 2010 (UTC)[reply]
March 27
Do topical vitamins still work if dietary intake is sufficient?
I know that a number of studies show topical vitamin A fights acne, and C and E rehydrate dry skin. Has it been studied whether these effects vary with dietary intake or bloodstream levels of these vitamins? That is, how do we know the topical vitamins aren't simply replacing missing dietary vitamins? NeonMerlin00:39, 27 March 2010 (UTC)[reply]
Just a guess - but most people get more than enough of those three vitamins in their diet - yet they still get dry skin and acne. If topical use of those vitamins actually has a measurable effect then they can't simply be making up for inadequate diet. SteveBaker (talk) 05:56, 27 March 2010 (UTC)[reply]
It is worth asking at the outset whether Vit A actually 'fights' acne and whether Vit C and E rehydrate the skin. My understanding is that acne is an infectious condition of the follicles of the skin and rehydration can only occur in the presence of added water. So, is vitamin A some form of antibiotic and how does vitamin C and E emulate water? I would check the source of the studies, the sale of cosmetics is riddled with dubious and outrageous science. Richard Avery (talk) 09:01, 27 March 2010 (UTC)[reply]
The hydration of skin occurs from the inside out as much as from the outside in, skin becomes dry because evaporation is occuring faster than replenishment from internally. Most "moisturizers" work not by rehydrating the skin, but by reducing evaporation from the skin. They are usually some sort of hydrophobic oil of some sort which prevents drying out. If the skin retains moisture, it doesn't dry out as fast. I have no idea if this is so, but Vitimins E & C may work by a similar mechanism. --Jayron3221:11, 27 March 2010 (UTC)[reply]
Animation of interference of waves coming from two point sources. There is strong nett radiation in directions where the source radiations add constructively and no radiation in the directions where they cancel. Cuddlyable3 (talk) 16:26, 27 March 2010 (UTC)[reply]
In that article, the description of the image to the right says: "As time progresses, the wave fronts would move outwards from the two centers, but the dark regions (destructive interference) stay fixed."
Is that true? 1) If the dark regions stay fixed, then how could the light regions move? 2) The animation further down the page does not have any stationary dark regions. Thanks 78.149.198.14 (talk) 01:19, 27 March 2010 (UTC)[reply]
The "dark regions" in the picture would be the cyan parts. There's several unmoving cyan lines radiating outward. There are also cyan regions moving along with the other-colored regions, but that's not what it's referring to. 67.172.112.226 (talk) 01:43, 27 March 2010 (UTC)[reply]
That incoherent caption is interfering with the propagation of my thoughts. When a wave from one source meets a wave (having the same frequency) from the other source, they destructive interfere at certain points. Those particular waves then move outward, but following waves will destructively interfere at exactly the same spots, so those locations will always be dark. Clarityfiend (talk) 04:27, 27 March 2010 (UTC)[reply]
"There's several unmoving cyan lines radiating outward." How can they radiate outwards and be unmoving at the same time? A contradiction. 89.243.43.75 (talk) 15:16, 27 March 2010 (UTC)[reply]
Radiating in the sense of being a line from the centre to the edge (like a radius of a circle), not radiating in the sense of radiation. The lines are static. There are 6 in the top half and 6 in the bottom half of the image. --Tango (talk) 16:48, 27 March 2010 (UTC)[reply]
Thanks for the clarification. So if for example you had two nearby radio transmitters, and were driving around with a car radio, then if you drove along one of the lines described above, you would not hear anything? Would this also apply if the radio transmitters were transmitting signals at a different frequency AND/OR a different phase? Thanks 78.144.250.185 (talk) 15:59, 28 March 2010 (UTC)[reply]
When two transmitters radiate identical signals then yes, cancellation causes nulls in some directions where a car radio would receive little or nothing. The directions of the nulls change if one source is phase shifted relative to the other. That is a simple example of a Phased array. If the sources are at different frequencies then the radiation pattern is not stable; if the frequencies are close the result is the same as a constantly changing phase shift. That is useful if one wants a transmitter beam to sweep across a given angle e.g. for a radar. Cuddlyable3 (talk) 22:45, 28 March 2010 (UTC)[reply]
Technically, you'd always hear something, as one of the transmitters would be further away, and wouldn't quite counter out the other. Incidentally, since sound is waves, you could just use two speakers so you don't need the radio. — DanielLC06:35, 29 March 2010 (UTC)[reply]
Obviously we are speaking technically. It is only necessary for the cancellation to reduce the signal strength below the noise threshold of the receiver. Practical factors such as the transmitters not radiating identical powers will change the depth of the nulls. The reason cancellation can occur is because the distances from transmitters to receiver are unequal. Sound wave cancellation can also occur but it is difficult to demonstrate in the presence of indoor acoustics, the separation of human ears, and the difficulty of producing omnidirectional point sound sources. It is also unpleasant to listen to the prolonged sinewave tone that would be necessary. Cuddlyable3 (talk) 15:26, 29 March 2010 (UTC)[reply]
gravity
Although of higher local intensity or density throughout the universe is gravity the equivalent or substitute for atheists what God is represents for theists? 71.100.6.104 (talk) 03:09, 27 March 2010 (UTC)[reply]
Gravity is the weakest of the fundamental forces! If I ever feel the need to worship one of the forces holding the universe together, I'm going to worship the strong force, and death to all the infidels that worship the Weak nuclear force.
Seriously, though, I'm not a hundred percent sure what you're asking, but I'm pretty sure that the answer is "no." APL (talk) 03:26, 27 March 2010 (UTC)[reply]
When you talk about God creating the Universe to people who believe that the universe was created by the laws of physics then you get the impression that they have a substitute for God. Since one theory about the end of the universe is that the universe will end when the last Black Hole explodes and gravity is deemed responsible for holding a Black Hole together then it seems like gravity in that context must or might be their "god."
It may be a minor or perhaps frivolous question to some but I'm trying to phantom what it is that such persons find to be a plausible explanation for the basis or creation of the universe other than God. 71.100.3.207 (talk) 03:51, 27 March 2010 (UTC)[reply]
Atheist usually don't see god as a plausible explanation for the basis or creation of the universe at all. I don't see why you chose gravity out of all things as a possible substitute for god. Dauto (talk) 03:55, 27 March 2010 (UTC)[reply]
(after edit conflict)
Oh. I think most atheists are OK with not knowing exactly how the universe came to be. I can't speak for anyone but myself, but I feel smarter acknowledging that I don't know than I would feel if I believed a creation myth. APL (talk) 03:56, 27 March 2010 (UTC)[reply]
Incidentally, I don't just feel this way about the GREAT MYSTERIES OF LIFE, I feel this way in general. If someone asks me for directions to the post office, I would prefer to tell them that I don't know where the post office is, rather than give them some made up directions that are probably wrong. APL (talk) 03:58, 27 March 2010 (UTC)[reply]
Buddhism, a religion that does not teach the existence of any gods with the exception of a few sects (when do religions ever agree on anything?), teaches that there was no creation of this universe or of any living things. Everything is impermanent and mutates from one form to another. 67.243.7.245 (talk) 04:10, 27 March 2010 (UTC)[reply]
Not to preach here but to reconcile any problems I have in terms of fantastical belief I see God as the entity that fits a certain logical definition of omni-everything. Sort of like playing a game of mime and people try to guess what it is you are miming. When no one can guess, finally they give up and you give a name to whatever it was. 71.100.3.207 (talk) 04:23, 27 March 2010 (UTC)[reply]
The problem with your idea is that God represents a great many things to theists. Atheists might believe some alternative cause for the creation of the universe (big bang theory, for example) - but they don't use that as a 'replacement' for all of the other things that theists use their "god hypothesis" for. Atheists don't worship anything - so right there, neither gravity nor anything else is equivalent to gods for theists. Also, most atheists would be immediately happy to replace the big-bang theory with some other competing idea if it proved to be a better fit for the observable facts. It takes a lot more than mere facts to persuade theists to change their views...which is likely the only reason they are theists in the first place. SteveBaker (talk) 05:54, 27 March 2010 (UTC)[reply]
Did you miss the sentence "atheists don't worship anything"? Claiming the contrary is a popular sport (at least among theists), but I've never understood why. Algebraist10:54, 27 March 2010 (UTC)[reply]
Another problem is--if God is real, that is, part of reality, then how did reality come to exist? This topic tends to lead to an infinite regress--and while i enjoy infinity as much as the next guy, I can't see how it supposts the notionon of God. Pfly (talk) 10:15, 27 March 2010 (UTC)[reply]
I think the closest you're going to get is that many but not all atheists have a reverence for the idea of the universe as a whole—with all its long history, its complexity, its improbability, and its incomprehensible vastness—that might come at least somewhat close to the reverence that many religious believers have for the idea of God. I would say (as an atheist of sorts, and one who has spent a long time associating with a wide spectrum of self-described atheists) that such is not a totally ridiculous statement. But atheists don't think the universe will help them (usually much to the contrary—the universe cares not for your problems!), or is a cosmic "self" that has any true intelligence, or that the universe dictates any kind of straightforward moral code.
Moreover, atheists are not organized. Some think Richard Dawkins is keen, some think he is an idiot. There is nothing more cantankerous and argumentative than a meeting of atheists. They agree on almost nothing (other than their lack of belief in the validity of religion), much less the proper basis for being an atheist. Trying to find a "substitute religion" is kind of a fruitless endeavor. I think you will do better in understanding "the other side" by just trying to make sense of it on its own terms—its beliefs, its values, its totems—than just trying to Find/Replace religious concepts with atheistic ones.
Anyway, no, "gravity" does not mean God to an atheist, it is not even close. Gravity is no more awe-inspiring than magnetism. (I think the latter is more spooky on a day-to-day level, to be honest!) It's neat that we can extrapolate from forces we experience here on Earth and make sense of the movement of planets, stars, galaxies, the entire universe as a whole! But an atheist is going to see that as a testament to human ingenuity, nothing more. Perhaps that is the more profound point: if you want to see what an atheist substitutes for "God", look in a mirror! ;-) --Mr.98 (talk) 12:41, 27 March 2010 (UTC)[reply]
Note: the question appears to be related to a quote attributed to Carl Sagan: "The idea that God is an oversized white male with a flowing beard who sits in the sky and tallies the fall of every sparrow is ludicrous. But if by God one means the set of physical laws that govern the universe, then clearly there is such a God. This God is emotionally unsatisfying... it does not make much sense to pray to the law of gravity." It's not listed on his Wikiquote page, but you can find it all over the web (and never directly referenced, at least in my brief search). Matt Deres (talk) 16:00, 27 March 2010 (UTC)[reply]
Another example is that pantheists believe that God is the Universe, or that there are ancient religions that regard God as a "spaceship" or an Anunnaki. Another proposition is that God created the Big Bang in an attempt to intertwine religion with modern science. ~AH1(TCU)16:37, 28 March 2010 (UTC)[reply]
dark matter
is it possible that dark matter,is/are,neutrally oscillating strings that are vibrating at the same frequency/pattern as the Universe itself.By this I mean,can strings blend into the background harmonic of an oscillating Universe?My thought is that such a string would only be evidenced through it's mass/ gravitational profile! —Preceding unsigned comment added by 122.107.53.106 (talk) 03:58, 27 March 2010 (UTC)[reply]
The question refers to strings with a t, not springs. I assume this is intended to be analogous to the strings on a stringed musical instrument. Dolphin51 (talk) 08:03, 27 March 2010 (UTC)[reply]
Maybe it means the oscillations cancel each other out, or they're vibrating at the background frequency of the universe (not sure how that would work)? ~AH1(TCU)16:30, 28 March 2010 (UTC)[reply]
F-CENTERS IN CRYSTAL DEFECTS
WHEN COMPOUNDS SUCH AS Nacl LiH are heated with high energy radiatin they became deficient in negative ion?
