Wikipedia:Reference desk/Archives/Science/2007 May 30

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May 30

History of Nursing in the Caribbean

I have been unable to find any references to Nursing in the Caribbean as it applies to the actually development of the profession. The only references have been to Dame Nita Barrow of Barbados and Mary Secole of Jamaica. I would appreciate any information that may be supplied in relation to this topic.

Dimensions > 3

Humans can only see the 0th, 1st, 2nd, & 3rd Dimension. We can not see past the 3rd Dimension. If we had bigger brains, would we be able to? Would aliens be able to?68.148.164.134 06:24, 30 May 2007 (UTC)

Light travels within three dimensions, taking some small amount of time to do so. Eyes use light to see with, so that is why humans, superhumans or aliens can only see three dimensions. The way the eye works it projects a two dimensional image on the retina, so you are only experiencing a two dimensional version. It could be theoretically possible to detect light holographically or over the space in a volume, to truly see three dimensions. However it si possible to imagine 4 dimensions, or to project it onto three or two dimensions. See Fourth dimension for an example of this. The size of the brain has little to do with this, but some geometrical or mathematical skill may help in imagnination of it! GB 06:46, 30 May 2007 (UTC)

Thanks, but what do you mean by '...holographically...'?129.128.67.22 17:43, 30 May 2007 (UTC)

Holographically means that the phase as well as the brightness of the light is detected, and this is done at microscopic sizes smaller than the wavelength of light. If this was possible to do, the shape of the wave form entering the detector can be reconstructed. This can then determine the depth of the visible object, and if the detector is big enough can see in different directions at the object. Usually holograms require special illumination and a long time to produce.

As Graeme says we only "see" two dimensions. We interpret visible light information and, using the slightly different information transmitted from either eyes, our brain is able to gauge depth, permitting us perceive the world in three dimensions. So unless whatever is "past the 3rd Dimension" generates light, a bigger brain would not help us "see" it. Instead we would need a different type of visual detection system. Some snakes can "see" in infrared, leaving some to suggest they can essentially "see heat". Depending on your interpretation, you could say this is seeing in another dimension. A more traditional concept of the 4th dimension is time. If we could "see" time we should be able to a limited time into the future or the past. There is no biological mechanism known for doing such a thing, and therefore it would be wild speculation to suggest aliens could do this. Rockpocket 06:50, 30 May 2007 (UTC)

Ok, our eyes actually each see a 2D world, but putting two together adds an appreciation of the 3rd dimension. Would adding more eye balls add more dimension, or would you need to add another dimension to our eye balls? Plus, you might have to add a 4th dimension to our world? Or is my first sentence wrong? Aaadddaaammm 08:19, 30 May 2007 (UTC)

Each eyeball only sees a 2D image. But since we are in a 3D world, both your eyes will disagree with eachother. This, along with a number of other effects mentioned in depth perception allows your brain to extrapolate how things are arranged in the actual 3D world. Since the world is not 4th dimensional (ignoring silly silly time), there are no 4th dimensional effects on your eyes for your brain to extrapolate a 4D image. No matter how many eyeballs you have, there is no 4th dimension to perceive. If you did suddenly find yourself in 4 dimensional space (or 5 dimensional spacetime, or whathaveyou) you'd find that your eyes didn't work all that well, and your body would likely shatter. Imagine introducing a 2 dimensional being into our universe (it would only be one molecule thick!). Someguy1221 08:29, 30 May 2007 (UTC)

Just a note: A being that can only perceive 2D is not necessarily one molecule thick. It can exist in all 3 dimensions, but only comprehend two of them - never knowing how it exists in 3D. This is similar to our inability to understand how we exist in a 4th dimension. What if we are all connected as one 4D being, but we only see separate 3D projections of the unified whole? --Kainaw ^(talk) 13:27, 30 May 2007 (UTC)

