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I've noticed that occasionally there are lots of dead flies on my outdoors plants (I've noticed them on basil, oregano, and thyme in particular). I occasionally just found one and thought, "well, I guess this just happens to be where he kicked the bucket," but the other day I found dozens of them. They look like normal houseflies, more or less, and are just sitting on the plants, clinging to them, dead. They aren't wrapped in a web or anything so obvious. What causes this? Is this some sort of fly life-cycle thing? Or some odd predator? Or what? We don't use pesticides on the plants. Any thoughts? Imagine a bunch of flies just sitting on a plant. Now imagine they happen to be dead. That's what it looks like. --98.217.14.211 (talk) 00:38, 5 September 2009 (UTC)[reply]
Houseflies and blowflies do not feed on green parts of live plants, but they may feed on the honeydew secretions of aphids if those are present on the plants. There may be some kind of pathogen (virus, bacterium, fungus, parasitic worm eggs, whatever) present in the honeydew and infecting the flies. I've never seen such a thing, so I have no good guesses as to what it may be. If it is a fungus, you would see fruiting bodies emerging from the dead flies. Otherwise, you'd just see dead flies. Sorry. --Dr Dima (talk) 02:59, 5 September 2009 (UTC)[reply]
OK, found a site with good images of flies killed by fungus, here. Does it look any similar? Note that Entomophthora is not the only fungus that can do that. We have an article on Entomopathogenic fungi, and it provides many links. --Dr Dima (talk) 05:01, 5 September 2009 (UTC)[reply]
Hmm, maybe. I'll take a closer look today. Sounds plausible enough.--98.217.14.211 (talk) 13:05, 5 September 2009 (UTC)[reply]
What is the offical Moon landing time? And if using USA time (20July) does that mean it actually happened on 19 July (using Australia/NZ time zones)? —Preceding unsigned comment added by Rosetoohey (talk • contribs) 00:42, 5 September 2009 (UTC)[reply]
"Actually happened" means nothing here, you're just asking about time zones, which are all about your reference point. From our article, "At 02:39 UTC on Monday July 21 (10:39pm EDT, Sunday July 20), 1969, Armstrong opened the hatch, and at 02:51 UTC began his descent to the Moon's surface." As always, it was a different day somewhere in the world... --98.217.14.211 (talk) 00:54, 5 September 2009 (UTC)[reply]
The internationally agreed-upon time in space is UTC, so the above times are technically the "official" Moon landing time. -RunningOnBrains(talk) 01:06, 5 September 2009 (UTC)[reply]
I'm surprised the second time zone given is EDT. Shouldn't it be Houston time, which would be CDT? --Trovatore (talk) 01:09, 5 September 2009 (UTC)[reply]
Well, that is the timezone it took off in, while they were "just" communicating with Houston. Besides, most of the country's power (namely Washington, D.C. and NYC) are in there as well. ~ Amory(user • talk • contribs) 02:21, 5 September 2009 (UTC)[reply]
The country is run from Los Angeles. Washington and New York are just being humored. --Trovatore (talk) 02:24, 5 September 2009 (UTC)[reply]
Of course, the real answer is that there is no time in space. That is to say, there is no chronology which may be calibrated. --Trovatore (talk) 01:13, 5 September 2009 (UTC)[reply]
OK, no one's picked up on this thread for more than an hour, and a Google search for "there is no time in space" shows a shocking number of people who appear to be serious, so I guess I'd better spoil my own joke here. It's a line from a very famous episode of The Twilight Zone, the one where the aliens' guiding book, To Serve Man, turns out to be a cookbook. --Trovatore (talk) 02:22, 5 September 2009 (UTC)[reply]
As to the original question, if it is July 20 in the US, it can be July 20 or 21 in Australia or New Zealand, but not July 19. Eastern Standard Time in the US, for example, is 15 hours behind Eastern Standard Time in Australia. In all Australian and New Zealand time zones both the landing and the moon walk took place on July 21. --Anonymous, 05:22 UTC, September 5, 2009 (it's September 4 in western US time zones, but September 5 in Australia and New Zealand).
At the beginning the Sun radiated about 30% less energy and continues to radiate more and more. What causes this to happen? --Halcatalyst (talk) 00:54, 5 September 2009 (UTC)[reply]
The accumulation of helium in the core causes a gradual increase in the rate of fusion and gravitational self-compression, as helium is denser than hydrogen. Higher temperatures must be attained to resist this increase in gravitational compression and to maintain a steady state.
