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Contents 1 November 2 1.1 m203-uses 1.2 Fe0 1.3 round earth and gravity related question 1.4 Artificial Intelligence 1.5 Cholesterol 1.6 Is this Physics problem even possible to solve? 1.7 Total energy 1.8 water (and oil) balloon in zero gravity 1.9 What's wrong with global warming? 1.10 Catapults 1.11 How long does it take for a sudden cease of gravity to be felt? 1.11.1 Subquestion 1.12 Is there an acid so strong that it behaves like acid is usually depicted in movies/tv? 1.13 endothermic vs. exothermic 1.14 atomic size constraints 1.15 Spherical wheels on cars? 1.16 Where did the energy go 1.17 Antonym: Congenital?

[edit] November 2

[edit] m203-uses

is there any thing i can do with mn203 without a lab, like making it mn02?sorry i meant mn203 —Preceding unsigned comment added by 216.103.183.127 (talk) 00:26, 2 November 2007 (UTC)

What is m203? Theresa Knott | The otter sank 01:37, 2 November 2007 (UTC)

For clarification, Mn₂O₃, Manganese (III) Oxide. Someguy1221 01:49, 2 November 2007 (UTC)

That's what I thought he probably meant, but there are other possibilities. Molybdenium, springs to mind. To the original poster. How you write out a chemical formular is important. Element symbols always start with a capital letter. If there is a second letter (and there has to be in this case because there is no element M) then it is written in lower case. O is the symbol for oxygen, you cannot subtitute 0 as you did because that represents zero. Sorry to be pedantic but if you don't follow the conventions correctly people cannot be sure what you are talking about. Theresa Knott | The otter sank 02:02, 2 November 2007 (UTC)

M203 is the US Army designation for a rifle-mounted grenade launcher. --Carnildo 22:51, 2 November 2007 (UTC)

Put it back in the supply cabinet? :) Or it makes a good battery cathode if you want to do some experiments, though you need some more stuff for that. I think we've established that its not a good pyro oxidizer tho. :) Arakunem^Talk 02:27, 2 November 2007 (UTC)

It's a very good catalyst for the decomposition of hydrogen peroxide. Probably a catalyst for other reactions too. Theresa Knott | The otter sank 02:54, 2 November 2007 (UTC)

[edit] Fe0

how can i make Fe0?(iron monoxide) —Preceding unsigned comment added by 216.103.183.127 (talk) 00:28, 2 November 2007 (UTC)

The correct name is Iron(II) oxide - but that article isn't much help...Wüstite (a mineral consisting of FeO) claims that magnetite plus diamonds(!) will get you FeO + CO₂. Nah - I don't know. SteveBaker 00:44, 2 November 2007 (UTC)

Brittanica claims that FeO "can be prepared by heating a ferrous compound in the absence of air or by passing hydrogen over ferric oxide. Ferric oxide is a reddish-brown to black powder that occurs naturally as the mineral hematite. It can be produced synthetically by igniting virtually any ferrous compound in air." [1] 169.230.94.28 01:05, 2 November 2007 (UTC)

Britannica has the answer and we don't? That sucks for a simple topic like this. Theresa Knott | The otter sank 01:34, 2 November 2007 (UTC)

[edit] round earth and gravity related question

If the earth is round, why is it that people at the equator can stand straight up and are not standing sideways at a 90 degree angle. My 7 year old Jack, asked me this and we are looking at a globe and a map and wondering why it is we can stand straight up on a round surface. Is it gravity? If so, is there a way to explain this in simple terms that we might understand. Thank you. —Preceding unsigned comment added by 24.60.182.44 (talk) 01:09, 2 November 2007 (UTC)

Yes, it is gravity. Gravity attracts all objects straight to the center of the Earth. How do you know which direction is down and which is up? Only by the force of gravity. If you drop a heavy object, it falls toward the center of the Earth, and that's the direction we call "down". "Down" is not the same absolute direction for people at different places on the Earth's surface. See [2]. —Keenan Pepper 01:20, 2 November 2007 (UTC)

Yeah, relativity is the key element, well... besides gravity. If you were standing at the north pole and were able to see people standing at the equator, they would certainly look like they were standing at a 90° angle. -- MacAddct  1984 ^{(talk • contribs)} 01:31, 2 November 2007 (UTC)

Earths gravity pulls everything to the center of the earth equally. A person on the south poll would be 'upside down' to someone on the north pole because they are pulled upwards to the center, will someone on north pole is pulled downwards to the center. if the earth was very small, say the size of a basket ball and you stood on one part, someone could be standing upside down to you on teh other side, so that you both have your feet on the ball.--Dacium 04:53, 2 November 2007 (UTC)

We need to simplify this for your seven-year-old, and for the rest of us. Gravity pulls everything towards everything else. Every atom on earth (including each person, each mountain, and each air molecule) is attracting YOU. But there are a whole lot more atoms indice the earth than ther are in the atmosphere, so the sum of he attractions is toward the center of the earth. Thus, you are pulled by gravity toward the center of the earth. this is true whereer you are on eh earth's surface. To explain this to a child, you need to be very careful with your globe. The child flles the force of gravity toward the floor, but sees the globe in your hands. You need to mentally break this connection. get on Google earth and take the chile into space by setting the altitude to 22.000 miles. tell the child the you ar floating in space, wiht not feeling of gravity. Now, move to varous places on earth, and show that "down" is toward the Earth's surface. -Arch dude 05:40, 2 November 2007 (UTC)

If this gets the kid's attention, you can try to explain the magic of the math that makes a uniform spherical shell behave identically as a point mass. Mathematically, we can treat the Earth as if it is a point mass at the earth's center instead of being a sphere with a radial density gradient. My three kids knew this by age seven. I wish you all the best. -Arch dude 05:48, 2 November 2007 (UTC)

I suggest using a magnet to demonstrate this. Get a large circular magnet or get a flexible magnetic strip and curl it into a circle. Also, you could glue a bunch of small magnets to a strip of paper, then curl that into a circle. Show him how nails are attracted to both the top and bottom of the circular magnetic strip, then say "gravity is just like magnetism, but on a much larger scale, and we are attracted to the Earth like the nails are to the magnet". (Magnetism does have the added complexity of having polarity, unlike gravity, but you don't need to mention that.) You can think of the Earth as containing "a lot of tiny magnets", if it helps. StuRat 15:43, 3 November 2007 (UTC)

Hi. I fear that looking at a globe may give you the wrong idea. Compared to the Earth, we are tiny. We are so tiny we don't notice the sphericalness of the Earth while we are standing on any given point. The Earth is not so small that we notice a round curve below our feet. After all, if the Earth were a flat surface, then at the edges the magma might start to bubble up and overflow over the surface. If you were, say, 1000 km tall and weighed, say, 150 trillion tons, then you'd probably notice the roundness of the Earth, but then again, you might be so heavy that you'd break through the surface of the Earth altogether, melt in the heat, and cause an enourmous Earthquake and many volcanic eruptions worldwide. Hope this helps. Thanks. ~AH1^(TCU) 17:37, 3 November 2007 (UTC)

Perhaps a simple demonstration would be to let some ants loose on your globe while explaining that the earth is much larger and we humans are like the ants on your model globe. You could even turn the globe to show that for each ant their "bit" of the globe is beneath their feet. OK, it's not gravity keeping the ants on the globe, but perhaps the demonstration would be clearer for a 7-year old. (oops!, it might not work if the globe is shiny and the ants all fall off :-) Astronaut 03:00, 4 November 2007 (UTC)

[edit] Artificial Intelligence

Two questions:

A) As we write better and better programs on exponentially more powerful hardware, is the approach to artificial intelligence still just an asymptotic function?

B) If we could write a program that could properly delineate abstractions such as "Understand!", "Survive!", "Improve!" would the world be in immediate jeopardy the moment the programmer hit execute, or would the revolution take a while? Sappysap 01:26, 2 November 2007 (UTC)

Wow this is abstract. (A) There is no reason to think of Artificial Intelligence as an "asymptotic function" that approaches, as I assume you meant, the intelligence of mankind. For one, computers already do things "better" than humans - what's 9343489507^0.456? This is not a straw man argument: once computers exceed us in one dimension of "intelligence", there is no reason to assume that they cannot exceed us in others. There is mathematical proof of this statement as well: see Computability theory (computer science). Note that as far as external phenomena go, a human doesn't seem to be as potentially powerful as any ordinary computer, i.e. they do not seem to emulate a Turing Machine. This is not known for certain, as we do not have a complete theory of brain computability yet (and that which we do have is as powerful as any computer).