Another very interesting question. It has been suggested by some string physicists that the universe is like this in some manner (but no evidence, of course, silly string physicists). If we are only seeing 3 dimensions but exist in 4, then if molecules can enter and exit this fourth dimension just as easily as all the others, it would be easily detectable (the mathematically calculated value for the relative volumes of solids suddenly wouldn't work). However, if our molecules are confined to a 3D "surface" of some larger 4D world, then you might see some pretty interesting phenomena. Spacetime curvature can only be accurately imagined in four dimensions, and indeed, spacetime is four dimensional (but "locally" three dimensional). Let's go back to the 2D case. Let's say the world is the surface of sphere. All particles are confined to travel along this sphere, so even light just bends around it and there's no way of looking away from the sphere, no way of seeing there's more to the world than this surface. However, assumining the sphere is small enough, you could easily detect some paradoxes. For example, two adventurers set out in parallel paths some hundreds of miles away, and have some neat gizmo to ensure they stay parallel. From their mastery of geometry, they know they should never meet, but nonetheless, they do! Enterprising mathematicians would point out that parallel lines do meet on the surface of a some curved shapes, so perhaps there is a curvature to their plane-world that is only visible in three dimensions. For our own universe, however, no one has ever found any curvature not explained by general relativity (that I know of). Someguy1221 16:06, 30 May 2007 (UTC)

Thanks, but I don't understand.129.128.67.22

Let's try that again. String theorists claim that the extra dimensions (beyond three) are very tightly curved such that the maximum size of anything in the curled up dimension is very, very small indeed - much smaller than an atom. Hence 'leakage' of water out of a bucket because of it flowing into the extra dimensions where there presumably is no bucket to stop it is not a reasonable possibility. This also blows away your alternative methods of detecting the extra dimensions. The best way to imagine this is to pretend you are a being who lives in a 1D world where all of life exists along a single infinitely thin line. The 1D scientists have just discovered that you actually live on an infinitely long but very narrow 2D sheet that's rolled up into a cylinder. So in addition to moving long distances in your familiar forwards and backwards way, you can actually move sideways in this "extra" second dimension. Now imagine that the diameter of the cylinder was a gazillionth of a picometer - vastly smaller than an atom. Everything would behave just like it does in their 'normal' 1D world even though they would actually live in a tightly curved 2D space. If string theory is correct then we live in a 15 dimensional world (or however many it is this week!) but only three of the dimensions are sufficiently uncurled to be noticable - even to exceedingly careful experiments. To probe distances that short requires vast amounts of energy - and we simply don't have the tools to do that. SteveBaker 18:25, 30 May 2007 (UTC)

Technically, shouldn't 4D vision let you see many instances of time at the same time, and all sides of a building at the same time? Some people had claimed to be able to picture things in 4D, but I wonder how they did. It'd look worse than Picasso. --Wirbelwindヴィルヴェルヴィント (talk) 08:37, 30 May 2007 (UTC)

You would see not only every wall, but also every internal detail of it. Any volume is essentially flat and infinitely thin in 4D, just as (from a mathematical perspective) any area has infinitely small volume. As to whether you'd see many instances of time, you have to decide whether the 4th dimension you want to see is time-like or space-like. If it's space-like, you see the wierdness I've described. If time-like, you'd see something akin to a world line. Someguy1221 08:49, 30 May 2007 (UTC)

(Thanks, but I don't understand.129.128.67.22)

For 129.128.67.22, I suggest playing with the "penny-man" example. Lay a sheet of paper on your table. Put a couple pennies (or any coin - or even anything flat) on it. Each penny is a penny-man that lives in a two dimensional world. They can see left-right (width) and near-far (distance). They have no concept of looking up or down. The sheet of paper is their world. Now, lay a pencil or something on the paper (or just draw a line on it). This is a two-dimensional wall. It has length, but not height. When penny-man approches it, he can see the wall, but has no concept of looking over the wall. To him, it is an impassable object. When we look at the situation, we know that penny-man could just hop over it if he could see in three dimensions. Now, put the other penny-man on the other side of the wall. They cannot see each other because they cannot look over the wall. However, we can see both penny-men easily. This is pretty easy to work with. You can even make a 2-D house by drawing a box (make sure to give them a little door to walk through). Inside the house, they feel enclosed. We, however, can easily see both inside and outside the house. After playing with this a bit, you can start to think about what our 3 dimensional world looks like to a 4 dimensional being. Then, you can imagine what it would be like if you picked up one of the penny-men and moved him around in 3 dimensions. What would he see? How does that relate to what we would see if we were lifted into a fourth dimension? It is all theoretical, so there's no wrong answer. --Kainaw ^(talk) 00:38, 31 May 2007 (UTC)

Candlestick or Faces?

