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February 24

Dual and quad satellite LNBs

Is there any relation between the dual and quad satellite LNBs in use these days and horizontal and vertical polarisation? Rich Farmbrough, 00:56, 24 February 2012 (UTC).

LNB here stands for Low-noise block downconverter. -- ToE 02:35, 24 February 2012 (UTC)

the quad LNB should be able to receive both polarizations and switch bands independently based on signalling on each of the four lines. Since you ask "these days" you will not be talking about the old ones that had separate polarization and band outputs on each line, which then went to a switch that selected one of them for the satellite receiver. So you should be able to feed 4 receivers, and each can select polarization. (for dual two receivers) Perhaps one only will have to supply a DC voltage on the coax to power the LNB. eg the Acer LNBFQUW1 can drive two PVRs (presumably needing two cables input each). Graeme Bartlett (talk) 10:32, 24 February 2012 (UTC)

Thanks, indeed "these days" is the key, since the current set-up did not make sense with what I remembered from the dim and distant past. All is now clear. Rich Farmbrough, 17:00, 24 February 2012 (UTC).

100% efficient engine violates the 2nd law of thermodynamics?

I was told this, but after reading the article on the 2nd law, I saw no where that suggested this. It says why a heat engine can not be 100% efficient but nothing about non-heat engines. From my understanding 100% efficiency can not be attained due to friction. But that really has nothing to do with the 2nd law. I think. Can someone clarify? ScienceApe (talk) 03:27, 24 February 2012 (UTC)

What's a non-heat engine? Vespine (talk) 03:57, 24 February 2012 (UTC)

Also, it's not friction that prevents 100% efficiency, this article discusses the reason, as you correctly state it applies to heat engines, but is there any other kind? Vespine (talk) 04:01, 24 February 2012 (UTC)

It's alot easier to explain and understand once you involve the concept of entropy. Plasmic Physics (talk) 07:25, 24 February 2012 (UTC)

Can you do anything with the 2nd law of thermodynamics without involving the concept on entropy? That's what the 2nd law is all about. --Tango (talk) 13:23, 24 February 2012 (UTC)

....Well I thought this was obvious but a fuel cell is not a heat engine. Therefore it can bypass the carnot efficiency limit. A Fission-fragment rocket or a Fission fragment reactor is not a heat engine. Just some examples. A heat engine is any engine that converts heat into doing useful work. In the case of a fuel cell, it converts chemical energy directly into electrical energy. Fission fragment reactor direct conversion of high energy ions into electricity. ScienceApe (talk) 14:46, 24 February 2012 (UTC)

It wouldn't violate the first law (which basically says you can't get more energy out of something than you put in, so rules out greater than 100% efficiency, but doesn't have a problem with 100%). It does violate the second law, though, which isn't really about energy efficiency, but it rather about entropy (basically, entropy/disorder almost always increases - not just doesn't decrease but actually has to increase whenever something happens). A 100% efficient engine would have to leave entropy constant, which is why it can't exist. --Tango (talk) 13:23, 24 February 2012 (UTC)

I thought that, but barring friction, what's stopping an fission fragment reactor from being 100% efficient? Seems to me that friction is the only thing keeping it from 100% efficiency. ScienceApe (talk) 15:00, 24 February 2012 (UTC)

Maybe a "non-heat" engine would be something like an electric motor? Just a guess. Dismas|^(talk) 14:32, 24 February 2012 (UTC)

I'm actually shocked that you guys never heard of engines that don't need to convert heat in order to do work... A machine that runs on solar cells for example directly converts light into electricity, it's not a heat engine... ScienceApe (talk) 14:50, 24 February 2012 (UTC)

The Second Law is more general in nature; it does not apply only to heat engines. Entropy is a broader concept, and the efficiency of heat engines is just one way of looking at it. Lynch7 15:04, 24 February 2012 (UTC)

Yeah I know that it doesn't only apply to heat engines. I'm trying to explain to someone why a space ship would always give off heat. Then I explained the carnot efficiency limit, and he said that it doesn't apply to non-heat engines, which is true. Then I said that all engines must give off waste heat due to the 2nd law, but he said that this is only due to friction and that there's nothing in the 2nd law that mandates that a frictionless engine can't exist. I wasn't sure what to say to that because barring friction, I'm not sure what else keeps a non-heat engine from being 100% efficient. ScienceApe (talk) 15:32, 24 February 2012 (UTC)

Particles emited from the fissile material are emitted in all directions, not just where you want to direct the thrust. This will limit the efficiency in the same way that the non-directional expansion of gas inside the cylinder of an internal combustion (carnot) engine limits its efficiency. 203.27.72.5 (talk) 23:23, 28 February 2012 (UTC)

Basically it boils down to the fact that information cannot be erased. Suppose you have a system A which contains some amount of energy, and you want to take that energy and perform work on another system B. The initial state is then system A being in some state with some energy content, the final state consist of system A in some other state containing less energy, and system B containing more energy, such that the total energy is the same.

Now, this transfer of energy from A to B will have to work regardless of the specific details of the state A is in. Obviously, there are a huge number of possible states that A can be in. However, when you transfer energy to B as work, then the state of B changes by only one degree of freedom. E.g. if you lift a macroscopic pobject against the force of gravity, you only have to specify the height of the object to specify how much work was performed against gravity. Such changes are reversible, you can lower the object and put the system back in the original state, the released energy can be put back to where it came from.

Suppse then that you can extract energy from A and use that to perform work on B without any restrictions. The initial state of A evolves to some final state with less energy and the change of B can be described by specifying one degree of freedom (like the change of the height of weight, or the change in the velocity of a spaceship). But had A been in a different state with the same energy, then B would have ended up in the same final state, but A would have to go to a different final state. A cannot evolve to the same final state as in the first case, because then the information about the initial state would have been lost. Each different initial state has to evolve to a different final state. But this is impossible, because at higher energy, there are more possible states for A than at lower energy. So, it is then inevitable that two different initial states would evolve to the same final state.

The only way out of this problem is to dump some energy in the form of heat to another system. So, some part of the randomness in the initial state of A must be transferred to another system, this makes room for different initial states to always be able to evolve to different final states. Count Iblis (talk) 17:02, 24 February 2012 (UTC)

This discussion is reminding me that I really need to learn thermodynamics. I've never understood it.

That said... the thermodynamic upper bound on heat engine efficiency is 1 − T_C / T_H, where T_C and T_H are the temperatures of the cold and hot baths. I think ScienceApe's question boils down to what, if anything, replaces that bound in cases where there's no apparent T_C or T_H. I haven't seen an answer to that so far (and I don't think I know the answer). A lot of people seem to be responding as though the limit on heat engines was just "not exactly 100%". It's much stronger than that.

Count Iblis, what you seem to be saying is that if you lose information, you emit heat. I believe that (isn't it a definition of heat?) but then the question is why an "engine" (whatever that is) has to lose information. You haven't explained why the information removed from A can't be stored in B, or for that matter in the environment, in a recoverable form. -- BenRG (talk) 02:19, 25 February 2012 (UTC)

When I lose information, I tend to emit swear words. Come to think of it, I seem to be losing (or forgetting) more of it lately. I'm evidently fighting a losing battle with entropy myself. :O AndyTheGrump (talk) 02:25, 25 February 2012 (UTC)

Re the examples of "non heat engines", I suppose I just didn't think of fuel cells as "engines" and I'd never heard of Fission Fragment rockets before. But I would have thought those have obvious reasons why they can't be 100% efficient. Fission releases energy as heat which would not contribute to propulsion. The combined mass of the fission products in a fission reaction is NOT equal to the total mass fissile material consumed. As for fuel cells, they also "heat up" due to the chemical reactions which means energy lost not converted to electricity. From the fuel cell article, 2H2 + O2 -> 2H2O reaction is highly exothermic, . If you tried to "capture" that heat and also make it do work, like one of the posts above suggests, then you run into the 2nd law heat engine problem, where capturing the heat it self requires work. Vespine (talk) 22:51, 26 February 2012 (UTC)

Ok I've re-read some of the definitions and I think I've cleared up my misunderstandings. Fuel and solar cells and fission fragment reactors them selves are not engines at all, but neither can they achieve 100% efficiency anyway, for the reasons mentioned above. They CAN be used to power electric motors which I do believe fall into the category of "non heat engine", hover efficiency of electric motors can also never reach 100%, even discounting friction, for reasons discussed on this site. Vespine (talk) 03:15, 27 February 2012 (UTC)

Tidal acceleration and the stopping of the Earth

I understand that, long before the Earth would stop rotating due to Tidal acceleration it will be destroyed by the expanding sun, BUT if the sun never expanded, how long would it take for the Earth to completely stop its day/night cycle? --ManyMonkeyMikes (talk) 11:49, 24 February 2012 (UTC)

It wouldn't. Keep in mind that the moon, which is already tidally locked, has a month-long solar day. As to when earth would get to that point, find a source for the rate of slowdown of Earth's rotation, and then you can compute how many years until the almost-24-hours day becomes 28-plus days. ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:02, 24 February 2012 (UTC)

Trying to find a reliable source for the slowdown is proving to be tricky. Wikipedia does not give an exact number. Much searching on the Internet gives a value of "1.4 milliseconds per solar day per century" which seems to come from this site (of dubious validity) claiming it came from The Sub-bureau for Rapid Service and Predictions of Earth Orientation Parameters of the International Earth Rotation Service at the US Naval Observatory, but a search of their site turns up nothing. If anyone can find a reliable source for the slowdown of the Earth's rotation it would be much appreciated. --ManyMonkeyMikes (talk) 13:24, 24 February 2012 (UTC)

I believe that the Moon will be pushed into higher orbit (and indeed lost) long before either the Earth is phase locked or engulphed. Rich Farmbrough, 17:25, 24 February 2012 (UTC).

Even after the Earth-Moon system is tidally locked, the solar tides will slow down the rotation of the Earth further, until the Earth is tidally locked to the Sun. I really don't know what will happen to the Moon in that case, and thinking about it makes my head hurt today... --Stephan Schulz (talk) 18:28, 24 February 2012 (UTC)

I believe the Moon would have to be lost or move into a distant orbit for that to happen. Otherwise, the Sun would try to make the Earth rotate once every 365 1/4 days, while the Moon tried to make it rotate every 28 days. The Moon, at least in it's present location, has the stronger effect, so would win this battle. StuRat (talk) 22:48, 24 February 2012 (UTC)

Similarly the sun should induce a slow rotation in the moon (more than countered by the Earth's effect), these two effects should be putting the Earth-Moon system into higher orbit with respect to the Sun. Sometime in the next billion years someone should run the numbers. Rich Farmbrough, 19:22, 25 February 2012 (UTC).

According to this book it would take 10¹¹ years for the Earth to become tidally locked to the Moon. Tidal locking to the Sun will obviously take much longer. There is a formula you can use at Tidal locking#Timescale but in reality it will probably never happen because Earth is outside the tidal locking radius of the Sun. SpinningSpark 19:47, 27 February 2012 (UTC)

psych question

do people with low self esteem regarding their bodies, even with "objective" reasons (severe obesity, for example), on some level feel flattered at the same time as they might feel violated by unwanted advances, even in the extreme? Applies to both genders and is a question of psychology. --80.99.254.208 (talk) 14:01, 24 February 2012 (UTC)

Some will, some won't. Everyone is different; I don't see how one can generalize or usefully speculate about the feelings of people with low self-esteem.--Shantavira|^{feed me} 17:28, 24 February 2012 (UTC)

I believe a portion of the morbidly obese are that way because of sexual abuse as children. That is, they found that they were abused less if they became unattractive, due to weight gain. Combine that with the comfort one finds in eating, and you have a deadly combo. StuRat (talk) 22:43, 24 February 2012 (UTC)

Wow! That's amazing. Any chance of a citation??? Also, if this is true, then what could be done about it?? Does the person need to feel very protected or what? Also it's an ongoing issue: even if that person drops a lot of weight TODAY, they will still get all sorts of unwanted attention. --80.99.254.208 (talk) 10:12, 25 February 2012 (UTC)

Sturat, this is a breakthrough contribution (no sarcasm) and I would like to hear more, or greater elaboration of your thoughts. --80.99.254.208 (talk) 10:12, 25 February 2012 (UTC)

Here are some studies: [1], [2], [3]. See more studies here: [4]. Note that only forced sexual relations (by threats, physical force, etc.) are likely to cause this reaction. Consensual incest or statutory rape would not be expected to have the same result. In fact, these forms may have the opposite effect on the victims, who then become sexually promiscuous, with large numbers becoming prostitutes. Also note that if the victim remains obese long into adulthood, then that weight and those eating habits become normal for them, and any attempt to lose weight may be resisted. Therefore, getting the victims out of that environment and having psychiatric treatment early on would be most beneficial. StuRat (talk) 20:41, 25 February 2012 (UTC)

See Obesogen, Infectobesity. Psychiatric explanations for disease are the last refuge of the clueless - I think anyone who remembers all the pop psychology about stress causing stomach ulcer before the discovery of Helicobacter will be very, very skeptical of them. Wnt (talk) 10:38, 25 February 2012 (UTC)

Note that the discovery of a bacteria doesn't eliminate stress as a contributing factor, any more than the virus causing colds eliminates cold, dry weather as a contributing factor. StuRat (talk) 21:02, 25 February 2012 (UTC)

The trouble with pop-psychology is that the plural of anecdotes is not data. We actually know a lot about causes of overeating and obesity, and they are many and varied, hence also the successful treatments are. For psychological causes CBT is reasonably successful. Rich Farmbrough, 19:28, 25 February 2012 (UTC).