This seems to make sense, but this is un-referenced, and I am not very familiar with the subject.-RunningOnBrains(talk) 01:04, 5 September 2009 (UTC)[reply]
It makes sense - to get two hydrogen nuclei to fuse you have to overcome their mutual repulsion. In the sun, it's gravity & pressure that does that - so there is some region in which the suns gravity is strong enough and some other region where it's not. As the sun transforms hydrogen into denser helium - it gets gradually denser - the region throughout which hydrogen fusion can occur gets bigger - so more hydrogen fuses and you get more energy. I don't have a reference - but it sure makes good sense. SteveBaker (talk) 02:46, 5 September 2009 (UTC)[reply]
It's not just that there is more volume that can fuse, the part of that volume that was fusing before is now fusing faster because it is hotter/denser. --Tango (talk) 10:42, 5 September 2009 (UTC)[reply]
Here's a response I got from Yahoo Answers: "As the core of the sun converts hydrogen to helium, it becomes more hydrogen-depleted. This forces the core to contract and increase pressure in order to fuse enough hydrogen to counteract gravitational collapse. As it contracts, however, the density gets higher, increasing the gravitational pressure; so the core must also get hotter by fusing even more hydrogen than before in order to maintain the equilibrium. So the sun gets hotter over time. As you say, it is 30% hotter now then when the earth first formed." --Halcatalyst (talk) 14:14, 5 September 2009 (UTC)[reply]
...and that's why it's unwise to ask questions at Yahoo Answers. SteveBaker (talk) 15:30, 5 September 2009 (UTC)[reply]
But that's the right answer... --Tango (talk) 16:00, 5 September 2009 (UTC)[reply]
It's incomprehensible! Why (to pick just one part of this so-called explanations) does becoming more hydrogen-depleted force the core to contract? We're told it does that in order to fuse enough hydrogen to counteract gravitational collapse! Why? Is it sentient or something? That's not an explanation - it's a mess. SteveBaker (talk) 18:47, 5 September 2009 (UTC)[reply]
It is pretty common to explain things using anthropomorphic language. The only problem with the explanation is that it isn't very detailed, but that isn't necessarily a bad thing - you have to get the level of detail right for your intended audience. The OP probably doesn't know much astrophysics and probably doesn't want to read the equivalent of a textbook on the subject, so you need to miss a few things out. If you want a slightly more complete version of the sequence of events (they are actually all happening at once, of course), here goes: The hydrogen gets depleted, which reduces the amount of fusion happening, that reduces the radiation pressure so gravity is stronger and the star contracts. As with any gas, it heats up as it contracts. That extra heat and density increases the amount of fusion taking place, increasing the radiation pressure and bringing the star into a new equilibrium. This new equilibrium position is hotter and, therefore, brighter. --Tango (talk) 19:02, 5 September 2009 (UTC)[reply]
It doesn't seem right to me intuitively that a reduction in the reaction rate would lead to a higher equilibrium reaction rate, although I guess it's not impossible. I would think the increase in the molar mass is mostly what drives the volume down and the temperature up. Rckrone (talk) 05:50, 6 September 2009 (UTC)[reply]
I'm not entirely sure about that bit either, that is why I kind of glossed over it. --Tango (talk) 14:43, 6 September 2009 (UTC)[reply]
Thank you all for your responses. --Halcatalyst (talk) 21:40, 7 September 2009 (UTC)[reply]
I'm sure it's less about hydrogen depletion and more about helium formation. Increased gravitational pressure merely alleviates hydrogen depletion. John Riemann Soong (talk) 04:15, 8 September 2009 (UTC)[reply]
My question is this: how can the atomic mass of oxygen-16 be 15.9949146 u when both a proton and a neutron have masses greater than 1 u? Shouldn't the mass be greater than 16 u? Nkot (talk) 01:56, 5 September 2009 (UTC)[reply]
See Binding energy#Mass deficit for an excellent explanation of the phenomenon. Come back and ask again if that article doesn't make sense... --Jayron32 02:01, 5 September 2009 (UTC)[reply]
Black hoes can evaporate due to hawking radiation...so would it be possible for a particle to pass the event horizion but have the black hoe evaporate before the particle is absorbed into the singularity? like the particle gets sucked past the "point of no return" but the black hoe evaporates right after?