Over the last two days I have had the remarkable pleasure of hearing the thoughts of Richard Axel on the neurobiology of perception. He had some very interesting things to say about visual perception v. reality. He actually said that "vision is a hypothesis", and is no more than an approximation of the reality of our surrounding space. Indeed, what we see in any dimension is a complex top down integration (in our brain) of a very simple bottom up signal (light). That what we understand as "seeing" is part genetic hardwiring, part experience and part expectation, therefore we all "see" our world differently, not because it is different, but because we all have differences in the parameters we use to interpret it. Optical illusions are a good example of demonstrating this. For example some people will first see the faces and others will see the candlestick in the Rubin vase. Why is that, since the light reaching our eyes should be the same? The reason is because we have different genetic hardwiring and experience and our brains integrate all of this with the expectation of what we think we are seeing and tells us it is a candlestick or two faces in profile. It usually doesn't take long to "see" the other image, since our brains are pretty quick at information processing. No-one, as far as we know, can encompass both interpretations simultaneously, our brains don't seem to work that way. It is implicit in the assertion that "vision is a hypothesis", that the 3 dimensions we "see" are not necessarily the limit of the world around us. So, to apply Axel's interpretation to your question. While having a bigger brain wouldn't help us "see" another dimension (if it existed), having a different type of brain might. Rockpocket 01:24, 31 May 2007 (UTC)

Dryers vs. handTowels

I notice that many dryers now claim that they are "more environmentally friendly than handtowels" - is this true? are their any studies that compare the two? --Fredrick day 11:14, 30 May 2007 (UTC)

Although I can't provide evidence, i'm pretty sure that neither are more environmentally friendly than my prefered method of giving my hands a good shake and wiping them on my trousers. Esspecially since my trousers are used in a multitude of other ways too. 213.48.15.234 11:20, 30 May 2007 (UTC)

There is some information at hand dryer, but note also the comments on that article's talk page.--Shantavira|^{feed me} 11:41, 30 May 2007 (UTC)

It will be a very difficult comparison on a worldwide level. For the towels, you have to consider where they come from (virgin forest or a tree farm), how did they get transported (floated in a stream or driven on a truck), how were they processed, what where they packaged in, how did they get to the facility where you used them, how are they disposed of? For the dryer, you have to consider the raw materials to make it (steel and aluminum) and the electricity. Is it coal power, nuclear power, wind power? How much electricity does it actually use? Does it produce greenhouse gasses as a byproduct of producing heat? Overall, the dryer is most likely the most eco-friendly way to go, but you can never really get your hands dry with one. --Kainaw ^(talk) 13:23, 30 May 2007 (UTC)

I thought most paper towels were recycled paper Coolotter88 00:40, 31 May 2007 (UTC)

That may or may not be a good thing. It does take quite a bit of energy to ship all that paper back and forth and recycle it. --67.110.213.253 23:22, 31 May 2007 (UTC)

Design considerations for a water bridge?

Water bridge is a bridge that carries water. This helps in navigation of boats, ships etc. Now while we design a water bridge do we consider the weight of the ship that floats across the bridge? Or do we just consider the weight of water alone the bridge has to carry? deostroll 14:04, 30 May 2007 (UTC)

Time to learn about buoyancy. When a boat is present, exactly how much mass of water will be displaced (i.e., "not in the water bridge due to the presence of the draft of the boat") compared to the mass of that boat? DMacks 14:34, 30 May 2007 (UTC)

Preface: this is not a legal Professional Engineer's opinion. You don't need to consider the ships.

1. A ship displaces a mass of water equal to its own mass. Reasonably free-flowing water means that the total mass on the bridge is equal under various conditions (more ship mass means less water mass).
2. Ship mass is negligible relative to water mass anyway.