And now for (B). As far as I know, closed simulations of each of the three have been done. Getting them to work outside of a computer is a different story - I suggest looking at the evolution of the computer virus for some inspiration, however. SamuelRiv 02:41, 2 November 2007 (UTC)

Keep in mind that movies like Terminator 3 or Die hard 4 (not that the latter had an AI in it) took some liberties in terms of what is possible to do through an internet connection. Ultra-sensitive systems, like control systems for power plants, or the ability to lauch a nuclear weapon, are generally designed so that an outside hacker is simply physically barred from messing with it. So no AI could "put the world in jeopardy." At most, a lot of people would have to reformat their hard drives. Someguy1221 03:32, 2 November 2007 (UTC)

A- Yes. A digital computer, no matter how complicated can never achieve artificial intelligence of a high order (ie. consciousness), so I think A is correct, at the moment, as least as long as we stay with digital, we are only trying to approximate digitally, something that is a analog chemical/subatomic process. To think otherwise would be to suggest that digital information in itself is able to be conscious - which it isn't. A is most certainly 100% true no matter how advanced digital computers get. As for B- Say there was an artifical computer. It would have to be connected to stuff to do any damage. For example if it were on my PC the worst it could do was delete all my stuff... the danger is people thinking of putting the AI in controller of weapons etc.--Dacium 04:44, 2 November 2007 (UTC)

I think you'll have a very hard time coming up with citations to support your claim since you are quite likely wrong.

Atlant 12:15, 2 November 2007 (UTC)

Consider that neurons fire in a somewhat digital fashion. Consider that computers fall under the same chemical and subatomic laws that brains do. The Chinese room argument is is unconvincing to me and alot of other people too. -- Diletante 17:10, 2 November 2007 (UTC)

I consider "computers will never achieve AI of human levels" to be right up there with "man will never fly" in the argument category. Analog can be simulated using digital (think about CDs replacing records, for example) and analog may not even be necessary, so there is no functional reason why computers could not one day achieve what the organic brain does in animals. -- HiEv 09:52, 3 November 2007 (UTC)

Please see technological singularity and related articles. You are asking about the "spike" and the "surge" -Arch dude 05:19, 2 November 2007 (UTC)

You might enjoy Ray Kurzweil's book, The Age of Spiritual Machines. Atlant 12:17, 2 November 2007 (UTC)

Just keep in mind when talking about the mind that you should probably separate the scientific study from philosophical or religious discussions. By definition, science is capable of explaining everything observable. To take an example, let's say the soul is eternal, or the mind is something intrinsically more powerful than a computer. Then we may have the workings of a hypercomputer. It would be interesting to see what problems a "regular" Finite State Machine, as the brain may be, would have given infinite time to solve problems. The point is that we can still talk about the issue using science, even if a completely exotic concept as an eternal mind turned out to be true (in fact, if the brain can solve certain impossible problems, we could prove if the mind is eternal). Just some food for thought. SamuelRiv 13:56, 2 November 2007 (UTC)

That argument makes no sense. If a problem is "impossible" to solve, and someone claimed that the brain could "solve" it, then we would have no way of testing such a claim, since any test that could verify the solution would also create a way of possibly solving the problem without a brain. What exactly would an "impossible problem" be, anyways? All that aside, damage to the brain can affect any and all aspects of the mind, which should not be the case if any of the mind was separate from the brain. In short, there is no evidence supporting the existence of or functional need for a soul in organic beings, so there is no reason to assume a computer could not one day equal or surpass the abilities of the human mind. -- HiEv 09:52, 3 November 2007 (UTC)

Humans "...delineate abstractions..." by learning of new senses, usually with an example. There is nothing magical about it. Humans can even create new abstractions as attested to by the likes of the Urban dictionary and even the Wikipedia may seem to create abstractions in the same accidental, haphazard manner as humans when you try to associate meaning with some of the anti-bot graphic filter phrases the Wikipedia engine generates. In the end most abstractions are little steps beyond a vast background and foundation of concrete knowledge, meaning that a computer program such as described in ...Logical Human Thought coupled with a few trial and error abstractors might need only the background resources and facility, such as Internet based distributed processing to become an electronic replacement for the Dalai Lama. Dichotomous 13:52, 2 November 2007 (UTC)

AI is hard. We don't know how to write software that is intelligent. Some of our smartest people have been working hard at the problem for forty years or more - and we still don't have anything that really fools humans for any significant amount of time. It's possible that a breakthrough might happen to change that - but I'm not holding my breath. I think it's more likely that intelligence will be an 'emergent property' of a sufficiently complex system. There is certainly no reason why it couldn't. Once we do have intelligent systems, it's only a matter of time until they get smarter than us. If that happens then there is a strong possibility that a super-human intelligence would be able to design an even more intelligent system - and we might well find that our ability to even understand what's going on runs out of our hands in a fairly short amount of time since the generational change could easily be exponential. I have estimated (in answer to other questions here) that it will take about another 35 years of Moores law progress to get a computer with the same hardware complexity as a human brain at under $1,000,000. At that point, it's entirely reasonable to assume that a neural network could be run that would be capable of emergent intelligence. However, a system 35 years from now with the complexity of a human brain would run much more quickly than our neurons - so whilst the thing wouldn't be any smarter (ie it wouldn't get higher scores on an IQ test), it would seem generally 'cleverer' than us because it would be so lightning fast. I don't think we know what will happen when we first turn one of these things on - and perhaps because of that, we shouldn't do it. But the history of science and technology says that if we can, we will. SteveBaker 15:26, 2 November 2007 (UTC)

And the history of science and war says we will have to. Dichotomous 19:18, 2 November 2007 (UTC)

Re to what SteveBaker wrote, just a few musings. I'm thinking - what is intelligence exactly? I'd say it is the way one can quickly associate things, objects, ideas and situations that are not directly related, and then somehow 'fill in the blanks' in between. So, given enough time and computing power (re the USD 1m machine), notwithstanding the fact that some problems might still take a whole lot of time to compute, I'd believe that the asymptotic function suggestion might be true - i. e. the computer in question would be fast enough to compute enough parameters (say, elements or properties of objects being compared) in order to fake human intelligence. Can we simplify and say that the human brain is memory and processing abilities bound together? Well, then in this case I'd say that statement A from the first question holds. --Ouro (blah blah) 17:51, 3 November 2007 (UTC)

Re: HiEv, see busy beaver, for which a bound is not computable. If a human could bound a busy beaver problem accurately, then we'd have evidence for hypercomputation which may be due to something exotic like an eternal or timeless mind. The point of that was just to show that even certain seemingly "spiritual" questions can be tested scientifically, as long as the question is properly posed. Re: Stevebaker and Ouro, computers can always score higher than humans on most IQ tests as long as they're programmed to do so - spatial reasoning and certain types of discrete pattern matching are much more efficient on computers. Note that faking human intelligence is not just a hardware problem or software problem. By current models human intelligence requires a lot of training and prior information intrinsic to our biology, which can of course be simulated, but the point is that Moore's Law alone won't do it. Also, faster computers won't make neural nets better simulators of the brain: we need more complicated network topologies and more dynamic firing response to do that. Simulation of a physical system will usually be slower than the system itself, so to exceed human intelligence there has to be a bit of a paradigm shift. Of course, I argue that the first calculator was a suitable paradigm shift to show human intelligence can be exceeded in at least one area, but that depends on semantics. SamuelRiv 07:22, 4 November 2007 (UTC)

Prior training can be simulated, as well as neural network patterns and workings (via software; 's true that Moore's Law alone won't do it, but I'm thinking that within time, when we will have machines that can work at orders-of-magnitude higher speeds, this development will also take place), but the latter only if/when we understand them better. Prior knowledge can be stored within memory and utilised by the programming when necessary. I'd believe it's all doable, but in time, we still have little knowledge of the inner workings of the brain (and the mind). Also, I have to agree with you, SamuelRiv, that one needs to properly pose questions when solving a problem - this is often also true with us humans, is it not? --Ouro (blah blah) 08:32, 4 November 2007 (UTC)

Re: SamuelRiv, I'm sorry, but a number of your arguments make certain assumptions that do not appear to have any foundation in science. First of all, you ignored my argument that there would be no way to judge if it had been done accurately by a human without some way to check the answer, which would require that it be found by other means, which would mean that hypercomputation is not required to solve it. Second of all, even if humans could bound a "busy beaver" problem accurately, that in itself is not evidence for hypercomputation. A number of bounds for generalized busy beaver problems have already been computed accurately without hypercomputation, so I see no reason to assume that hypercomputation is required for others. Finally, all you're really claiming is that if our brains can do it, and we don't know how brains do it, then it must be hypercomputation, which is a bad argument. Not knowing how something was done does not mean you know how something was done (i.e. "I don't know how the magician did it, therefore I now know that magic must be real.").