Also, if you have two black hoes close enough to each other such that their event horizons overlap at a single point...(perhaps the black hoes could be orbiting one another, so they dont fall into each other), what would happen to a particle that traverses a path where the gravitational forces are neutralized by the opposing black hoe--that is the black hoes are possitioned such that their event horizions are both tanget ( at the same point) as the path of the particle. What would happen to the particle, since it is effectively has passed the event horizions of both black hoes? XM (talk) 04:54, 5 September 2009 (UTC)[reply]
It's not often you see such a persistent typo. APL (talk) 05:22, 5 September 2009 (UTC)[reply]
In my experience, hoes don't evaporate, absorb particles, or neutralize gravity, regardless of their color. They also don't have event horizons. Or did you mean black ho's? If you did, I have to tell you that they don't do any of these things either. :-D 98.234.126.251 (talk) 05:37, 5 September 2009 (UTC)[reply]
At first I thought XM's L key didn't work, but then it would have been back hoe. The question seems legitimate aside from the... typo?... so I guess I'll answer it. The short answer to all of the questions is that anything that passes the event horizon will hit the singularity because that's how the event horizon is defined. If you escape then by definition you didn't cross the event horizon. So the real question is whether an evaporating hole has an event horizon at all, and the answer to that seems to be yes, at least if you believe the Penrose diagram that was in Hawking's original paper (reproduced here). Hawking himself seems to have decided that he was wrong and there is no horizon or singularity really, but that view doesn't seem to be very popular right now. If the event horizons of two black holes overlap then you really have one black hole with one event horizon, and the no-hair theorem implies that it will quickly become spherical. There's only one singularity so there's no ambiguity about where the particle ends up. -- BenRG (talk) 18:45, 5 September 2009 (UTC)[reply]
Forget about the amusing if racist typographical error! This user has asked a question that is so profound I wonder why I didn't think of it myself! Because if the hole (presumably a very small one) evaporated quickly enough, the particle would have "seen inside" the black hole and then "escaped" as the event horizon moved past it, and this would violate "cosmic censorship" which states that not even information about the inside of the black hole can escape-some physicist think this is just as well since new laws of physics might enter our universe should a naked singularity ever occur.
My answer is that since the particle would effectively be moving near the speed of light (or above(?)) when inside the event horizon, and that the horizon shrinks according to how much hawking radiation it can shed, then there is no reason to suppose that the horizon can catch up with the particle. I could be wrong-can someone do the maths?80.2.195.218 (talk) 13:24, 10 September 2009 (UTC)[reply]
Hey, that's a fascinating possibility -- that information could actually escape from a shrinking black hole. Unfortunately I can't crunch the numbers on this -- my area of expertise is hydrocarbon/petroleum chemistry, not astrophysics -- but it does make sense intuitively that if a black hole is shrinking, then matter/information could escape from it. Note also that the "cosmic censorship" hypothesis has so far only been definitively proved for non-shrinking black holes. (I wonder if Hawking is logged in right now, maybe he could crunch the numbers for you, he's a genius at this kind of stuff.) Well, clear skies to you! 146.74.230.106 (talk) 21:29, 10 September 2009 (UTC)[reply]
leaving groups, chemical attacks, MOs, and stuff.[edit]
Sometimes I get the feeling that in some electron movement mechanisms, incoming electron density "pushes" existing electron density in the other direction ... so say in esterification an alkoxy oxygen's lone pair is "pushing" away the pi electrons onto the carbonyl oxygen, and the pi electrons happen to be mobile. So in fact, before that carbon converts to sp3, it's like the alkoxy oxygen's lone pair is participating in the pi system and pushing the existing bonding electrons away. (Okay this might be an unacceptable classical picture though.)
It was mentioned to me in passing half a year ago how nucleophile electrons attack antibonding orbitals ... and in fact I had totally forgotten about antibonding orbitals (as a mechanism for nucleophilic attack) until it was mentioned again. I guess the classical picture is kind of misleading, because the typical picture of SN2 seems to be of say, the better nucleophile "pushing" away the leaving group through backside attack, forming and breaking bonds simultaneously that way. Do those nucleophile electrons in fact, contribute to an antibonding orbital, that somehow negates the bonding between the LG and the carbon? If so, how does an antibonding orbital convert into a bonding orbital (presumably that as the LG leaves a bond must be formed simultaneously)? Is bond breaking and formation truly simultaneous? Will one aspect be slightly ahead?
Another thing that doesn't get discussed much is sp2-sp3 (and sp2->sp3) transitions ... it almost seems to be a factor (if you could somehow get stuff to happen during the transition, and how much it affects equilibrium. It seems to me that possibly in such transitions you could have stereochemistry-affecting effects. What does the hybridisation of the two central carbons look like when the bromonium ion is attackihg an alkene for instance? 1/2 sp3, 1/2sp2?
Also, is acid catalysis always a first order reaction? It seems to me that for example, in acid-catalysed esterification of carboxylic acids where the C=O bond is being restored and kicking out the LG, that the electron density (coming out from oxygen's previous lone pairs) pushes electron density away from the C-OH bond, and meanwhile, this also pushes electron density from the OH (part of COOH) onto a nearby "proton". That is, I'm thinking you don't have to wait for acid to protonate the leaving group, and THEN wait for bond displacement to occur -- acidic protons act like terminal electron acceptors that stabilise the LG. Which kind of mechanism would acid catalysis support? (I'm referring to protonation, such as OH protonation, to create a good LG, not protonation to increase the carbonyl carbon's electrophilicity.) John Riemann Soong (talk) 07:44, 5 September 2009 (UTC)[reply]
First of all I'm going to recommend another book if you haven't already read it, it's "Mechanism and Theory in Organic Chemistry" (Thomas H. Lowry, Kathleen Schueller Richardson) which covers the first three paragraphs of your question in very gory details.
also Transition state - should be the best search term - eg for google books.