Difficulty with understanding

I have been looking at the induction coil and the transformer articles and cant understand the difference in operation as it says the induction coil is a type of transformer but the transformer works by induction into the secondary coil. So what is the actual differnce in workings?--92.28.64.112 (talk) 14:22, 24 February 2012 (UTC)

The induction coil also works by induction. As far as I can see, the only special feature of an induction coil is that it requires an interrupter so that it can transform DC to AC, whereas a transformer that is transforming AC to AC (and just changing voltage) does not need an interrupter. Gandalf61 (talk) 14:36, 24 February 2012 (UTC)

The basic difference is that the induction coil does not have the secondary winding; the transformer has the primary and secondary windings connected by common magnetic flux. If the secondary winding of the transformer would be removed then it becomes just an inductor. In fact, if the secondary winding is just left unconnected then the primary also works much like an inductor. — Preceding unsigned comment added by Sivullinen (talk • contribs) 14:50, 24 February 2012 (UTC)

An inductor doesn't have a secondary winding, but an induction coil has both primary and secondary windings (according to our article). Gandalf61 (talk) 14:59, 24 February 2012 (UTC)

Correct - both an induction coil and a transformer are essentially the same - both have primary and secondary windings; both have voltage across both windings due to the magnetic field, the field being created by the current in the primary winding. Fundamentally, they are named different not according to principles of operation but according to application. The application of an induction coil is to create a high voltage with little concern over load carrying ability; a transformer application is the conversion to a convenient voltage with effective load carrying ability. The physical geometry and materials are chosen in each case to optimise for each application. Similar to a variable resistor: if it installed to vary voltage it is called a potentiometer; if it is installed to vary current it is called a rheostat. Keit124.178.176.225 (talk) 15:34, 24 February 2012 (UTC)

Yeah but in the transformer article it says the pimary current creates a secondary voltage by induction. So if you connected something to the secondary, you would get a current- Yes? So does that secondary current act backwards to create more primary voltage by induction or have i got it wrong?--92.28.64.112 (talk) 16:32, 24 February 2012 (UTC)

In transformer (that is used with AC) it rather works so that when you connect something to the secondary then the primary will take current from the source connected to the primary - quite natural. But the induction coil (I indeed mixed it up with inductor in my previous post) is typically used so that it is connected to DC for a short duration, then the circuit opens and the energy stored in the inductance of the coil creates high voltage 'spike' - both in primary and secondary. — Preceding unsigned comment added by Sivullinen (talk • contribs) 17:55, 24 February 2012 (UTC)

So what's all this stuff about magnetic flux. Does the secondary current create flux or is it only the primary? --92.28.74.5 (talk) 23:44, 24 February 2012 (UTC)

The bit about connecting to DC for a short duration is a red herring, as succesive pulses of DC is, as far as an induction coil / transformer is concerned, the same as AC. What happens is this: The primary current creates a magnetic field which creates a proportional voltage in both primary and secondary. This induced primary voltage always opposes the driving EMF (the source) If there is a load on the secondary, the resulting secondary current creates a magnetic field in the OPPOSITE polarity to that created by the primary current, i.e., the secondary current field tends to cancel out the primary current field. This reduces the voltage induced in the primary, allowing the source EMF to force more primary current to flow, maintaining the magnetic field strength. The magnetic field created by secondary current cannot AID the field established by the primary - if it did, you would have a perpetual motion machine. Keit124.178.44.47 (talk) 00:18, 25 February 2012 (UTC)

So you are saying that the secondary flux cancels out the primary flux??--92.28.74.5 (talk) 01:38, 25 February 2012 (UTC)

Yes. In an ideal transformer, regardless of whether no load on the secondary, full load, or somewhere in between, the net flux remains the almost same, being the amount of flux required to approximate the EMF applied to the primary, even though both primary current and secondary current increase proportaionately with load, and, each on their own, would produce a very much larger flux. If the secondary flux added to the primary flux instead of substantually cancelling it, then you could (forgeting saturation), for any actual load, have flux build up from nothing to infinity - a truely stunning perpetual energy source: infinite ouput with no input.

This principle of a result opposing what caused it is an important principle in electrical theory, and is refered to as Lenz's Law by electrical technologists, analogous to the Newton's Third Law in dynamics, and Le Chatelier's Law in chemistry. Keit121.221.76.24 (talk) 02:16, 25 February 2012 (UTC)

So the diagram of the ideal transformer in the basic principles section of the transformer article is wrong? It shows just one flux that the text says is due to the primary. This does not agree with your version.92.28.74.5 (talk) 11:47, 25 February 2012 (UTC)

The Wikipedia transformer article is not entirely wrong in this regard, just simplistic and not explained very well, and can mislead. If you read thru my explanation carefully, you should find it is obviously correct. But if you were to use an instrument that measures magnetic flux, you would obviously get only one reading, even though there are two components of the flux, and that reading (the sum of two fluxes so to speak) would be in aproximate proportion to the EMF applied to the primary, as I said. The value of Wikipedia is a) it can help you think thru a subject of interest, and b) it points you in the direction of good references. You should not rely much on Wikipedia content on its own, or get too distressed if you find it wrong. Keit60.230.195.53 (talk) 12:28, 25 February 2012 (UTC)

Yes it has mislead me and maybe others too! So if i was to measure the flux in the core of a transformer under differnt conditions of load current, what would i find? Would the flux increse with increasing load current? You say it only depends on the primary voltage. So it doesnt change with the amopunt of power transmited thro the transformer? The article shows only one flux created by the primary but you say thers are two fluxes?--92.29.192.13 (talk) 20:30, 25 February 2012 (UTC)

I've already explained this. Correct: the net flux (being the sum of both fluxes) does not significantly change with the amount of power transmitted. There are two fluxes notionally, but of course as the two fluxes are in the same magnetic path, you can only measure the sum of the two, and that sum APPEARS to be dependent on source EMF.

The key to understanding it is the opposition to primary curent by the induced primary voltage, this voltage not to be confused with the EMF from the power source. On zero load, you will have a certain flux amplitude, being that amount of flux large egough to make the primary voltage just large enough to oppose the source EMF, so that the primary current is that amount of current that produces the flux. When load is added to secondary, the resulting secondary current causes a flux that tries to cancel the flux established by the primary current. This would mean that the induced primary voltage drops, so that the source EMF is less opposed. As the source EMF is less opposed, it forces in more primary current, increasing the primary flux to compensate for the secondary flux. In a practical transformer, this compensation is very effective, so you see little change in net flux. But if you increase the source EMF, it will force more primary current and increase the flux. Keit121.221.230.136 (talk) 01:14, 26 February 2012 (UTC)

So this flux that is proportinal to the primarry voltage only , is that the magnetising flux?--92.28.74.149 (talk) 01:33, 26 February 2012 (UTC)

I'm not sure how to answer this in a way that is meaningfull for you. the term "magnetising flux" is not used in any of the textbooks I have. Rather, a MAGNETISING CURRENT causes a MAGNETIC FLUX, or, more correctly, a magnetic field. Magnetic flux is what you get when you have magnetised something. I guess if you used the term magnetising flux when talking to an electrical engineer, he would assume you meant the flux established by the primary current with no load on the secondary. But he might assume you meant the total (sum) flux under load, as it is practically the value, as I've said. Keit120.145.3.209 (talk) 02:51, 26 February 2012 (UTC)

Earth proportion to blue whale

how big would the earth be if a human was the same proportion as a blue whale is to earth as it exists? — Preceding unsigned comment added by 165.212.189.187 (talk) 14:30, 24 February 2012 (UTC) I.E. Earth:blue Whale::X:Human — Preceding unsigned comment added by 165.212.189.187 (talk) 14:31, 24 February 2012 (UTC) Nevermind — Preceding unsigned comment added by 165.212.189.187 (talk) 14:35, 24 February 2012 (UTC)

1)Decide which aspect of a human's size you want to use in the comparison (height or mass perhaps)

2)Use the Human Being and Blue Whale articles to find out how much bigger the whale is in this respect

3)Multiply the size of the Earth by this factor.

Done! Rojomoke (talk) 14:48, 24 February 2012 (UTC)

If you don't want a specific answer, the Earth would still be really really big in comparison. The Earth is so much bigger than either a human or a whale that the difference wouldn't be noticeable, say, to a suddenly-whale-sized human. Staecker (talk) 15:53, 24 February 2012 (UTC)

But the earth does not have enough place to hold the resources (food, energy needed for industrial consumption, petroleum etc.) for 7 billion whale-sized humans. When a human suddenly becomes whale-sized, he will need a lot more food. The vehicles that will transport the whale-sized humans from one place to another, be it an aircraft, or a car, will need a lot more fuel. --SupernovaExplosion ^Talk 16:31, 24 February 2012 (UTC)

On the other hand if the earth shrank and the humans stay the same, we would need less energy for transport, as things would be closer together, and gravity would be considerably less. A human weighs c. 100 k, a blue whale c 200,000k, the Earth 6 x 10 ²⁴ k so "Whale Earth" would be 3x10²¹ about the mass of Titania ( see this list if you want to calculate other massses or sizes). Rich Farmbrough, 17:43, 24 February 2012 (UTC).

Now that I think about it volume or equator circumference would be the best way to envision it. what would those be??

I would stick my neck out and say "go with Titania again". 2 billion cubic km, 1756 π km circumference (1/8 of Earth's) (on pure numbers it would be about 1/12 but of course density varies). I'm not sure this is a useful visualisation though "if we were blue whales the world would seem smaller" the important thing that makes the world small for blue whales is that they travel almost all of it, and they can apparently communicate across a fair percentage of it. That and the relatively low world population, who may be mostly known to each other. Rich Farmbrough, 19:39, 25 February 2012 (UTC).

For a favourable interpretation of the question, the earth would have a diameter of 4.4 km. Using wolfram alpha, the oddest questions can be answered quite straightforwardly: [5]. A slightly less nonsensical answer is that a mini-earth with the same mass ratio to the original earth as a human to a blue whale, would have a diameter 8% that of the original earth [6]. Wolframalpha is no good with parenthesis, but nothing a bit of copy and paste can't solve. EverGreg (talk) 11:52, 27 February 2012 (UTC)

Paintings and time

When an old painting, like the Mona Lisa, gets yellowish or black, what process causes that? 88.14.192.178 (talk) 15:48, 24 February 2012 (UTC)

it's called aging. It's explained on this page, which also shows some ways to fake it: http://www.faux-painting-techniques.com/ageing_and_distressing.html --80.99.254.208 (talk) 16:36, 24 February 2012 (UTC)

I doubt the OP wanted to know that it's called aging. I suppose the OP wants the chemical process that makes paintings look different. — Preceding unsigned comment added by 80.31.146.38 (talk) 17:21, 24 February 2012 (UTC)

I believe a lot of the blackening is accumulated soot and dust, whereas yellowing is due to changes in the pigments or varnish, caused by sunlight or chemical reactions. Someone more knowledgeable will be along in a minute ... Rich Farmbrough, 17:54, 24 February 2012 (UTC).

One of the causes colors changing in paintings is due to the pigments chemically reacting. Many pigments are inorganic, transition metal compounds, which will often react with each other, or with stuff in the air. Perhaps most notably, white lead is pretty notorious for darkening over time due to the formation of lead(II) sulfide. Red Lead was popular in some cave paintings, and also tends to darken over time. I think I've read that this is why some of the buddhist figures in the Mogao Caves have dark faces, though I can't find that now, so I might be wrong. Buddy431 (talk) 19:27, 24 February 2012 (UTC)

young's modulus

Please i greatly need help in this question!!

A string has a length of 2.0m and a density of 8000kg/m³.When the string is vibrating in the fundamental mode with a frequency of 200Hz the tension in the string produces a strain of 2%. Calculate the young's modulus for the string. — Preceding unsigned comment added by 41.205.4.4 (talk) 18:18, 24 February 2012 (UTC)

Please do your own homework.

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. Red Act (talk) 18:36, 24 February 2012 (UTC)

I've moved this misplaced reply from the OP to the correct section. IP 41: please use the 'edit' buttons to make sure your responses go in the right place. SemanticMantis (talk) 20:28, 24 February 2012 (UTC)

• This is not a homework assignment, rather it is a past advanced level question. i solved it with my friends and we had difficulties then we decided to ask it to the teachers separately, but surprisingly, the two teachers had different solutions!!!!. So now i don't know what to do — Preceding unsigned comment added by 41.205.4.6 (talk) 20:16, 24 February 2012 (UTC)

Young's modulus is stress over strain, as given by our article. You have the strain. How do you find the stress? Specifically, how is the stress related to the frequency of the fundamental mode, the density, and the string length? --140.180.9.36 (talk) 21:54, 24 February 2012 (UTC)

This is not a valid question, as insufficient data is supplied. The amplitude given (a strain of 2%) is not determined soley by the length, frequency, and mass, it is determined by the energy, which is related to the intial strain. As a though experiment, visualize plucking the string with various amounts of pull. The frequency remains the resonant frequency (determined by length, mass, and young's modulus), but the amplitude (the strain) will vary. Unless your teachers were incompetent, you have remembered the question incorrectly. Keit124.178.44.47 (talk) 00:05, 25 February 2012 (UTC)

2% is not the amplitude of the oscillation; it's the change in the string's equilibrium length due to the tension. This problem is perfectly solvable with the information given. --140.180.9.36 (talk) 02:30, 25 February 2012 (UTC)

Strictly speaking, you are correct. However the end to end tension is (for reasonable amplitudes) virtually directly proportional to oscillation amplitude. And amplitude can be any magnitude. Keit60.230.195.53 (talk) 06:02, 25 February 2012 (UTC)

Please what is my solution then!!! — Preceding unsigned comment added by 41.205.4.99 (talk) 11:01, 25 February 2012 (UTC)

Maybe if you show how you arrived to your solution and what your teachers said, that would convice us that you have spent effort to solve the question. As said above, we don't like to solve homework questions if you don't show any effort in trying to solve it yourself. – b_jonas 17:40, 25 February 2012 (UTC)