08:51, 27 March 2010 (UTC)Souravraaj (talk)
Air pressure drop
Could anybody point me in the direction of some of the general principles I'd need to work out the pressure drop of a gas through a hole in a sphere into a vacuum? Something designed for pedagogical purposes would be ideal. --superioridad(discusión)10:37, 27 March 2010 (UTC)[reply]
Obviously the pressure drop of the gas passing through the hole is from the starting pressure to zero (unless the vacuum is in another container small enough to be significantly pressurized by the inflow of gas). Are you asking about how quickly the pressure in the sphere will drop to zero ? If this is a rigid sphere, then the shape won't much matter (unless it was maybe a really long narrow tube). The shape, smoothness, and material of the hole could make a big difference, though. Laminar flow will be quicker than the turbulent flow which could result from a rough or uneven hole. StuRat (talk) 13:29, 27 March 2010 (UTC)[reply]
Well, the question I've been given asks me to calculate the time for a particular percentage pressure drop, given an air-containing sphere in space that has a hole of known diameter punched in it. So, yes, rate of pressure drop is the one. Damned if don't have the least clue where to start, though. --superioridad(discusión)00:48, 28 March 2010 (UTC)[reply]
Heya, sorry for the late response. This link might help, bottom of page 18 and the top of page 19 (about effusion). They're lecture notes from the first-year Chemistry course at Oxford; you'll get the most out of it if you know a bit of calculus but some formulae in the middle of p19 are calc-free. kB is the Boltzmann constant. Hope this is of use. Brammers (talk) 13:24, 28 March 2010 (UTC)[reply]
Right, got it! Using equation 9.12, I think that if you set p0 to 1 and p to whichever percentage you desire (where 1 is 100%, 0 is 0% and 0.5 is 50% etc), then plug in the right values to give you tau, you can solve for time. Have fun! Brammers (talk) 18:44, 28 March 2010 (UTC)[reply]
Well, the problem specifies that the hole is rather large (d=2 mm), so close, but no cigar (i.e. not an effusion problem). Link should be helpful, though, cheers. --superioridad(discusión)22:34, 28 March 2010 (UTC)[reply]
In case it's not obvious, I should state that the flow rate through the hole will start out high, then gradually drop down asymptotically to zero. The pressure inside will be reduced in a similar fashion. StuRat (talk) 15:38, 28 March 2010 (UTC)[reply]
How do we make a bathtub full of lather and bubbles like how they show in the movies?
Have tried several times and failed, even after using the top end bath salts and shampoos I never never get a bathtub filled with lather and bubbles like how it is meant to be... whenever I try I can get scanty lather but not the real copious whitish romantic look.
Is there a secret or am I missing something?
I tried hot water, warm water, cold water, even tried increasing the lather by skimming the water with a mug, all this produces mere minimal increase and the water is good water, it isnt hard water or anything... have tried it even at the star hotels and failed....
Then asked room service to do it and they did a fabulous job but was too embarrassed to ask them how they did it.
This anonymous behind the computer shelter gives me the neceessary courage to ask this question which has always been in my mind.
And whats the scientific reason for me not getting anough lather? —Preceding unsigned comment added by Fragrantforever (talk • contribs) 12:55, 27 March 2010 (UTC)[reply]
It sounds like you're using the wrong stuff. Bath salts aren't meant to create suds, they are to soften the water. And shampoo creates minimal lather, as it's main purpose is to clean your hair, not make bubbles. What you need is called Bubble Bath: [14]. If you can't find any, then liquid dish-washing detergent is the next best thing. Pour it right where the faucet empties, then turn on the water full blast. The more you use, the hotter the water is, and the more turbulent, the more suds you will get. Also, avoid bath oils, as those will retard the suds. StuRat (talk) 13:03, 27 March 2010 (UTC)[reply]
That article starts off talking about bubbles in the water, as in a soda pop. Then they talk about a "foam bath", which is what we want, bubbles on top of the water. StuRat (talk) 13:20, 27 March 2010 (UTC)[reply]
I remember seeing a modelling session on tv where foam was being produced in a bath with an electric kitchen wisk or beater. Bubbling air through the water may also work. 89.243.43.75 (talk) 15:28, 27 March 2010 (UTC)[reply]
Water spraying at high velocity will create more foam than water running smoothly from the tap. A spray hose with a nozzle would thus do a better job than running the tap. Edison (talk) 19:11, 29 March 2010 (UTC)[reply]
What is the spin rate of a black hole?
We have 'clocked' pulsars rotation or spin rate. I have not found any reports of detecting a black hole's rate of spin, and I guess this may be because they emit nothing we can detect as yet. I was told that a black hole is a quasi-stellar object(Quasar? -Idunno) and so could not be clocked. By analysis of probable mass and the surrounding environment(a galaxy?) have we been able to take an educated guess at the spin rate of any black holeHebegb2 (talk) 14:55, 27 March 2010 (UTC)[reply]
While the objects falling into the black hole may spin in, I'm not sure the singularity itself has a spin. Since it's a single point, with no "lever arm", how would a spin be defined ? StuRat (talk) 15:42, 27 March 2010 (UTC)[reply]
Elementary particles have spin, despite being considered point particles. See rotating black hole. Conservation of angular momentum means a rotating star cannot collapse into a non-rotating black hole. --Tango (talk) 16:39, 27 March 2010 (UTC)[reply]
I thought elementary particles where more correctly modeled as "probability functions" than as point singularities. Maybe the same is true of black holes, since they have a certain uncertainty in position (which allows them to evaporate). Is that the explanation, or am I just applying Euclidean geometry and Newtonian physics to a realm where neither applies ? StuRat (talk) 17:53, 27 March 2010 (UTC)[reply]
I think the short answer is that it is complicated! In the Standard Model, elementary particles are considered point particles, although mathematically they are modelled using Quantum field theory where the concept of a particle doesn't really occur. I don't think that is relevant to the question of how they spin, though. The angular momentum of a particle is, like the angular momentum of a black hole, just an arbitrary constant in the equations that happens to obey the same conservation laws as angular momentum in other situations, so we call it angular momentum. (This apparent coincidence probably has an explanation, but it isn't part of our theories as far as I know - hopefully we'll work it out sooner or later.) Hawking radiation isn't really to do with uncertainty in the black hole's position (incidentally, I don't think the position of the singularity is really defined - it is just "in the future" - it is the position of the event horizon that we talk about), it's more to do with uncertainty in the energy density (I may be using the wrong term there, but in my intuition it is energy density) just outside the event horizon which allows particle-antiparticle pairs to be spontaneously created and then fail to annihilate. --Tango (talk) 23:39, 27 March 2010 (UTC)[reply]
I have heard that before about spin being only a theoretical concept when it comes to subatomic particles, not actual physical rotation of the particle. But, that does bring up the Q about what happens to the conservation of angular momentum in a black hole's singularity. If it's an infinitely small point, then even if it spins infinitely fast it still wouldn't have any angular momentum. Sounds like an argument for it not to be an infinitely small point, to me. StuRat (talk) 15:29, 28 March 2010 (UTC)[reply]
The black hole is not infinitely small. The event horizon has a radius. The angular speed people are talking about here is the top angular speed of the ergosphere which BTW doesn't spin as a solid body. see Kerr metric#Frame dragging. Dauto (talk) 18:51, 28 March 2010 (UTC)[reply]
But we're talking about the singularity, not the whole hole. If there's any mass to the singularity, then how is conservation of angular momentum handled there ? StuRat (talk) 19:46, 28 March 2010 (UTC)[reply]
That looks to be exactly what I was talking about, an alteration of the point singularity model to allow for angular momentum. StuRat (talk) 22:43, 28 March 2010 (UTC)[reply]
A claimed measure of the ergosphere of a particular black hole has been done recently. If it stands up, this would be the first measurement of the angular momentum of a black hole ever done. 20 points to the person who can locate the reference of which I'm speaking. ScienceApologist (talk) 03:16, 30 March 2010 (UTC)[reply]
Flying Black Holes
Assuming Black Holes are the only cosmic devices that eject matter and energy at their North and South poles presumably in exactly the same amounts and these beams of energy could be diverted at 90 degrees to the same perpendicular would the beams create a thrust that would send the Black Hole flying? 71.100.3.207 (talk) 16:52, 27 March 2010 (UTC)[reply]
Your premise is impossible, so your question is meaningless. The jets have to go in opposite directions in order for momentum to be conserved. It is conservation of momentum that makes rockets work (which seems to be what you are basing your result on). You can't ignore conservation of momentum when creating the jets and then use it for propulsion. Your physics would be inconsistent. If you have an inconsistent theory, anything can happen. --Tango (talk) 16:58, 27 March 2010 (UTC)[reply]
You are misunderstanding the question (my fault). The jets would be diverted to facing in the same direction like two side by side rocket engines. 71.100.3.207 (talk) 17:14, 27 March 2010 (UTC)[reply]
In that case, I think the black hole would stay still, and whatever device you used to divert the jets would get blown in the opposite direction as the redirected jets. StuRat (talk) 17:37, 27 March 2010 (UTC)[reply]
No, that's what I understood. It's impossible. That's just not what the jets do and they can't do that without violating the laws of physics. Unless you mean diverting them after they have left the black hole, in which case it won't have any effect on the black hole since there is no connection to it (but whatever is doing the diverting will be pushed away diagonally away from the black hole and in the opposite direction than the rays end up).
The jets are sent at opposite directions not because of momentum, but because the object is spinning, and the jets emit from the poles. Momentum can be easily conserved no matter the direction of the jets because the material being emitted acts as the other half of the conservation law. If you had some sort of method of attaching a redirector to the black hole (which you don't, but if you did) then you could send the black hole flying through space. But it wouldn't fly very fast because the black hole is heavy and the jets are pretty light (pun accidental, but enjoyed). Ariel. (talk) 01:25, 28 March 2010 (UTC)[reply]
To try and make the fault in the question clearer... The "deflectors" the questioner is suggesting will be off in space somewhere. So, this is like asking if you squirt a water gun at an angled wall, will it make the water gun fly out of your hand? Once the exhaust leaves the black hole, redirecting the exhaust will not affect the black hole - with the exception of redirecting it back into the black hole. What is required is a means of attaching the deflectors to the black hole. Then, the deflectors will attempt to move, but being attached to the black hole, will end up moving the black hole also. As mentioned, the black hole is too massive to move, but simply attaching deflectors to one will be hard enough - no need to worry about moving something so massive. -- kainaw™02:26, 28 March 2010 (UTC)[reply]
My God can noone understand him? Yes, if you had some devices deflecting the beams of matter at 90 degrees, and those devices where somehow attached to the black hole, then it would move. Although that's not possible.--92.251.136.245 (talk) 22:14, 28 March 2010 (UTC)[reply]
You can't connect something to a black hole. It's a logical impossiblity. Nothing crosses the event horizon. You could just keep the jets a little in front of the black hole and move it gravitationally, or divert one of the jets back into the black hole to move it. — DanielLC06:32, 29 March 2010 (UTC)[reply]
As I acknowldeged. Have you no imagination? I will never be overlord of the whole world, that doesn't stop me talking about what I would do if I was.--92.251.191.108 (talk) 00:27, 31 March 2010 (UTC)[reply]
It works for sugar as well (although sweet garlic would be pretty nasty). The general principle is that a flavor consists of an odor combined with a taste. Most of the information is in the odor component, but the flavor system needs to avoid being activated by pure odors, because pure odors frequently don't come from things in the mouth. The mechanism that makes this work is a strong nonlinear interaction between tastes and odors -- the presence of a taste strongly potentiates responses to odors. (Sweet and salty have stronger effects than sour and bitter, though.) This interaction is very well understood by cooks, who are systematically taught how to use salt and sugar to enhance the flavor of herbs and spices, but it's hard to find straightforward accounts of the science behind it. Looie496 (talk) 18:54, 27 March 2010 (UTC)[reply]
If I'm understanding this right, could I stick my nose in some crushed garlic, lick a small quantity of salt, and by doing this taste garlic? Vimescarrot (talk) 20:00, 27 March 2010 (UTC)[reply]
Yeah, though you probably don't need the salt at all. Your taste buds only receive info on whether something is sweet, salty, sour, savory, or bitter; what you perceive as "garlicky" is a small amount of sweetness mixed with aromatic compounds you sense in your nose. See flavor and taste for a fuller explanation. Matt Deres (talk) 20:18, 27 March 2010 (UTC)[reply]
Well, if it's tricks you want... Take an apple and cut out a small chunk of it - say 1cm x 1cm x 1cm or something. Cut out a similarly sized piece of onion (or even a garlic clove). Blindfold your friend and pinch his nose shut with a clothespin or something similar. Have him open his mouth and stick out his tongue and place one of the chunks right on there. By removing the "aroma" portion of the "flavour", your friend will have to use only "taste" to determine what's sitting on his tongue - and onions and apples "taste" roughly the same - only the aroma part is significantly different. Chances are good he won't be able to tell the chunks apart. Matt Deres (talk) 00:12, 28 March 2010 (UTC)[reply]
Ooh, and here's a somewhat related... well, not a "trick" per se, but a neat tip. Next time you make up a cup of hot chocolate, shake in a couple of sprinkles of cayenne pepper. Just a shake or two will do it. Oddly, the beverage won't taste hot at all, but rather chocolate-yier. The capsaicin in the cayenne bonds with your taste buds rather indiscriminately, boosting the flavour. The compounds in the cocoa (the milk solids) prevent the drink from getting spicy (assuming you don't dump a whole lump of cayenne in there), though it may get slightly bitter if you put too much in there. I know, it sounds nuts, and until I tried it, I didn't believe it either, but it really does work and it will improve the flavour of your cocoa. Matt Deres (talk) 00:23, 28 March 2010 (UTC)[reply]
Google scholar gives to the question:'Hydrogenation of vinyl chloride -poly the answer Vinyl chloride is cleanly hydrogenated to ethane at 120" and 3 atm pressure in a PtCl2 solution in molten (C2H&N SnC13. in doi:10.1021/ja00780a013 --Stone (talk) 18:26, 27 March 2010 (UTC)[reply]
@24.150.18.30: I am trying to eliminate insaturation of VCM, producing preferably chloroethane, but the production of ethane (or other substances) it's however better than don't reduce VCM at all.