First paragraph - the alkoxy (why is it not alchoxy, alchemy, alchohol, k?emistry) O lone pair does just what you say - interacting with the pi system - specifically the C p orbitals. For RC(=O)OR' + R''O- the transition state is close to sp2 C bonded to O- , R , R' with the R''O bonding via the p (pi) orbital.
I don't think your use of the term 'antibonding orbital' is right in all the contexts you use it specifically the ester example Doh. It is right, silly me (deleted stuff below)
Second paragraph:
In a pure Sn2 reaction bond breaking and formation is simultaneous, however in practice if one of the leaving groups is better than the other, asymmetry may develop.
Both views are right. ('classical' and molecular orbital - they're describing the same thing, the same 'movement' of electrons (or electron density) )
Yes - the nucleophile feeds electrons into the leaving groups antibonding orbital - weakening the bond. As the bond is weakened the leaving group will move away from the molecule (electron repulsion).
Curiously the antibonding orbtital for the leaving group is the bonding orbital for the attacking nucleophile (and vice versa) - it's like a see-saw.
The antibonding orbital for the leaving group is the mirror image of the bonding orbital, this might need a little more explanation if you're not familiar with it, both can be anti, or bonding orbitals depending on which side a group is attached.. In the absense of groups on either side both orbitals can be considered plain p orbitals of the same energy, did that make any sense?
More: once a leaving group has left (in Sn2) the bonding orbital it was using, becomes the new antibonding orbital - it relates to the mirror image thing.
Third paragraph:
Roughly yes, someway between sp2 and sp3. For the stereochemistry the vaguely similar epoxide ring opening reactions are a good source of data (epoxides being more stable)
Fourth paragraph - do you mean that reaction of the intermediate in esterification is a concerted reaction in which the OH leaves as it is protonated (1 step) rather than being protonated then leaving (2 steps) - this is quite likely (though depends on absolute conditions)
R R
| \
R'O-C-O-H >>> C=0+-H
| /
HO: H+ R'O +H2O
I can't draw the arrows for electrons, but you can assume all the above happens in one step—Preceding unsigned comment added by 83.100.250.79 (talk) 11:10, 5 September 2009 (UTC)[reply]
Yeah, that's what I meant. Basically I'm wondering whether this will allow protonation to occur more often, even though the pKa of a protonated alcohol is kinda low. I wonder this cuz I'm interested in improving the efficiency of longer-term esterification, in weakly or moderately acidic solutions [with acetic or phosphoric acid for example] that must be edible later on. Thus, if I had a pH of 2.7 (or say I somehow got my hand on food-grade phosphoric acid and lowered it to 2 or something), would protonation of the OH group occur more often than the general 2-step depiction would indicate? That is, as the carbonyl's pi system electrons are entering the antibonding orbital (is my conception correct? sp2 bond is being reformed though) of the C-OH bond, this makes the OH group more basic ... making it a terrible leaving group, of course. But wait! There's a hydronium ion nearby (as well as various other protons on HOH's) which basically neutralise the OH group's basicity as it forms, in a sort of concerted reaction. John Riemann Soong (talk) 15:35, 5 September 2009 (UTC)[reply]
Basically yes. (at least there's nothing above that is obviously wrong)
To expand a bit there's a good example of trans-esterification where a chemical buffer (acid base) mixture is used (in dehydrating conditions of course) - in an even more concerted reaction the acid part of the buffer "stands by" to protonate the leaving group (-OR) as it leaves, and the basic part of the buffer "stands by" to deprotonate the attacking alcohol as it attaches to the carbonyl - higher concentrations of buffer help here - but the reaction goes suprisingly well for what may be an effectively neutral solution. Note how both the presence of the acid and base help here. This helps to reduce the activation energy - though the chances of getting all the molecules in the right place mean the reaction is quite slow nevertheless (High concentrations of buffer help). The same method can be used for straight esterifiaction.
The only tricky with the above method is choosing weak acids and bases that will not get involved in the esterification reaction (ie not acetate, or boric acid) Triethyl amine (does not react with carbonyl permanently) is a good choice for the base. Have you heard of tertiary amine carbonium ions? see
R3N+C(=O)R'
They're useful (and often overlooked) intermediates in esterification reactions, especially when you want to avoid any strong acid - you can prepare them before hand from the amine and acid chloride in suitable conditions.