You can get the amplitude value using the value of the strain given (making some approximations of the shape of the wave). There is no information about the forces acting, and I don't really know how to use the frequency. With the given data, and as far as my knowledge goes, the best thing that can be done is to use the bending and deflection equation (Its a second order differential equation). Since we know the amount of deflection (the amplitude that is), and making some reasonable assumption of the shape of the cross section of the wire (in order to calculate the second moment of area), we can arrive at the value of the Young's modulus. But I don't know where to plug in the frequency. Lynch7 18:07, 25 February 2012 (UTC)

I've realised my initial response was wrong. You get that! Amplitude can be deduced from the strain, using trig. We now have amplitude and frequency, so we can nd get acceleration. With acceleration and the mass (given), we can use Newton to calculate peak force and relate it back to strain. So we now have both stress and strain and calculate Young's Modulus. Sorry teachers. Keit121.221.230.136 (talk) 01:01, 26 February 2012 (UTC)

Extra moons and tides

A bit of a silly question:

I'm starting up a new DnD campaign where a rogue intelligent planet invades the solar system and gets close enough for atmospheres to touch so that dragons can fly orks over to invade the realm. It has the power to not destroy orbits so things won't crash into the sun or fly out into space unless it wants to, but I was wondering how bodies of water would act in a situation like this? If another earth parked in the sky so that the atmospheres were kissing, would all of the oceans slowly shift to one side of the planet? Would it be enough to trigger plate tectonic shifting and cataclysmic earthquakes? Thanks for any insight 142.244.35.91 (talk) 18:22, 24 February 2012 (UTC)

If the new planet still has gravity, then the occeans would shift, but not slowly, and not just the occeans. The planets would violently crash into each other and pretty much be destroyed. If this is somehow magically being prevented, then you have to consider: what things on the old planet are affected by the gravity of the new planet and what things aren't? Anything on the old planet that is subject to the new planet's gravity will be violently stripped off, including occeans, atmosphere, tectonic plates, people, dragons, etc. If the old planet is not affected by the gravity of the new planet, well then of course nothing will happen. Rckrone (talk) 18:42, 24 February 2012 (UTC)

(ec) For two planets to be in such close contact, either their relative orbital speeds will have to be immense, or they will collide. (In other words, the angular momentum has to be huge in order to avoid falling into each other). This is part of the reason, conceptually, why the three body problem tends to be unstable. Furthermore, since you've described atmospheric contact, gas drag is going to be a pretty huge factor as well - so even if the orbit began stably, it will quickly decay, and the planets will collide. Since you're already using a lot of imagination, it's probably worth closing this one as "magical physics," and ignore the consequences. Nimur (talk) 18:46, 24 February 2012 (UTC)

This sounds very all or nothing, but I did get a cool idea of having daring navies riding a massive waterspout to the other planet. Awesome.142.244.35.91 (talk) 19:14, 24 February 2012 (UTC)

The Science Fiction writer Bob Shaw's Land and Overland trilogy was set on a double planet with a common atmosphere, enabling flight from one to the other: you might want to read it to see how he avoided mentioning the problems that would, in reality, ensue. He used to claim that it was set in an alternative universe where the value of π was different, allowing the setup to exist, but when asked what that value was, said (as I recall) "Whatever it takes to make my planets work."

For inter-planetary waterspouts, you might also want to check out Robert Forward's Rocheworld and sequels, where two planets mutually orbit so close as to periodically exchange seawater. Forward being a researcher in gravitation, his underlying maths are probably more rigorous than BoSh's. {The poster formerly known as 87.81.230.195} 90.197.66.60 (talk) 21:06, 24 February 2012 (UTC)

How about if instead of making it a new planet, you make it something smaller, like a comet, which was frozen in the outer reaches of the solar system for thousands or millions of years, but occasionally comes close to the star, causing it to temporarily melt and develop an atmosphere, which would be lost to space in time, but it has the near miss before it gets the chance. The much lower mass of the comet would mean it's effect on the planet would be far less. You could also make it skip off the planet's atmosphere, by striking at a shallow angle. StuRat (talk) 22:37, 24 February 2012 (UTC)

Are you sure you understand how close this is? Here's a picture of Earth's atmosphere to scale. Note that while 100 km is the nominal "thickness of the atmosphere", it's not breathable (to humans) above about 3 km. -- BenRG (talk) 23:17, 24 February 2012 (UTC)

? 3km is only 9842 feet, 6 and 15 ⁄ 64th of an inch. List of highest towns by country in feet.--Aspro (talk) 23:52, 24 February 2012 (UTC)

Since you do have the means to bring in an alien planet and other wonderful things, for its atmosphere to touch ours and prevent atmospheric drag, you will want to maintain these planets in a very precise and mutual geostationary orbit. Currently, this orbit is at about an 22,000 miles altitude. A much larger, but far less dense planet of equal mass might work at this distance (with a radius five and a half times ours, its surface gravity is thirty times less so its upper atmosphere would need to be considerably thicker to obtain an atmospheric pressure), but for adding another Earth-like planet, the geostationary orbiting distance could be reduced by doubling the Earth's rotational velocity.

Given a stable configuration (by means of magic or not, for the orbit is similar in configuration to the proposed space elevator), there will be less gravity experienced on Earth near the other planet, as well as an 11% increase or less in gravity on the opposite sides (each of the more distant planet's center is at least three times more distant and gravity drops off by the inverse square law). On Earth, the directions of the gravitational field will still point in the general direction towards the ground (some will notice a small eastward or westward shift in direction). Between the planets where their gravities become more or less equal, each will oppose one another so our pond water and atmosphere there will weigh considerably less. Less weight means less pressure, thus the atmospheres will expand outward to build pressure (which in turns means that the planets' orbit altitude can also be considerably higher).

In the exact middle between these planets (in the shared portion of their upper atmosphere should the other planet have a similar radius), your dragons' and the DnD characters will be completely weightless, so make sure they have some fun with that scenario.

Our ocean at the center will still have a small weight due to it being on Earth's surface, but weighing substantially less it will bulge or increase to a large height that will depend on the exact altitude chosen. Thus you may want to park the new planet over the vast Pacific Ocean after first evacuating the islands there perhaps. I would certainly expect some new plate shifting dynamics, tides etc, but with a DnD game you don't really need a computer and lots of data to sort all that out though. --Modocc (talk) 00:17, 25 February 2012 (UTC)

You seem to be all out of paragraph breaks, so have some, on me: ¶¶¶. :-) StuRat (talk) 00:25, 25 February 2012 (UTC)

Done. :-) Yep. I wrote up more than I intended or thought I had. Sometimes, I need to step back and revise for readability, but when I'm in thick of it and adding details, that does not always happen. Thanks. --Modocc (talk) 01:50, 25 February 2012 (UTC)

Hmmm, since this is fantasy, how about saying that the intelligent comes from another dimension with "incompatible gravity"? Whatever means of propulsion the planet uses to get next to yours, it could just sit there without affecting or being affected by your gravity. (It might have to match the course of your planet around the sun... unless it's the sun that revolves...) Earth material would fall to Earth, Planet material would fall to planet, so if you rode a dragon up to their surface you'd still feel "upside down" until you drank enough of their water; maybe your heavy Planet-made armor would hold you down in the meanwhile. ;) Wnt (talk) 17:00, 25 February 2012 (UTC)

Jules Verne's novel Off on a Comet considers a comet that temporarily gets very close to the surface of Earth but luckily doesn't go under the surface. The comet does steal some water from Earth. – b_jonas 17:38, 25 February 2012 (UTC)

Unfortunately, Jules Verne's science was rather iffy, even for his day. He had Martians launched to Earth in giant cannons, for example, which would kill the occupants both when launched and when they struck Earth, due to massive g's. StuRat (talk) 20:26, 25 February 2012 (UTC)

Well, there's no guarantee the Martians aren't sturdier, and actually, a giant cannon might not have that high an acceleration. I'm also not entirely unwilling to allow them some parachuting or aerobraking to their pods, since after all, if you can land a Space Shuttle safely I'd think the Martians could think of something.

The weirdest idea though is something I remember from one of those Earth to the Moon movies, where drastic acceleration from a gun was countered by spinning people around really fast in little tubes. My feeling is that the centrifugal force should rapidly become worse than the acceleration from the rocket, yet I haven't entirely convinced myself that this couldn't work. But I don't think this was suggested in the original story. Wnt (talk) 19:46, 26 February 2012 (UTC)

The portrayal of the ships leaving craters where they struck the Earth makes it clear they were going entirely too fast. As for countering gravity with centrifugal force, I don't see how, since centrifugal force constantly changes direction. At best, they wouldn't effect each other at all, while, at worst, the centrifugal force would add to the gravity at one point in the spin. I suspect they are misinterpreting the common carnival ride, where people spun inside a cylinder "stick to the walls" instead of falling down when the floor lowers. This isn't because the force of gravity is reduced, but because the "cylinder" is more of a conical shape, which narrows at the bottom, so the centrifugal force pushes them upward. However, the total g force isn't reduced, it's just in a different direction. StuRat (talk) 20:03, 26 February 2012 (UTC)

Well, what I'm thinking is, if the gravitational force pools blood to one side of the body, pulls the bones toward the other side, and then you change orientation 180 degrees, the process reverses. Sort of like running an electrophoresis gel in reverse to get all the bands to come together again - but more like a sedimentation gradient. ;) But - the gravity also probably very quickly applies tension/pressure to some cell membranes which could burst, and a 180 degree rotation won't fix that. Also, if the spin is too slow, the bones have already ripped off the meat before you pull them the other way. I don't know from first principles how important all such processes are. Wnt (talk) 23:31, 26 February 2012 (UTC)

Rapidly reversing your g-forces sounds very unhealthy, to me, resulting in nausea, at the very least. StuRat (talk) 23:35, 26 February 2012 (UTC)

Um, where does Verne consider Martians launched to Earth in giant cannons? I don't think he does. What he does write, however, is Earthmen launched towards the Moon from a giant cannon in From the Earth to the Moon. Here he describes a mechanism for reducing acceleration forces that obviously wouldn't work in real life (see Chapter XXIII in an English translation). – b_jonas 10:23, 28 February 2012 (UTC)

Pressure and the nonmetal-to-metal transition

I was just doing some editing on a topic that I don't know very well (yes, that's bad, but it can be interesting ;) - Nonmetal#Metallic allotropes. The table is all new; if you see something stupid, many thanks for singing out! But I have some more general blue-sky questions about this.

I'm thinking that our planet's atmospheric pressure is arbitrary, so where we see the boundary between metals and nonmetals should be arbitrary. If it were 40-20000 times denser, we'd have fewer than 18 nonmetals. But does that apply in the opposite direction? If you do all your metallurgy at 0.0001 atmosphere, do many of the poor metals form nonmetallic allotropes? Aside from some obvious ones like alpha-tin I haven't had so much luck finding these, nor did I find clear sources linking them with low pressure - I don't actually know that the metals don't just straight sublime to gas, even if you work at lower temperatures.

Is there any distinction between an allotrope of an element and a "phase" of the element, as commonly used to apply to forms of the element with different crystal structures or chemical formulas?

If a high-pressure phase of an element is shown to be a superconducting metal, at a few kelvins, does that guarantee that above the T_c it is still a metal, or could it go straight to being something else without a change in structure?

Some properties of a metal are defined in terms of the properties of compounds, and others in terms of the physical properties of pure allotropes. These are not interchangeable, and the allotropes depend on the environmental pressure. Does that mean that in some broad philosophical sense, for those who can move freely between any environments in the universe, being a "metal" is actually two entirely different concepts, one dependent on pressure, one not?

Wnt (talk) 22:33, 24 February 2012 (UTC)

There is a difference between phase and allotrope, look at sulfur - one of its allotropes can exist as either solid or liquid. Plasmic Physics (talk) 04:41, 25 February 2012 (UTC)

Good point. The situation here can be surprising - to quote one source:^[7] "Gallium melts at 30^oC and can be easily supercooled to ~3^oC below the melting point. On melting, the covalent bond and the well-defined lattice of α-Ga are destroyed and the DOS minimum disappears, resulting in more metallic properties and free-electron behaviour for the liquid [Hafner90]. Gallium’s behaviour does not seem to change after subsequent meltings and solidifications, but does depend on the annealing temperature that it is heated to. That is, the liquid seems to have a memory of the α phase in its structure, but annealing it above 45oC changes this structure towards a more β-like one, resulting in a significant decrease in the temperature at which it solidifies [Kofman79, Wolny86]." Wnt (talk) 22:04, 25 February 2012 (UTC)

The main effect that you would see at very low pressures is that some metals would be considered as a gas, for example mercury is a gas at 0.8 Pa at 40°C but that is below your stated low pressure. At even lower pressures your alkali metals will be gas, eg potassium at 13 microPascals at 40°C. Graeme Bartlett (talk) 20:35, 27 February 2012 (UTC)

I'm more interested in what that gas might condense into, if it is cooled enough. Are we really at the very bottom of all possible pressures in the cosmos, so that with higher pressure nonmetals become metals, but at lower pressures metals never become nonmetals? Wnt (talk) 05:13, 28 February 2012 (UTC)

The chemical bond energy is much greater than atmospheric pressure, as you can look at the pressure needed to drush something, or the negative pressure to stretch a material. It gets much less likely that there are two allotropes extremely close in energy to each other that would be changed by a low pressure. Even if there was a change in temperature would probably change the equalibrium point any way, and change may be slow, for example with tin. There are likely to be large negative pressure regions that have not yet been explored. Graeme Bartlett (talk) 06:04, 28 February 2012 (UTC)

An astronaut on a very small moon...