A vinyl chloride is not a good Michael acceptor: chloride cannot easily stabilize a lone-pair/anion at the carbon to which it is attached. For example, the haloform reaction does not release the carbanion until you get three halides on the carbon, and attempts to deprotonate halocarbons often lead to carbene-like structures (α-elimination). Vinyl chloride is fairly inert to hydride attack, stable enough that NaBH4 can reduce other things in its presence quite selectively (U.S. patent 5,068,474). DMacks (talk) 14:16, 29 March 2010 (UTC)[reply]
Re cost -- in my experience hydride donors are cheaper to implement at the laboratory scale than pressurised H2 with expensive metal catalysts. Of course, if you want economy of scale at the industrial level, you want catalytic hydrogenation, but still. Thanks for the heads-up, DMacks. John Riemann Soong (talk) 02:59, 30 March 2010 (UTC)[reply]
The glottal stop and /h/ would also be still pronouncable. Since nasality is triggered by the lowering of the velum (rather than the raising of the tongue towards the velum), /m/ would be one of the possible labial sounds. — Ƶ§œš¹[aɪm ˈfɹ̠ˤʷɛ̃ɾ̃ˡi]03:59, 29 March 2010 (UTC)[reply]
A likely explanation would be that when you're run down and tired, your immune system isn't as effective as it is when you're completely refreshed, and thus you are more likely to catch a cold which presents with symptoms (i.e. not killed in time to avoid the symptomatic stages of the common cold). It's not so much that yawning shows that a cold is coming, but instead that fatigue can increase the likelihood of you not removing a cold from your body before the symptomatic period. Regards, --—Cyclonenim | Chat 23:54, 27 March 2010 (UTC)[reply]
That's possible, but it could be tiredness caused by the cold. It takes several days after you catch a cold before you develop cold symptoms, but your immune system is working hard in that time fighting the cold. It seems plausible to me (and fits my own experience) that your immune system working hard can make you tired. --Tango (talk) 00:00, 28 March 2010 (UTC)[reply]
Not sure what you mean by "missed heartbeats". There are several different conditions which could be described as such. See tachycardia, bradycardia, Palpitation, or Cardiac dysrhythmia for articles dealing with abnormalities in heartbeat. The article Cardiac cycle discusses the normal heartbeat. You should be able to find the information you need in those articles. If you are personally concerned about your own heartbeat, find a physician who can see you in person and find out what is wrong with you; random strangers on teh intrewebz are unlikely to be able to provide a good diagnosis. --Jayron3223:41, 27 March 2010 (UTC)[reply]
While missed heartbeats may not mean much, they may also be an indicator for AV block. More info would be medical advice. This situation should be dealt with a physician in person. Nobody can or has the right to make a diagnosis on a message board. 88.242.231.192 (talk) 10:42, 30 March 2010 (UTC)[reply]
March 28
Stomach Cancer Genome
Our article on Stomach cancer includes an un-cited sentence: "China being member of International Cancer Genome Consortium is leading efforts to map stomach cancer's complete genome." (last sentence; causes)
What does this mean? I thought most cancers were uncontrolled growths forming as a result of genetic damage. This sentence seems to imply that this type of cancer is either of a specific genetic form, or similar to a virus. NByz (talk) 04:00, 28 March 2010 (UTC)[reply]
Cancer has many genetic causes, but they're probably looking for oncogenes that are linked to those particular cancers. Sometimes there's a genetic predisposition to cancer, and I'm guessing the ICGC is trying to find either oncogenes or proto-oncogenes in the human genome. SDY (talk) 09:15, 28 March 2010 (UTC)[reply]
[15] which seems high up on a Google (albeit not Bing) search for 'cancer genome' should be of relevance, particularly the part on research as would International Cancer Genome Consortium and [16]. Incidentally, the wiki-article is already linked to from the stomach cancer article (as has been for at least a week) and the ICGC homepage is already linked to from the ICGC article. While they don't go into detail on China's stomach cancer efforts, they do provide a general idea of what the ICGC's efforts entail and what is generally meant by a cancer genome project in general which appears to be the source of confusion/uncertainty here.
As an example of some of the work that's part of the Chinese efforts, you may want to take a look at the work of You-Yong Lu[17] who is listed as one of China's representatives to the ICGC. Also Henry Yang who is probably this guy [18][19]. An investigation of the BGIwebsite[20][21] may also yield information on relevant research.
The basic idea is that there are some genetic mutations which are often associated with particular cancers and large scale genome sequencing of cancer cells followed by comparison to the reference human genome will allow the "mutational spectrum" of different cancers to be defined. This could lead to better diagnostic categories based on cancer genotype rather than merely phenotype (histology etc.), paving the way for treatments targeted at particular mutations. Looking at the mutations associated with particular cancers in lots of samples will also hopefully give some insight into which mutations are causative or "driver" mutations, and which are merely the result of the cancer disease process ("passenger" mutations). Equisetum (talk) 10:30, 28 March 2010 (UTC)[reply]
Where to start with extreme theoretical physics?
"Okay, what is the definition of a blackhole? Okay it has an event horizon- what's that?" So I click the link titled event horizon, "An event horizon is where you have to be going faster than the speed of light to break free of the gravitational pull of a black hole- or gravitational or quantum singularity... What's a normal singularity?" etc etc etc...
Where do I start? Rather than back tracking to what i want, why can't i go forward? Rather than asking about everything on the page, how about understanding it and taking the next step so that i can understand things from the beginning? Here's where wikipedia goes wrong- you have to click a billion and one links before you get down to understandable parts of a complex topic... the beginning. So what would i have to search to begin with the very basics of say, a Riemannian Manifold? Or a black-body? Singularity? Euclidean Space? Don't make me keep listing them...
I'm 14 so no i can't just take a course at uni, believe me i would if i could! —Preceding unsigned comment added by SamitaEJ (talk • contribs) 10:58, 28 March 2010 (UTC)[reply]
The problem is that Wikipedia is an encyclopedia, not a textbook. It doesn't necessarily have great ways to go through topics in the same way you would a book. Personally, I recommend getting a book! There are many excellent books on physics out there that lead you through this stuff in any interesting way, at any level of detail you desire. --Mr.98 (talk) 13:23, 28 March 2010 (UTC)[reply]
If you really want to get to grips with a subject, do it the same way you would at school. Start from the basics (in the UK we have GCSEs, then A levels, then university) so you could read individual guides for the first two levels which can give you a good basic understanding of the topic, then go on to read books which expand on the details. Regards, --—Cyclonenim | Chat 14:52, 28 March 2010 (UTC)[reply]
There is a problem with technical articles at Wikipedia, that they tend to be written by, and for, experts in the field. Some concepts are inherently complex, so there's no easy way to explain them. But others aren't all that complex. A black hole can be described quite simply: "A black hole is where so much matter is packed into such a small space that the gravity causes it to pull itself down to a single point, called a singularity, and nothing, even light, can escape from a roughly spherical region around it, called the event horizon.". Now, our black hole article isn't bad, but the opening paragraph does go into the "general theory of relativity", the "deformation of spacetime", a "perfect black body in thermodynamics", the "theory of quantum mechanics", and "Hawking radiation". That's probably a a bit much for the average person to take in at once. I'd leave those details until later. StuRat (talk) 15:07, 28 March 2010 (UTC)[reply]
If you're prepared to struggle a bit, Roger Penrose's The Road to Reality is worth looking at. The first few hundred pages cover the mathematical background for the rest, which aims to be a pretty complete introduction to modern physics. At 14 the mathematics will probably be very difficult for you, nevertheless you might get a lot out of it, especially if you've got somewhere you can go for help (e.g. the ref desk!). I wish I'd had the book when I was that age. Tinfoilcat (talk) 17:05, 28 March 2010 (UTC)[reply]
It doesn't seem anything like a true story, to me. The subjects obviously would have died with their internal organs removed like that. Also, there have been plenty of real sleep deprivation studies to know what happens. The paranoia part seems right, but not the rest. StuRat (talk) 14:43, 28 March 2010 (UTC)[reply]
But why would you think your subconscious mind is suicidal ? It evolved, just like the conscious mind, based on the survival of the genes most likely to bring about the survival of the genes. Thus, being suicidal isn't a winning strategy. I can imagine the higher reasoning part of the brain ceasing to function, leaving you in an animal state, but those "animals" would still have a sense of self-preservation. StuRat (talk) 19:42, 28 March 2010 (UTC)[reply]
The subtitle is "The best short story I've read" and the opening blurb says "Don’t know who wrote this or where it came from. My brother emailed me a link to it and I was completely immersed in it. Kudos to the author. If you know the author or any other works by this person, please leave me links in the comments, I would love to read more from this person." The only real question is what could possibly have made you think there was even the slightest chance this was real, let alone a reputable source to base articles on? No offence, but it says flat out that it's a story; why is there any confusion? Matt Deres (talk) 18:41, 28 March 2010 (UTC)[reply]
What a load of complete bullshit. Honestly - not one single part of that rings true or has even fractional basis in known science. Zombies created in Russian death-camps. Sheesh! It's obviously fiction. Ignore it...and DEFINITELY don't write a Wikipedia article about it without a heck of a lot of reliable-source references. SteveBaker (talk) 23:06, 28 March 2010 (UTC)[reply]
Obviously fiction, and the writing's not stellar (The story would be a lot punchier if they left out that explanation in the next to last paragraph.) , but it's still a fun story. It comes from a long tradition of sci-fi/horror about what happens to people when they stay awake for ridiculous periods of time.