Any reasonable engineering safety margins for weight-of-water will account for any and all deviations from theoretical ideals noted above. — Lomn 14:35, 30 May 2007 (UTC)

So long as all of your ships are floating - you're OK. If one ever sank though - you could be in trouble! You'd probably also have to consider the gradual build-up of silt at the bottom of the water - silt is heavier than water so I'd certainly expect the weight to slowly creep up over the years if the bridge isn't dredged periodically. SteveBaker 16:44, 30 May 2007 (UTC)

An interesting extension of this issue is a boat lift: the lifted basin always weighs the same, boat or not, so two basins always nearly perfectly counterbalance each other, unlike an elevator, in which the lift motor has to work against the weight of the occupants or mis-balanced counterweights. DMacks 17:05, 30 May 2007 (UTC)

Also consider rain, especially in form of monsuns. :) 81.93.102.185 17:38, 30 May 2007 (UTC)

Unless the river gets deeper - rain doesn't matter - the incoming raindrops displace water just like the boat does. But obviously if the river can flood then you have to take account of the deepest the water can ever be. SteveBaker 19:03, 30 May 2007 (UTC)

A ship also imposes certain dynamic forces that differ from the dymanic forces of flowing water with no ship or those of standing water. I suspect that these are negligible, but the engineer had better check. Harbors and canals have speed restrictions to minimize damage due to these dynamic effects. -Arch dude 23:27, 30 May 2007 (UTC)

Mesons

Do mesons every really "exist" for any substantial amount of time? Mesons are made of one quark, one antiquark; but wouldn't they just immediately annihilate (and turn into virtual bosons obviously) instead of exchanging gluons and forming a hadron? --Codell^«T» 15:10, 30 May 2007 (UTC)

Mesons are not made of one quark and *its* antiquark. They are made of, say, up and anti-down, not up and anti-up. The quarks cannot just annihilate each other.

So annihilations can only happen with the same types of quarks/antiquarks? Ex. down, anti-down? If so, then wouldn't this just result in an annihilation instead of a quarkonium meson? What really is the difference between quarkoniums and annihilations? --Codell^«T» 15:58, 30 May 2007 (UTC)

What annihilation really means is decay into photons. So annihilation is just a special form of decay. Every decay must obey a set of conservation laws. Decay into photons for an up and anti-up quark for example would violate charge conservation I think I should better go to bed right now... Now when the possiblity of annihilation exists, it does not automatically mean that annihilation happens. For example Positronium is quite stable (for some nanoseconds, but that is long in this case). Compare that to 1.2E-20 seconds for the Upsilon particle. It hardly exists at all.

π^± are long-lived mesons, at 2.6×10^-8 s. The two quarks do annihilate to form a (virtual) W^±, which then produces leptons. That's how Fermilab makes a beam of neutrinos. The π⁰ is neutral, a superposition of uū and d-dbar. The quark and antiquark can annihilate to form two gammas, with a much shorter lifetime. So you're exactly right, mesons don't last for a long time because the quarks can annihilate. But their lifetimes are long enough that they do "really exist" within nuclei. The speed of light is a foot per nanosecond, so a charged pion could propagate up to ~2.6 feet - or even farther, because time dilation stretches their lifetimes. So they're plenty real on nuclear scales. On larger scales, a particle accelerator can produce a beam of pions, then focus them in a magnetic "horn" before they decay. That's how they focus the neutrino beam for MINOS. The K-long meson has a longer lifetime, 5.2×10^-8 s. Its decay is more complicated. --Reuben 18:13, 30 May 2007 (UTC)

Okay, thank you. I now understand. --Codell^«T» 15:58, 30 May 2007 (UTC)

MCAT Prep Question

I am reading through a comprehensive review book for the MCAT and have come across a question I need help with. It is a passage question and the passage states that students are leaning out of a classroom windows (h=5.2m) and dropping different balls from the same height. They measure the velocity of the balls instantaneously before impact in order to determine the value of gravity (g). They ignore air resistance in their calculations.