• An astronaut standing on Thebe (moon) - would he be able to push himself off (jump off) the surface to "escape" the gravitational pull of the mini-moon?
• Asked the other way round: What would be the smallest size moon (assuming density 3 g/cm3) where this would be possible? Thanks! 213.169.162.159 (talk) 23:00, 24 February 2012 (UTC)

The Escape velocity in that article is reckoned to be some 20–30 m/s. Even a Soviet athlete on steroids would have trouble achieving that. So I'd say no. This is a bit of an odd question to my mind. Why are you asking. Are you writing a sci-fi novel or sommit? It would help us focus our response --Aspro (talk) 23:08, 24 February 2012 (UTC)

There are no odd questions, only odd answers :-) Background: At the surface points closest to and furthest from Jupiter, the surface is thought to be near the edge of the Roche lobe, where Thebe's gravity is only slightly larger than the centrifugal force. As a result, the escape velocity in these two points is very small, thus allowing dust to escape easily after meteorite impacts, and ejecting it into the Thebe Gossamer Ring. So - even there "he" would not be able to "jump away" ?? Grey Geezer 09:15, 25 February 2012 (UTC)

That's different. Both tidal forces and centrifugal force (yes, it does exist, despite what lots of ignorant pedants may have told you!) can cancel out some, or all, of the gravitational force. There are irregularly shaped asteroids where the bits furthest away from the centre are rotating so fast that surface gravity is effectively negative - you would have to hold on tight to avoid being thrown off into space. The formulae we've been talking about are for escape velocity taking into account only the gravity of the object you are standing on. Once you take other factors into account, it can get very complicated. --Tango (talk) 14:19, 25 February 2012 (UTC)

As for how small it would have to be, it's not that straightforward. You might think a moon with 1/10th the mass would have 1/10 the escape velocity, but it's not so, since the mass near you has far more effect, and with a smaller moon more of the mass is close to you. The angle also matters, since mass out to the sides of you doesn't pull you down much. And the density also changes, with smaller masses having generally denser elements, because they don't have the gravity to keep the lighter elements from blowing away, but also being packed less densely. Finally, irregular shapes must be considered. In the mass range we're talking about you no longer get a spherical shape. Considering all this, a moon with 1/100th the mass of Thebe should be safe to kick off from. Also note that Thebe has a density around 0.86 g/cm³, so your assumption of 3 g/cm³ seems a bit high. StuRat (talk) 23:21, 24 February 2012 (UTC)

Thebe has a density around 0.86 g/cm³, so your assumption of 3 g/cm³ seems a bit high. NASA is wrong? 213.169.162.159 (talk) 09:09, 25 February 2012 (UTC)

I wonder why our article says the lower value, then. StuRat (talk) 20:19, 25 February 2012 (UTC)

What angle matters? The angle your trajectory makes with the ground? That doesn't matter at all, as long as it is doesn't make your path to infinity go through the asteroid. You can see that the angle doesn't matter by thinking in terms of energy. Whether you can escape or not depends on the sum of your gravitational potential energy (which depends only on height) and your kinetic energy (which depends only on scalar speed), so angles don't come into it. (It gets a little more complicated if you take things like rotation or irregular shapes into account.) --Tango (talk) 00:39, 25 February 2012 (UTC)

I refer to the angle of the ground beneath you. On a large body, much of the mass near you is almost out to the side, so it contributes little to the downward pull of gravity. This is one of several factors making the escape velocity not proportional to the mass of the body. StuRat (talk) 01:55, 25 February 2012 (UTC)

I guess that's one way of thinking about it, but it isn't a very useful way. If you are standing on the surface of a 2m radius sphere with mass 100kg or you are hovering 1m above the surface of a 1m radius sphere with mass 100kg then the gravity you feel is the same (see shell theorem). More of the 1m radius is directly below you, but it is further away. Those two factors exactly cancel out. The reason escape velocity is not proportional to mass is simply that you are further away from the centre when standing on the surface of a larger object. (Of course, if the object is sufficiently massive then adding more matter to it would actually cause it to shrink due to increase pressure, so surface escape velocity would increase by more than the increase in mass. See Jupiter#Mass for one example. Neutron stars are another example.) --Tango (talk) 14:19, 25 February 2012 (UTC)

I'm more interested in why it can be modeled as a point mass, and the angle of the ground beneath you is one of the reasons. StuRat (talk) 20:21, 25 February 2012 (UTC)

E.g. on Deimos (moon), it should be possible, you just need 6 m/s to escape. --Roentgenium111 (talk) 23:54, 24 February 2012 (UTC)

Escape velocity is straight-forward to calculate. As the article mentions, the formula is $v_{e}={\sqrt {2GM/r}}$ for a spherically symmetric body. If your density is ρ = 300 kg/m³, then M = ρ(4/3)πr³, so in terms of radius $v_{e}=r{\sqrt {8\pi G\rho /3}}\approx r\cdot 0.0004\;\mathrm {s} ^{-1}$. The part I'm missing is that I don't know what a reasonable velocity from a jump is. Rckrone (talk) 17:00, 25 February 2012 (UTC)

Does the speed you can jump at depend on gravity? I'm not really sure... Assuming it doesn't, then we can use our experience of jumping on Earth. A typical person can probably jump roughly 0.5m in the air (that's increase in centre of gravity - you can get greater clearance from the ground by lifting your legs up, but that's not useful to us). As we know, $v^{2}=u^{2}+2as$. Substituting in v=0 (you are at rest at the peak of your jump), s=0.5m and a=-9.8m/s/s, we get u=3.13m/s. So, let's take a jumping speed of 3m/s (the 6m/s Roentgenium mentions for Deimos is looking a little high - that would require an Earth jumping height of 2m, while the world record for high jump is 2.45 metres, using a technique that doesn't involve lifting your centre of gravity over the bar and wasn't done wearing a spacesuit!). That would give r=7,500m, compared to Deimos' mean radius of 6,200m, (but its density is only 1.5g/cm³, so that's doesn't make sense... ah, your density is wrong - 3g/cm³=3,000kg/m³. That means the radius for 3g/cm³ would be more like 2,400m.). --Tango (talk) 17:54, 25 February 2012 (UTC)

Thanks for your calculations. I was assuming that a human can reach a similar vertical speed as he can reach a horizontal speed (>10 m/s for the world record, and presumably more without air friction), but apparently that's not possible. Still, you could probably reach an orbit just by running fast... --Roentgenium111 (talk) 19:39, 25 February 2012 (UTC)

Of course you would have trouble gaining the required traction and would probably leave the surface due to curvature, before obtaining escape velocity, only to fall back with a less useful velocity... But if you could find a suitable staircase to run up, you might be able to do it. Rich Farmbrough, 20:04, 25 February 2012 (UTC).

Yes, leaving the surface due to curvature would definitely be a problem. Orbital velocity is ${\frac {1}{sqrt{2}}}$ times escape velocity, so you would reach orbital velocity (and leave the surface, although you would come back the surface after one full orbit so you could try and gain a little more speed at that point). That's assuming a spherical planet, though. An irregularly shaped asteroid could be a lot like the staircase you mention (you don't need steps, just an inclined ramp should be enough). If you are moving towards a bit of the asteroid where the surface is further away from the centre, then you are effectively moving up hill. Whether that could ever be enough for you to reach escape velocity, I'm not sure... the rotation of the asteroid would be important to consider. --Tango (talk) 21:10, 25 February 2012 (UTC)

Given that such a small moon's gravity is very small, I'm thinking even a steep incline is inadequate because of the Earth-based tendency to push away from the surface and so one's inertia will not be falling back towards the incline or stairs very rapidly. A suitably long concave incline might work well though. --Modocc (talk) 22:22, 25 February 2012 (UTC)

I'm skeptical of this. I think of one person running up a staircase on Earth and not stopping at the top, versus the other standing on the top step and tensing for as high a jump as possible, and I just can't picture the one running making the higher jump. Wnt (talk) 22:28, 25 February 2012 (UTC)

There is practically no gravity to work with here. For Thebe its only .004g or only four thousands of our gravity. The gain with running is similar to pole vaulting, because running permits the accumulation of kinetic energy. Its counter-intuitive with inclines though, because we lose this energy when running up steps here due to our stronger gravity. Because the g force is so small, the incline does not much effect the outcome other than it is the general direction that you want to be heading... which is away from the rock. One could even run the interior rim of a shallow crater in circles only to exit the crater at the last moment. --Modocc (talk) 22:46, 25 February 2012 (UTC)

I haven't thought too much about it, but I think neither curvature nor the inertia in falling back would be much of a problem: As long as you haven't reached escape velocity, you will get back to the ground to make another "step" in your running. Each "step" may ultimately take you a kilometre or so to make, but you keep your horizontal speed during this kilometre due to the lack of air friction, and can accelerate with each step just as you would in an Earthly running, until you finally (after maybe ten steps) leave the surface continuously for a complete orbit, and then push yourself off the "takeoff" point a bit higher when reaching the ground after completing the first orbit (as described by Tango), to stay in the orbit indefinitely... --Roentgenium111 (talk) 22:49, 25 February 2012 (UTC)

True, although obstacles such as running into boulders or crevices might present this runner with some difficulty. I'm assuming that the OP has some important place or ship that some stranded guy or guys must reach on their own. With any such attempt, one may end up in a perpetual orbit about the rock that takes a very very long time to decay and that could be a lonely place to be. Maybe they just need the exercise, or more likely, working on a mineral excavation. --Modocc (talk) 23:36, 25 February 2012 (UTC)

Some of the details described in Arthur C. Clarke's 1949 short story 'Hide-and-Seek' may be of some interest and relevance here. {The poster formerly known as 87.81.230.195} 90.197.66.166 (talk) 00:20, 26 February 2012 (UTC)

Seems like I read that story, although its been decades, so I don't remember the details. One problem with trying to simply "run" off the moon as suggested above is that its more likely that one will simply reach some arbitrary orbital velocity, but not the desired escape velocity because the difference between these is too large. When "running" along the surface with progressively longer strides one will get very close to a sustained orbital speed which then only allows for a single last "jump" or "step" to either an orbital or escape velocity with no second chance of increasing one's speed further. But if one runs the rim of a crater and spirals outward, they can check their speed to ensure the velocity they need is attained (should they know what this is of course),and they can even stop at any point due to any problems with their equipment or intended trajectory. --Modocc (talk) 01:24, 26 February 2012 (UTC)

There isn't a problem there - you would always come back to the ground unless you reached escape velocity. An orbit always goes through the point where you last had any thrust, so if your only thrust occurs when in contact with the surface your orbit must always intersect the surface at at least one point. If you pushed off every time you reached that point, you would gradually extend of apoapsis (furthest point in your orbit) while your periapsis (closest point) stays the same. Eventually, you would achieve escape velocity and that is when you would stop coming back to the surface. --Tango (talk) 01:53, 26 February 2012 (UTC)

I see and when reflecting back on the previous comments, they make more sense now, and I may want to study up on the reason for this dynamic which I'm sure has everything to do with the conservation laws. Since the OP's only request was whether or not an escape velocity could be reached, the running off bit works fine then. Of course, if one needs a particular velocity that exceeds the escape velocity to perhaps get to a nearby ship then a controlled run makes more sense. --Modocc (talk) 02:43, 26 February 2012 (UTC)

This method of running seems sufficiently difficult to turn the OP into a QWOP. -- ToE 12:50, 26 February 2012 (UTC)

(Outdent) Orbital mechanics à la Newton's cannon becomes complex in two-body interaction problems, where a larger planet (say, Jupiter)'s Roche boundaries influence the local gravity on a smaller moon. You'd need to take into account the Hill sphere around the smaller object. Maybe the person jumping off will assume a horseshoe orbit with respect to the moon and its larger planet, or interact with the Lagrangian points in some way - the end result appears similar to a set of resonance frequencies by which the astronaut will have an easier or harder time jumping off said moon at different angles and different times. ~AH1 ^(discuss!) 21:27, 26 February 2012 (UTC)

February 25

Hermitian condition in Dirac notation

A Hermitian operator satisfies

$\int \operatorname {d} x\left[(A\psi )^{*}\psi \right]=\int \operatorname {d} x\left[\psi ^{*}A\psi \right]$

How do you write this in bra-ket notation? --superioridad ^(discusión) 02:01, 25 February 2012 (UTC)

Is this homework? You need to use the fact that $(A\psi )^{*}=\psi ^{*}A^{*}\!$. -- BenRG (talk) 02:24, 25 February 2012 (UTC)

This is not homework. So is it then just $\left\langle \psi \right|A^{*}\left|\psi \right\rangle =\left\langle \psi \right|A\left|\psi \right\rangle$? --superioridad ^(discusión) 02:42, 25 February 2012 (UTC)

Yes. Though you could also write $(A|\psi \rangle )^{*}|\psi \rangle$ for the left hand side. You don't have to use $(A\psi )^{*}=\psi ^{*}A^{*}\!$, it just makes it look nicer. If this were homework they would probably want the nicer looking version. -- BenRG (talk) 02:47, 25 February 2012 (UTC)

Electric trains and rails

Since electric trains put the return current to the running rails, is there any danger of shock if someone touches the running rails with a train nearby? Does the return current go to the ground directly under the track, or does it travel along the running rails? 169.234.111.180 (talk) 02:37, 25 February 2012 (UTC)

Yes, there is danger. See Third rail. RudolfRed (talk) 04:33, 25 February 2012 (UTC)

Assuming there's a third rail present, that would be dangerous. Inadequately-bonded running rails might present a danger, but I don't know what difference in potential might normally be present between the running rails and ground. Probably not much, if any, given that people walk over tram rails all the time in wet weather, or across rails at station platforms. I have no definitive information to offer, though. Acroterion (talk) 04:39, 25 February 2012 (UTC)

So just to clear things up, the current from the third rail/overhead lines is picked up by the train, and is returned through the wheels of the train, to the rail, and then to the ground; and the current doesn't have to return to the power station through the running rails, right? 169.234.111.180 (talk) 04:57, 25 February 2012 (UTC)

From this paper, it sounds like most modern systems (DC at least) do use the running rails to return the current, which can lead to dangerous shocks, but the lines are equipped with special devices which measure the potential between the rails and the earth, and open a connection if the potential gets dangerously high. Smurrayinchester 11:30, 25 February 2012 (UTC)