Spoilers for short story linked above : I didn't catch this until I saw it pointed out in the comments, but check it out : Of the original five test subjects, one is killed by the other subjects, one dies when his spleen is stepped on, one died on the operating table when he was anesthetized, one of them dies mysteriously when he falls asleep, and two of them are shot by a researcher.
Why when I stir a drink mix into water does the sound of the stirring lower in tone until it reached an equilibrium? Actually, I'm not sure the drink mix has anything to do with it, though I haven't tried it with plain water. My guess is that as you're stirring, air is being forced into the liquid and this changing the tone until no more air can be incorporated. Just a guess. Looking for a real answer.--162.84.128.153 (talk) 17:36, 28 March 2010 (UTC)[reply]
Hmmm. This is very strange. First, there's all this stuff in the past discussion about hot liquid. This is an everyday phenomenon for me and cold liquid is at issue (well tap water temperature). No need for anything hot. Second, no tapping on the outside is going on. Just the sound of the stirring and the spoon making noise off the inside of the glass. This happens whenever I am mixing (I am completely addicted to Crystal Light) and it doesn't matter one wit what vessel I use. Second, people kept taking about the tone rising. The tone lowers. I just tried an experiment with interesting results. I tried it with Crystal Light, then I tried it without. With the mix there was the normal effect. With just tap water, the same effect was present but not as strong. And here's where it gets interesting, if I fill the glass and let it sit for a few minutes, the effect goes away entirely, regardless of whether I use the drink mix or not. So, there certainly is much of this that is related to the gas present as I fill the glass from the tap coming out of solution (the opposite of what I suspected when I first posted).--162.84.128.153 (talk) 18:46, 28 March 2010 (UTC)[reply]
You beat me to it. That's just what I was going to say about air bubbling out as it sits or is stirred. This is especially true if you have an aerator on your faucet. StuRat (talk) 19:35, 28 March 2010 (UTC)[reply]
Personally, I don't believe the "air bubbling out" theory - it doesn't fit all of the facts we've heard over the years - and I can't reproduce the conditions where the effect "goes away" after these bubbles have supposedly escaped. I suspect (without proof) that we're hearing the result of doppler shift in the sound due to the motion of the liquid. But this question has come up at least three times in the last year - and there is no solid science to explain the effect that I'm aware of. This would make a great study for someone with access the the right equipment. SteveBaker (talk) 22:50, 28 March 2010 (UTC)[reply]
Could it be that the frequency of the stirring sound is related to the relative speed of the stirrer in the liquid? The liquid was stationary when you first started stirring it. After a while, its has caught up with the stirring so the stirrer is no longer moving as fast relative to the liquid. --173.49.14.137 (talk) 11:43, 29 March 2010 (UTC)[reply]
I think that the properties of the solution are changing as the sugar, flavoring, etc. are dissolved, so it transmits sound differently. --98.221.179.18 (talk) 20:18, 29 March 2010 (UTC)[reply]
will bleach oxidise hemiacetals?
Suppose that all the OH groups on a sugar ring have been protected except the OH group on the anomeric carbon (since it easily undergoes exchange). It possible to oxidise this selectively unprotected OH group to an ketone, with mild conditions and bleach? I am hoping the anomeric effect will actually promote the reaction and make the OH group more reactive than normal, allowing me to use bleach under very mild conditions without oxidising anything else. Ideally my yield would be >70%. John Riemann Soong (talk) 18:11, 28 March 2010 (UTC)[reply]
Bleach (especially under acidic conditions) can oxidize secondary alcohols. Not sure if being anomeric helps or hurts: for the normal (not hemiacetal in equilibrium with an aldehyde), I think the mechanism is E2-like, with from anomeric H coming off and its σ bond becoming the carbonyl π). That mechanism doesn't work unless there is an exchangeable H on the O, so an acetal or other ether would not react rapidly.DMacks (talk) 19:00, 28 March 2010 (UTC)[reply]
Oops, I meant hemiacetals. Bleach is too weak of an oxidant to oxidise C-H bonds, right? What about the potentially activated C-H bonds on the other side of the acetal-ethoxy oxygen? Also the largest issue I'm worried about is the bleach oxidising the other alcohol in the straight-chain form... which would give me a dialdehyde...
That's actually the question I answered:) Your real problem is keeping the sugar-ring in the closed (hemiacetal) form. If it opens, you risk oxidizing the aldehyde to acid, and that is a very easy process (basis for the many chemical tests for reducing sugars). Bleach doesn't touch alkanes, and I do not think (based on mechanism) that it would touch a carbinol hydrogen (the H on a C, where that C is attached to an O as ether or hydroxyl)--except when it is an OH group. DMacks (talk) 14:09, 29 March 2010 (UTC)[reply]
No, I don't know if bleach does oxidize aldehydes. But if it does, your major problem is not "ring opens, primary alcohol oxidizes to aldehyde" but instead "ring opens, aldehyde oxidizes to acid". Either way, it's a problem. But it's a different problem. One way, you have to consider primary vs secondary-alcohol/hemiacetal oxidation rates, other way you have to consider secondary-alcohol/hemiacetal vs aldehyde oxidation rates. In general, aldehyde→acid without also oxidizing alcohols is very easy to do. And the specific bleach oxidation mechanism looks to me like it could work well on aldehydes also. Try this experiment: take some simple aldehyde, some simple secondary alcohol, and some simple primary alcohol, and try to oxidize each with bleach under the same experimental conditions. See if each one reacts, and approximately how rapidly. DMacks (talk) 15:06, 30 March 2010 (UTC)[reply]
Anyone know about Victoria building and guttering?
Is it possible that the downpipes carrying rainwater off the roofs of our house simply sink about 3 metres vertically into the ground and then soak away directly into the foundations/ watertable? Did they do that? --BozMotalk19:55, 28 March 2010 (UTC)[reply]
Unlikely. A downspout concentrates the volume of rainfall collected from a large area into a single spot. The permeability of the ground to water, however, is limited. If you merely buried the downspout in the ground (whether 3 meters deep or half a meter or whatever), the water would simply back up in the downspout until the gutter overflowed, thus defeating the purpose.
If you ran the downspout down into a dry well you'd constructed, on the other hand, it would work quite well. Such wells tend to get clogged with silt and other material after a while, however, and stop working. After 100 years I can imagine that it might be largely indistinguishable from soil. —Steve Summit (talk) 20:57, 28 March 2010 (UTC)[reply]
Ok, thanks. There has been a problem with ground water flooding into the cellars in the last couple of decades. I assumed it was to at least below the cellar level originally. I guess a silted up dry well is a possible cause. I could divert the downpipes fairly easily so perhaps I should just do it. --BozMotalk21:21, 28 March 2010 (UTC)[reply]
(ec)It depends a lot on the type of ground conditions you have whether such a soakaway (as we call dry wells in the UK) would be effective long term. I live on river gravels over chalk and I reckon that the water table is a long way down, as we have a completely dry unlined cellar. I think that under those conditions it would work (the down side is that it makes growing stuff in the garden during dry summers a pain). Mikenorton (talk) 21:25, 28 March 2010 (UTC)[reply]
Yes, this arrangement is common in the area where I live, but the water is usually ducted away from the foundations before being allowed to sink into the ground. As Steve says above, silting-up is a problem, and it is possible that the original underground drain has collapsed. Dbfirs21:30, 28 March 2010 (UTC)[reply]
We are on real Suffolk clay (chalky boulder clay to geologists). We live right at the top of a hill but the moat and pond in our garden stay full and have been 5 m apart with 1m difference in water depth since about 1150 AD without one draining into the other (the moat isn't around our house, it is the site of a medieval hall). But the cellar used to be dry: indeed even thirty years ago it had electrics and central heating fitted which is not sensible today. So something has silted or blocked I guess. --BozMotalk21:33, 28 March 2010 (UTC)[reply]
Sounds like a great place. You must have a sanitary sewer or a septic tank, and you could always direct the water there (although it might not be legal to drain it into the sewer in some places, and it might overflow a tank). The most obvious solution is just to have some maintenance done to remove the generations of silt built up in the dry wells. StuRat (talk) 22:01, 28 March 2010 (UTC)[reply]
As far as I know, one should never divert rainwater or other runoff water into either a septic tank or a sanitary sewer. The former is a significantly bad idea; the latter is (as Stu mentioned) illegal in some/many/most jurisdictions. —Steve Summit (talk) 13:46, 29 March 2010 (UTC)[reply]
Well, there is the combined sewer. While I agree that they are a seriously bad idea, they seem to be widespread in certain areas, anyway. In South-Eastern Michigan we pretty much have to close all the lakes and beaches when there's a heavy rain, and the basements of all homes in low lying areas flood with sewage, because of combined sewage overflow: Combined_sewer#Mitigation_of_CSO_Impacts_in_United_States. StuRat (talk) 18:41, 30 March 2010 (UTC)[reply]
Some options for what to do with rainwater coming out of downspouts:
You could put a French drain under the downspout which directed the water elsewhere. A french drain is basically a trench filled with gravel. The large pore space between the gravel makes an effective means of containing and/or diverting water. Some French drains also contain a semi-permiable pipe to direct the collected water away towards a more desirable area.
You can bury some flexible piping and use that to direct your water, used by itself or in connection with a French drain, it can be very effective. Years ago I had a problem with my downspouts washing away a flower garden; so I dug a trench across my yard to an overgrown area, and installed something like this corrugated black plastic pipe, covered it back again with dirt and grass, and and it works great. Its been in the ground almost 7 years, and it has never clogged or backed up at all. I have three downspouts (roughly 1/4th of the water from my roof) all directing their water into a pipe 4 inches in diameter, and it runs underground down a slope that has about a 1-2 foot drop over about 30 feet, and its plenty to direct all of the water away from the house. The entire project took a Saturday, a buddy, and a case of beer.
You can collect the water and reuse if for gardening. The article Rainwater tank contains some examples of rainwater collection barrels and tanks.
Some buildings take the water to the gutter of a nearby road. But roads can get built up over time blocking the outlet. (I have excavated near an old road and found the old piping no longer going to the road). Polypipe Wrangler (talk) 01:15, 30 March 2010 (UTC)[reply]
Ok lets say that the Earth (or some other large solid planet) had a solid core, and we drilled a great big hole from one side of the Earth, through the centre, and out the opposite side. What would happen if you jumped down this hole?--92.251.136.245 (talk) 22:20, 28 March 2010 (UTC)[reply]
Without air resistance, you'd speed up, falling to the center, and then start slowing down again, making it just far enough to poke your head out the other side (assuming the Earth is a perfect sphere here), and the process would repeat forever. With air resistance, you'd quickly reach terminal velocity, so you wouldn't make it very far past the center. However, since air resistance is a bigger deal the faster you move, you'd probably oscillate back and forth for a long way near the center. PaulStansifer22:36, 28 March 2010 (UTC)[reply]
If we were to build a colony on an asteroid or small moon, that might actually be an efficient way to transfer items from one side to the other. As a practical matter, you might want to build an elevator shaft with a magnetically "levitated" car, so the people or objects wouldn't keep hitting the sides. StuRat (talk) 22:40, 28 March 2010 (UTC)[reply]
There have been many discussions of this in the past - and the consensus is that outcome depends entirely on the assumptions you make.
From most realistic assumptions to least:
The hole would fill up with molten lava and other materials faster than you could dig it. So you can't construct such a hole.
In order to prevent this collapse, we'd have to line the tunnel with something - but there is no material we have that could survive the temperatures and pressures involved.
Even if we imagine such an 'unobtainium' material to make a liner with, the hole would fill up with air. At the pressures you'd get towards the center of the hole, the air would liquify and block the hole that way.
So you'd jump in and as the air density increases towards the center of the earth, then either:
You'd be crushed by the pressure.
If you wore some kind of pressure-resistant suit then you'd eventually float like a balloon at some distance before you got to the center.
If you put enough weights onto the suit then you'd slowly sink to the bottom of the liquified air and arrive at the center of the earth - where you'd feel no gravity at all.