And perception, to bridge the gap between physiological sensation and psychological awareness of sensation. Nimur (talk) 17:28, 5 September 2009 (UTC)[reply]
For an overview have a look here: >> Sense << Ostracon (talk) 18:53, 5 September 2009 (UTC)[reply]
Symbols used in this type of molecular diagram[edit]
I would like to learn a bit more about the symbols used in this type of molecular diagram, example In this particular example, the wriggly line is the symbol that I don't know about. There surely is an article on wikipedia, but I can't locate it. Can somebody please supply a link? Thanks! --TrogWoolley (talk) 10:24, 5 September 2009 (UTC)[reply]
It relates to the stereochemistry, the atom attactched to squigly line can be in one of two positions (up or down with respect to the paper) - it's a variation of the Natta projection
In this case a squigly line means that it is either, ie that it is not defined which of the two types it is.83.100.250.79 (talk) 11:02, 5 September 2009 (UTC)[reply]
Here's an example with the two types ('forward', and 'back') - they are the triangular line , and the dotted lines, if a squicgly line was used it would mean that it could be either.
The term for indeterminate stereochemistry, if you would like to read more, is racemate. --Jayron32 11:46, 5 September 2009 (UTC)[reply]
Talking of such things - if you see a percentage above or near the squiggle especially something like "40% R" , or "30% S" this gives the ratio of the two enantiomers - labled "R", and "S" (the R and S should be explained somewhere in one of the linked articles)83.100.250.79 (talk) 19:06, 5 September 2009 (UTC)[reply]
This advises a saltwater mouthwash, as does the leaflet I was given after my lignocaine-assisted tooth extraction. Our article claims that saltwater mouthwash does not kill bacteria, though this information was added unsourced by a frequently-warned IP. Still, the dessicant effect of salt would, I assume be negated by the moist conditions of the mouth. Why use saltwater as opposed to plain water? And why not simply use Listerine? Vimescarrot (talk) 13:00, 5 September 2009 (UTC)[reply]
I'll tell you what. Give yourself a half-inch cut on your arm and pour Listerine into it. Then cut your other arm and pour salt water into it. Tell us which one hurts the most! --TammyMoet (talk) 15:04, 5 September 2009 (UTC)[reply]
Even the effect of Listerine in the mouth is only temporary. Saltwater as a mouthwash may not be as potent as Listerine, but it will make your mouth and throat feel better for a little while, certainly better than plain water will. Baseball BugsWhat's up, Doc?carrots 16:57, 5 September 2009 (UTC)[reply]
My mouth and throat both feel fine. Does this negate the need for mouthwash? Vimescarrot (talk) 17:08, 5 September 2009 (UTC)[reply]
Sorry, quick update on that. My mouth and throat felt fine until I used the saltwater mouthwash. Vimescarrot (talk) 17:09, 5 September 2009 (UTC)[reply]
Salt water has applications to oral hygiene other than germ-killing. The salt water interacts with the gums and keeps them in an isotonic equilibrium, minimizing fluid loss. Listerine, or other antibacterial fluid, probably does not satisfy this property. Nimur (talk) 17:31, 5 September 2009 (UTC)[reply]
You're probably right. The anti-bacterial component usually consists of alcohol, although some brands pointedly do not use it. 69.201.150.69 (talk) 13:44, 6 September 2009 (UTC)[reply]
I think a salt water mouthwash would help kill bacteria. In the past food was preserved by salting. I imagine that salt shrivels up the germs by osmosis. It may, I guess, be less harmful to the healing socket wound than the more aggressive chemicals in mouthwash. I think, for example, that when you use an antiseptic cream on a wound, that it kells off a layer or two of healthy cells. 78.146.76.67 (talk) 11:44, 6 September 2009 (UTC)[reply]
Eating a packet of Ready Salted doesn't sterilise my mouth; is there any real evidence that a saltwater mouthwash would? Especially since the salt will be gone within minutes (unlike preserved food, which remained salted)? I'm not really convinced. Vimescarrot (talk) 17:11, 6 September 2009 (UTC)[reply]
Be sure not to use salt substitute which is potassium chloride instead of sodium chloride. Potassium chloride will damage cells and keep the wound from healing. DISCLAIMER: This is not medical advice. Please consult your doctor to obtain medical advice. --Taxa (talk) 17:14, 6 September 2009 (UTC)[reply]
Also, mixing potassium chloride with sugar and giving it a tap with a hammer can cause fire, so if you do use salt substitute, make sure you clean your mouth out before you eat anything with sugar in it. I think it was potassium chloride, anyway. The stuff you get in water purification tablets.....? --KageTora - (영호 (影虎)) (talk) 11:52, 7 September 2009 (UTC)[reply]
Saltwater rinse recommendations are empirical -- they are not shown to reduce bacterial load any more than tap water, and especially not in the fraction of a minute that rinses are typically used. Even chlorhexidine requires a concentration of 2% and 10 minutes to kill some of the more terrible periodontal pathogens. DRosenbach(Talk | Contribs) 01:34, 8 September 2009 (UTC)[reply]
IS ANY OBJECT (ARTIFICIAL OR NATURAL SATELLITES) IN THE STATE OF FREE FALL IN GRAVITY FREE SPACE IF YES THEN WHY? —Preceding unsigned comment added by 115.178.96.7 (talk) 13:40, 5 September 2009 (UTC)[reply]
Please do not write in block capitals, it is harder to read and is interpreted as shouting. I'm afraid I do not understand the question. "Free space" is an idealised concept, it doesn't actually exist in reality. --Tango (talk) 13:46, 5 September 2009 (UTC)[reply]
I think he meant "gravity free space", which also doesn't exist. (I think.) Vimescarrot (talk) 13:56, 5 September 2009 (UTC)[reply]