In DC traction systems, it is a lot harder to keep the current in the rails and out of the earth, than it is for AC, but it is still the aim. DC systems are a nightmare for causing corrosion in any metal (rails, pipes, buildings, whatever), and causing all manner of problems and safety issues in nearby AC and low voltage systems. DC Traction is a very old and obsolete system devised when the only way to vary the speed of an electric traction motor without punishing loss of efficiency was the Ward-Leonard motor-generator-motor system http://en.wikipedia.org/wiki/Ward_Leonard_motor_control_system. As soon as magnetic amplifiers were developed (1930's to 1950's) AC could be used at lower cost, and DC traction became obsolete, much to the relief of engineers everywhere, and the distress of litigation lawyers. From the 1960's onwards, semiconductor control (TRIAC/SCR) was available, giving extremely high reliability and efficiency with AC power. Keit60.230.195.53 (talk) 12:50, 25 February 2012 (UTC)

DC traction is certainly an "old" system, dating back to about 1895, but hardly an "obsolete" one, since many of the larges urban rail systems use DC, and plan to retain them for the indefinite future. See List of current systems for electric rail traction. Agreed, AC is the typical choice for new electrification, especially in less dense population areas and for long distance transportation. The simple regenerative braking of DC allows a train slowing down or going down a grade to power a different train on the system which is accelerating or going up a grade, rather than just dumping the energy into resistor banks on the train or into heating up brake linings. AC regeneration is far more complicated.Edison (talk) 23:02, 26 February 2012 (UTC)

Competent electrical engineers never design systems so that significant current returns to the power station or substation via the earth - earth current only occurs during certain fault conditions. This is because earth current causes a wide range of problems, including interferance and hazardson telephone lines (even if overhead and not buried) and hazardous conditions on metal pipelines. The rails thru which the locomotive return current flows are connected at regular interavls to a return cable system, which "drains" off the current. In the overhead feed system, transformers are installed each couple of km or so to force the return current out of the rails. With overhead systems, it is not in the least dangerous (electric shock wise) to touch the rails - of course being run over by the train is another matter. With 3rd rail systems, yes - touch the 3rd rail (which is the energising rail), and you can die. Fortunately 3rd rail systems are not very common in most countries, due to performance issues, problems with interference to other services due to the necessarily low voltage & high current, the need to securely fence off and gate the railtrack, and residual serious safety issues (what happens if the train breaks down? Usually passengers get fed up after a while and want to get off - Don't ever do that) 3rd rail systems are used most often in coutries like Britain, whose focus of safety and reliability, due to cultural standards & industrial history, is lower. In countries with a greater focus on performance and safety (eg Australia), 3rd rail feeding is not allowed. Keit60.230.195.53 (talk) 05:55, 25 February 2012 (UTC)

I think that you will struggle to find evidence that there is a lower focus on safety and reliability in Britain than Australia. The main issue with both the original electrification of urban railways in Britain and converting the existing network is bridges and tunnels - the cost of rebuilding these would be enormous. Mikenorton (talk) 14:50, 25 February 2012 (UTC)

There might be other reasons than "a greater focus on performance and safety" for Australia not having the third rail systems found in the US and Britain, such as not having dense urban areas needing electrification of rail systems as early on. The assertions sounds a bit chauvinistic. As for side effects of the current used to power rail systems, I agree that DC leakage promotes corrosion, but AC causes inductive interference with phone and other signal systems. "Competent engineers" realize induction is also a problem with AC powered trains, especially with thyristor or other solid state controls, since the rails are used as returns and transposition is not the simple affair it is with overhead power transmission systems. Edison (talk) 23:02, 26 February 2012 (UTC)

There's heaps of evidence that reliability and safety is less a priority in Britain than in Australia, for cultural and historical reasons. One of the more well known examples known to electrical and electronic engineers in Australia is colour TV. Britain was first, Australia didn't go colour until 1974. There was immense pent up demand. To meet it some companies decided to import British-made sets. These didn't last long in the market - they were rubbish compared to Australian and Japanese sets - too many faults, too many safety compromises, too expensive to repair due to rough construction & old-fashioned circuitry. One set, "Decca", shortly after market release in Australia, was banned by the Authorities, because not only was it liable to give electric shocks if only minor wear & tear defects occurred, it had a transformerless half-wave rectifier power supply, causing DC in your house earth, which causes corrosion and earth system failure in the street MEN earthing system. A truely disgusting thing - yet it was a major volume seller in Britain. Those of us of my generation remember British cars, before they were driven out of the Aust market when Japanese designs began to be imported and made locally in Australia in the 1960's. British cars in comparison to Jap and American - sourced cars were cheap, nasty, unsafe, had high failure rates, and poor finishes that didn't last. Before WW2, Britain had a huge export market. After WW2, little exported. Why? Because by pre-war standards, British products were on a par, but during WW2, other countries used their experience in making high quality war materiel to improve quality, but Britain was economically damaged, and resumed with pre-war standards and expectations. Another good example: motorcycles: In the 1960's Japanese bikes (Honda, Yamaha, etc) became available in Australia. That was the end of British bikes (BSA, Triumph). British bikes had poor finish, high failure rates, and vibrated. Jap bikes had high standard of finish, better handling and brakes, very low failure rates, and were smooth. I could go on and on with many many more examples. Australia has, ever since WW2, been economically much stronger than Britain, with a higher standard of living. That has allowed Australia to spend the money to do things right. Yes, it cost a lot of money to put in Overhead feed AC rail traction. Australia just went ahead and did what was necessary for a reliable safe system. Keit60.230.195.53 (talk) 16:19, 25 February 2012 (UTC)

The vast majority of UK electric stock that does not run underground uses overhead (Great Northern Electrics uses overhead and switches to 3rd rail underground - a matter of a few miles). Safety is a complex issue and the comparison between two countries can't be dealt with easily (there is some interesting background comparing DV/AC and UK US on Talk:Vacuum_tube). The sad fact is that the three pin system, for example, used in the UK and Australia (with different plugs) is being undermined by two-wire devices suitable for the US and mainland Europe, which are significantly less safe. Although the British manufactures post war certainly took a much smaller share of the world market, and quality was a problem (which was mainly about the Far Eastern improvement in quality, as you say) none of these things generalise that well. We see, for example, cycles in the competitiveness and quality of other producers, and we see an increase in the demands of standards bodies such as the IEEE regulations. Earth bonding requirements in the UK, for example, have changed radically in the last 50 years. Rich Farmbrough, 20:30, 25 February 2012 (UTC).

In Australia at least, if the plug does not have an earth pin, the appliance must conform to the requirements of "double insulated", and must display the symbal. This makes it more safe than an earthed appliance, not less. The reasoning for the 3-wire eath system is that if appliance metal work thru a fault becomes live, the user is protected from electric shock because the appliance metalwork is as low in voltage as anything else nearby. But this assumes that the appliance and the house wiring is not otherwise faulty, and the house is multiple earthed per code requirements. If this is not so, or the source is a portable generator (which makes earthing rather a theoretical exercise), the user can still be electrocuted. There are other, admittedly quite uncommon, circumstances too complex to explain here, where you can be electrocuted with earth-pin/3-wire systems that are not faulty. With double insulated appliances, it is virtually impossible for such faults to occur, and thus virtually impossible for users to be electrocuted. In any case, in Australia, the use of ELCB's for ALL wall outlets installed for portable or mobile appliances is mandatory. Lastly, IEEE is an American professional body for engineers (broadly similar to IET in Britain), and has nothing to do with the UK. Keit121.221.230.136 (talk) 00:36, 26 February 2012 (UTC)

The IEEE article seems to disagree with your characterization of it. 75.41.110.52 (talk) 21:31, 26 February 2012 (UTC)

Despite what you may have or have not misread or read in either my posts above or in any Wiki article, it is a fact that the IEEE, as an AMERICAN association of professionals, cannot and does not control how things are done in Britain (or any other country) any more than the BRITISH standards bodies can expect to control how things are done in America - that is the point I made. Basically, both are associations of professionals extablished to maintain professional standards by dissemination of information, however they differ in detail as to how they do that, and what they do. If you work in Britain, you will have to conform to any applicable British mandatory standards, which will in many aspects require you to violate some aspects of American standards. This does not mean either standard is wrong, as they will have been written in different contexts. The same with working in America - you will have to conform to NEC etc, not British standards. This is not to say you cannot use the standards of another country as a GUIDE to good practice, should a local or international standard to cover the subject not exist. Generally, if there is a local standard, comply with that, if no local standard comply with an applicable International (eg IEC) Standard, if no international standard, comply with an applicable voluntary industry code of practice, if no industry code, use a subject applicable standard or code from another country if in your profesional judgement it seems fit, if not, just use your best professional judgement. Keit124.182.21.71 (talk) 02:03, 27 February 2012 (UTC)

"more than 400,000 members in more than 160 countries, about 55% of whom reside in the United States" from the article with two references. If you have better references please improve the article. 75.41.109.190 (talk) 20:08, 27 February 2012 (UTC)

The data on membership is totally non-relevant to both the OP's question and the points I've made, however the reasons for the membership is very interesting. As a practicing consulting engineer based in Australia but with some limited international work, I am a member of both the IEEE and the IET. Neither have any "control" in the countries my customers are located in, but a) membership gives me access to their journals so I can keep current on scientific developments etc, and b) membership gives me some international credibility. The equivalent Australian institution is the IEAust. Keit124.182.55.62 (talk) 09:02, 28 February 2012 (UTC)

I removed the excess "E" from my post, and refer the hon. IP to the IEE/IET regs BS 7671. I note that these are also the standards for "Mauritius, St Lucia, Saint Vincent and the Grenadines, Sierra Leone, Sri Lanka, Trinidad and Tobago, Uganda, Cyprus". Also that they have a requirement for RCDs on all household sockets. Double insulation is no match for a properly earthed device, there is no reason, of course, that a properly earthed device can't also be properly insulated. Rich Farmbrough, 02:56, 27 February 2012 (UTC).

Incorrect - a double insulated appliance, which may have exposed metal work, SHALL NOT HAVE THIS METAL WORK EARTHED, as that defeats the whole idea of double insulation. If you earth it, you will reduce safety, not increase it, as earthing it allows the user to be electrified due to the house or whatever earth/nuetral wiring being at voltage above earth either due to faults or in circumstances where a good MEN system earth is not possible. Double insulation, as difined in standards, is, atleast with the Standards I am familaier with (eg Australian Standards) is not a matter of being "properly insulated" or good insulation, it is a set of construction and testing requirments that make an appliance fault that electifies exposed metal work, or metal parts that can be exposed without the use of a tool, virtually impossible. For example, the complete failure of any single insulation item shall not be possible or shall not enable such electrification. See my post on this above. Note: I have added the word "otherwise" which was inadvertantly left out on a previous post of mine. BS7671 would be a British Standard, not an IET publication. In any case I cannot access it without paying money, which I will not do just for a RefDesk debate - can you post the title of the standard and the relevant paras, should you feel I've got something wrong somewhere? Keit58.170.170.136 (talk) 03:36, 27 February 2012 (UTC)

The OP may be interested that in the New York City subway there are what looks like short snips of limp metal cable connecting all the expansion gaps in the regular rails. I've also seen things spaced at regular intervals on the running rails which appears to be where the power goes after passing through the train. You must be rather close to the train for all three rails to be electrified. I've heard on the news of people electrocuting themeselves on the third rail. However, I don't think I've ever heard of someone being electrocuted by the running rails, even being saved in the nick of time from the train by a Good Samaritan, which happens sometimes (you know, very large population, a crazy bum tries to kill themeselves, even other people). Unless you're under the influence of a substance it's likely intentional if you get electrocuted as the TR's covered on 5 out of 6 sides. (and Mythbusters has busted that you can get electrocuted from peeing on it) Sagittarian Milky Way (talk) 00:32, 28 February 2012 (UTC)

"Things spaced at regular intervals" may be the drain system I mentioned in my post near the start of this thread. The drain system is designed to take the return current out of the rails so that it doesn't normally try and return via the earth - that ensures that the return rails stay at very low voltage compared to earth, and you can't be electrocuted. However short snips of cable as described across the rail expansion gaps are required in any case to ensure end-to-end good rail conductivity within sections for signalling purposes - when the train is near, and passing a road crossing, it electrically shorts the rails together and this is detected by the signalling controller box to bring the barriers down & turn the lights and bells on, so that the cars & trucks halt to let the train thru. Keit124.178.178.1 (talk) 06:56, 28 February 2012 (UTC)

Extreme oxidation states

Is the +9 oxidation state (or any other oxidation state higher than +8) possible? (Talk:Potassium nonahydridorhenate suggests CoH₉, RhH₉ and IrH₉.) What about oxidation states lower than -4? Double sharp (talk) 07:09, 25 February 2012 (UTC)

The highest I've ever heard of is +8 (Xe(VIII)), but my ignorant mind is certainly not all-knowing. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 14:31, 25 February 2012 (UTC)

No idea about what's possible, but the Oxidation state article mentions +8 and –4 as the currently-known extremes. DMacks (talk) 15:50, 25 February 2012 (UTC)

Well, with enough energy and the right equipment you can zap any number of electrons from an atom (see Ionization energy) - so your maximum "oxidation state" there is +Z. However that's quite an artificial set-up - I guess you're thinking about states found in actual, stable molecules. LukeSurl ^{t c} 00:08, 26 February 2012 (UTC)

Yes, I am thinking about states that could (hypothetically) be found in a stable molecule. How stable would Ir(IX) probably be? (Co(IX) and Rh(IX) would probably be less stable, but what about Am(IX) and Mt(IX)?) Double sharp (talk) 04:34, 26 February 2012 (UTC)

This paper thinks Mt can have a +9 oxidation state: Himmel, Daniel; Knapp, Carsten; Patzschke, Michael; Riedel, Sebastian (2010). "How Far Can We Go? Quantum-Chemical Investigations of Oxidation State +IX". ChemPhysChem 11 (4): 865–9. doi:10.1002/cphc.200900910. PMID 20127784. Ratbone60.230.199.111 (talk) 13:17, 29 February 2012 (UTC)