If you could pump the air out to make a nice hard vacuum throughout the hole - then you'd obviously suffocate on the way down...so you need to wear a space-suit.
If you get rid of all of these annoying issues, then you'd still be bouncing off the sides of the hole all the way down due to coriolis forces - so the friction would slow you down to an eventual stop at the center of the earth - where you'd float freely and feel no gravity.
To solve that problem, we'd have to drill through the axis of rotation (ie from pole to pole).
With only gravity acting on you, you'd have to enter the hole at a very precise vertical angle to avoid hitting the sides on the way down.
Assuming you get around ALL of those issues, then there is no friction in the hole - so what would happen is this:
In your spacesuit, you jump into the hole at the North Pole. Since you are falling, you feel no gravity from the instant your feet leave the ground - you fall all the way towards the center of the earth, getting faster and faster - so fast in fact that you overshoot the center and start to slow down. You finally stop just as your head pops up above the surface at the south pole - then you fall all the way back again, through the center and back out to the North Pole. This happens over and over again until someone or something intervenes to helps you out!
How would one calculate the pressure at the center of the Earth in an Unobtanium tube exposed to the atmosphere? It seems counterintuitive that air pressure gets higher and higher as you descend below the surface, since gravity should decrease steadily, due to some of the mass being above you and a smaller amount below you. At the same time, though I admit that the "column of air" above the point would get higher and higher, which would seem to increase the pressure. And wouldn't the temperature have to be lowered to get air to liquify?Edison (talk) 16:36, 30 March 2010 (UTC)[reply]
Only if there was some friction acting on you. For instance if you were bouncing off the sides, or there was some air. ... But, as Steve mentioned, if there was air, it would increase in pressure until some parts of the tunnel were actually filled with liquid air. APL (talk) 23:04, 28 March 2010 (UTC)[reply]
You'd need a lot more air to liquify in the center than you would to slow you down a bit with each pass. If we pumped out all the air we could, to get as close to a vacuum as possible, we would still be slowed appreciably at those speeds we would reach near the center. StuRat (talk) 23:54, 28 March 2010 (UTC)[reply]
I think a more interesting question involves a lined (with unobtainium) perfectly straight and regular cylindrical hole through the center of the Earth containing absolutely nothing (a vacuum). A lead sphere is released from beneath the cap of one end of the tunnel (at the surface of the Earth). Would the sphere oscillate indefinitely or would it eventually come to rest at the center of the Earth? Let us say that the tunnel was drilled from pole to pole to eliminate the Coriolis forces that SteveBaker presciently pointed out. Bus stop (talk) 00:24, 29 March 2010 (UTC)[reply]
Well, the axis moves around a bit, so I guess if you waited long enough the ball would end up hitting the sides and slowing down. --Tango (talk) 00:30, 29 March 2010 (UTC)[reply]
Precession of the equinoxes takes tens of thousands of year, though. I suspect that before that effect you would see something similar to tidal forces, whereby the passing lead sphere would warp the shaft inward, ever so slightly increasing the pull of gravity towards the sphere. As the sphere would be continuously moving past this bulge, it would tend to slow the sphere down, over time. StuRat (talk) 00:35, 29 March 2010 (UTC)[reply]
Entering at the equator one's momentum would be conserved. Comparable to ice skaters who bring their arms to their bodies to increase their rate of spin one will rotate progressively faster than the Earth as one falls. A tunnel must have the shape shown for one not to touch its sides. Cuddlyable3 (talk) 14:44, 29 March 2010 (UTC)[reply]
Imagine you moving back and forth as the Earth spins. The path you draw on the Earth would be the shape of the curve. It looks kind of like a flower. Since a day isn't perfectly divisible by the time it takes to make one lap, the "petals" of the flower would never quite line up, so you'd have to keep digging until you eventually just hollow out the entire equator. — DanielLC06:25, 29 March 2010 (UTC)[reply]
I'll bet that you could make it evenly divisible by starting the passenger some distance above ground level. APL (talk) 16:24, 29 March 2010 (UTC)[reply]
Regarding the picture, that can't be right. I think your path would be some kind of oval. You wouldn't spiral towards the center. You'd pass around it. Compare it to an elliptical orbit. — DanielLC07:36, 30 March 2010 (UTC)[reply]
You're right; in an inertial reference frame, your path would just be an elliptical orbit. The path using the rotating earth as a reference frame would be different; it probably would look a bit like a spiral. The picture is certainly not correct: the path would not hit the center of the earth, and it would begin perpendicular to the surface of the earth. (The picture may be accurate for something dropped straight into the earth that wasn't traveling along with the earth's rotation at the surface.)71.240.196.243 (talk) 14:03, 30 March 2010 (UTC)[reply]
Yes, there are applications. For instance, Fermi's theory does not account for neutral currents which allow neutrinos to scatter ellastically off of matter. That fact was skillfuly used at snolab to clinch the proof that solar neutrinos oscillate into different kinds of neutrinos along their path towards the earth. Dauto (talk) 20:52, 29 March 2010 (UTC)[reply]
Space Radiation and the Earth's Atmosphere —how are high-frequency light rays affected?
I have a question about how exactly the Earth's atmosphere protects us from space radiation?
As far as I'm aware, high-frequency radiation (gamma rays, x-rays, and most ultraviolet rays) are in some way "filtered" by the atmosphere and thus never gets to the surface. Whereas lower-frequency radiation (some ultraviolet rays, visible light, infrared waves, microwaves, and radio waves) penetrate the atmosphere and are felt down here.
My question is twofold:
(1) How does the atmosphere interact with the higher-frequency radiation, exactly? Does it reflect the gamma rays, etc. back into outer space; decrease the frequency of it so as to render it into visible light, etc; or some other process?
and
(2) Is this done by the radiation of the atmosphere, the chemical composition, or something else?
Would exploding a hydrogen bomb on the planet Jupiter or any of the other gas giants turn them into a second sun? And if so, would there be any effects on the Earth?80.1.88.25 (talk) 12:33, 29 March 2010 (UTC)[Trevor Loughlin][reply]
The size and heat of even the largest H-bombs ever set off on Earth is still quite small on a planetary scale even if they are large on a human scale. The amount of energy that is released when, say, an asteroid hits a planet is vastly larger (though not always concentrated in the same area). In any case, even if a significant fusion reaction was set off inside the planet (where the material would be of highest density), without there being more mass and gravity, it would not be self-sustaining like reactions in the Sun. (For the same reason, setting off an H-bomb on Earth does not begin a fusion reaction in the nitrogen in the air or the water in the sea, even though in theory both could undergo fusion under the right conditions. But Earth does not have the right conditions for that.) --Mr.98 (talk) 12:42, 29 March 2010 (UTC)[reply]
Bombs cause explosions, which move things outwards. To turn Jupiter into a second sun you would need to compress it to increase the temperature and density to a point where fusion can occur. However, once that fusion did occur the radiation pressure would cause the planet to expand again and the fusion would stop. You would need some way to compress it and keep it compressed, and I don't think there is any practical way of doing that, even assuming futuristic technology. I think any attempt to compress it would end up using more energy than would be generated. --Tango (talk) 13:07, 29 March 2010 (UTC)[reply]
Agree with Mr.98 and Tango. None of the gas giants are large enough to sustain fusion reactions at their core - otherwise they would be stars already. The minimum theoretical mass for a star with a similar composition to our Sun is about 75 times the mass of Jupiter, and the least massive known star, AB Doradus C, is 93 Jupiter masses - see star. Jupiter became a star at the end of 2010: Odyssey Two, but only with the help of advanced alien technology.
To answer the second part of your question, if Jupiter did somehow become a star, but without changing its size, mass or orbit, I think there would be a minimal impact on the Earth. Jupiter is 1/10 of the diameter of the Sun, and even at its closest, it is four times as far away from Earth as the Sun. This means that its apparent area in the sky is at most about 1/1600 that of the Sun. If we assume a Jupiter-star would emit the same amount of radiation per unit surface area as the Sun, then it would increase the solar radiation on Earth by less than a tenth of 1%. This is the same order of magnitude as the natural solar variation that we experience as the Sun's output changes over its 11-year solar cycle. Gandalf61 (talk) 13:10, 29 March 2010 (UTC)[reply]
To the contrary, I think it would be devastating if Jupiter ever became a sun. Much of Earth's night would be gone, which would wreak havoc on nocturnal animals and mess up the circadian rhythms of the diurnal animals. --206.130.23.67 (talk) 14:39, 29 March 2010 (UTC)[reply]
If Jupiter (at the current size) were burning like a star, it would not eradicate nights on Earth. Your statement appears to be based on the idea that it will shed enough light every night to cause problems. First, it is 1/1047 the volume of the sun. It is more than 4 times further away from Earth than the sun. (Please check my math - I am doing this with a crying baby on my lap, so not concentrating well.) Combined, the luminosity will be minuscule compared with the Sun (or even a full moon). It will be very visible, but mainly as a very bright star, not something that lights up the night. Further, that is only if it is in the night sky. Jupiter is not always in the night sky. -- kainaw™14:50, 29 March 2010 (UTC)[reply]
Since we don't know any way of turning Jupiter into a star, it is difficult to guess what luminosity it would have. A key thing to keep in mind is that human light perception is logarithmic. A full moon doesn't seem that much dimmer than the sun - it's plenty to see by. That is despite the Sun being 449,000 times brighter than a full moon. If Gandalf61 is correct (his guess is as good as anyone's) and Jupiter would be only 1000 times dimmer than the Sun, it would be many many times brighter than the full moon. You wouldn't really notice that it was dimmer than the sun at all. --Tango (talk) 15:09, 29 March 2010 (UTC)[reply]
Using the basic luminosity function L=4πR2σT4, you can see that the radius of Jupiter is directly related to the luminosity. Since Jupiter is 1/10th the radius of the Sun, it has 1/100 less luminosity (squared produces 100 instead of 10). If we assume T to be the same, Jupiter would be 1/100th the luminosity of the Sun. That would be the apparent luminosity if Jupiter were the same distance from Earth as the Sun. It is more than 4x further away, so it would have an apparent luminosity less than 1/100th that of the sun. -- kainaw™15:34, 29 March 2010 (UTC)[reply]
That is consistent with Gandalf's guess (I think you did the same calculation). As I say, being 1/1600 the brightness of the Sun wouldn't be very noticeable to humans (your pupils would dilate a little more, and then everything would seem the same). That formula doesn't really apply, though, since we're assuming something weird is happening to make Jupiter a fusor. --Tango (talk) 15:56, 29 March 2010 (UTC)[reply]
OK - now that I can work out equations and type without little fingers trying to press all the keys on the keyboard... The apparent luminosity (usually referred to as "magnitude") will be approximately 1/1600 that of the Sun if Jupiter has the same temperature as the Sun. But, that will not be a reasonable assumption. Instead, consider a very small star, still larger than Jupiter, but damn small. VB 10 is the best I can find as a reference. It has an absolute luminosity that is 0.0008 that of the Sun. Therefore, the apparent luminosity of Jupiter, if it was as bright as VB 10, would be about 0.00005 that of the sun (1/20000, which is significantly dimmer than the 1/1600 previously quoted). Further, this would not an object the size of the Sun in the sky. It would be a tiny dot that is visibly brighter than the other stars. So, it is not reasonable, in my opinion, to claim that it will disrupt all nightlife on Earth - which is the point I was attempting to make much earlier. The best I can explain it is comparing lighting up a room with a flood light or with a tiny little LED. Even if the tiny little LED is 10 times brighter than the flood light, it won't do a good job of lighting up the room. In our case, the tiny little LED is 20000 times dimmer than the flood light. -- kainaw™17:34, 29 March 2010 (UTC)[reply]
According to apparent magnitude, yes. It is also a given since we know we can see the moon during the day and it is around 450,000 times dimmer than the Sun. So, an object 20,000 times dimmer will be visible. Jupiter would be a tiny little bright dot in the sky. -- kainaw™19:11, 29 March 2010 (UTC)[reply]
The angular diameter is irrelevant, it is brightness (what you are calling apparent luminosity) that matters. How spread out it is will determine whether you can look at Jupiter without blinding yourself, but that's about it. A flood light and an LED of equal brightness will illuminate a room equally well. 1/20,000 times the brightness of the Sun is 22 times brighter than a full moon. Basically, it would like having a full moon much more often than we do now. That may not have too big an impact, but it would have one. --Tango (talk) 19:22, 29 March 2010 (UTC)[reply]