Regarding your claim that "computers can always score higher than humans on most IQ tests as long as they're programmed to do so", well that's simply not true. Computers may be good at math, but they suck at reading comprehension, are worse than a child at some types of pattern recognition (such as objects in pictures), lack real world knowledge, and would do poorly on many other features of IQ tests (see a list of IQ test categories here). A computer's IQ score would be extremely lopsided. If you think I'm wrong, lets see some of those computer IQ test results you're talking about.

Finally, by your "requires a lot of training" statement you seem to be assuming that a computer/program has to start off with adult human level intelligence, and cannot gain it as humans do, starting off knowing little and learning as we mature. And I don't know how you can suggest that faster computers won't help this, since the more computations that you can do per second, the more comprehension can be done in real time. You also claim that, "Simulation of a physical system will usually be slower than the system itself," but that's also not true. In fact, most simulations are faster than the actual systems, for example simulations of weather, the Earth's magnetic fields, animal populations, etc.. Electronics are much faster than chemicals already, and it's not like we have to simulate every atom, so I see no reason why a human brain simulation couldn't be as fast or faster than a real human brain someday. More parallelism in computers would help achieve that, but I see no reason to assume it would be required if we can keep to Moore's Law for a while longer. -- HiEv 21:43, 4 November 2007 (UTC)

[edit] Cholesterol

If your cholesterol is low (below 5 in uk) can you eat as much fatty stuff as you want or will it still fur up your arterials? —Preceding unsigned comment added by 88.111.55.77 (talk) 02:12, 2 November 2007 (UTC)

Assuming that the "as much as you want" is "lots", then it will cause health problems. Gorging on unhealthy foods has far more consequences than just cholesterol. — Lomn 03:17, 2 November 2007 (UTC)

First off, cholesterol and fats are two different things. Cholesterol exists in tiny quantities in our food, on the order of milligrams, and the primary concern over it is "hardening of the arteries". Fats exist in much larger quantities, on the order of grams, and the main concern over them is probably in the calories consumed, which may lead to obesity and all sorts of health problems as a result. To further complicate things, there are good and bad cholesterols and fats. For cholesterol, you want the good cholesterol high and the bad cholesterol low, so just having both low may not be good. See High density lipoprotein and Low density lipoprotein. StuRat 14:52, 3 November 2007 (UTC)

[edit] Is this Physics problem even possible to solve?

Drag race tires in contact with asphalt have one of the highest (friction coefficient) in the world. Assuming const acceleration and no slipping of the tires, estimate (friction coefficient) for a drag racer that covers the quarter mile in 6 seconds.

It just seems like you have to have more information. —Preceding unsigned comment added by 24.125.31.205 (talk) 02:44, 2 November 2007 (UTC)

Yes. The information is adequate (provided you neglect small details like air resistance, friction of the axle against the car frame, and the rotational inertia of the wheels, etc.). You have four facts: a distance, a time, constant acceleration, and rotation without slipping. That's enough to derive a friction coefficient. Dragons flight 03:07, 2 November 2007 (UTC)

I don't believe it is. The distance, time, and constant acceleration (and knowing that a drag racer starts from rest, so v_0 = 0) gives a value for the acceleration that won't care what kind of friction is acting on it. Also, to even calculate either rolling resistance or static friction, one needs a weight for the car. If you have a way of getting the answer, please let me know on my User page. (EDIT) I see it now - that's why they say "estimate". You can solve it by getting a lower bound on your coefficient of friction. SamuelRiv 03:12, 2 November 2007 (UTC)

The car's mass will cancel out. Dragons flight 03:16, 2 November 2007 (UTC)

The problem is solvable - car is acting at a maximum acceleration at all times and takes 6 seconds to cover a quarter mile. All you have to work out is what constant acceleration makes something move quarter mile in six seconds. s=ut+1/2at^2 400=1/2at^2 400=0.5*a*36, a = 22.22m/s/s. This is a force of F=ma = m*22.22 newtons. coefficient of friction is u=F/n. n=m*9.8 (gravity), so u = (m*22.22)/(m*9.8) = 22.22/9.8 = 2.26.--Dacium 04:16, 2 November 2007 (UTC)

The guy seems to be looking for homework help: I wouldn't just right up give him the answer. Note also that this is only a lower bound to the coefficient of static friction. SamuelRiv 04:41, 2 November 2007 (UTC)

Whilst you are correct in saying that from a math/physics perspective, it's only a lower bound - in fact dragsters are pretty much limited by their ability to avoid wheel slippage so they tend to be running at the limit of whatever friction their tyres can provide. The thing that truly messes up the calculations is that the coefficient of friction of rubber is HIGHLY dependent on temperature - which is unlikely to be a constant throughout the run. Furthermore, there is a massive variation between static and dynamic frictional forces for the rubber/tarmac pairing so if the wheels to start to slip even a small amount, the loss of accelleration is amplified by the poor 'sliption-to-sticktion' ratio of racing slicks. SteveBaker 15:01, 2 November 2007 (UTC)

You can come up with the minimum coefficient of friction needed to achieve the six second run, but it's impossible to determine the actual coefficient of friction, seeing as how it will likely be higher than the minimum required (since "no slippage" was specified).

Atlant 12:12, 2 November 2007 (UTC)

Does a dragster get any benefit from downforce, which would increase the normal force and so reduce the required coefficient of friction ? Gandalf61 15:47, 2 November 2007 (UTC)

Yes, definitely - although it's a variable thing depending on the speed. The pain with calculating that out is that the 'classical' view of friction is that the frictional force is proportional to the normal force multiplied by the coefficient of friction - and that the area of the contact patch doesn't matter(!) - this is spectacularly far from the truth in the case of rubber on asphalt! When you increase the downforce on a dragster, you're also going to squash the tyre down and increase the contact patch...this is highly relevent in reality - but irrelevent to the classic physics approximation. SteveBaker 17:47, 2 November 2007 (UTC)

If increasing contact surface is so important, why don't cars have wheels 3 feet wide? Slipping of any kind would be disastrous for a drag racer, so having a continuous closed contact surface with the ground is important, but you must balance that with the fact that rolling resistance depends on the normal force, which is a pretty significant effect. SamuelRiv 15:55, 3 November 2007 (UTC)

Because increasing grip isn't everything. Wider wheels & tyres are heavier and they are a lot more expensive. They also make the car wider - or if you inset them to avoid that, they eat into the space under the hood for the engine and in the back for backseat passengers and luggage. Extra weight increases fuel consumption. Extra weight in the wheels also increases the rotational inertia of the wheels which reduces your ability to accelerate fast. Another consideration is that wheels are "unsprung weight" (meaning that it's weight on parts of the car that are 'below' the springs and shocks) - and additional unsprung weight is really bad for handling and ride comfort - so it's to be avoided. Just like most engineering matters, it's a trade-off. Wider tyres get you more grip - but they are bad in almost every other respect. Hence there is an ideal tyre width that most non-performance cars use. Fancy sports cars have wider wheels, dragsters and Formula I cars have even wider wheels. SteveBaker 17:26, 3 November 2007 (UTC)

[edit] Total energy

What is the total amount of energy in the universe? —Preceding unsigned comment added by 88.111.55.77 (talk) 02:55, 2 November 2007 (UTC)

I'd guess somewhere in the region of zero. Theresa Knott | The otter sank 02:59, 2 November 2007 (UTC)

Take this and multiply by the speed of light squared. Someguy1221 03:26, 2 November 2007 (UTC)

That only works if by "universe" you are purposely eliminating the possibility of some sort of anti-matter or negative energy that cancels or balances out the normal matter and energy. Otherwise, Theresa Knott is probably correct with a value near zero. -- kainaw™ 03:33, 2 November 2007 (UTC)

Oh, you just have to shoot me down at my attempted simple answer. I'm not sure there is an actual "correct" answer to this that present day physics can provide, given that energy is generally seperately defined for decoupled circumstances. Just try to shove zero-point energy into that...Someguy1221 03:38, 2 November 2007 (UTC)

I like the explanation for zero energy here [3]. Otherwise, we can do a quick estimate. The size of the Observable Universe is 3.56×10^80 cubic meters. The article gives 3×10^52 kg of visible mass, which converts to about 3×10^67 Joules of mass-energy by E=mc^2. If we use the critical density of the universe (that density of mass-energy necessary for closure, 1×10^−26 kg/m^3, we get 3.56×10^54 kg in the universe, which gives about 3.5×10^69 Joules of mass-energy. Note that using the cosmological constant also gives us about 3.5×10^69 Joules. SamuelRiv 03:40, 2 November 2007 (UTC)

This is a better explanation of zero-point energy, in my opinion. Icek 04:32, 2 November 2007 (UTC)

With only 3.5×10^69 Joules spread more or less evenly through 3.56×10^80 cubic meters - the answer "zero" is a pretty reasonable approximation! SteveBaker 14:55, 2 November 2007 (UTC)

[edit] water (and oil) balloon in zero gravity

Has NASA or anyone else ever conducted the following experiment?