Yes, if there was no gravity there could be no 'free fall' as there would be no gravity causing the object to 'fall'. --KageTora - (영호 (影虎)) (talk) 14:02, 5 September 2009 (UTC)[reply]
I believe your question should have been worded as 'why is it that when an object is in orbit, why is it in a permanent state of freefall?'. The reason is that the Earth's gravitational pull is dragging on the object bit-by-bit, so that every time the object goes around the Earth it gets just that little bit lower. This is why the Shuttle et al., have to compensate every now and then by firing off their booster rockets to get it back into position. There is no such thing as a permanent orbit. Either something starts coming back down or it starts floating away. Even the moon is getting further and further away from us. --KageTora - (영호 (影虎)) (talk) 14:10, 5 September 2009 (UTC)[reply]
I am sorry, but that is completely wrong. The need to boost is because of atmospheric drag. If you are in a high enough orbit that there is no significant atmosphere, orbits are essentially permanent. The Earth's orbit around the Sun, for instance, will last essentially unchanged until the Sun dies. The Moon is moving away because of tidal forces, but it will never leave orbit entirely. Orbits work because the gravitational pull is always towards the centre of the body and the object orbiting is moving perpendicular to that direction. This means gravity changes the direction of the motion without changing the speed, so it ends up going around the body at constant speed. (This is for a circular orbit, a general elliptical orbit is a little more complicated, but the basis concept is the same.) --Tango (talk) 14:18, 5 September 2009 (UTC)[reply]
OK, Tango. In that case I join the OP in asking the question I reworded for him/her. --KageTora - (영호 (影虎)) (talk) 14:22, 5 September 2009 (UTC)[reply]
Because free fall does not mean "rushing towards the ground". It means "Accelerating due only to gravity" - which is what orbiting bodies do. Vimescarrot (talk) 14:27, 5 September 2009 (UTC)[reply]
You may be making the common mistake of thinking that astronauts are in microgravity because they're out in space, which is 90% wrong: they would be subject to about 90% of the amount of gravity as they are on the ground if they were just standing still at their usual altitude. The reason they're in microgravity is because they're in free fall, as above. Newton's cannonball is an intuitive way to understand that astronauts are continually falling. --Sean 15:15, 5 September 2009 (UTC)[reply]
In fact, when it's time to return to the ground, the spacecraft fires a rocket tangent to the orbit (adding no momentum towards the earth). After losing some of the angular momentum, the spacecraft just "falls" back to the ground, indicating that it is still completely under the control of Earth's gravity. Nimur (talk) 17:36, 5 September 2009 (UTC)[reply]
I'll offer some further explanation - astronauts appear weightless because they are "falling" at exactly the same rate as their spacecraft, so relative to each other they don't move at all, hence appearing to float. This is because all objects in a given gravitational field accelerate the same amount. The gravitational force is proportional to mass, but acceleration is given by F=ma (which implies a=F/m). That means the mass cancels out when you calculate acceleration, meaning acceleration does not depend on mass. --Tango (talk) 15:31, 5 September 2009 (UTC)[reply]
To be perfectly clear, "Free fall" does not necessarily imply that the spacecraft is plummeting towards earth. Plummeting is only one kind of free fall. If your "sideways" speed is high enough you can "fall" for ever without ever getting any closer to the ground. (You can sort of visualize this as falling towards the earth, but going sideways enough that you miss it entirely. ) APL (talk) 04:15, 6 September 2009 (UTC)[reply]
That is why I put "falling" in quotes. We call it falling even though it isn't actually moving downwards. --Tango (talk) 14:44, 6 September 2009 (UTC)[reply]
How come a Blue bottle fly makes so much noise while a housefly is completely silent? I know the former is slightly larger, but can that alone make so much difference?--Shantavira|feed me 16:36, 5 September 2009 (UTC)[reply]
Houseflies aren't silent. They're quite noisy. But given that their flying speed is almost the same, any difference in their loudness would probably be attributed to their larger wingspan. Dougcard (talk) 04:14, 6 September 2009 (UTC)[reply]
There may be more to it than just size... It's true that a hornet is much more noisy than a wasp or a bee. On the other hand, the average fly is much larger--but in many cases more silent--than a typical mosquito. To confuse things further, mosquitoes (and probably many if not most flying insects) can apparently operate in more than one flight mode, some of which may be more silent or noisy than others. Just think of a helicopter that generates different noise levels while moving or in hovering position. Also, many sources of noise, in particular from oscillating or rotating objects are quite directional, so the perception of noise may depend on the relative position of the observer with respect to the emitter. Last but not least, we may be more sensitive to noises that are perceived as threatening (fear of mosquito bite) than to those that are considered benign or inconsequential. Michel M Verstraete (talk) 20:32, 9 September 2009 (UTC).[reply]
I has spare computer smps. can I make switching audio power amp from it? How? —Preceding unsigned comment added by 79.75.72.226 (talk) 17:38, 5 September 2009 (UTC)[reply]
Maybe - key factors here are the sampling frequency, and possibly you will need a separate comparator.
