Can a marble be suspended in the exact center of a volume of fluid if enough pressure is applied to the fluid?
90516
This can be done with a density gradient column.  It is made of 2 fluids that are carefully mixed so that the density of the mixture varies from the top of the container to the bottom.  An object will then come to rest where its density matches the mixture.  This is actually a very accurate way of determining the density of materials.  However, over time the fluids in the column mix by diffusion and will eventually cause the suspended objects to sink or float.
695604
The only way I can see this being done is to have a piston or other way to separate the water and air; otherwise, the air will simply bubble up through the water.\n\nAssuming there is a piston or other separation device and the water will overflow at the top, the initial pressure needed to begin the displacement of the water is the height of the water in inches  times its density in pounds per cubic inch. Taking the density of water as 62.4 pounds per cubic foot and converting to pounds per cubic inch the density becomes .036111. The height is of course 100 ft time 12 inches per foot or 1200 inches. This yields an initial pressure of 1200 time .036111 or 43.33 psi. As the piston moves upward, the required pressure will decrease directly in proportion to its height or .4333 psi per 12 inches of rise.
867059
A curious result is that if you put a charged object anywhere inside a spherically symmetric charged hollow sphere, the net force on that object is zero.  It doesn't have to be in the centre.  This is a consequence of Gauss' Law.
60322
Archimedes principle on bouyancy states that a body submerged in a fluis is subjected by an unbalance force called bouyant force equal to the weight of the displaced fluid. By formula, Fb = Vd\n;Fb = bouyant force\nV = volume of te dispaced fluid\nd = density of the fluid\n\nDimensional Analysis,\nFb must had a unit of force; Kg, N, lb\nV is volume; cu.m, cu.ft\nd is density; kg/cu.m. and so on.\n\nFb(Kg) = V(cu.m.)*d(Kg/cu.m.)\n\nu'll get Kg.
883842
the pressure will be higher but the density won't change, it can't possibly change if it's a sealed, rigid container, density is mass/volume and you aren't changing the mass or the volume, therefore the density will be unchanged
317039
Imagine the stretching of the surface of the drum as the stretching of a spring. use hooke's law to do calculation. u will need to know about the strength of material and etc. very very very complicated
922775
If everything were lined up exactly inline and marble 1 hits 2 which hits 3, then both marbles 2 and 3 would move together at the same rate as marble 1.    (1~~>2~~>3)   You have probably seen that demonstrated with the newton's pendulum toy that has 5 chrome balls hanging by strings.   When you pull 1 of the balls back and let it swing into the other 4, 1 ball swings off the other end.  2 balls cause 2 to swing off the other end.\n\nIn your experiment, the 1st marble will hit the other 2 and release all of it's energy.  They will then go at the same rate as the first marble.\n\nV1 = 0\nV2 = 15.5 in a direction 60 degrees off to left.\nV3 = 15.5 in a direction 60 degrees off to the right.
188175
it is possible but it would take an extreme amount of pressure and you might not be able to manually pump it full enough, you would have to use an electric pump and make sure the air needle can't pop out.
288682
The most common way to think of that would be, by lifting the marble higher.  By lifting it you give it potentiol energy that can be released by dropping it.  You could also push it into a spring.  As the spring gets tighter you are storing Potential enrgy that can be released by letting go of the marble.  A sling shot would do the same thing.
860910
Pour the liquid in the graduated cylinder and there will usually be mL markings along the side.  This measurement is volume
97560
Are you referring to a fluid mechanics problem, or?\n\nCan you restate the question?
979511
a) If the subject was a gas, you can use Boyle's Law to calculate the change in volume. The Boyle's Law states that the pressure of gas is inversely proportionate to its volume, and hence P1V1 = P2V2.\nHowever, Boyles Law only covers the case of gas, and liquids are only VERY SLIGHTLY compressible. the volume of water doesnt really changes.\n\n***Anyway, how do u even carry a litre of water in an ocean? u can use a bubble, and you cant use any use any hard cased material cos tt will prevent you from finding the change in volume!\n\nb) ???\nUnless you are so sure you can find the change in volume, the density remains the same. I don't know any formula similar to Boyles Law that can explain the relation between the pressure of water and its volume.