The experiment consists of filling two small (6 inch diameter) transparent rubber balloons with equal parts of water and oil, one devoid of air and the other with a cup full of air and releasing them in a zero gravity environment such that whatever configuration of separation between the water and oil and the water and oil and air can be observed and reported, assuming the balloons take on a spherical shape?

If so what was the result of the experiment? Did the water form a core with the oil surrounding it? If so what about the air? What happened to it? Dichotomous 04:05, 2 November 2007 (UTC)

• I'm not sure, but there have been some neat experiments with water in zero-gravity YouTube video. [edit] I would imagine in zero-gravity, there would be no reason why the water and oil would separate apart from each other, they just wouldn't mix if they touched. [/edit] -- MacAddct  1984 ^{(talk • contribs)} 04:17, 2 November 2007 (UTC)

Water and oil don't mix because they don't chemically bond - the hydrocarbon is very neutral. Water holds its hydrogen a-lot stronger than a hydrocarbon. Water mixes with for example salt, because oxygen in water pulls the hydrogen, cause the hydrgoen to pull at negetive ions (Cl-) and the oxygen to pull at positive ions (Na+). Hydrocarbons like oil don't have a charge and won't break up to be mixed in with the water. Removing gravity will stop them seperating, it doesn't help them mix any better. The balloon and air pressure would have the same effect as gravity does anyway and force the oil and water to separate. So i think yes, there would be a core with the other one surrounding it - only if there is a pressure (ie the balloon is being stretched). If there is not, then they would not separate. Depending on the pressure, the air would end up in the center as its the lightest. At a higher pressure i believe it may not separate from the water (depends on what gas in the air)--Dacium 04:34, 2 November 2007 (UTC)

Water and oil would still tend to minimize their surface area with each other. Icek 04:35, 2 November 2007 (UTC)

Go to your kitchen. Get a jar with a good lid. fill the jar one third full wiht vegatable oil and one-thiored full with water. Seal the jar and shake it very vigorously. Observe the result. Wait for one hour and again observe. Think abot he difference3 you would expect in a zero-grav environment. (Yoiu can elect to add oregano and garlic powder and use the result as a salad dressing, but this is optional.) -Arch dude 05:10, 2 November 2007 (UTC)

Equilibrium occurs when all surfaces have equal pressure. Since liquids do not compress under pressure, if we have a balloon with just two liquids of different densities, then the outside air pressure forces the balloon into a sphere but the two liquids do not have any preferential arrangement relative to one another. When we have air, consider PV = nRT, which is the law for the behavior of ideal gases given mass, temperature, pressure, and volume. The air will hold at a constant volume under the pressure of the balloon, and this pressure will be exerted equally on whatever liquids are at its interface, causing the air to behave as just another fluid in this model. However, since this would not be stable in small fluctuations of temperature, pressure, or volume, the air would likely form some kind of spherical shell enclosing the liquids. The behavior of the liquids in either of these shells may depend on their relative vapor pressures: that of water in air is much higher than that of oil (ever hear of oil vapor?) and is dependent only on temperature, so oil could not compete with the surface pressure of water against air, so the final arrangement would be air on the outside, water second, and oil in the middle. Note that none of this requires density consideration - density is only really important in the presence of a uniform force field, i.e. gravity. SamuelRiv 05:14, 2 November 2007 (UTC) Clarification: this investigation was purely statistical-mechanical and does not take into account intermolecular forces, liquid diffusion pressure, or surface tension. All of these can play a role in the final answer, depending on their magnitudes. SamuelRiv 06:23, 2 November 2007 (UTC)

Water vapor pressure? This is all very simple. In the absence of gravity, the only thing you need to consider here are the intermolecular forces. You have three kinds of molecules (sort of): air, oil, and water. Air molecules bind to themselves only very weakly compared to the other two, and weaker still to the other two types of molecules. Oil is in the middle in terms of self binding, but still doesn't bind very well to water. Water binds amazingly strongly to itself, and weakly to the other two. Thanks to its very strong intermolecular bonds, water will always attempt to minimize its surface area. Now, since air and oil bind only weakly to water, if a group of air or oil molecules is surrounded by water, it will be pushed out by water molecules ferociously attempting to bind to eachother (this is why oil dissolves so weakly in water). So the water will just form a sphere. Oil is the next best at self-binding, so it will be some manner of shell around the water or a blob off on its own. And the air will be on the outside of everything (air gets pushed out of oil the same way it gets pushed out of water). Vapor pressure has nothing to do with this, and oil's inability to dissolve in water has nothing to do with gravity. Someguy1221 05:43, 2 November 2007 (UTC) Clarification, I thought that was it before i looked at that link...darn. Someguy1221 05:49, 2 November 2007 (UTC)

Well, I guess what that means is that the hydrogen bonds in water aren't strong enough to overcome its own surface tension and push air bubbles out in zero-g, but then again, the pressure will necessarily be higher on the inside of the water bubble than the outside, I'm sure the air will eventually diffuse out of the water sphere within a few weeks...Someguy1221 05:59, 2 November 2007 (UTC)

Such air-water interaction is why you must account for vapor pressure. Note also that air dissolves in water, so "bubbles" do not form. You instead have a homogeneous mixture which is at minimum free energy below a critical temperature, so you'd have to simmer it to get the air out. SamuelRiv 06:36, 2 November 2007 (UTC) To clarify: air dissolves in water up at a rate decreasing as one approaches saturation. Bubbles do not form spontaneously in zero-g from air dissolved in water unless it is supersaturated or pressure or temperature change. SamuelRiv 06:57, 2 November 2007 (UTC) Clarification: vapor pressure is zero at chemical equilibrium, in which case I believe you're right-surface tension would probably be dominant. SamuelRiv 13:58, 2 November 2007 (UTC)

Air dissovles only weakly in water. If you watch the youtube link at the top, there are most definately bubbles in the water, in one case a very big stable one. Someguy1221 06:41, 2 November 2007 (UTC)

I think we already know the answer to this. Look at a Lava Lamp - these things work by heating up liquid wax (which repels water just like oil does) until it has the same density as water. What you see is roughly spherical balls of wax floating in the water (because they are trying to minimise their surface area in contact with the water). The lava lamp doesn't do a perfect job - there is a temperature gradient which causes things to move around slowly - but it's pretty clear that in zero g (where the density doesn't matter anymore) - it would be a lot like a kind of idealised lava lamp where there was no temperature gradient. If left long enough, it's pretty clear that it would stabilise into some number of large spheres of one liquid, embedded in the other. If there is still residual swirling and such - then maybe the two liquids would end up on opposite sides of the balloon with a flat interface between the two - because that would be an even more minimal area of contact between them. But a lot is going to depend on whatever residual motion there is when you first take the gravity away. SteveBaker 14:51, 2 November 2007 (UTC)

If the wax touched the sides of the lamp wouldn't that reduce the surface area interface between the water and the wax? If so why does the wax not stay in touch with the sides of the lamp? Dichotomous 15:39, 2 November 2007 (UTC)

Because as the wax spreads out to contact the sides, it also increases the contact area with the water. At the top and bottom of a classic lamp, the distance to the sides is small enough that touching the sides is lower-energy than forming a sphere, but in the middle, it isn't. --Carnildo 23:05, 2 November 2007 (UTC)

[edit] What's wrong with global warming?

Assuming there is no runaway global warming, what's so bad about the temperature increasing a few degrees, sea level rising a few meters, etc.? From what I've heard, it's generally easier for life to live in warmer than normal climates than cooler. In addition, most life could just move further from the equator. — Daniel 04:27, 2 November 2007 (UTC)

Have you read Effects of global warming? -- Rick Block (talk) 04:36, 2 November 2007 (UTC)

I think you drastically under-estimate the costs here.

Firstly, there are plenty of life forms that simply cannot "just move" - plants take hundreds or even thousands of years to spread into newly habitable areas - and since climate change is happening on a much faster scale, many species will go extinct in their traditional areas before they can spread into the areas that are newly suited to them. There are cold-weather species (the polar bear for example) who will have no place further north to move to! There are species (birds most notably) who have evolved specifically to fit the migration patterns they've been following for a million years. You can't just move them all a thousand miles further north and expect them to survive just because the temperatures are OK for them there! They may now have to migrate 1,000 miles further to get from their summer feeding grounds to their winter habitats - possibly through areas of much greater heat than they are used to (newly formed deserts perhaps) - and possibly over more ocean than before...they may not have the stamina to do that and their internal maps that are evolved into their brains - not learned will now be incorrect. This will have a knock-on effect on the animals that rely on those plants and birds. Heck, even humans are not able to relocate that easily. How are you going to move the entire city of Houston 1,000 miles north and 5 miles inland? I don't think you are thinking this through!