The power supply will as standard match a single DC voltage, but for audio you want it to match an AC voltage.
What the thing does is monitor the output voltage, and compare it to a standard, if the voltage is less than the standard then the supply is switched on. What you need to do is disconnect the wires that compare the output sample the with the standard, and connect it so that the output sample is compared with an audio input (suitably scaled) - this is where the comparator comes it - the comparator compares the two voltage and supplies a switching voltage to the power transistors.
There's a chance that the whole set up is so well integrated that you won't be able to access the comparator voltage input, (ie a single chip solution) - most likely you will be able to use the recitfied power supply, and switching transistors, but will need a separate comparator. You will probably need to disconnect any smoothing capacitors from the output, (but not imput).
Also note that the switching frequency of the amp will limit the frequency responce, and that there might be a huge amount of switching noise. (a small capacitor in the output stage helps fix this, as does any inductance a loudspeaker has)
Some one who knows more about the specifics of computer smpa supplies will know how adaptable they are.
This should be a useful read [1] - note it says the switching frequency is typically 33kHz - that isn't really enough for high quality audio, but should be good for a subwoofer, or active woofer.83.100.250.79 (talk) 18:33, 5 September 2009 (UTC)[reply]
Another alternative is to use the powersupply as a powersupply for a conventional amp (replacing the old transformer) - extra capacitors in the output stage are a good idea here, the average computer power supply has the potential to outperform the average audio amp power supply if you can isolate the switching noise.83.100.250.79 (talk) 18:41, 5 September 2009 (UTC)[reply]
I totally forgot that the power above method would only produce half wave AC - that complicates things a bit..83.100.250.79 (talk) 16:59, 6 September 2009 (UTC)[reply]
The smps is designed to deliver current at a given DC voltage. An AC current output will be needed to drive a loudspeaker i.e. positive and negative currents with average zero. Connect the smps output via a resistor to ground. Calculate the resistor value R ohms to drain I amps = half the available maximum current. THe resistor will dissipate IxIxR watts. Connect a large value electrolytic capacitor, typically 470µF 50V or more, from the smps output to the loudspeaker; this blocks DC current into the speaker. Inside the smps you need to access the reference voltage which is often given by a zener diode. Your AC audio signal will be added to the DC reference voltage. Depending on the circuit you can probably arrange that using a resistor and an input coupling capacitor. Cuddlyable3 (talk) 21:39, 5 September 2009 (UTC)[reply]
That sounds like it would be wasting energy (maybe I misread) - there's an alternative - use a push pull formation of two identical ouptuts...83.100.250.79 (talk) 17:04, 6 September 2009 (UTC)[reply]
You read correctly. The arrangement described is a Class A driver that wastes power. A push-pull or Class B driver is difficult to construct with a smps that is designed only to "push". Cuddlyable3 (talk) 17:57, 6 September 2009 (UTC)[reply]
How about if there are two 12V supplies (I believe there are usually 2 12V rails, but not sure if they are powered separately, or have the same rating), both fixed to 6V (for O input) then one can go up whilst the other down, the speaker connected across the two outputs, I suppose the outputs would need to be connected to ground via a capacitor - getting complex..83.100.250.79 (talk) 18:12, 6 September 2009 (UTC)[reply]
Why is it possible to have a giant neutron (as in neutron star or Black Hole) but not a giant proton and electron as products of a giant neutron decay? --Taxa (talk) 19:03, 5 September 2009 (UTC)[reply]
A neutron star is not a giant neutron, but many neutrons, with a skin of atomic matter. Protons repel each other, so it is very hard to make a collection of tons of pure protons. In fact the energy needed to bring them together will exceed their rest mass. I think there was a reference desk Question about this a year ago. Normally you will get a combination of electrons and protons - that is hydrogen, after all a neutron decay makes protons and electrons as well as the neutrino. Graeme Bartlett (talk) 22:13, 5 September 2009 (UTC)[reply]
Okay, so a behive of neutrons... What happens to the neutrons that decay? Do they turn into protons and electrons and if so where do they go? Do they follow the path of least resistance and get shot out in jets someplace? --Taxa (talk) 17:03, 6 September 2009 (UTC)[reply]
The pressure from gravity forces electrons and protons together into neutrons since neutrons can pack much closer than normal atoms can, which is how a neutron star forms. Neutrons in the core of a neutron star are much more stable than free neutrons and if they do decay the resulting protons and electrons would likely soon be forced back into neutrons. Neutron_star#Structure mentions the core might have electrons and protons mixed in. In the crust where the pressure is lower and normal matter can exist there's probably some transition back and forth. Rckrone (talk) 20:31, 6 September 2009 (UTC)[reply]
I've been trying to solve this using (unnormalised) formula similar to
Wavefunction = eΣan rn
And ignoring the angular component (eg taking δ2/δθ2 fn(r,θ,μ) to be zero ) ( because I'm not very good/lazy)
Which works and gives me solutions, mostly of the form
an+2(n+2)(n+3)=k1an+1+k2an
where k1,k2 are numbers (I believe there is a name for polynomials of this type but have forgotten - can anyone remember?)