The question states: "Which of the following would change the measured value of g in this Experiment?"
I. Increasing the mass of the Earth
II. Using balls having a different mass but the same volume
III.Throwing the balls horizontally instead of dropping them vertically

A: I only / B: III Only / C: I and II only / D: II and III only

Correct Answer given: A
Answer I gave: C

Now, I can immediately eliminate III as a possible choice and understand why. Also, I understand why increasing the mass of the Earth would increase the value of g. However, it seems as though II was hastily skipped over by the designers of the question. The explanation given as to why II is a wrong choice was this: "Now we need to consider Statement II. It says that using balls with different masses but the same volume would change g. We know that in this experiment the students ignored air resistance, so the volume of the balls is not relevant. What about the mass? The value of g is of the object's mass. So Statement II is false."

OK, this seems like something you learn in high school physics, day 1. If you drop a 1 pound weight and a 10 pound weight they hit the ground at the same time. I got that concept. The question asks which of the following would change the measured value of g. It seems that objects with the same volume but different masses would be affected differently by air resistance. Theoretically, the object with the lesser mass would be more affected on the way down (when colliding with air particles) than would the more massive object, due to the differences in their momentum. Thus, the lighter object would be moving slower the instant before impact and the measured value of g would be different. Now, the explanation of why II is not correct says that the students ignored air resistance, so volume wouldn't matter. However, this does not make sense to me. If they ignored air resistance, but still performed the measurements in air...then there was still air resistance affecting their measurements. So, it seems as though the fact that they ignored air resistance would be a contributing factor to the difference in their measured values of g based on the conditions in Statement II. It seems as though only if they took air resistance into account would balls of different masses and the same volume produce the same measured value of g.

Am I right? I don't know if my reasoning here is correct. Mrdeath5493 17:32, 30 May 2007 (UTC)

The reasoning is very simple. The question is "What can change the Earth's gravity?" The answer: "Change the Earth's mass." It doesn't matter how your throw the ball or what size/mass the ball has. That won't change the Earth's gravity. --Kainaw ^(talk) 17:35, 30 May 2007 (UTC)

I don't think so. the question asks "What would change the measured value of g?" Mrdeath5493 17:40, 30 May 2007 (UTC)

"g" is the "Earth's Gravity". The question is: What would change the measured value of Earth's Gravity? --Kainaw ^(talk) 17:41, 30 May 2007 (UTC)

OK I agree with that. The problem comes in the measurement. If you dropped a feather and a bowling ball in air at a height of 5m and then measure their instantaneous velocity before impact, are you suggesting that the measured value of g would be the same in both cases? Mrdeath5493 17:44, 30 May 2007 (UTC)

So in short you are telling me never to go on a holiday in the US when I am sick? I can even give an argument for III: The pressure gradient of the atmosphere will generate a net lift on a ball flying sideways at high speed.

You were correct to begin with. The actual value of g is unchanged by changing the mass of the balls. The measured value of g is subject to air resistance and to how the calculation is performed. Someguy1221 17:54, 30 May 2007 (UTC)

The question stated that air resistance is not an issue. See Galileo Galilei. He did this experiment in the 1500's I believe. It demonstrated that the velocity is the same if the height is the same, regardless of mass. --Kainaw ^(talk) 17:56, 30 May 2007 (UTC)

With no air resistance, yes the final velocity should be the same regardless, according to V_f = V₀ + at. With initial velocity equaling zero since you're dropping them. Of course, Answer choice III (Throwing the balls horizontally instead of dropping them vertically), is somewhat misleading, since if you threw the balls at a high enough speed, you could technically throw them into orbit, but I don't think they are considering that, also since it's completely off topic and wouldn't change g ;)--GTPoompt 18:02, 30 May 2007 (UTC)

The question did not state that air resistance is not a problem, it states that the students ignored air resistance in their calculations. It's a poorly worded problem. Someguy1221 18:04, 30 May 2007 (UTC)