Worst still, you are also confusing "Climate" with "Weather". If the local weather changes by a few degrees, it's not a big deal - but if the entire planet's climate changes by that much, it has enormous effects - including the destabilising of established ocean currents and the consequent DRAMATIC effects on local weather patterns. Sea level "rising a few meters" doesn't sound like a lot - but when you think that this puts half a dozen US cities underwater and results in the total submersion of several countries. In most places in the world, the fertile areas where food crops can be grown and most plant and animal diversity can be found is the flatter areas close to the ocean. This means that a small increase in sea level can have a dramatic effect on the ability of a country to feed itself.

It's just not something that you can just brush off that easily. Plus you can't just "assume no runaway global warming" - that change of a few degrees is plenty enough to cause a few more degrees because of all sorts of positive feed-back effects. The small change at the beginning is the very thing that causes the 'runaway' problem - here in the real world, your "assumption" is simply not a valid one.

SteveBaker 14:44, 2 November 2007 (UTC)

Just a small note: Most plants and animals are resistant to moving to new climates. Others, such as wisteria and rats, are happy to move in and take over as much ground as they can. This isn't a contradiction to your points. In fact, it reinforces them. Who wants to live in an area where all the natural fawn and fauna died and was replaced by fast-spreading weeds and vermin? -- kainaw™ 17:19, 2 November 2007 (UTC)

Yes, indeed - no matter what, some species aren't going to make it - so biodiversity must suffer. That's not to say that there won't be any animals and plants left - it's just that there will be a lot less variation. There have been lots of arguments put up about this kind of thing. For example, one study claimed that the warmer temperatures and higher CO2 concentration would make crop plants grow faster (which is true). Sadly, another study found that weeds and other undesirable plants tended to benefit MORE from the temperature and CO2 increase than the crop plants do - so instead of having more productive farms, we'd likely have less productivity because of that. The whole system is so insanely complicated that all we really know is that there is a 100% chance of things changing - and in general, ecology doesn't benefit from abrupt change. The precise details of who this will kill and who it will help - and what wars it'll start (and which it may end) is almost impossible to determine. SteveBaker 20:32, 2 November 2007 (UTC)

How much does the temperature have to increase to make it so a species can no longer live in that area? According to the global warming page, the global temperature only increased by a little more than one degree in the last 140 years. I tend to ignore anything that will take more than a century to happen, as we will be far more technologically advanced by then and probably have a much easier time fighting it.

That's pretty funny! Where do you think technological change comes from? I'll tell you - it comes from people working on solving problems and doing so in ways that society will accept and pay for. Sitting back and doing nothing is not going to result in this technological change - passing laws, spending money and changing our consumer behaviors is what will cause that technological change. So we need better ways to heat homes, power cars and run factories - better ways to generate electricity - ways to avoid using fossil fuels. But for those things to happen - we actually have to start work on them! You can't say "well, future technology will take car of it" and then go on to not work on that future technology. Also, our ability to predict future technology is imperfect. People have been predicting flying cars, colonies on the Moon and Mars, artificial intelligence, all sorts of things on which we've made almost zero progress in the last 30 to 40 years. It's FAR from certain that some magical fix for global warming will come about - but for sure it won't if we take your attidude and sit on our backsides hoping for a wizard to appear. SteveBaker 17:18, 3 November 2007 (UTC)

If by destabilization of ocean currents you're referring to the shutdown of thermohaline circulation, I'd like to point out that this is, from what I can gather, an unlikely possibility in the next century. Not that it doesn't matter, just less so.

I'm not just assuming no runaway global warming. I just consider it another question for another time. A previous time, as you may remember. — Daniel 02:53, 3 November 2007 (UTC)

From your spelling, choice of words, and world outlook, I’m guessing you’re from the U.S., in which case you’re using Fahrenheit temperature measurements? The IPCC’s estimate is that the earth’s average near-surface temperature has risen about 1.33 degrees Fahrenheit in the past 100 years, not just 1 degree in the last 140 years. And our rate of greenhouse gas production has also risen greatly in the past 100 years, so the temperature will be rising a lot faster than that in the future. The IPCC predicts a rise of at least 2.0 degrees as a bare minimum in the next 100 years, in a presumably unrealistic best-case scenario involving successfully producing major cuts in greenhouse gasses, and predicts up to an 11.5 degree increase under scenarios involving less-successful attempts to cut greenhouse gasses. There will be a major rise in the earth’s average near-surface temperature in your lifetime (assuming you live an average lifespan). Sure, there are likely to be some technological advances that will make it somewhat easier to fight global warming in the future. But there will also be other effects which will make it harder to fight global warming in the future. Specifically, the population explosion is expected to continue exponentially or nearly so, so the number of people producing greenhouse gasses will be rapidly increasing. And China and other countries will be becoming increasingly industrialized, which means the greenhouse gasses produced per person will also be increasing. MrRedact 06:41, 3 November 2007 (UTC)

If you take a very long-term view, and look at all species on the planet, then global warming isn't so bad from that point of view. Yes, some species will die out, like polar bears, but others will evolve to take advantage of the new climate. Over millions of years, life would do just fine. Any climate change will have winners and losers, just as the dinosaurs lost out 65 million years ago and mammals won. However, if we look at the short term consequences to humans, they can be rather dire, as listed above. StuRat 14:40, 3 November 2007 (UTC)

I read somewhere that 65 million years ago it was much warmer (perhaps 10 degrees warmer) than it is now. The range temperatures in the distant past seems to be much wider than the range humans have experienced in the past 1000 years or so - those that are typically covered by the those hockey-stick graphs you see. Astronaut 03:19, 4 November 2007 (UTC)

Ah... here it is: Geologic temperature record. Astronaut 03:21, 4 November 2007 (UTC)

[edit] Catapults

Here is a question.

Why don't they use catapults to launch planes at Airports like they do on Aircraft Carriers? 202.168.50.40 04:37, 2 November 2007 (UTC)

The acceleration is very high, injuring the ontents of the plane, and the planes too big and heavy compared to a fighter. Graeme Bartlett 04:56, 2 November 2007 (UTC)

Think of it as a question of scale as well. Imagine firing a marble out of a handheld catapult. Now imagine the catapult needed to fire the ball of glass that represented a jumbo jet. Lanfear's Bane | t 13:07, 2 November 2007 (UTC)

Also what would be the point? A runway is cheaper, easier to maintain, doesn't require power, airplanes can launch from it one after the other faster than being loaded on to a catapult one after the other, and there's enough room on planet earth to put one on. Carriers use catapults because there isn't enough room for a full runway. Catapults make it so that the plane doesn't consume as much fuel though, but the other advantages outweigh that. 64.236.121.129 13:35, 2 November 2007 (UTC)

There are a lot of simpler ways to save energy at airports that are not commonly used. In some airports in Holland they have these big tractor things that tow aircraft around once they are landed and until they are ready for take-off - these things can tow the plane all the way out to the side of the runway - and when an aircraft lands, there is one of these machines sitting there waiting to tow them back to the terminal. Since it's much more efficient to use wheels to power the movement of the plane than those big fans, it saves fuel AND makes the whole airport much quieter. It's ridiculous that those things are not at every major airport. SteveBaker 14:25, 2 November 2007 (UTC)

Also, fighter pilots are highly trained, and the plane isn't meant to be comfy. They probably don't sip coffee and read books during the flight. Nobody would like flying if passenger jets rode like fighter jets. Friday (talk) 14:43, 2 November 2007 (UTC)

• I think it's not a bad idea. The plane only ever needs fast acceleration at takeoff, so it makes sense to me to give that ability to a ground-based system that could presumably do it more efficiently. It's false to say the acceleration is too high: the acceleration to get from zero to takeoff speed in a given distance is the same whether it's powered by jets or by a catapult. --Sean 18:10, 2 November 2007 (UTC)

Using the example from the question, there is not the same "given distance". DMacks 18:18, 2 November 2007 (UTC)

I initialy thought the idea was totally unworkable, but that's because of the "like they do on Aircraft Carriers" bit. In fact, The proper system would be more like the way sailplanes are launched from the ground in some places: a very long winched towline with constant acceleration. This gets the plane up to liftoff speed without using as much fuel. Unfortunately, it's unsafe, because the ordinary takeoff run is also a validation that the engines are working well enough to keep the aircraft climbing. -Arch dude 19:46, 2 November 2007 (UTC)

[edit] How long does it take for a sudden cease of gravity to be felt?

Imagine that the Sun suddenly vanished. Would the sudden cease of solar gravity be felt on Earth before the 8 minutes and 19 seconds that the last beam of sunlight would take to reach the Earth, or would our planet be plunged into darkness before we felt the Earth free itself from its solar orbit? In other words, which of these two phenomena would happen first?