Transistor question

for a transistor the current amplification factor is 0.8 ,when transistor is connected in CE configuration . calculate the change in the collector current when base current changes by 6mA? — Preceding unsigned comment added by 117.252.66.67 (talk) 15:23, 25 February 2012 (UTC)

No. You calculate it...it's your homework not ours. DMacks (talk) 15:49, 25 February 2012 (UTC)

Can you multiply 6 by 0.8?--92.29.192.13 (talk) 20:35, 25 February 2012 (UTC)

I know that this is just a homework problem, and the OP needs to run with the figures given, but does a current gain of 0.8 make any sense in a common emitter circuit? Transistor beta states: "It is typically greater than 100 for small-signal transistors but can be smaller in transistors designed for high-power applications." In practice, would you ever see a CE current gain of less than unity? -- ToE 09:12, 26 February 2012 (UTC)

does the Q mean the alpha of the transistor insted of the beta?--92.29.200.31 (talk) 11:57, 26 February 2012 (UTC)

It took me a moment to realize that by "Q", you meant Q, not Q. -- ToE 12:19, 26 February 2012 (UTC)

The terms "alpha" and "beta" (although beta is still in common use) have been obsolete for decades - Alpha was a useful property with the very earliest of transistors. Yes, the current gain in CE mode (termed hfe) is usually much greater than unity (may be as much as 100,000 or more for very small "supergain" transistors, up to 1000 for small discretes intended for audio premaps), but it can be significantly less than unity for very high voltage switching power transistors. So, the question should probably be taken as written. Unless the correct standardised symbols are used (hfe in this case), it is impossible to be certain. Keit124.182.18.249 (talk) 12:32, 26 February 2012 (UTC)

What's the point of the "being" portion of Human Being

I know "human" refers to our genus, Homo, but I don't know what the point of the "being" is. It doesn't refer to our species, sapiens, so what's the point? I know you can refer to other animals in the genus Homo as human, but are they "Human Beings" as well? ScienceApe (talk) 19:29, 25 February 2012 (UTC)

Well, it's not a scientific term. I would hazard to say it is simply idiomatic — that is, the "being" doesn't mean much other than "I am implying this individual human has a soul of some sort," or something similarly meant to imply a notion of "dignity." The OED says that a "human being" is simply "a person, a member of the human race; a man, woman, or child." This implies to me that while you could drop the "being," it's that last part that matters — you're implying personhood, you're implying some specificity (man, woman, child), you're implying that they're alive (or invoking their formerly living status explicitly), you're implying something more than a pure zoological classification. The term dates back to at least the 17th century. --Mr.98 (talk) 19:36, 25 February 2012 (UTC)

Back to the days when adjectives were not usually used as nouns. "Human" has become a noun, but it was originally an adjective describing a type of creature, or being. We still sometimes hear the expression "alien being", but it's usually abbreviated to "alien". Same thing with human creatures or beings. "Creature" has a mainly pejorative use, so "being" is preferred. -- Jack of Oz ^{[your turn]} 20:52, 25 February 2012 (UTC)

Some folk think of human as an adjective only, like Latin humanus (although the OED cites substantive uses back to the 16th century), so they need to use "human being" for the noun (Latin homo). Deor (talk) 21:07, 25 February 2012 (UTC)

I will just note that the use of human as a noun goes back to the early 16th century. So it's not that recent, though of course usage patterns vary. Ngrams suggests using human as a noun was not uncommon but far less common than using "human being".[8][9] (just two variants I used to try and gauge relative frequency) --Mr.98 (talk) 21:43, 25 February 2012 (UTC)

You have anologies in other languages, too: human <=> human being; Mensch <=> menschliches Wesen; humain <=> un être humain. As above "pre-scientific" notation. 213.169.162.159 (talk) 09:16, 26 February 2012 (UTC)

...and 'ser humano' in Spanish. Richard Avery (talk) 14:08, 26 February 2012 (UTC)

Being. ~AH1 ^(discuss!) 18:53, 26 February 2012 (UTC)

And "istota ludzka" in Polish. Equivalent terms exist in practically all European languages. They are all calques for Latin philosophical terms from Aquinas that are in turn calques for Greek terms from Aristotle. Dominus Vobisdu (talk) 18:59, 26 February 2012 (UTC)

Actually, isn't "being" a verb? Maybe we should take this to the Language desk. ;) Wnt (talk) 19:42, 26 February 2012 (UTC)

No. It's a noun derived from a verb. Specifically, a Gerund. Dominus Vobisdu (talk) 19:45, 26 February 2012 (UTC)

It just sounds better, wouldn't you say? Vranak (talk) 03:11, 1 March 2012 (UTC)

Oxosulphuric acid

What is the name for the analog of sulphuric acid with oxygen in place of sulphur (H₂O₅)? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 20:15, 25 February 2012 (UTC)

Sulphuric acid is H2SO4.--92.29.192.13 (talk) 21:19, 25 February 2012 (UTC)

Super oxygenated water is all I can find. Apparently, it is not very stable.--92.29.192.13 (talk) 21:23, 25 February 2012 (UTC)

Oxygen doesn't form that sort of structure ("O with four bonds to atoms"). H₂O₅ is all linearly attached "hydrogen pentaoxide", not a "sulfate-like core". DMacks (talk) 21:32, 25 February 2012 (UTC)

That acid can't exist, so I assume it is a hypothetical acid, in which case you'll have to invent a new name, possibly peroxygenic acid. Plasmic Physics (talk) 21:36, 25 February 2012 (UTC)

February 26

Molality calculation

A mass of 168 g of manganese dibromide is dissolved in 225 g of water. What is the molality of the solution — Preceding unsigned comment added by Nanceninja (talk • contribs) 02:28, 26 February 2012 (UTC)

Please see Molality and Molar mass. If you still have trouble with this homework problem, show us your work so far, and point out where you are stuck. -- ToE 02:45, 26 February 2012 (UTC) (I also shortened the title of this section.)

..., and is manganese dibromide a synonym for Manganese(II) bromide, MnBr2? -- ToE 02:49, 26 February 2012 (UTC)

... apparently so (NIST MML). I went ahead and created the redirect: manganese dibromide. If a chem-head tells me it's wrong, I'll speedy WP:G7 it. -- ToE 05:26, 26 February 2012 (UTC)

Lightning in Hawai'i

A friend of mine who lives in Hawai'i says that he rarely sees lightning there. Is there data to support this and if it's true that there is little lightning there, what's the cause? Dismas|^(talk) 07:56, 26 February 2012 (UTC)

There are very few lightning storms over the ocean and Pacific islands in general. Quoting from NASA, The ocean surface doesn't warm up as much as land does during the day because of water's higher heat capacity. Heating of low-lying air is crucial for storm formation, so the oceans don't experience as many thunderstorms. The Hawaiian islands are presumably small enough that the weather there is more ocean-like than land-like. Someguy1221 (talk) 10:12, 26 February 2012 (UTC)

The areas of the world that see the most lightning (I'm going to use Florida, USA, which is the state with the most annual lightning strikes, as an example) experience it so often because they are prone in their respective warm seasons to produce localized "garden variety"-type storm cells—Someguy's link describes the upwards motion needed to initiate convective weather activity. Synoptic weather events (large-scale low pressure systems, cold fronts, tropical cyclones) can also produce lightning, but because they occur much less often in a given location, regions where pop-up thunderstorm cells are rare have to rely on these synoptic systems for their fill of electric storms. This is the case for Hawaii, but interestingly, due to the tropical environment surrounding the islands, storm systems that are capable of producing lightning there often go all-out; see here, for example, which discusses such events as 21,000 lightning strikes in five hours. Juliancolton (talk) 20:24, 27 February 2012 (UTC)

How do they measure genetic proximity for unrelated individuals?

In Steven Pinker's The Blank Slate, there is a fairly lengthy discussion of the contribution of genetics, family environment, and "unique" environment to an individual's personality and abilities. The contribution of each is measured according to its role in explaining the variance: more fully, the proportion of variance of the response variable of the sample that is attributable to that characteristic. Note that the three contributions add to 1 (the third is determined by subtraction, I think). I can make easy sense of this when we are dealing with measures that vary continuously, which would include the response variables, since personality measures can be made continuous. But surely the same has to apply for the independent variables, or else we need a categorical variable for each individual. How do they measure, in particular, genetic proximity as an independent variable? I know there are measures of consanguinity, like the coefficient of relationship, but you can only measure a small handful of people on the same scale like this. Furthermore, this is only defined by relationship, not in any absolute sense, so what reference point do you use? What do you do when you are using a large sample of unrelated people? For that matter, how do they measure the family environment? Is it just two categories, "same" and "different"? IBE (talk) 22:17, 26 February 2012 (UTC)

I'm having a little trouble parsing your question (which is probably why nobody else has answered it, as well), but to your basic question of isolating genetic relatedness as an independent variable, isn't the answer to this simply by using twin studies? --Mr.98 (talk) 22:35, 27 February 2012 (UTC)

Is that what the OP was on about? My first reaction was "Eh? wot? wot langwidge is dis?". My second reaction was: This is some kind of joke ripping us off, or the OP just likes the sound of his own words. Wickwack121.215.46.219 (talk) 00:25, 28 February 2012 (UTC)

Twin studies does give a good overview from what I can tell. The language of my post, imho, is that of mathematics, expressed in words, by someone who doesn't fully understand the concepts involved, hence the question. I also claim to be way ahead of my time, which is why Wickwack can't understand me ;). As for the maths, I can understand when it's presented, but I'm used to one type of study (simple regression, from my undergrad days), and I could see that twin studies didn't fit that particular mould. Twin_studies#Methods has the main particulars, so I'll get my head around that and then I'll ask stuff again if still needed. Thanks for the link - I should have checked first, but I'm not used to finding such good details on methodologies, even knowing the WHAA.. principle. IBE (talk) 04:29, 28 February 2012 (UTC)

February 27

Time travel

I have heard it's possible to travel in time if we can somehow travel faster than light. It has been proven scientifically and mathematically. But i don't understand how it all works? Can someone explain it in the way that any ordinary people can understand it? Thanks!Pendragon5 (talk) 04:37, 27 February 2012 (UTC)

"Can someone explain it in the way that any ordinary people can understand it?" Probably not. Then again, it has also been "proven scientifically and mathematically" that we can't travel faster than the speed of light in the first place. Why do you think the universe should be understandable by "ordinary people"? It doesn't seem to make that much sense to those that spend their entire careers trying to understand it... ;-) AndyTheGrump (talk) 04:44, 27 February 2012 (UTC)

(edit conflict) It has not been proven scientificially, in the sense that science doesn't really "prove" anything, it provides evidence that it happens. Insofar as no experiment has ever been done to confirm time-travel, and so-far only one tenuous experiment has possibly confirmed FTL travel at all (the Faster-than-light neutrino anomaly) I would say that the empirical evidence of either faster-than-light travel or time travel is simply not there for it to be "scientifically proven" (again, for whatever you take "proven" to mean, which science doesn't really do). Mathematically, sure, there are all sorts of ways to manipulate equations to show that time travel is possible, if faster-than-light travel is possible, but empircal evidence for the latter condition is super tenuous, and the connection between FTL travel and time travel is unexplored (except mathematically), I would say that the jury is not just out on time travel, I'd say there hasn't even been an arraignment... --Jayron32 04:46, 27 February 2012 (UTC)

What do you mean by that science doesn't prove anything?? Science is always a tool for human to prove things, it's by far the most acceptable method around the world to prove stuffs.Pendragon5 (talk) 19:37, 27 February 2012 (UTC)

Strictly speaking, Jayron is correct. Science helps us develop models to predict the outcomes of experiments, but since there are always multiple models to explain any occurrence, nothing is strictly proven correct, although certain models will inevitably be disproven. Someguy1221 (talk) 10:22, 28 February 2012 (UTC)

Also (edit conflict). I'm not sure if you could say it's been "proven", it's probably safer to say that it fits certain interpretations using current models. We have an article which addresses what you are talking about Time_travel#Via_faster-than-light_.28FTL.29_travel Vespine (talk) 04:50, 27 February 2012 (UTC)

(edit conflict) It has never been proven, in that no one has actually done it. Mathematically and theoretically, it comes down to the relativity of simultaneity (you can also read Special_relativity#Causality_and_prohibition_of_motion_faster_than_light). Basically, as you'll surmise from the first article I linked, the order in which events appear to have occurred is not sacrosanct in special relativity (i.e. reality). There are situations in which the order that events occurred is different for different reference frames. But if you allow things to move faster than the speed of light, you can get some ridiculous observations, mainly effects occurring before their cause. My college special relativity text used the example of a projectile being sucked out of its target and back into the cannon that fired it. Such things can be interpreted as objects, such as that projectile, traveling backwards through time. Someguy1221 (talk) 04:51, 27 February 2012 (UTC)

It's not really true. The "proofs" you've seen start from unstated assumptions that are inconsistent with faster-than-light travel, and proceed to assume the existence of faster-than-light travel. From that, you can prove anything (ex falso quodlibet). The reason people write about amazing things that physics "proves" is the same reason they write about miracle diets and angel sightings. There's a market for it. It's not because of any scientific merit. -- BenRG (talk) 07:14, 27 February 2012 (UTC)

I don't think there's anything in the derivation of the tachyonic antitelephone that's logically inconsistent with faster-than-light travel. If there is, please enlighten me.