No! (C) *IS* the correct answer - Galileo was wrong! Mathematically II is also correct - the masses of the balls does have an effect on the time it takes for them to hit the ground. Remember that not only does the earth exert a gravitational pull on the ball - but the ball also exerts it's own (very, very tiny) gravitational field on the earth. So in theory the heavier ball does hit the ground infinitesimally faster than the lighter one because the earth moves towards the ball by tiny amount. However the ratio of the mass of the earth to the mass of either of the balls is truly phenomenal and there is no practical measurement you could do that would show the difference. But if we are ignoring such major effects as air resistance then we should probably ignore this effect too. Note that the question doesn't ask what would change the value of 'g' (this wouldn't) it asks what would change the value as measured by this experiment. Ask your physics teacher to explain what would happen to the value of 'g' the students would calculate if one of the balls had a teaspoonful of neutron-star material inside - you'd certainly notice it fall to the ground a heck of a lot faster than a ball made of rubber or something. So if you are DESPERATE for that extra point to make a grade then it's worth trying to appeal on those grounds! (Good luck with that!) SteveBaker 18:42, 30 May 2007 (UTC)

You know what - forget the ball made of neutron star stuff. The question says that the students do an experiment to measure 'g'. They ignore the effects of air resistance (meaning that they time the balls falling in air and measure their impact speeds and do the simple calculation to get 'g' - stupidly ignoring the effect of air resistance - and therefore they get an incorrect value for 'g'). Now the question doesn't ask what would change the true value of 'g' - it asks what changes the value of 'g' in the Experiment - since a heavier ball will hit the ground faster than a lighter one because of air resistance, changing the mass of the ball will affect the value of 'g' in the experiment. They'll get a totally different value depending on the weight of the ball they use. So - yeah - you were correct - the answer is definitely (C) - without any appeal to quirky arguments.

SteveBaker 19:00, 30 May 2007 (UTC)

I would seriously reconsider if I even *want* to be in an education institution that chooses it applicants by such a bad test. Maybe not if that question was the only negative thing, but if the whole test and application process makes a similar impression, what does that tell you about the quality of the institiution that chose the test? I would expect a lot of mindless rote learning in that case.

See Earth's gravity - especially the section on the measured value of "g". The mass of the falling object is not part of the equation for "g". And, the question is really just asking if you know what "g" is. It may as well ask, "Did you pay attention in class when they talked about Galileo?" --Kainaw ^(talk) 19:09, 30 May 2007 (UTC)

Not to get too off the point, but MCAT prep books are notorious for some of the terrible questions their writers come up with. --David Iberri (talk) 01:52, 31 May 2007 (UTC)

If you understand gravity thoroughly, and accept that the MCAT is only testing a very basic understanding of physics, you can infer that they will not be convoluting the problem with complex mathematics required for drag and air resistance. Rather than try to answer this question from the standpoint of a thorough physics explanation, I recommend you accept this at face value - it is an elementary physics question in a multiple-choice medical school qualification exam. Advanced physics explanations are unlikely to be relevant to this scenario. When the scope is thus reduced, it is evident what they want the answer to be. Nimur 06:14, 31 May 2007 (UTC)

Wouldn't they both hit the ground at the same time and the same speed, because their speed is proportional to g divided by their mass, as it takes more force to move a heavier object than a lighter one to the same speed :) HS7 13:27, 31 May 2007 (UTC)

The full form of the gravitational equation is that the force exerted by one object onto another is proportional to the product of their masses divided by the square of the distance between them. So the gravitational force exerted on the lighter sphere is much less than that on the heavier one. However, accelleration is force divided by mass - so the heavier object takes more force to accellerate it at the same rate as the lighter object. If you multiply out the two equations, the mass of the ball cancels out - so the accelleration of the ball due to the earth is the same no matter how much it weighs - so 'g' is a constant unless the mass of the earth changes (or you get closer or further away from it). That's the simplistic view that this exam question is trying to get at. However, there is a deeper truth that comes from realising that the ball also attracts the earth - and the force with which it does it is exactly the same as that of the earth pulling on the ball. But the earth is insanely heavy - so the rate that the earth accellerates towards the ball is a truly tiny number! However, that number is different for the light ball than for the heavy one. So the time it takes the ball to hit the earth does depend on the mass of the ball - but for 'normal' balls - the difference between a 1 ounce ball and a hundred ton ball would be far too small to measure. However, if the heavier ball were made of something like neutron star material - then you'd definitely notice the heavier ball hitting the ground sooner than the lighter one because the earth would be pulled upwards with the same force as the ball going downwards. But - as has been pointed out - that's clearly not the answer the physics paper was looking for - however correct it may be! SteveBaker 15:22, 31 May 2007 (UTC)