I pity the poor baby having to put up with Kainaw discussing Jupiter sun's when he/she's supposed to be the centre of Kainaw's attention :-P Nil Einne (talk) 06:53, 30 March 2010 (UTC)[reply]
Also, nuking Jupiter will not turn it into a non-star fireball, because to have a fire you need heat (supplied by nuke), fuel (Jupiter has quite a bit of hydrogen), and Oxygen (technically Chlorine might do), which is lacking, so no fire will be maintained. Googlemeister (talk) 13:54, 29 March 2010 (UTC)[reply]
It's also worth pointing out that the fusion matterial in the hydrogen bomb is either a mixture of deuterium and tritium or a lithium deuteride. Hydrogen itself leads to a slow nuclear burn instead of an explosive one. Thanks god, otherwise the sun would have exploded a long time ago :-) Dauto (talk) 14:49, 29 March 2010 (UTC)[reply]
I've looked and all I keep getting is frosted glass which is not what I mean.I wanted to know why ,when windows frost up,the pattern looks like ferns or flowers? —Preceding unsigned comment added by 88.96.226.6 (talk) 13:47, 29 March 2010 (UTC)[reply]
The article titled (unsurprisingly) Frost contains a section on window frost. Ice forms a a hexagonalcrystals. The frost patterns you see result from a propagation of this hexagonal structure across the surface of the glass, much as in a snowflake. The angles involved in the branching are all generally 120o (the internal angle of a hexagon), which gives the pattern you are familiar with. --Jayron3214:24, 29 March 2010 (UTC)[reply]
Other than the information given in sinusitis, we can't really give any more information specific to your case because we don't offer medical advice. You should consult a medical professional if you need to know more. —Akrabbimtalk20:14, 29 March 2010 (UTC)[reply]
Well, Matt asked for details and I gave it. I just wanted to know if there were any cases where someone infected someone else with sinusitis. --Reticuli88 (talk) 20:30, 29 March 2010 (UTC)[reply]
It was a rhetorical question. You got the infection from somewhere, therefore it is contagious. How contagious, as in how worried should you be if you sneeze, is a more complicated question that depends on a lot of variables that we wouldn't be able to evaluate here. —Akrabbimtalk20:58, 29 March 2010 (UTC)[reply]
Sinusitis and a cold could be the same organism. Or it could be an opportunistic infection with bacteria after a cold (which is a virus). So you can infect other people, but that doesn't mean it will turn into sinusitis for them - it might simply be a cold. A bacteria sinusitis is unlikely to be infectious. If you are contagious you might transmit the germ, but you can't control in which part of the body it will cause illness in the other person. Ariel. (talk) 21:03, 29 March 2010 (UTC)[reply]
The textbook I use derives the relationship τ = dL/dt in the following manner: dL/dt = d/dt(r x p) = dr/dt x p + r x dp/dt = v x mv + r x F = r x F = τ. But doesn't this assume that the position vector for the torque and the particle are the same (ie that the force is being applied directly on the particle)? 173.179.59.66 (talk) 17:30, 29 March 2010 (UTC)[reply]
Whatcha talking about? By definition the coordinate of the point of action of force on a point particle (and the torque as well) is the coordinate of the particle itself. Dauto (talk) 17:53, 29 March 2010 (UTC)[reply]
Well, what if we were looking at a massless lever with a large mass somewhere in the middle. The the force can be applied at the end of the lever, while the particle would have a different r than the force. 173.179.59.66 (talk) 21:41, 29 March 2010 (UTC)[reply]
No, a lever is made of many particles, the external force acts on a massless particle at one end of the massless lever and the action gets transmited to the other end by internal forces between the massless particles used to make the lever. Lots of forces and each one of them act exactly at the coordinate of the particle they are acting on. Dauto (talk) 22:16, 29 March 2010 (UTC)[reply]
Agreed, but then how would you calculate the torque on the massive particle? I would imagine that you would have to use τ = rF x F, and this would equal d/dt(Lm) = d/dt(rm x p), with rF not equal to rm. So what gives? 173.179.59.66 (talk) 01:12, 30 March 2010 (UTC)[reply]
I guess I'm missing something lol, maybe an example with numbers would help me. Imagine we have a 2 metre rod of negligible mass with a small rock with a mass of 10 kg attached at the middle (the rod will be rotated about one end). A force of 3 Newtons is applied to the end of the rod. What's the angular acceleration? τ = 6Nm, and τ = d/dt(rmmv) = d/dt(Iω) = (mrm^2)α = (100kg.m^2)α, so α = 0.06rads/s. Here, rF didn't equal rm. I agree that the internal forces of the rod will ensure that a force with a different magnitude than F will act on the particel at r = rm, but it remains to be shown that dL/dt still equals rF x F, which I don't think my book adequately proves. If I'm not making sense, let me know. 173.179.59.66 (talk) 02:38, 30 March 2010 (UTC)[reply]
Everything you said above is correct except for "it remains to be shown that dL/dt still equals rF x F". Yes, for that example rF and rm are different but that's only possible because you are dealing with an extended object. The book's proof is intended for a set of point particles. But since extended objects are made of point particles, at the end of the day the proof also applies for extended objects. The internal forces will be whatever they have to be in order for dL/dt to be equal to rF x F. Either that happens, or the massless bar will bend and brake. Dauto (talk) 04:49, 30 March 2010 (UTC)[reply]
How do you know the internal forces will arrange themselves so that dL/dt = rF x F? I would think it would have to be justified on the grounds of mechanical advantage or something, which my book doesn't seem to address. It seems to me that the proof doesn't apply to extended objects for that reason. 173.179.59.66 (talk) 06:41, 30 March 2010 (UTC)[reply]
Yes, there are some missing steps in the proof but they are almost trivial. Denoting the extended object angular momentum by and its external torque by , we want to prove that . But we already know that . So the only step missing in the proof is to show that . This comes about because the internal forces come in action-reaction pairs of oposing forces acting along a common axis which leads to pairs of oposing torques that cancel pairwise identically. Dauto (talk) 15:12, 30 March 2010 (UTC)[reply]
I don't know what you mean by a central torque. But I know that the reason that they cancel pairwise is that the pair of forces act along the line connecting the particles. May be that's what you mean by central torque? That's true because this is the only way to preserve the axial symmetry. There is only one important exception. Electric charge particles in motion produce magnetic fields that generate forces on othe moving charges. The magnetic force depends on the velocity of the particles as well as on their positions and can generate non-axial forces. Heck, The magnetic forces may not even opose each other, violating Newton's third law. But all is well since in those situations there will be electromagnetic radiation being produced and if the energy, momentum, and angular momentum of the radiation is also included in the calculation than the usual laws (including ) will still apply. This situation is clearly outside of the intended scope of the proof presented by your book. Dauto (talk) 19:52, 30 March 2010 (UTC)[reply]
Defrosting and re-freezing
If a product has a caducity of some days in the fridge (7 C) or several months frozen (-15 C), what is the problem of defrosting and re-freezing it? You buy it frozen, let it defrost 1 day and then re-freeze it and defrost it again in another further day. That just makes 2-3 days spoiling, is that a huge deal?
--Quest09 (talk) 17:49, 29 March 2010 (UTC)[reply]
If you are defrosting it in the fridge, you should be fine (although you may damage the food, particularly the texture). If you are defrosting it to room temperature, it could be a problem. --Tango (talk) 19:23, 29 March 2010 (UTC)[reply]
When a product is frozen commercially they (often) use flash freezing, i.e. freeze it so fast that there is no time of ice crystals to form. This preserves the texture of the product. If you refreeze it, you will allow ice crystals to form, and burst the cells of the food, so it will be limp and have a poor appearance. Ariel. (talk) 21:15, 29 March 2010 (UTC)[reply]
This is an issue that often comes up with frost-free freezers, which do, at times, melt the food then refreeze it. This absolutely ruins some types of foods, such as bread, by driving all the moisture to one end. Thus, you get part that's hard as a rock, and part that's slime. Meat can suffer from "freezer burn" for a similar reason. Ice cream also forms large ice crystals when it refreezes, making it no longer smooth and creamy. Some foods do quite well when melted and refrozen, like juices and soups, just as long as they are fully thawed and stirred before you consume them. StuRat (talk) 22:11, 29 March 2010 (UTC)[reply]
Grasping the EPT
Can I jump from a helicopter and safely grasp the electric power transmission line to hang for a few seconds, like birds that commonly sit onto it? 213.154.8.70 (talk) 21:33, 29 March 2010 (UTC)[reply]
(ec) There are at least two cables in a power line, one taking the electricity away, the other bringing it back. If a circuit is made, and you are a part of it, you will be fried to a crisp. With low power cables, they are insulated, but with high power cables, the bare wire is usually exposed, because insulation wouldn't do anything anyways. As long as only one cable is touched, you should be fine. If two uninsulated cables, however, are touched simultaneously, a circuit will be formed, and you will be turned into a crispy critter. --The HighFinSpermWhale21:40, 29 March 2010 (UTC)[reply]
Original poster: Note that the above answer is not to be interpreted as some sort of guarantee. That is, don't jump from a helicopter and say Wikipedia said you could. Gabbe (talk) 21:45, 29 March 2010 (UTC)[reply]
I would advice against the jumping part (It would likely be to high up to survive a fall and to close to the ground to be able to open a parachute), But they DO, on a routine basis, use helicopters as a working platform for work on LIVE extra high voltage power grid lines. (i.e. without turning off the power!). So if what you are yearning for is to touch a live wire (and survive), then that experience is definitely within reach (pun intended!)(Of course Provided that you have sufficient money to bribe the working crew :-) Unfortunately I am not able to remember what kind of protection, if any, one needs to have on the helicopter for it to fly safely in the (I presume) unusual ammount of static electricity. --Seren-dipper (talk) 22:15, 29 March 2010 (UTC)[reply]
I assume no liability for the following statement. It would be doubtful you could jump from a helicopter and grab even a non-electric rope without falling to your death. Utilities commonly have "live line work" done by personnel carried by helicopters. The personnel wear a metallic garment, and a probe absorbs the arc as the helicopter assumes the potential of the high voltage line (345 kv or 765 kv, for example). Then the worker installs a line separator or whatever, at the line potential, but isolated from ground potential. So for information only, and not proffered as advice, I assume that if I jumped from a helicopter and somehow grasped a high voltage line, without falling to my death, there would be a surge of electric current when I passed near the wire, causing an electric arc which might cause injury or death, which might cause my hands to grasp the wire, or to be unable to grasp the wire. I have seen photos of high voltage line maintenance in China in which the worker shot a nonconductive line across a transmission wire, then pulled herself up to the conductor, then attached a little wheeled cart which she used to travel to the line separator needing service. (By no means should you attempt this). The trick is to be at the line potential, without a path to ground or a different phase. Once there, corona discharge might be painful or disabling, even without arcing to ground or a different phase. The higher the voltage, the more difficult it would be to hand on. Birds land and perch on distribution lines, but not on transmission lines. Edison (talk) 01:59, 30 March 2010 (UTC)[reply]
There was a programme on the BBC nor long ago - Richard Hammond's Invisible Worlds - which showed an engineer sitting on a power line. I don't think it said whether it was a distribution or transmission line, but they used a special camera to show the electric field around him. It has to be said he was wearing a special metal mesh suit as well! --TammyMoet (talk) 08:19, 30 March 2010 (UTC)[reply]
The question is unclear and difficult to answer. If I touch my toes with my fingers, then yes, there is a circuit through the body, and some current could flow, if something is creating a voltage difference in the body. It is not clear why there would be a difference of voltage. Don't try the experiment. There is the likelihood of electrocution due to touching two conductors or a conductor and ground (the wood of a utility pole counts as ground), and arcing can occur across some distance of air, perhaps several feet, depending on the voltage. At high voltages, there is corona discharge even without a connection to ground or a different phase. This is due to the air being ionized near the sharper points connected to the conductor. The shape of a conductor affects the location and amount of corona discharge. Edison (talk) 16:25, 30 March 2010 (UTC)[reply]
What is the psychological term denoting the strong reaction (weeping, sobbing) when divulging a traumatic experience?