The assumptions may be inconsistent with faster-than-light travel plus currently understood physics, but that's quite a different matter. --Trovatore (talk) 02:13, 28 February 2012 (UTC)

Well, we have managed time travel! Ok, perhaps not quite, but this experiment was very interesting. http://en.wikipedia.org/wiki/Hafele%E2%80%93Keating_experiment They managed to get a clock to travel nearly 40 nanoseconds in to the past/future. Ok, it's not time travel as such... Zzubnik (talk) 10:48, 27 February 2012 (UTC)

That's "time travel into the future", aka the twin effect/paradox. It's a totally different beast from traveling to the past. -- BenRG (talk) 19:45, 27 February 2012 (UTC)

Hi BenRG. I'm aware of that. I thought it might be an interesting read to the OP. Personally, I think that time dilation like this is the closest that we will ever get to "travelling in time". — Preceding unsigned comment added by Zzubnik (talk • contribs) 10:17, 28 February 2012 (UTC)

If I'm playing snooker ball and move it back as it moves backward in time its an example . Thanks Water Nosfim — Preceding unsigned comment added by 81.218.91.170 (talk) 13:33, 27 February 2012 (UTC)

Here's a simple way of thinking about it. As you accelerate, time starts to slow down for you. At everyday speeds, this is entirely imperceptible, but once you get up near the speed of light, it ramps up and becomes noticeable (from your point of view, time will pass normally, though). As you go faster and faster, approaching the speed of light, time slows further still, and if you were able to travel at exactly the speed of light, time would stop completely. What this implies... is absurd, and just one of the many reasons why you cannot travel at the speed of light. But if you could then accelerate even more, time would begin to go backwards. The faster you went, the faster back through time you would go. Again, from your point of view, time would proceed normally. Goodbye Galaxy (talk) 15:21, 27 February 2012 (UTC)

I meant it has been proven scientifically and mathematically if we can somehow manage to travel faster than light. And yep as with our technology today, we are no where close to the speed of light. A lot of people today would say travel faster than time is impossible. But impossibilities always have changed over time, it is impossible right now doesn't mean it will be impossible in the future. A thousand years ago, there were a tons of stuffs that consider as impossible but we can do them with ease today. I'm not saying certainly time travel is possible but we don't know for sure it's impossible either. So basically theoritically it's possible if we can travel faster than light but practical is we will see in the future or our descendants will??Pendragon5 (talk) 19:37, 27 February 2012 (UTC)

@Goodbye galaxy: Why the faster you move = the slower than time is? I don't understand the concept that0 if you travel faster than light then why is it that time is running backward?Pendragon5 (talk) 19:37, 27 February 2012 (UTC)

Goodbye galaxy is wrong about that (see my reply below). As for the rest I don't know what to say except what I said above. -- BenRG (talk) 19:50, 27 February 2012 (UTC)

The time dilation factor is ${\sqrt {1-(v/c)^{2}}}$. This does go to zero as v → c, but it goes imaginary, not negative, when v > c. In fact, negative values of the factor correspond to speeds less than c; it's only by convention that the positive square root is used. Special relativity, without additional assumptions, has no concept of going "forward" or "backward" in time. -- BenRG (talk) 19:33, 27 February 2012 (UTC)

LOL then we just make an assumption that time connected to speed? I don't see how they connected, time has nothing to do with speed. Time is something always happening no matter what we do. As we travel faster and faster, how can it possible effect the flow of time? Time is just a concept that humans have came up with, it's arguable of what time actually is. As i can understand, we still have a poorly understand about time. We still have long way to go if time travel is possible. But the time travel concept many famous scientists have came up with really confused me.Pendragon5 (talk) 20:00, 27 February 2012 (UTC)

"As we travel faster and faster, how can it possible effect the flow of time?": Few if any people understand it completely, but it does. This is part of Einstein's theory and it has been confirmed by accurate clocks in spaceships. They travel at great speed, and their clocks go a little slower than clocks on earth. The theory of relativity is mind-boggling, but it does work. It makes accurate predictions. You are right that real time travel (i.e. backwards, because we always travel forwards anyway) is usually based on poor understanding of the theory. -- Lindert (talk) 20:21, 27 February 2012 (UTC)

Well the concept of the faster we go = the slower the time flows is just mind crunching, i don't think i can make any sense out of it. I see no connection between the two, no matter of faster you travel, it is only get you to one place to the other faster not travel in time. Even if we can travel faster than light, i don't see how it can possibly travel back in time. Perhaps there is something else is effecting it and we don't even have any knowledge about it yet, such as dark matter and a bunch of other predicting particles that scientists think it must be there so the whole thing can make sense. As i can see, we still have a lot of assumptions. I don't mean to say human race is stupid, we are the smartest species in the Earth indeed but i think we are still very stupid "compare" to the universe as the whole. I feel like as it is right now i only consider that we barely know anything about the universe yet, there are still a lot of things left unexplained or doesn't make sense to human's conscious. Plus what we think is right for now may be proven wrong in the future. It always have been like that, that's the natural of science. I wonder if there is an absolute truth that will forever never be proven wrong nor can it be improve one day.Pendragon5 (talk) 23:41, 27 February 2012 (UTC)

You're thinking about time in the wrong way. You're thinking of time (change) as more primitive than everything else in the world. That's why you don't understand how the laws of physics can "affect time". Actually, time is an aspect of the world on the same footing as distance. Clocks tick off the seconds, and people think and perceive the world and grow older, by means of physical interactions between their various parts. If you think of time in that way then there's nothing strange about special relativity at all. As someone who does understand it, I assure you that it's not nearly as amazing, or interesting, as it might seem at first. -- BenRG (talk) 01:34, 28 February 2012 (UTC)

Along that line, I used the phrase relativity is "a basic consequence of geometry" when a question came up a week or two back. If you work out the math to its logical conclusions, you'll see that there's no other way things could work - there'd be an inconsistency somewhere. That's why we have things like length contraction or time dilation. As far as physicists are concerned, conundrums like the barn door paradox are less inconsistent than the consequences if we did not make relativistic corrections. For example, if the speed of light were not constant, as observed from any reference frame, the natural consequence would be that some reference frame is the canonical one, and everyone else is wrong; the laws of physics would change from place to place and person to person. Not only is the idea of a canonical universal "correct" reference frame a profoundly unsettling concept for many physicists, it's also not what we measure. We use relativity (in both special and general forms) because it is internally consistent (the math all works out) and externally consistent (it matches our experiments). All the equations and the mathematical abstractions about relativity are simply tools to help us formally explore the consequences of our observations. Nimur (talk) 02:08, 28 February 2012 (UTC)

Hopefully these explanations have disabused the OP of any notion that the highly questionable (at best) nature of the concept of time travel is in any way comparable to the well understood and experimentally verified nature of the feature of our universe that is time dilation. The OP may wish to read Poul Anderson's classic hard science fiction novel Tau Zero, which prominently features, explains, and is titled after time dilation (to the extreme). If they have sufficient self-discipline, they should postpone reading our article until they have completed the novel; it is a good read. The OP may also find our Twin paradox article interesting. -- ToE 13:23, 28 February 2012 (UTC)

The truly strange thing is that the speed of light is the same for all observers, regardless of their velocities with respect to one another. Once this is accepted as experimentally proven, then length contraction and time dilation follow as natural consequences; it's the only way it all makes sense. If the OP does choose to major in astronomy at university, they will work simple special relativity problems in Freshmen-level Physics. A fuller understanding will come in their Junior year when they study classical electrodynamics. (The "classical" here distinguishes it from quantum electrodynamics.) Having mastered a text such a Griffiths', the student will understand the derivation of special relativity, and should be able to easily work most relativistic problems of motion, but it won't be until they study a graduate level text, such as Jackson's, that they will be in a position to work more complicated relativistic electrodynamic problems. General relativity is a much different story. Its consequences will be discussed in many an astronomy course, but its fundamental mathematics isn't really tackled until graduate level physics.

I'm no science historian, but I suspect that most physics graduate students will understand special relativity as well as Einstein, and those specializing in general relativity will likely understand it better than Einstein ever did. The profession marches on, and long past are the days when only a handful of people on Earth understood Einstein's work.

Finally, the OP should not assume that an undergraduate degree in astronomy is the best route towards graduate studies and a profession in that field. In my experience at university, I found that the astronomy majors were much less proficient at math than the physics majors. Both took the same Junior-level Classical Electrodynamics and Classical Mechanics classes, and grades were bimodal. I believe that the students who either double-majored or majored in physics and minored in astronomy had a much better chance of being accepted into a good graduate program, and those who were most interested in cosmology were often applying to physics departments, where that sort of work is mostly done. -- ToE 22:09, 27 February 2012 (UTC)

Just as an historical note: special relativity wasn't the one that required heavy mathematical work. That was the one where people just didn't like the conclusions. One of the main criticisms was that it was mathematically simplistic — a trick of algebra — as compared with, say, the kind of math you needed to calculate vortex stability in luminiferous ether flows. General relativity was the one that required some more mathematical ability than your average physicist had at the time, and was first embraced by heavily mathematical physicists in England primarily for this reason. (The British had a stronger mathematical physics tradition at the time; the Germans were mostly about very careful experiments.) Later generations of physicists would become increasingly mathematical and theoretical. (Even they didn't work on GR, though. GR more or less wasn't even taught from the 1920s through the 1970s, or something along those lines — I can't recall the exact dates. It wasn't a hip area of physics.) --Mr.98 (talk) 02:12, 28 February 2012 (UTC)

Many times the particle goes back and forth in time. When he starts moving faster, the average of time begins to run down, and see the parallel universes which is related only to the speed of C, what remained of the particles that move backward and forward in time remains to one parallel universe . Thanks Water Nosfim — Preceding unsigned comment added by 81.218.91.170 (talk) 05:39, 28 February 2012 (UTC)

Binocular vision and microscopes

Is there a known condition whereby one cannot look through the microscope with both eyes simultaneously? How is this phenomenon called? What are the treatments if any? Thank you. Gidip (talk) 11:10, 27 February 2012 (UTC)

The same issues should apply to both microscopes and binoculars. Therefore, what you are looking for is the section Disorders of binocular vision in the binocular vision article. It lists several relevant disorders of the eye. As for treatment, the reference desk is not allowed to give medical advice. You would have to consult a professional Orthoptist. EverGreg (talk) 11:32, 27 February 2012 (UTC)

There are too many disorders listed there (I browsed this page before posting my question). I can't figure out which disorder is the relevant one. If someone can suggest one or a few specific disorders that would really help. Thanks. Gidip (talk) 11:57, 27 February 2012 (UTC)

Before we get into medical disorders, consider that it might just be improper setting of the microscope. The eyepieces have to be exactly the right distance apart or you have to move one way and the other to see through them. If you can't see through the microscope with both eyes at the same time, obviously you don't know that the current setting is correct. I'd say, keep trying. Wnt (talk) 12:53, 27 February 2012 (UTC)

He fathered seven point five children in one night and about Genghis Khan he only laughed

• http://www.youtube.com/watch?v=pzmI3vAIhbE

Er zeugte sieben Kinder in einer Nacht		He fathered seven children in one night
Und über seine Feinde hat er nur gelacht		And about his enemies he only laughed
Denn seiner Kraft konnt keiner widerstehn		Because nobody could resist his strength
Hu, ha...		Hu, ha...
		-- Dschinghis Khan by Dschinghis Khan

http://www.dailymail.co.uk/sciencetech/article-2106776/No-wonder-theres-violence-world-Hundreds-millions-people-related-despots-claims-study.html

... One historian believes that Yangdi, the 6th-century Sui dynasty emperor, for example, had children with 100,000 women.

The Emperor Yang of Sui (569-618) died when he was 49 years old. When he was 12, his father established the short-lived Sui Dynasty. So he was a prince when he was ready for having babies. He must have slept with at least 7.5 women each night for 37 years to father 100,000 kids. I mean from age 12 to age 49 and not a single miss.

Who is that historian?

Emperor Yang of Sui was 5 centuries before Genghis Khan. I guess he was unable to laugh at a man who could only father seven children each night. -- Toytoy (talk) 11:24, 27 February 2012 (UTC)

The Daily Mail references the anthropologist Laura Betzig. She mentions Yang in this article: "Yangdi, the Sui Dynasty emperor who built the Grand Canal and rebuilt the Great Wall, was credited by an official historian with 100,000 women in his palace at Yangzhou, alone, and when the Yuan dynasty founder, Kublai Khan, put up a capital at Beijing, he left a summer palace at Xanadu behind, with room in the excavations for another 100,000 (Wei, 2008)." So, he kept an enormous harem, but there is no claim that he had children or even slept with all of these women. --Wrongfilter (talk) 11:59, 27 February 2012 (UTC)

And it probably should be noted that estimates of population sizes (e.g. army sizes, harem sizes, city populations) by ancient historians are often off by an order of magnitude or so, either out of error or for dramatic/propagandistic effect. (Heck, we still do it today when estimating death tolls — the estimates before bodies have been counted in any systematic way, in either war or disaster, are often an order of magnitude off of the actual counts, in my anecdotal experience.) --Mr.98 (talk) 13:09, 27 February 2012 (UTC)

It is probably impossible to do that many women... to have many kids. I have read many Vietnamese and Chinese histories. Hundreds of children are typical for emperor but i never ever heard about any emperor has reached a thousand children mark yet. So 100,000 children = not even close to be possible. There were always some people who try to exaggerate stuffs about the kings so the king can favor them. Perhaps that emperor likes people to think he does a lot of woman. And as the emperor in any where of the world back there then they can basically say whatever they want and forced people to believe it. If you don't believe in it, the consequence will be really bad... Possibly deaths.Pendragon5 (talk) 19:44, 27 February 2012 (UTC)

Are you sure this is a literal description, rather than a gullibility of speech? After all, they say an emperor built a pyramid and nobody ever saw him hauling any bricks. Can't he have 100,000 children the same way? Wnt (talk) 00:57, 28 February 2012 (UTC)

I know nothing about 6th century Chinese medicine. But as to Chinese medical belief of early 20th century, I think many people, under Taoism influence, believed that one drop of semen equals to maybe 10 drops of blood or so. As a result, a playboy may be "sucked up dry" by women if he sleeps with too many of them for too many nights. I think many athletes still keep themselves from women before they are going have a big fight.