My understanding is the force of gravitational pull is applied to both objects (the earth and the ball) at the same time. Assume you dropped a ball 1/10 the mass of Earth. 10% of the force will be applied to Earth and 90% will be applied to the ball (being rather simplistic and ignoring radius of the objects). If you dropped a neutron ball 1000 times the mass of Earth, the neutron ball will remain rather still and almost all the force will be applied to the Earth. In your explanation, you appear to be creating extra force - the full force of pull on the ball and some more force pulling on the Earth. I believe that is incorrect. The force is, in total, based on the mass of the ball, the mass of the Earth, and the distance between them. So, force between ball 1 and the Earth is m1*me/dd where m1 is the mass of ball 1, me is the mass of the earth, and dd is the distance squared. The force between ball 2 and the Earth is m2*me/dd. The first force is applied to the ball and the Earth, creating an overall acceleration of f1/(m1+me). Considering the second acceleration being f2/(m2+me), you can go back and see that m1 and m2 are cancelled out and the acceleration remains constant. That is a relative acceleration (the acceleration at which the distance shrinks to zero). Because we are referencing this with the surface of the Earth, we say that the ball is falling to the Earth, even if the ball is so massive that the Earth actually moves to the ball. --Kainaw ^(talk) 15:46, 31 May 2007 (UTC)

The force is in no way divided up. Each object feels the exact same force, in opposite directions, Newton's third law.

Exactly - that is my point. The force between the ball and the Earth is one force. It pulls on both objects. In our experience, a very small amount of the pull affects the Earth. Most of the pull affects the ball. If you make the ball much more massive than the Earth, the ball will be affected very little by the pull and the Earth will be mostly affected. In SteveBaker's explanation, the force pulls down on the extremely massive ball and then there's some additional force that pulls on the Earth. Please don't confuse what I said with a claim that there is some division of force. There is a division of the applied force - or the effect of the force. For example, if you grab your desk with one hand and your chair with another and pull, the force is the same measured from either hand. However, the effect of the force on the chair (assuming it is much lighter than your desk) is greater. The chair moves and the desk does not. You can play semantics all you want, but I don't see how you can claim some additional mystery force slips in to have an extra affect on the desk. --Kainaw ^(talk) 17:17, 31 May 2007 (UTC)

I'm not sure exactly what you're arguing against here. SteveBaker is right. Even if the ball you dropped is a small chunk of neutron star that is more massive than the Earth, it will still accelerate at the same rate as any other ball (for this, ignore general relativistic effects). The only difference is that now the Earth's motion from gravity will be visible. There is no "mystery force" at work here. You know the form of Newton's law of gravity, think about for yourself. If you make the ball 10²⁶ times heavier than it was, but keep it the same size and shape, the force on it will increase by a factor of 10²⁶. So will the force on the Earth. Someguy1221 17:51, 31 May 2007 (UTC)

Someguy1221 17:07, 31 May 2007 (UTC)

I would like to clarify that this was not a question on the actual exam. It was from a publisher that publishes many different test prep books. The Association of American Medical Colleges (which makes the actual exam) is very thorough in designing it's questions from what I can see from retired tests. In response to the argument that this question was trying to test a basic physics concept and that I should have inferred that my reasoning was "too convoluted" to get the correct answer: I realized that II was correct within 20 seconds of reading the question. I went through the entire chain of reasoning outlined above in that time. Is this something a test for medical school should punish or reward? Applied in a medical context, I would hope my doctor could do the same. The reason I posted the question was to make sure my reasoning was correct since the book disagreed with me. I hope to be an ER Physician one day and assume that such fast thinking might be something I need to practice. Mrdeath5493 16:41, 31 May 2007 (UTC)

Million decibel march

While discussing the consequence of very high sounds, my brother mentioned the "Million decibel march"-tour by Metallica. This lead to a debate of the actual consequences of such a powerful sound. Would it be powerful enough to tear the Earth apart? 213.237.10.244 18:14, 30 May 2007 (UTC)

Well, a million decibel is a number, not measurement for sound wave energy, so the question makes no sense if taken literally. A sufficiently powerful soundwave could destroy the earth in theory. It would probably degenerate into a system of shockwaves and not keep its frequency. It is more like a huge number of disassociated extremely fast nitrogen atoms on a rampage. This might interest you: http://qntm.org/destroy.