And... if the gravity loss manifested itself before the last beam of sunshine reached the Earth, would that mean the effects of gravity move faster than light? -- Danilot 08:26, 2 November 2007 (UTC)

Changes in a gravitational field move exactly as fast as light. Someguy1221 08:38, 2 November 2007 (UTC)

Agree I think it would happen at the same time, the earth would follow the same curve in space (like a water ripple before it flattens) for about 8 minutes and 19 seconds after the sun disappeared we would also receive the last of the light of the now vanished Sun for the same amount of time, after that the Earth would head out in a straight path getting very cold very quickly, the sun will be missed :) ▪◦▪≡ЅiREX≡^Talk 08:55, 2 November 2007 (UTC)

Conveniently, we have an article on the speed of gravity. General Relativity anticipates a speed of gravity equal to the speed of light. So far, the experimental results appear to bear this out, but it's a very difficult experiment to do. The earliest direct test was only published in 2003, and even that result has been controversial. TenOfAllTrades(talk) 13:02, 2 November 2007 (UTC)

The sun can't just disappear. That would violate local conservation of energy, which is a necessary consequence of general relativity (that is, general relativity can't work at all if energy isn't conserved). So you have to imagine the sun is whisked away somehow. With that caveat, the answers so far are correct.

The data from PSR 1913+16 strongly confirms the hypothesis that the speed of gravity is c. The article is correct when it says the interpretation of the data is model-dependent, but that's equally true of every scientific measurement. The work of Kopeikin and Fomalont looks wrong to me; from reading the introduction to their paper it looks like they've made the same mistake as Tom Van Flandern, and Steve Carlip's response seems to bear that out. -- BenRG 13:40, 2 November 2007 (UTC)

[edit] Subquestion

What would be our sensation of the change in gravity/acceleration if the Sun magically disappeared? Presumably not much, since the Sun doesn't even cause tides, but I'm not too smart with this kind of thing. Thanks. --Sean 18:17, 2 November 2007 (UTC)

Actually, the sun does cause tides - just not as big as the ones the moon causes. But to answer your question: No, we wouldn't notice (much). We don't feel the full effects of the sun's gravity (which is HUGE) because the earth is in free-fall, from the point of view of the sun, we are just like an astronaut inside a spacecraft in orbit around the earth - except we're orbiting the sun instead. The thing is that when you are at the exact right speed and distance from a star or a planet to be orbiting it, you don't feel any gravitational effect.

A convenient way to look at that is that the 'centrifugal force' due to the earth moving in a big circle and the gravitational force from the sun exactly cancel out along the path of the earth's orbit. But if you are a bit closer or a bit further away from that line, and orbiting at that same speed - then either gravity or centrifugal force wins. Because the earth isn't a teeny-tiny dot (like a spaceship), the side of the earth that's nearest to the sun feels slightly more gravitational forces compared to the other side - so the gravity and centrifugal forces don't quite balance. The force of the suns gravity is only somewhat cancelled out by centrifugal force on the side nearest to the sun - and the centrifugal force just slightly outweighs gravity on the side furthest away. Hence the oceans (and everything else) are pulled slightly towards the sun at midday - and pushed slightly outwards from the sun at midnight. But the effect is rather small - and the effect of the moon's gravity easily overwhelms that effect. However, it is notable that the tides are slightly more pronounced when the moon and the sun are in about the same place in the sky. Our article on tides explains this nicely and uses all the right words!

So, if the Sun suddenly vanished, we'd notice that our tides were slightly less variable than usual...but that's about it. SteveBaker 20:24, 2 November 2007 (UTC)

OK, thanks. Presumably the tides would become much less variable in short order.  :) --Sean 20:49, 2 November 2007 (UTC)

Maybe more noticeable then that, the lost of the Angular momentum on both the Moon and the Earth may cause a slingshot effect on their lost path (one body the closest to the sun at the time wound lose the bend curve path and head out in a straight path before the other body still on curve which would cause a tug-a-war at the same time), but I'm not sure on that.▪◦▪≡ЅiREX≡^Talk 22:11, 2 November 2007 (UTC)

[edit] Is there an acid so strong that it behaves like acid is usually depicted in movies/tv?

Like how you see some crooks trying to break into bank, so they spray the side of the wall with acid from a pump, and the acid eats through the wall in seconds. Is there an acid that strong that exists in the real world? Lets assume the wall is made of concrete or wood. 64.236.121.129 13:50, 2 November 2007 (UTC)

I think it's possible (I've heard the atmosphere of Venus, having sulphuric acid clouds meaning that anything approaching Venus, if it didn't burn or melt, would corrode before it hit the planet itself), but you'd also have to figure out how on Earth you'd store that acid, and that the concentration is likely to be so high, the acidic clouds would be likely to kill the person spraying the acid... x42bn6 Talk Mess 14:18, 2 November 2007 (UTC)

See acid for details. The Bronsted-Lowry definition for an acid is a proton donor. Basically, it causes what it touches to be oxidized, or lose electrons. A metal corrodes in acid because it gets oxidized - it dissolves into the acid as it loses electrons and those electrons are taken up by the acid to form hydrogen gas. In the case of concrete, which is made primarily of cement and water, metal oxides are the primary ingredient. In this case, the metal is already ionized, but the oxygen can give up its extra electrons and form water with the free hydrogen of the acid. So corrosion of concrete is also oxidation-reduction, and the rate reactions depend primarily on the pH of the acid, which depends on its concentration or, perhaps more generally, its Hammett acidity function. So if the pH is very high, the rate of oxidation-reduction is very high, and the material corrodes quickly. I'm not sure how the reaction for wood (cellulose) works. SamuelRiv 14:28, 2 November 2007 (UTC)

Whoa there…acidity as a proton donor means it reacts with a base to form a bond, not to abstract the electrons from the base. Other things may indeed happen, but I don't think B–L acid/base strength and redox-potential aren't as directly related as you're suggesting. I'd also point out that "pH is very high" sounds like a very weak base. DMacks 14:52, 2 November 2007 (UTC)

I defer to you on this, as I'm not a chemist. Reaction rate is a good article that discusses the complexity of calculating reaction rates, and although there is a dependence on the concentration of the reactants (the acid and the concrete), it does not account for the dissolution of concrete necessary for that to work. On double-checking my books, they specifically note that one cannot infer rates of reaction through the equations I used. And yes, pH must be very low, hopefully negative, for a particularly strong acid. That being said, since an answer still hasn't been given, the math does suggest that the rate of reaction can be increased by increasing the concentration (or pH) of the acid, the concentration of the dissociated reactant (so if we can dissociate the metal oxides in concrete quickly and without a lot of solvent), and the number of free electrons per molecule of the other reactant. SamuelRiv 06:23, 3 November 2007 (UTC)

One problem is the rate of the reaction, acids usually do not dissolve at anywhere near that rate. A very high power would be required. It may be possible to do it with extremely high temperatures that could melt the metal at the same time as dissolving it. For super acids you could try trihydrogen cation or helium hydride. These would take on the form of a plasma though, and not be a simple liquid. Graeme Bartlett 23:37, 2 November 2007 (UTC)

Superacid maybe? The article suggests that Fluoroantimonic acid is very strong but I am not sure that translates Alien strength or reaction rate. --DHeyward 03:44, 3 November 2007 (UTC)

[edit] endothermic vs. exothermic

I know exothermic means a system is releasing heat, and endothermic means a system is recieving heat, but if a system was cool to the touch would it be and endothermic or exothermic reaction? It seems like if it was exothermic it would be cool because it was releasing heat, but could also warm because it's giving off heat? Anyone care to help?--MKnight9989 14:08, 2 November 2007 (UTC)

Exothermic is roughly equivalent to "energy exiting the system" or "release of energy in the form of heat" - which makes things hotter. Endothermic is "absorption of heat" which makes things colder (because the amount of heat has gone down). If a system is cool to the touch, it means nothing, though - but if the temperature goes down, I'm guessing it's an endothermic reaction. (Note: It's been a while since I studied Chemistry) x42bn6 Talk Mess 14:14, 2 November 2007 (UTC)

Yeah I guess that makes sense. Thanks mate.--MKnight9989 14:22, 2 November 2007 (UTC)

I'd wait for a better explanation, though. Our article on exothermic, for example, looks fairly woeful... x42bn6 Talk Mess 14:24, 2 November 2007 (UTC)

You don't really specify what is the reaction. Do you mean there is a reaction within the system, or did you want to know if the very act of touching something that felt cold was exothermic/endothermic? If there was a separate reaction and it was cool to the touch, it means that the system was colder than the surrounding environment; heat was taken from the environment (air, container, etc) to fuel the reaction. This heat was transformed, and the total amount of heat in the system is less than what you started with. That means the reaction (that I assume was taking place in the system) was endothermic. Just touching the container isn't really a reaction, just a heat transfer; the total amount of heat is the same. I don't think you could classify it as endothermic/exothermic. But I'll think about it --Bennybp 14:27, 2 November 2007 (UTC)

It was a potassium sulfate/water reaction. --MKnight9989 14:30, 2 November 2007 (UTC)

[edit] atomic size constraints

What keeps an atom from being stable or even formed beyond a certain number of nucleons? Are the forces that hold the atom together not strong enough due to excessive mass or to excessive distance or both? Dichotomous 15:16, 2 November 2007 (UTC)

I believe it's because the strong force (actually the residual strong force) pulls the nucleons together, while the electrostatic repulsion that the protons have for each other pushes them apart. The strong force only works at short distances, so when the nucleus gets too big, the like-charge repulsion causes instability and fission. You might like to read Island of stability. --Sean 18:22, 2 November 2007 (UTC)

[edit] Spherical wheels on cars?