Sex is a very important subject in Chinese medicine and philosophy. Today's Taoists still practice a way to keep themselves from ejaculation during an orgasm. They trained not to "shoot" in order to enjoy sex and preserve their valuable bodily fluid. They did it to promote themselves to a higher degree of consciousness. An imaginary goal for some mysticism practioners is to "dominate" 100 women for a night without losing a drop of semen ("御百女而不洩"). I don't think there was feminism 1500 years ago, sorry.

There were very high ranked officials to decide who should sleep with the emperor each night. I don't think an emperor was allowed to waste his "dragon's essence" by sleeping with 100,000 women. If you play so much, you'll end up like a zombie! An emperor might have 100,000 women in his harem. However, the majority of these women were living in so-called "Cold Palace" ("冷宮"). They were not allowed to have sex with the emperor unless they were chosen. They end up marry an eunuch when they got too old to serve the emperor. The emperor could have several women with him each night. He was not allowed to ejaculate to all of them. On the other hand, if he ejaculated to the specially chosen woman, an eunuch outside was ordered to register the exact time of ejaculation in order to have the time reviewed by an astrologer. At least it was the practice of China's last Dynasty, Qing Dynasty.

Wang Zhaojun (50BC?-?) was a extremely beautiful lady-in-waiting of Han Dynasty. It is said that she refused to bribe a painter. As a result, she was painted as an ugly woman. The emperor did not know her before she was given to the Huns. ...... Well, the painter was sentenced death and she was sent far north .... Never been enjoyed by the emperor ......

I don't want to blame a scientist for her ignorance of Chinese history and philosophy, even though having 100,000 kids certainly is beyond science, logic and imagination. Not even a NBA basketball player could come even close without artificial insemination. I just want to know who's that historian? Ha! Ha! Ha! -- Toytoy (talk) 07:38, 28 February 2012 (UTC)

Can a Silicon atom get split into smaller atoms, like a carbon and oxygen atoms?

Hi, Is there any way to split a silicon atom (14) into a carbon atom (6) and an oxygen atom (8)?

Or is the fusion of lighter atoms (< 26) a one-way process towards iron (26) only?

Thanks --InverseSubstance (talk) 23:27, 27 February 2012 (UTC)

This kinda rings a bell. Have you been reading something like this:Biological transmutation. Neutron bombardment can transmute Si to phosphorous but as to the other theories I want to keep clear of them.--Aspro (talk) 23:51, 27 February 2012 (UTC)

Ah. Nothing sucks-seeds like a toothless biggie. With the aid of my dowsing pendulum I have found this... [10]. Well, it seems one learns something new every day ;-) --Aspro (talk) 00:06, 28 February 2012 (UTC)

Thanks - Not sure if the above is to be taken seriously? I not looking for a wacko nut job's outrageous claim. It seems physics basically says, once an atom has fused, it can never get defused.

Phosphorous (15) is higher than Silicon (14). I'm just wondering if light atoms can only ever become heavier atoms. That is, they can't ever migrate backwards towards lighter atoms. Just wondering. --InverseSubstance (talk) 00:21, 28 February 2012 (UTC)

Bombarding a nucleus with energetic particles (e.g. cosmic rays) can sometimes cause even relatively light nuclei to break apart. For example, the traces of Beryllium-10 found on Earth are mostly caused by cosmic rays breaking apart nitrogen or oxygen in the atmosphere. Dragons flight (talk) 00:42, 28 February 2012 (UTC)

For reference, the process is called Cosmic ray spallation Vespine (talk) 01:21, 28 February 2012 (UTC)

The Discovery_of_fission section of our Nuclear fission article alludes to "splittings" or fissioning of light elements achieved when the phenomenon was first investigated. Although Silicon is not explicitly mentioned, I can think of no reason why it should be any less fissionable than other light elements. {The poster formerly known as 87.81.230.195} 90.197.66.193 (talk) 02:09, 28 February 2012 (UTC)

All right! Thanks. Because now it seems possible to change sand into oil. Theoretically at least. I'll keep grinding away... --InverseSubstance (talk) 05:15, 28 February 2012 (UTC)

I don't think so. In theory, you can take any atom of your choice (eg silicon), and hit it with any atom or particle of choice (eg electron, gamma photon, thru to the largets atom possible). For each combination of traget atom and incident particle, you'll get smaller atom plus one or more particles and some energy radiated. Just what you get is governed by certain rules. More often than not, the smaller atom wil not be nuclear stable, and within microseconds to kiloyears, depending on the target and incident combination, decay into something else - this may continue several times. Problems: a) not all combinations have been tried, b) the technology to select and use the optimal combination may not exist, and c) we know enough to know what you get is seldom convenient. So, in practice transmuting silcon into carbon is not possible, and nor is the reverse. Ratbone124.178.178.1 (talk) 07:22, 28 February 2012 (UTC)

It's definitely possible to go from stable silicon to stable carbon, it just might not be possible to do directly. Regardless, the energy costs would be obscene compared to the energy you'd get from burning the resulting oil you plan on making. If you're already cool with the process being energy inefficient, you are much better off making oil out of carbon dioxide, which is easily doable. Someguy1221 (talk) 07:27, 28 February 2012 (UTC)

You can't make oil out of just carbon dioxide - you would need some hydrogen from somewhere. Water would be the obvious source. Then you're just doing the usual combustion reaction backwards. If combustion works, then combustion backwards should work, you just need to put enough energy into it (and somehow prevent the newly formed oil and oxygen from immeadiately combusting back into carbon dioxide and water - that would probably be the tricky bit). --Tango (talk) 12:34, 28 February 2012 (UTC)

February 28

can an exothermic reaction create ice?

could an exothermic reaction underwater create, as a byproduct, ice: however, doing so by decreasing the density of the affected area greatly.

I am thinking of a cubic centimeter of water that we mark so we can trace it. Then an explosion of that cubic centimeter into a cubic meter. The greatly reduced density could cause it to freeze, could it not, yet the explosion from cubic centimeter from cubic meter is, well, explosive. Which leads to the question posed above. Please let me know if I seem to indicate that I am laboring under false assumptions of either physics, chemistry, or thermodynamics. Thanks. (Note that I don't make any assumptions about the actual process involved! It could be a special chemical, a physical process, whatever...just that it's exothermic. I also don't care if, as a result, the water outside our new 'cubic meter' gets hotter/undergoes an increase in entropy: indeed I expect it to.). --80.99.254.208 (talk) 06:54, 28 February 2012 (UTC)

This is quite possible if pressure is sufficient. All you need is for the heat removed by the expansion to be greater than the heat needed to be removed as heat of sublimation. If the water is super critical, ie the pressure is above the critical pressure 22.064 MPa for water, the heat of sublimation is essentially mimimum (about 6 MJ/kmol) when the triple point temperature is reached. However, if the pressure is well above critical, the specific heat of the super critical phase is mimimised, making it harder to make ice. This is quite easy to see if you look at a plot of temperature versus internal energy. Ratbone124.178.178.1 (talk) 07:08, 28 February 2012 (UTC)

Semi conductors

Substances whose conductivity lies between that of a conductor and an insulator is called semi-conductor.But why isn't it called semi-insulator. — Preceding unsigned comment added by Aditi keerti (talk • contribs) 08:29, 28 February 2012 (UTC)

It could have been. Go find the person who chose that convention and shoot him with a semi-manual gun. StuRat (talk) 09:26, 28 February 2012 (UTC)

The definition of a conductor is that which has charge carriers (typically valence electrons fully mobile within crystals), can be ions in liquids), thereby permitting the carrying of current. You can thus have a material with not very freely available carriers (typically mid-valence electrons & missing electrons partially/semi mobile with crystals), so "semi-conductor" makes some sense in this case.

The definition of an insulator is that which has NO charge carriers, so cannot carry ANY current (in practice, with the right instruments, very minute currents may be detected, due to impurities, leakage thru contaminants on the surface, etc). It's just semantics, I guess, but semi-nothing doesn't make as much sense in English. Semi-something is still something; semi-nothing is an oxymoron at best. Ratbone124.182.55.62 (talk) 09:58, 28 February 2012 (UTC)

A ball rolling down an incline

When a ball rolls down a incline, how do we know how much of the potential energy gets converted into kinetic energy and how much gets converted into rotational energy? Widener (talk) 11:52, 28 February 2012 (UTC)

See Moment of inertia and List of moments of inertia. Dolphin (t) 12:08, 28 February 2012 (UTC)

If you know the radius of the ball, it is easy to calculate (if it isn't gliding, but rolling) what the relation is between the rotation- and movement velocities. Using the moment of inertia calculated for the ball, you can derive the ratio between rotational and kinetic energy. -- Lindert (talk) 12:37, 28 February 2012 (UTC)

(edit conflict) You also need take friction into account. Is there sufficient friction that the ball is purely rolling or will it also slide a bit? If there is no sliding, then you can easily relate the linear velocity to the rotational velocity (distance travelled in one rotation equals the circumference of the ball). Once you've got the velocities, you can work out the kinetic energy using E=1/2mv² and the rotational energy using the moments of inertia Dolphin links to. If there is sliding, then it gets rather more complicated (friction, generally, is complicated - there are massive simplications usually taught in schools that might get you somewhere, but to get it accurate is really hard). --Tango (talk) 12:40, 28 February 2012 (UTC)

What does Lidocaine have to do with wood?

Other names for the anesthetic lidocaine are "xylocaine" and "lignocaine". The prefixes xylo- and ligno suggest this has something to do with wood, but the Wikipedia article doesn't say anything about it being made from, or being chemically similar to, some substance extracted from wood. What, if anything, does lidocaine have to do with wood? -- Finlay McWalterჷTalk 13:16, 28 February 2012 (UTC)

According to the Oxford English Dictionary, 'xylocaine' got it's name because of the chemical relationship with xylene (and cocaine), which in turn is obtained from wood-spirit ('Crude methyl alcohol obtained from wood by destructive distillation'). It is also called 'lignocaine', because ligno- is the Latin equivalent of 'xylo'. -- Lindert (talk) 13:35, 28 February 2012 (UTC)

distance and time to orbital velocity at tourist-acceptable G's at sea level, assuming vacuum?

If the Earth were a perfect sphere and had no atmosphere, then a pod could accelerate fast enough to reach Orbit right at sea level, couldn't it? (well, a few inches/feet above sea level, let's say). So, at g's that are acceptable to tourists/passengers, how long would this acceleration take, both in terms of time taken and distance travelled in terms of our physical geography/Earth miles or kilometers? What if we increase the acceptable G's from tourist/passenger to whatever is the human maximum, however uncomfortable? Then how long would it take and what distance would be covered? Thanks. 188.6.78.231 (talk) 14:19, 28 February 2012 (UTC)

also could some confirm my premise itself, that at the right speed, orbit a few cm/feet above sea level in a perfect-sphere (uniform) earth without an atmosphere scenario is possible? 188.6.78.231 (talk) 14:27, 28 February 2012 (UTC)

Orbital speed for a circular orbit at zero altitude is about 7.9 km/s. If we assume a lateral acceleration of 5 m/s² (half a g, equivalent to the acceleration in a fast car) then it would take about 1600 seconds or about 27 minutes to reach this speed, in which time you would have travelled around 6,400 km, which is a little more than the distance from London to Chicago. Gandalf61 (talk) 15:00, 28 February 2012 (UTC)

Thank you! Do you think this is reasonable for a tourist/passenger to undergo for 27 minutes? (The acceleration of a fast car?) Would it be possible I don't know to have the pod turn toward the direction of travel, so everyone just feels a bit heavier instead of being pushed BACK (and not down) into their seats? Do you think this is a reasonable tourist experience? Thanks again for the calculations. 188.6.78.231 (talk) 15:17, 28 February 2012 (UTC)

On the other hand, at a crushing 10g, it would take 80 seconds and 3000 km. SpinningSpark 15:14, 28 February 2012 (UTC)

wow. Okay, here's one. How long would it take at a g so low that you can't notice that you're moving? Would you already go around the earth / several times before you reached orbital velocity at such a speed? 188.6.78.231 (talk) 15:17, 28 February 2012 (UTC)

Just as an aside: the slower you accelerate, the more energy you waste to gravity burn. Using the "high school algebra" formulation, as discussed in this excellent NASA page from Glenn Research Center, you need to modify your ideal rocket equation, yielding a loss term proportional to the time of the burn, "-g0*tb" - energy (or, fuel, or, dollars), that is purely wasted and gains absolutely nothing for the final orbit height or velocity. Nimur (talk) 18:00, 28 February 2012 (UTC)

To reach orbital velocity at sea level you'd likely use something like a maglev train in a vacuum tube, which should deal with gravity more efficiently than a rocket. OP should make plots of time vs horizontal g-forces (sqrt(radius of earth * gravitational acceleration) / time) vs distance-to-orbital-velocity (.5 * sqrt(radius of earth * gravitational acceleration) * time^2) to see what's plausible. --81.175.230.91 (talk) 18:44, 28 February 2012 (UTC)

Animal fat fuel

So I was cooking some ground beef the other day, and I drained the fat, and I looked at the slimy mess and I thought to myself, "Is it possible to fuel a car with this stuff?". So is it possible? ScienceApe (talk) 15:34, 28 February 2012 (UTC)

Animal fats can be turned into fatty acid esters (fatty acid methyl esters, in particular) that make suitable biodiesel. Using the fat directly would present practical difficulties. 148.177.1.210 (talk) 15:43, 28 February 2012 (UTC)

Slaughtering with a guillotine

What speaks against slaughtering animals with a guillotine? It seems more reliable, faster and less painful than stunning and cutting the throat. XPPaul (talk) 18:34, 28 February 2012 (UTC)

Probably not faster because it takes time to position the head correctly in the guillotine, adding to the distress the animal feels and increasing the risk of injury by slaughterhouse personnel. Also, the guillotine would have to be checked, cleaned and reset after each operation. Dominus Vobisdu (talk) 18:40, 28 February 2012 (UTC)