So, so far I have infinite solutions (using infinite range of a0)
But there is more because I want to normalise which I believe is "integral over all space of fn(r,θ,μ)fn*(r,θ,μ)" which I believe that this will only give non infinite integrals for certain values of a0 - and hence give "quantised solutions" (as wanted).
Can anyone give a clue on how to show which of the integrals will be finite. (Even though I have all the parameters from one of the equations above) - and hence can integrate the equation (giving another infinite polynomial) I haven't got a clue how to evaluate the integral... Thanks.83.100.250.79 (talk) 19:31, 5 September 2009 (UTC)[reply]
Can you be a little more specific about what you're trying to do? Is this energy eigenstates of the hydrogen atom? What's the integral you're trying to evaluate? Rckrone (talk) 05:34, 6 September 2009 (UTC)[reply]
As far as I can tell, the OP is asking for help in evaluating integrals of the type
Specifically, he suspects that the requirement that the above integral be finite will lead to only some allowed sets of . For physical reasons, the OP then hopes that some of these sets of will also satisfy the recursion relation he calculated initially and therefore produce allowed wavefunctions for his problem. Martlet1215 (talk) 10:56, 6 September 2009 (UTC)[reply]
That's it, unfortunately my notes are a bit mixed up and now I'm not sure the equations I wrote above are correct - the only way I'll be able to tell is to start again from the beginning. Nobody seems to have noticed an error though..83.100.250.79 (talk) 11:24, 6 September 2009 (UTC)[reply]
Let where for . Then , so . Unfortunately I think the integral will diverge unless . -- BenRG (talk) 11:17, 6 September 2009 (UTC)[reply]
Thanks, unfortunately I've been having some technical difficulties - meaning that a correction to the above recurrence relation didn't get inserted, then wikipedia wouldn't let me edit (overnight) - as a result the solution you saw lacked an (n+2)(n+3) term. To add insult to injury my notes are a mess, and as I tried various formula to solve the differential, I've lost track of which one I was using (basically I'm working from memory and a few scraps of paper). I'll need to start again to get this sorted out.
I should add a on hold tag to this.. However if someone is very familiar with the correct solution, they might still be able to help.83.100.250.79 (talk) 11:32, 6 September 2009 (UTC)[reply]
I still can't find the equation I used - I think it may have been similar to wavefunction =f(x)eax where f(x) is a (possibly finite) polynomial. Maybe someone (with knowledge of what works in the long run) could suggest a good choice for a generic function to start with to solving the differential equation.83.100.250.79 (talk) 16:58, 6 September 2009 (UTC)[reply]
Like Rckrone said initially, it would help if you explicitly told us what you are trying to do. Is this the problem of finding the radial part of the energy eigenstates of the Hydrogen atom? Martlet1215 (talk) 10:54, 7 September 2009 (UTC)[reply]
You guessed right in your first post - I can get solutions to the differential equation (radial only time independant schrodinger), using a variety of different formulations eg e^f(x) , f(x) , f(x)e^ax etc , but I can't work out a way to find out which solutions are not good, becuase the wave function cannot be normalised.
f(x)e^ax seems the most promising so far (a is negative) since if f(x) is a finite polynomial the the integratl will be finite. However I'm having difficulty showing the conditions under which f(x) is finite, and still stuck on rejecting other solutions which I can't integrate (normalise) - I probably need a way to tell which integrals will be infinte (since I can reject all those)>83.100.250.79 (talk) 15:03, 7 September 2009 (UTC)[reply]
In this particular example, the wriggly line is the symbol that I don't know about. There surely is an article on wikipedia, but I can't locate it. Can somebody please supply a link? Thanks! --TrogWoolley (talk) 10:24, 5 September 2009 (UTC)
Resolved