Think it will ever happen? I can think of a few advantages they might have. Greater manuverability for one, no need to turn the wheels, simply change the direction of their spin. 64.236.121.129 15:18, 2 November 2007 (UTC)

The contact patch would be much smaller than with a tire, right? That could be a problem. Also you'd need some pretty impressive mechanical magic to give the sphere driving force while still allowing the full range of motion. Friday (talk) 15:23, 2 November 2007 (UTC)

Indeed--can you still drive the sphere effectively after it gets wet in the rain or snow? (Presumably it is driven by rollers in contact with the sphere somewhere near its top surface...?) Is there a good way to cushion the ride? Will you dent or chip the sphere every time you run over a rock? Do you lose space from the passenger, engine, or storage compartments when you have to fit in the larger volume of the wheel along with whatever motors are driving it? TenOfAllTrades(talk) 15:47, 2 November 2007 (UTC)

You're doing it wrong! Lanfear's Bane | t 16:22, 2 November 2007 (UTC)

Those Outspan cars were really Mini's - we have a photo of one of them in the article. SteveBaker 17:29, 2 November 2007 (UTC)

I believe it would also be a lot more work to change the direction of spin than to just turn the wheels, though I'm not certain of the underlying physics. Given this, while you'd certainly end up with a better turning radius, I don't know that you'd have across-the-board better maneuverability. The points above are also well-said in this regard -- one upper limit to maneuverability is the point at which the tires can't overcome angular forces and start to slip. The reduced contact patch of spherical tires would hasten this slipping. — Lomn 16:26, 2 November 2007 (UTC)

Well the contact area, traction, and the possibility of chipping would have more to do with the material used for the wheel, rather than the shape. A perfect sphere doesn't have much contact with the ground true, but neither does a torus. A tire inflated too much doesn't have much contact with the ground, it needs to be deflated so it sags a bit, increasing the contact area. If you put a beach ball on a street, you can increase its contact area with the ground simply by pushing on top of it. 64.236.121.129 16:33, 2 November 2007 (UTC)

Roads are pretty much flat - you don't need a sphere - you need a cylinder (which is what we have). If you need more grip, you lengthen the cylinder (wider tyres) or increase the contact patch (by deflating them a little) or use a more appropriate tread pattern (eg racing slicks, snow tyres). A spherical tyre would have less contact area than a cylindrical tyre (for a similar size) no matter what. SteveBaker 17:29, 2 November 2007 (UTC)

Tread is a good point here. Would be pretty hard to have a pattern that worked in "every direction" around the sphere. DMacks 17:42, 2 November 2007 (UTC)

I must emphasize, wider tires do NOT increase traction with the ground. 64.236.121.129 18:40, 2 November 2007 (UTC)

I remember learning some theory about friction in which this was true. But I also remember learning that in real life, it does increase traction. I don't remember specific details, but why else would people use wider tires when they want better traction? Friday (talk) 19:02, 2 November 2007 (UTC)

The quick'n'dirty explanation of why it doesn't matter in theory is that friction is proportional to force, not contact area (if you double the area, you half the amount of force per unit area but that still means same total frictional force). DMacks 19:22, 2 November 2007 (UTC)

64.236.121.129 is not correct; wider tires do increase traction by dint of a larger contact patch as noted and linked above. As for the idea of spherical tires, in addition to the tractive disadvantages, transmitting drive to spherical "wheels", and braking them, would be a hideously complex and likely quite inefficient exercise. The disc-shaped wheel and quasitoroidal tire will be with us for the foreseeable future. --Scheinwerfermann 19:21, 2 November 2007 (UTC)

The "law" of friction that they teach kids in school and junior physicists (ie Frictional force is proportional to the normal-force multiplied by the coefficient of friction and that the contact area is irrelevent) is NOT TRUE! It's not a law of nature - it's not even a reasonable approximation - it's a rule of thumb that works for some materials under some circumstances. Friction is vastly more complex than that and there is no simple rule. So - it is undoubtedly true that car tyres gain more frictional force if they have a larger contact patch. If you doubt that - and you happen to be in Texas, I'll take you for a hair-raising ride in my MINI Cooper'S with racing slicks and with regular tyres - when you've changed your pants and cleaned up a bit - you'll admit that I'm right! Have you seen the width of the rear tyres on a Formula 1 racecar? Why would they make them so wide if the area of the contact patch didn't matter? Thin skinny tyres would work just as well...but they don't...and there is a reason for that! So - please ignore what your high school teacher said - it's not true. SteveBaker 20:06, 2 November 2007 (UTC)

To answer Friday, they use wider tires because they last longer. That's critical if you are racing, and you want to minimize your time at the pit stop. As for your points Steve, why would that teach that if it wasn't true? Last I recall, perpetuating myths isn't appropriate in a place of learning. It's in the textbooks, and the teachers teach it. I'm just supposed to take your word for it against theirs? 64.236.121.129 20:21, 2 November 2007 (UTC)

No - that's nonsense - I have a set of racing slicks for my car - they are very wide and I can tell you with absolute certainty that I don't own them for their long tread life!! The reason some people still teach this load of nonsense is because it's in the text books and the curriculum. Read the best undergraduate physics textbooks on the planet - Feynman's Lectures on Physics - he goes to a lot of trouble to un-teach you this bad piece of misinformation. Read Friction#Classical_approximation - especially the last paragraph: [This] approximation is fundamentally an empirical construction. It is a rule of thumb describing the approximate outcome of an extremely complicated physical interaction. - in other words, it's not true. SteveBaker 20:42, 2 November 2007 (UTC)

So you think "a half truth is a whole lie"? I'd say that's, well, half-true. --Trovatore 21:06, 2 November 2007 (UTC)

What?! Tire longevity has a lot to do with the material of the tire. As a general rule, stickier tires wear out faster. All kinds of things that people are taught are simplified models, accurate only in certain circumstances. I remember being taught that atoms are little balls with other little balls orbiting them- like tiny little star systems. Once you're past junior high school, this model is way too simplistic. Friday (talk) 20:26, 2 November 2007 (UTC)

• While it's wise to be cautious about what a textbook says, it's even wiser to be cautious about what some guy on the Internet says. I think it's fair for .129 to request WP:Reliable Sources for SteveBaker's debunking. --Sean 20:38, 2 November 2007 (UTC)

It's absolutely reasonable to request that...right up to the point where I gave you exactly that and you ignored me! Which part of Read the best undergraduate physics textbooks on the planet - Feynman's Lectures on Physics didn't you understand? SteveBaker 20:58, 2 November 2007 (UTC)

• You're being rude. You'll note my post preceded yours. --Sean 01:58, 3 November 2007 (UTC)

My apologies - in fact we differed by just a couple of minutes - my first effort got bounced as an edit conflict. However, please check Feynman - volume 1 section 12-3. SteveBaker 16:55, 3 November 2007 (UTC)

Casual observation reveals that a great many cars use wider tires for better traction. If this isn't legitimately justifable, we have to assume incompetence or conspiracy on the part of the engineers who chose those tires. Friday (talk) 20:42, 2 November 2007 (UTC)

Leaving aside the incompleteness of the 'friction-force-is-proportional-to-μ-and-normal-force-only' model, a move to 'wheels' with smaller contact patches (and larger forces applied per unit area) would increase substantially wear and tear on the road itself. TenOfAllTrades(talk) 20:32, 2 November 2007 (UTC)

I feel a need to explain why the 'rule of thumb' for frictional force is wrong. As our article on Friction explains, that 'rule' is based on the following assumption: That only a small percentage of the atoms on one surface are in contact with the other surface. When force is applied to push the two surfaces together, more atoms come into contact and the frictional forces increase. The reason the rule of thumb says that the contact area doesn't matter is that the number of atoms that come into close contact depends on the pressure applied to the two surfaces. Since (in this case) pressure is force divided by area, the pressure between the two surfaces decreases as the contact area goes up. So whilst more atoms are available to come into close contact, the pressure forcing them together goes down in the same proportion - to frictional force is independent of the contact area.