Resolved
When a patient for the first time divulges an until then repressed traumatic event, then many people (patients) experiences a (to them) surprisingly, and sometimes even frighteningly strong reaction of crying, sobbing, shaking etc. What is the specific psychological term for this (strong) reaction? --Seren-dipper (talk) 21:36, 29 March 2010 (UTC)[reply]
Most residential houses in the US appear to be supplied with 220-240 voltage which is split into 2 legs with each leg carrying 110-120 volts AC. Both legs seem to be connected by a neutral that is sometimes referred to as ground. Does this mean that one of the wires in a wall outlet has not difference in voltage that the ground outside or the floors in the house and that the bottom of the sin or cos wave for each leg is where neutral is actually at rather than at the zero point of the wave? 71.100.3.207 (talk) 02:24, 30 March 2010 (UTC)[reply]
To your first question: yes, one of the prongs is at ground potential.
To your second: it's hard to describe without drawing a picture. Each of the "hot" "phases" oscillates to a positive and negative voltage about that neutral. The scheme is described in our Split-phase electric power article. (Too bad there's not a picture on that page, either.) —Steve Summit (talk) 02:30, 30 March 2010 (UTC)[reply]
voltages of the two "hot" legs in Split-phase electric power Okay, here's a simple illustration. The black wire alternates from +120V to -120V with respect to the neutral (white) wire, which stays at ground potential (0V). Meanwhile, the red wire is alternating from -120V to +120V, 180 degrees out of phase. The voltage from either the black wire or the red wire with respect to the white wire is 120VAC. But if you connect a load across the black and red wires, you get 240VAC. (As you can see from the figure, at any given instant, the distance from the black curve to the red curve varies from +240 to -240.) This is how you can get 120V for ordinary household appliances, or 240V for high-power appliances such as electric stoves, ovens, and clothes dryers. (These are of course the voltages for North America; YVMV.) —Steve Summit (talk) 03:26, 30 March 2010 (UTC)[reply]
Please provide an English translation for "has not difference in voltage that the ground outside or the floors in the house and that the bottom of the sin or cos wave for each leg is where neutral is actually at rather than at the zero point of the wave?" Edison (talk) 05:09, 30 March 2010 (UTC)[reply]
I believe the first part has to do with the distinction between a floating ground and an earthed ground. That is, is neutral (zero) voltage in the electrical distribution system 0V with respect to a metal rod sunk into the ground. The answer to that, at least at the household level, is yes. The neutral bus bar in the electrical distribution panel in most US homes is directly tied to a metal rod sunk into the soil. The second part is asking about where the zero reference point is on the sinusoidal wave of the live conductor - does the alternating voltage swing from -120V to +120V (or whatever the numbers happen to be) or from 0V to 240V (with respect to the given neutral "zero"). In US household current it's the former. -- 174.31.194.126 (talk) 06:15, 30 March 2010 (UTC)[reply]
A U.S. utility is typically allowed to let the voltage (at the meter) vary +/- 6 volts from the nominal 120. With voltage drop along the wiring, this typically still supplies at least 110 volts at the outlet in the house. During peak loads there may be a "brownout" which causes voltage too low for some equipment to operate. The peak of a 120 RMS voltage is higher than 120 volts. To get the peak value, divide the RMS of a sinusoid by .707 and the result is 169.7 volts. A cheap (non-RMS) voltmeter will report a voltage other than the RMS value if the waveform is distorted, but will correspond to the RMS value for a pure sinusoid. So the voltage to neutral/earth from a "hot" wire varies typically from about -170 to about +170, 60 times per second (in the U.S.). If there is distortion (harmonics) in the waveform, the peak may vary substantially. The 240 volts in household supply is the RMS measurement between the two "hot" conductors. In 120/240 supply, the transformer low voltage winding is center-tapped for the neutral, so the two hot supply lines measure 120 from the neutral. The neutral is typically grounded to a substantial metallic low resistance earth ground at the transformer and at the neutral bus of the electric panel. The earth ground in the house might be the cold water pipe coming in from the street, or it might be a driven ground rod. If the neutral were ungrounded, such as in the case of the ground at the utility pole not being connected and the neutral at the house power panel not being grounded, there would still be 240 between the hot wires and 120 from either of them to the neutral wire, but with no earth ground there could be very high voltage, up to the primary voltage, between any of the supply wires and ground, a very hazardous situation. Edison (talk) 16:04, 30 March 2010 (UTC)[reply]
Role of the Pudendal Nerve in male sexual function?
I've been studying the role of the Sympathetic Ganglion in sexual intercourse, and I am stumped when it comes to the role of specific neurons in the male ejaculatory function.
As far as I can tell, the Prostatic Plexus and the Pelvic Splanchnic Nerves relay the sensory function —as well as increased heartbeat, blood pressure, body temperature, etc.— and the Pundendal Nerve relays the motor function (enabling ejaculation).
To wit, the Sympathetic Ganglion conveys the relexive arc from the genitals to the spinal cord (and back again), whereas the sensation of arousal/orgasm is received (some time later) by the brain.
My question, however, relates to the Pudendal nerve's function apropos ejaculation and the (subsequent) termination of sexual excitement —and the start of the refractatory period.
If —during the act of coitus— the Pudendal nerve were numbed or incapacitated, how would that affect a male's sexual performance? Would a man become able to feel the thrill of ejaculation for an extended period of time, without actually ejaculating (or losing his arousal)?
There is a technique, taught to couples where the men suffers premature ejaculation, where the perineum is pressed and held in just before ejaculation: this is said to delay ejaculation. I'm not an expert on human anatomy, but is this where this particular nerve is situated? --TammyMoet (talk) 08:16, 30 March 2010 (UTC)[reply]
PAC MAN MOON
I seen a quick article on the PAC MAN MOON shape (Saturn i think) It seemed glaringly obvious to me that the with the giant impact area that the energy transfered from the impact through the planet to other side and this was some sort of risidual engery left over from this. Ok maybe i am completly wrong but thats what struck me looking at it? So the question is i suppose could this be the case? —Preceding unsigned comment added by Chromagnum (talk • contribs) 05:18, 30 March 2010 (UTC)[reply]
Your description in terms of energy sounds pretty ropey. However, the idea that the creation of the Herschel crater could be linked to the variations in the observed temperature is suggested in the article you provided:
"The Cassini team says the creation of the crater itself might have played a key role in changing conditions across extensive regions of the moon's surface."
The Cassini-Huygens probe has an official website from NASA. Here's a little more technical information from the NASA/JPL Press Release: 1980s Video Icon Glows on Saturn Moon. The early assumption is that the temperature difference is due to surface texture changes, which affect the heating and cooling rate. "Even if surface texture variations are to blame, scientists are still trying to figure out why there are such sharp boundaries between the regions, Spencer said. It is possible that the impact that created Herschel Crater melted surface ice and spread water across the moon. That liquid may have flash-frozen into a hard surface. But it is hard to understand why this dense top layer would remain intact when meteorites and other space debris should have pulverized it by now, Spencer said."[24]. Nimur (talk) 14:48, 30 March 2010 (UTC)[reply]
Number of organs
I've read that humans have 22 or 23 organs depending on how you count.
How do you define an organ? Are cerebrum and cerebellum seperate organs? What about hypophysis? Is it an organ on its own or a mere extension of the brain? The eyes? If the stomach is granted the title of organ, what prevents the poor duodenum from being called one? Maybe the entire gastrointestinal tract is a single organ, then. Last thing I heard, salivary glands have been lobbying for independence. Seriously, unless the IAU decides what is an organ and what is a dwarf organ, I don't think it matters much. 88.242.231.192 (talk) 10:28, 30 March 2010 (UTC)[reply]
Maybe he was thinking of chromosomes? Humans have 23 pairs of chromosomes, often split into the 22 pairs of autosomes and one pair of sex chromosomes. His numbers match this fairly precisely. If he meant chromosomes, the answer is that the number of chromosomes varies wildly from species to species (note that the preceding link gives total number, not number of pairs, so humans show up with 46). —ShadowRanger(talk|stalk)12:21, 30 March 2010 (UTC)[reply]
No. I mean organs. Yes, I realise it's hard to quantify natural phenomena that fall on a gradient such as languages, planets, species, and organs. That does not mean that questions regarding them are not interesting or important. -Craig Pemberton13:22, 30 March 2010 (UTC)[reply]
Well, anyways, other animals, even other mammals, have organs that we do not share. Ruminants have a much more complex digestive system, for example. And I have it on good authority that many politicians manage without a neocortex. --Stephan Schulz (talk) 14:08, 30 March 2010 (UTC)[reply]
(ec)If you can't rigidly define an organ, then your question is meaningless. I have found at least one source that supports your "22 or 23" contention (where the 23rd is the skin, though it neglects to enumerate the rest). I can guarantee you that the number of organs in other animals would not match those of a human though, except by coincidence or in cases of extremely recent speciation (many of the apes may match us though). According to the page I'm referring to, the definition is "a relatively independent part of the body that carries out 1 or more special functions." Given that the appendix would probably not count as an organ under that definition, it would logically follow that any mammal with a functioning appendix (that actually serves the purpose of digesting cellulose) would not have a matching number of organs. Most herbivores have one, so the number wouldn't match. Similarly, many herbivores have multiple stomachs, which would screw up the calculation. And we haven't even left mammals yet. Birds, reptiles, etc., all have substantially different digestive tracts, reptiles, being cold blooded, don't have a liver in the same sense as humans, non-mammals wouldn't have a direct analogue to the mammalian reproductive system, etc. —ShadowRanger(talk|stalk)14:14, 30 March 2010 (UTC)[reply]
Similar to variations in the digestive system, how about variations in the scent glands of animals, such as skunks ? Venom producing/delivering organs would be another. Then some animals have other sensory organs, like extra UV and infrared-sensitive eyes and pits, and the echo-location "melon" on a porpoise or dolphin. Also, some animals have organs which provide the ability to sense weak, or create strong, electrical fields. Only (female) mammals have (milk producing) breasts and a uterus. StuRat (talk) 14:21, 30 March 2010 (UTC)[reply]
Evolution - E. coli
Greetings! Two questions:
How old - evolutionwise - is E. coli? (~ 300-400 Mill. years?)
As it is not found in the guts of cold-blooded animals (molluscs, fish) what ist the main type of gut bacteria in these species? Thanks Grey Geezer 07:26, 30 March 2010 (UTC) —Preceding unsigned comment added by Grey Geezer (talk • contribs)
i have a practical in course its aim is to determine the boiling point of the given sample of water.
in the setup of apparatus we are said to put some pumice stone in the boiling tube containing water.
tell me why that pumice stone is to be kept? {its not a HW question its indeed a doubt}
in precaution it is clearly mentioned that "the bulb of thermometer should not dip in liquid. it should be about 4-5 cm above the liquid surface."