Decibel can be taken as a unit of power per unit area, see Sound intensity level. A million decibels would correspond to about 10^{99 988} watts per square meter. Certainly enough to destroy the Earth, and over one square meter it exceeds the power output of the Sun by a factor of 2.61x10^{99 973}, which also happens to exceed the total output of all stars in the visible universe, but I'm not sure by how much. So...good luck making that loud a sound! Someguy1221 18:32, 30 May 2007 (UTC)

There has to be an upper limit to the amount of power that can be contained in a continuous sound wave - sound travels by compression and rarefaction of the air - at some point surely there would be a dead vacuum in the 'troughs' of the sound and the 'peaks' would be just maybe 2 atmospheres of pressure. You can't crank the volume any higher because the troughs can't get any more rarefied than a hard vacuum. You could have an impulse (a single 'peak') with more energy in it - but that's an explosion - not a 'sound' in the conventional sense. SteveBaker 18:48, 30 May 2007 (UTC)

I was skimming through the material on maximum amplitude, and at the max. amplitude, the material changes phase, or material properties. Thus, the nitrogen solidifies, or some such thing. As well, the 'speakers' would be subject to cavitation. --Zeizmic 20:27, 30 May 2007 (UTC)

If you imagine the classic sine wave sound/air pressure graph, the amplitude is the height of the graph right, the pressure wave has to travel from the top to the bottom of the curve so the higher the amplitude the further the pressure wave has to travel. The limiting factor therefore will be how far the pressure wave can travel which of course, is a factor of how fast it can travel, which would be, not surprisingly, the speed of sound. The speed of sound would be relatively constant therefore it would be inversely proportinal to the frequency of the sound, the higher the frequency the further the pressure wave has to travel therfore the less amplitude you can push it to. Which makes sense since it takes more energy to create a high frequency sound then to create then a low frequency sound of the same amplitude.Vespine 22:32, 30 May 2007 (UTC)

I'd love to direct you to the article about Disaster Area, but we don't seem to have such an article yet.

Atlant 13:07, 31 May 2007 (UTC)

See Hotblack Desiato. Laïka 13:28, 31 May 2007 (UTC)

"Human Information Processing" by Lindsay and Norman (1972) has a nice chapter on the physics of sound. The threshold of hearing at 1khz is 0 dB (relative to .0002 dynes/cm²). A whisper is about 30 dB, conversation is about 80 dB, shouting is about 100 dB, loud thunder or a rock band hits about 130 dB, the pain threshold is about 150 dB, and a manned spacecraft launch (think the Saturn V moon rocket) from 150 feet is about 180 dB. There does not appear to be a requirement that noise be sinusoidal to have a dB reading. I expect that for a nonsinusoidal noise, such as an explosion, one could look at the waveform and compare it to the peak of the sine wave. How many dB would the largest H bomb with a peak power of 5.4×10²⁴ watts present at its peak pressure intensity of the shock wave? How about the most powerful blast from a volcano? How about the Tunguska event? How about my neighbor's kids at 7 in the morning playing in their back yard? The article Effects of nuclear explosions says that nuke blasts are done to achieve 20 psi blast over the maximum area (rather than max blast per se). So how loud woud that be in dB? Edison 14:56, 31 May 2007 (UTC)

Using your 20 psi (138 kPa) as the peak value of a sinusoidal overpressure, then I make it 20 log10 ( [138 kPa / sqrt(2)] / [20 μPa] ) ~= 194 dB SPL.

You could also compare sound intensity levels to get a figure in dB. The reference power density for SIL is 10E-12 W/m^2. You need to assume a distance from the bomb, say 100 metres. Then your 5.4E24 watts would be spread over 126,000 m^2, giving you 4.30E19 W/m^2. 10 log one over the other gives ~ 316 dB SPI. Even if you hold the thing at arm's length, say 1 m away, that only adds 40 dB. --Heron 20:18, 31 May 2007 (UTC)