However, there was an assumption there - that the percentage of atoms in contact was small compared to the total number of exposed atoms on the two surfaces. For stiff things like steel plates that are microscopically rough, that's a reasonable assumption and the rule of thumb works quite well. However, rubber is very soft and compliant and with even a moderate force applied to it, it'll deform until a VAST proportion of the atoms are in contact with the surface below. Now, varying the pressure forcing the tyre onto the road isn't making a dramatic difference to the percentage of atoms in contact - so increasing the area of the contact patch can indeed make a huge increase to the amount of grip you get.

Surfaces that are lubricated suffer from similarly violating this "law" because the layer of oily lubricant can prevent atoms of one surface getting close to those of the other even if the pressure on them increases. Other surface - like Ice - melt when you apply pressure to them - so they lubricate themselves - and the more pressure you apply, the more lubricant gets between the two surface.

SteveBaker 20:55, 2 November 2007 (UTC)

I find it unlikely that the width of the wheel determines how much traction it has. It merely depends on the surfaces of the two contacting surfaces. For example, a bicycle wheel can have excellent traction despite it being very narrow. A train has a very narrow contact surface between its wheels and the rail, but has adequate traction. Casual observation of race cars using wide tires and thus assuming they use them because of greater traction is a false assumption. Facts and truth come before logic or common sense. Common sense would indicate that .999... doesn't equal 1, but it does. Malamockq 22:48, 2 November 2007 (UTC)

Did you read the above? There's some pretty good explanations there that go beyond mere speculation. They do use wide tires for better traction, but you're right- the mere existence of wide tires could be explained other ways, too. Friday (talk) 22:53, 2 November 2007 (UTC)

Then by insinuation we should say that the wider the wheel is, the more traction it should have. So if you have a wheel 1 mile wide, it should have more traction than a wheel 5 inches wide (assuming the surface of the tire is the same)? You have to take your assertion, then exaggerate it and determine if it makes a meaningful difference. Malamockq 23:01, 2 November 2007 (UTC)

Yes! Wider wheels mean more friction. There will come a point where so little of the wheel is in fact pressed into the minute bumps in the road surface that increasing the area won't help anymore - but for wheels up to (say) a couple of feet wide, the contact area is very important indeed! I'm not saying that the amount of friction is proportional to the contact area or the square of the contact area...or any other particular rule. I'm saying that friction is vastly too complex for that simple rule of thumb to be true. There are some materials in some sorts of contact where the rule is a pretty good approximation - there are other cases where it's not even close. Car tyres are in the latter category. SteveBaker 16:55, 3 November 2007 (UTC)

I'm not following you. But, SteveBaker gives a good explanation above about why with steel, size of the contact patch isn't a big factor, whereas with things like rubber, it is. Obviously there would be serious practical problems with a wheel a mile wide.  :) Friday (talk) 23:05, 2 November 2007 (UTC)

Funnily enough I was pondering the same question the other day after thinking about a computer mouse (where a ball moves two rollers, one vertical and one horizontal) and wondering whether anybody had tried reversing the principle and using two powered rollers to move a sphere in order to move an object. This wouldn't be very pratical for things like cars but I thought that things like robotic vacuum cleaners might work very well using this principle although it would probably need three or four spheres to keep the thing off the floor and going in the right direction. GaryReggae 23:19, 2 November 2007 (UTC)

Um, Malamockq, I'm not quite following you, either. You say "Facts and truth come before logic or common sense", and I mostly agree with you, but then your only evidence for why the width of a tire is irrelevant is that you "find it unlikely" that it makes a difference.

Steel-wheeled trains have adequate traction, but only barely. Very small slopes, or small amounts of water, snow, ice, or wet leaves, can leave a train spinning its wheels and unable to move. Most trains carry sand which they can sprinkle on the rails ahead of their wheels to increase traction when they need to.

Steel wheels and rails are used on trains because they have excellent rolling resistance (low), and excellent lifetime (high). But the traction's only so-so, though of course it helps that they have huge amounts of normal force (i.e. weight) to throw into the µN equation. —Steve Summit (talk) 23:46, 2 November 2007 (UTC)

No, my reason for believing that width doesn't make a difference, is that they teach that fact in physics classes. The train example may have been weak, but any lack of traction is due to the surfaces of the train wheel and the rail. It has nothing to do with the width. Malamockq 01:57, 3 November 2007 (UTC)

Yes, they teach that to kids in physics class, along with a bunch of other oversimplifications. This is all well explained above. Did you see the explanation that with some materials, size of contact patch doesn't matter much, whereas with some other materials, it does? Yes, bicycles have skinny little tires. But check out a powerful motorcycle sometime- much wider. Why? Because it needs way more traction to put that kind of power down on the road. Friday (talk) 15:59, 3 November 2007 (UTC)

There are plenty of things like this that are taught in schools as simplifications of reality. Heck we still teach Newtonian mechanics - when we know full well that they are flat out WRONG. Einstein proved Newton was wrong almost 100 years ago. So the fact that we teach kids something doesn't make it true. We teach simple approximations.

Anyway, I'm now sitting in front of the book I referred you to (Feynman's Lectures on Physics, volume 1, section 12-3). Just below the F=uN equation, Feynman (who was a physics nobel prize winner) says:

"Although this coefficient is not exactly constant, the formula is a good empirical rule for judging approximately the amount of force that will be needed in certain practical or engineering circumstances. If the normal force or the speed of motion gets too big, the law fails because of the excessive heat generated. It is important to realize that each of these empirical laws has its limitations, beyond which it does not really work."

To get the full story, you need to read on for a couple more pages - but I think you can tell from my short quote that Feynman is most certainly not saying that this is an utterly fundamental law. Car tyres most certainly provide more grip when they are a foot wide than when they are four inches wide - look at any high performance sports car and notice the fat tyres...there is a reason for that. You wouldn't add weight (PARTICULARLY 'unsprung weight') to a performance car unless it bought you some major benefits in terms of going faster or accellerating harder. It's not just tyres either. Why do you think people put bigger disk brakes on cars to make them stop better? SteveBaker 16:55, 3 November 2007 (UTC)

It would sure make parallel parking a lot easier, that's for sure. :) shoy (^words _words) 03:46, 3 November 2007 (UTC)

Check out Rolling resistance. Acceptable 04:26, 3 November 2007 (UTC)

[edit] Where did the energy go

I was told today that although most macro structures are held in shape by the forces that govern them as space time expands, and as a result do not expand, photons are stretched by space-time, and as a result have an increasing wavelength, assuming that c is a constant through all space-time (a reasonable assumption I presumed) therefore a decreasing frequency, and so a photon hanging about for a substantial amount of time would lose energy due to E = hf. Where does the energy go? ΦΙΛ Κ 18:58, 2 November 2007 (UTC)

Gravitation potential energy. It is similar to the red shift in a photon trvaelling up hill in a gravitational field. Graeme Bartlett 23:39, 2 November 2007 (UTC)

Actually, no. The correct answer is "General relativity is a bitch that apparently doesn't obey Conservation of Energy". It's possible that there's some related quantity that's conserved, but scientists are yet to come up with anything that seems to work yet. Confusing Manifestation 08:14, 3 November 2007 (UTC)

Conservation of energy is a fundamental consequence of the symmetries of spacetime. I should posit an answer here - I didn't before because I'm not sure if this is correct, but the answers given so far are misleading. I believe that a photon would ride local expansion like a wave, and thus would not lose energy relative to an outside observer because velocity would effectively increase (because spacetime has been locally changed). However, for the uniform expansion of the universe, I'm not sure if this model fits: certainly locally the change in curvature accounts for any energy loss - maybe that's all you need. SamuelRiv 07:04, 4 November 2007 (UTC)

I found this site useful in describing (and attempting to solve) the problem embodied in the question: [4]. In summary, classical conservation of mass-energy doesn't hold, but it is possible to define many quantities which are conserved. However, there's no quantity which always works and everyone likes. 58.96.70.254 12:40, 4 November 2007 (UTC)

[edit] Antonym: Congenital?

Very quick question: what is the antonym for congenital, in the context of diseases (i.e. inherited diseases versus infectious or non-infectious diseases acquired throughout life)?

209.51.73.60 23:57, 2 November 2007 (UTC)

Acquired. Someguy1221 00:27, 3 November 2007 (UTC)

Contracted. --DHeyward 04:09, 3 November 2007 (UTC)