You can measure the boiling point of water by measuring the temperature at which steam condenses (on the thermometer, for instance). Therefore the thermometer does not need to be immersed in the water. As for the pumice, its purpose is to assure that the water boils at its boiling point instead of superheating. If the water is quite pure, and the vessel in which it is heated is quite clean and smooth, then the water is liable to be heated above its boiling point and then violently erupt, yielding a false measurement (and possibly giving you a nasty burn as well). The porous pumice provides a rough surface on which the water will readily vaporize into bubbles. You can read more here, in considerable detail.--Rallette (talk) 09:27, 30 March 2010 (UTC)[reply]
See also boiling chip, if you ever need to buy some from a catalog. As Rallette notes, small, inert, irregularly-shaped chunks with a large surface area (for their volume) are used to discourage superheating and violent bubbling. TenOfAllTrades(talk) 13:07, 30 March 2010 (UTC)[reply]
transpiration
is it so that a tree on high altitudes will have more or less transpiration , assuming that temperature is same for it in comparison to another tree at lower altitude...thanx--Myownid420 (talk) 07:37, 30 March 2010 (UTC)[reply]
There seem to be quite a few studies on this. The abstract of the first one in my search says: "An important factor at high elevations is the reduced barometric pressure. It has long been recognized that the reduction of air pressure brings about an increase of evaporation, and the increase of plant transpiration in response to low air pressure has also been demonstrated under laboratory conditions." Other factors also seem to influence the rates of transpiration at different elevations; your best bet is to read some of the articles for yourself. Deor (talk) 11:52, 30 March 2010 (UTC)[reply]
Note that at that site, they mention the World Trade Center, Iraq conspiracies, and conspiracies around Tesla (this one actually has a bit of truth to it). It's surprising they don't mention the JFK assassination and the rise of Hitler. StuRat (talk) 15:47, 30 March 2010 (UTC)[reply]
I don't blame you for not liking that source, as it has all the hallmarks of crackpot pseudoscience.
Whenever someone says they've got an extraordinary new theory, which is unbeknownst to (and is perhaps being suppressed by) mainstream science, it's possible in some theoretical sense that they're right -- science does always consider the possibility that it's wrong, and it does always have more to learn -- but it's spectacularly, ludicrously, vanishingly improbable. Electromagnetic waves aren't an energy source anyway, so to suppose that these hidden, different kind of waves could somehow provide limitless "free" energy is just, I'm sorry, nonsense. —Steve Summit (talk) 13:14, 30 March 2010 (UTC)[reply]
According to the feedback column of New Scientist the "scalar energy" notion is an increasingly prevalent brand of what it calls fruitloopery. Consensus on Wikipedia, for what it's worth, said pretty much the same thing: Wikipedia:Articles for deletion/Scalar field theory (pseudoscience). However, unless I am missing something, I disagree with Steve Summit's assertion that electromagnetic waves aren't an energy source. See for example solar power. Steve - perhaps you mean that they aren't an "ultimate" energy source, in that they must have been generated by something else such as nuclear reactions within the sun, but under that interpretation kinetic energy wouldn't be an energy source in lots of real world situations either. Equisetum (talk) 14:01, 30 March 2010 (UTC)[reply]
I made the point a bit too hastily, but Stu caught my meaning. I think of electromagnetic fields and waves as being a means of transmitting energy, but not of storing it, let alone being a source of it. (Although it'd probably be hard, I concede, to rigorously distinguish between a "source" and a "storage", if it came to that.) —Steve Summit (talk) 16:57, 30 March 2010 (UTC)[reply]
Just curious; wouldn't it be more useful to have an article on scalar energy that says it's crap rather than just not have one at all? I'm not familiar with the WP deletion policies but common sense says that having information telling you it's highly improbable leads to less confusion than no information at all. I'll read into the guidelines further and perhaps I can answer my own question. -Pete5x5 (talk) 14:47, 30 March 2010 (UTC)[reply]
For very common ideas, like perpetual motion, we have an article explaining the common fallacies. Unfortunately, there is no limit to the number of incorrect theories - it's hard to have an article on all of them. We reserve our articles for only those most important pseudoscientific theories - the ones which are notable. Nimur (talk) 15:07, 30 March 2010 (UTC)[reply]
My last molar on the upper left is giving me trouble. I had a crown placed on it a few years ago, and several months ago, ¼ of it broke off. I haven't had the money or the insurance to get it fixed until now, and it hasn't given me much discomfort until the past week or so anyway. (This is not a request for medical advice, because I *know* what I need to do-- have the dentist fix my tooth. I have the appointment. It's in May.) The associated toothache comes and goes, but when it comes, it sometimes is blindingly bad.
I'm taking Ibuprofen and treating topically with Orajel, which helps some. However, I have also found that swishing slightly-below-room-temperature water around my mouth also helps ease the pain some.
Cooling the nerve may be helping you a bit. Cold brings a certain amount of numbness, and if your tooth is damaged, the nerve is exposed more directly to the cool water. If the tooth is infected, the cool water may also be reducing swelling a bit. You may also be reducing the degree of infection a bit (by washing away some of the pathogens), but that isn't likely to have a measurable effect on your pain, and they reproduce quickly enough and enough of them are likely unreachable by the water that the long term effect is negligible. —ShadowRanger(talk|stalk)14:18, 30 March 2010 (UTC)[reply]
Thanks. However, I had some slushy (partially frozen) lemonade yesterday, and five minutes later, I thought I was going to die. So, apparently not TOO cold? Kingsfold (talk) 14:40, 30 March 2010 (UTC)[reply]
Too cold is going to be a problem, but lemonade (regardless of temperature) would probably make it worse. Lemonade is quite acidic, and contains a lot of sugar (which bacteria in your mouth process, producing more acids). —ShadowRanger(talk|stalk)15:19, 30 March 2010 (UTC)[reply]
Might be the acid in the lemonade, but it was Crystal Light, so... no sugar. I'll try another flavor today (not frozen), and see if it still causes trouble. Kingsfold (talk) 15:48, 30 March 2010 (UTC)[reply]
May seems like a long time to wait while you are in such distress, can't you get an appointment with a dentist sooner ? StuRat (talk) 15:41, 30 March 2010 (UTC)[reply]
Great idea. Wish I had thought of that. (Wink.) I'm on the "call if someone cancels" list. I'm considering changing dentists.... Kingsfold (talk) 15:48, 30 March 2010 (UTC)[reply]
I had toothache that was relieved by swirling water. It was due to a large cavity between my teeth - due to moving town I had not had regular check-ups for some time - my fault for not signing up with another dentist and not the NHS's. Coincidence - it was the last tooth on my upper left as well! Luckily it was just a wisdom tooth, so it was removed without any cosmetic or functional effects. Am I right in guessing from the OP saying that they could not afford treatment that OP is from the USA? (Why dont Americans revolt and demand free health care from their government?) 78.146.180.118 (talk) 20:17, 30 March 2010 (UTC)[reply]
Did you read the post? He does have an appointment to get it fixed, and presumably can afford treatment. But his regular dentist is backed up, so he has to wait a little while (or find another dentist, which he has not yet tried to do). —ShadowRanger(talk|stalk)20:22, 30 March 2010 (UTC)[reply]
But I though this kind of delay was what Americans imagined happened with the UK NHS, and not what is supposed to happen in the US? I mean, May and he's in pain! Over a month away! This side of the Atlantic you'd probably get it done as an emergency appointment the same day, or the next day at most. 78.146.180.118 (talk) 20:26, 30 March 2010 (UTC)[reply]
Yes, and being cancer, there are a nearly infinite number of possible symptoms. That said, we cannot give medical advice at this reference desk. You should contact a doctor for specific advice. —ShadowRanger(talk|stalk)15:24, 30 March 2010 (UTC)[reply]
Here is a PDF of Canadian cancer statistics for 2008. It also has a special section of Childhood (Ages 0-14) cancers beginning on page 60. You can use that to look for articles on specific cancer types. -- Flyguy649talk15:50, 30 March 2010 (UTC)[reply]
number of sites
If Iran packed every square inch of land under its control with nuclear sites is there a limit in the amount of electricity it could sell to other countries without fear of invasion if Iran cut them off? 71.100.3.207 (talk) 17:24, 30 March 2010 (UTC)[reply]
You're going to have to clarify what you mean. How would Iran's nuclear sites affect its ability to sell electricity (do you mean nuclear power plants, as opposed to nuclear weapons or nuclear refining sites?), and how would selling electricity have anything to do with fear of invasion? Another country could theoretically become dependent on foreign electrical power, but they usually maintain enough homegrown generation capacity to keep their military functional. —ShadowRanger(talk|stalk)17:43, 30 March 2010 (UTC)[reply]
Lets see... after a certain distance the cost of transmission might become too high for electricity to sell. with that restriction what is the limit of power Iran could reasonably expect to sell, use or store without unreasonable risk of invasion by others to remedy an energy crisis?. 71.100.3.207 (talk) 18:16, 30 March 2010 (UTC)[reply]
The range for transmission would likely not be much of an issue. You do lose power over long distances, but given that there are serious (if unlikely to be fulfilled) proposals to use solar power in the American Southwest to power the whole country, it's clear that the losses aren't enough to make it completely impractical. Answering the question of how much power is "too" much, in the sense of attracting the attention of foreign powers looking to remedy an energy crisis, that's asking us to predict the future and human political psychology all in one. According to this chart of power generation capabilities, the U.S. produces as much power as China, Japan and Russia put together. Iran is producing roughly 10% of what Japan uses, so unless you assume that Iran's neighbors are more likely to launch an invasion than Japan's neighbors, Iran should be able to multiply it's power generation capabilities by 10x without risk. —ShadowRanger(talk|stalk)18:43, 30 March 2010 (UTC)[reply]
Land isn't the limiting factor. I would expect them to run out of nuclear fuel long before they ran out of space. --Tango (talk) 17:50, 30 March 2010 (UTC)[reply]
Breeder reactors aren't magic. They can turn certain thorium isotopes into usable reaction material, and they can use the initial uranium fuel more thoroughly, but they're still limited by the need for fairly uncommon materials. —ShadowRanger(talk|stalk)18:43, 30 March 2010 (UTC)[reply]
The transmission cost for electric power is mainly the one-time investment cost in infrastructure i.e. lines, pylons, switchgear and transformer stations. The selling price for electricity varies with market demand, which varies with season. Availability of cheap and reliable nuclear power will tend to stifle use of other forms of power, particularly if it is perceived as less polluting. There is no economical way to store electric power so it makes more sense to run nuclear stations at partial capacity and adjust output to suit demand. Neighbouring countries that might invade Iran for reasons that have nothing to do with electric power would seem less likely to do so if they depended on Iran for cheap power. But the OP makes a big assumption that the world powers will allow Iran to develop nuclear capabilities unchecked. Cuddlyable3 (talk) 19:17, 30 March 2010 (UTC)[reply]
Some things are clearly beyond anyone's (and the whole world's) capacity. Iran cannot acquire such potential overnight or in a year or in ten years. There's nothing to discuss here. They can build the system organically, one radius at a time, but it's a whole different scenario. NVO (talk) 19:21, 30 March 2010 (UTC)[reply]
calorie consumption
Is there a formula to calculate the number of food calories needed to fulfill both the BMR requirements and number of calories needed to ride a bicycle different distances and at different speeds? 71.100.3.207 (talk) 17:39, 30 March 2010 (UTC)[reply]
There are a million different calorie calculators online, just Google (or Bing) for them. This is one. They're all going to have a very limited amount of accuracy, due to the fairly substantial differences in metabolic rates between people. —ShadowRanger(talk|stalk)17:47, 30 March 2010 (UTC)[reply]
You'd have the same problem with a formula. Unless you spend a lot of time to determine your particular basal metabolic rate, you will just have to settle for an average, which won't be accurate enough to be of much value, unless you just happen to be a throughly average person. See basal_metabolic_rate#BMR_estimation_formulas. Also, the air temperature, humidity, hilliness of the path, clothes you wear, whether it's sunny or cloudy, etc., would all have an effect on the calories you use biking. But, if you want to ignore all that, here's a site which has the formula to calculate calories burnt bicycling: [26]. StuRat (talk) 18:14, 30 March 2010 (UTC)[reply]
Resolved