1093836
Centrifugal force and gravity.
182931
Lift under a liquid (or BOUYANCY FORCE) is defined as the weight of the volume of liquid displaced.  In other words, say a ball displaces 1 cubic meter of water.  The ball will be pushed back up with the same amount of force as the weight of that 1 cubic meter of water.\n\nSince water's density is 999 kg/cubic meter, we're looking at 999 kg of water (water is really heavy!).\n\nNow if we want the force of that water, we'd simply multiply it by the gravitational acceleration, 9.8 m/s/s:\n\n999 * 9.8m/s/s = 9790 Newtons (1 Newton = 1 (kg * m)/(s^2))\n\nSo the ball would be pushed back up with a force of 9790 Newtons.\n\nNow, the hard shell ball will (theoretically) not deform when subjected to the pressure of 10 feet of water.  So the *bouyance force* placed on the hard shell ball is easy to calculate, given the formula above.\n\nThe latex balloon, however, will shrink as it is exposed to increasing pressure, so the volume of water it displaces will be smaller. (I will not solve that problem here as it takes up quite a bit of space to show how much a balloon shrinks as it is subjected to increasing pressure.  Essentially it comes down to writing a spherical force balance and integrating over volume)\n\nConsequently, the bouyancy force exerted on the latex balloon will be smaller, and the ping pong ball would be forced back up through the water harder (and faster) than the balloon.\n\nAs an extreme, ever wondered what happens to helium filled balloons when they are released into the air and they float up until you can't see them anymore?  Since the atmospheric pressure is lower and lower the higher you go, the balloon expands until eventually it pops.  At that point, the popped balloon either falls back to earth or gets swept up into atmospheric currents, which can take it basically anywhere on earth.
858875
First off, much of the interior of the earth is molten after you get about 30K down, so you couldn't really drill a hole through it. But for arguments sake, let's say you built a super-strong tube out of some nano-material and inserted it through the center of the Earth from pole to pole. Now you drop your stone.\n\nThe stone would fall towards the center of gravity, the center of the Earth. It would accelerate as it fell, so it would presumably fall past the center and head towards the South Pole. As it headed towards the South Pole, the gravitation pull would slow it down. Would it manage to to escape the Earth's gravity? No.\n\nConsider that the stone being at the top of the North Pole represents "potential kinetic energy". You could have generated energy from its fall much like a dam creates energy from falling water. Instead, it fell freely towards the center of the Earth, and all its potential kinetic energy has been turned into velocity. That velocity can't be enough to go any further than it fell (that is, out of the South Pole) because where would the excess energy come from?\n\nNow when the stone is at the center of the Earth, it has no potential kinetic energy. It can't fall anywhere, because every direction is up. All of its potential kinetic energy has been turned into velocity and so it should "fall up" the distance that it fell down. When it reaches the South Pole, it will have no velocity left, but it will have the same potential kinetic energy as when it started to fall. So you might expect it to fall from pole to pole indefinitely.\n\nHowever, if the tube is full of air, we have a problem. Falling objects normally have a "terminal velocity", which is the speed at which the air resistance stops the object from accelerating further. The air pressure at the center of the Earth would be enormous (air would probably be a liquid at these pressures and temperatures), as it would be subject to the weight of the air both north and south of the hole pulled in towards the center. The increasing density of air would slow the object as it fell to where it had little momentum as it approached the center. It might overshoot somewhat, but friction would continue to slow it until it was drawn to the center of gravity.\n\nNow if the tube contained a perfect vaccuum, the object should continue to fall from pole to pole indefinitely...except there's the problem that the stone would hit the sides of the hole due to the rotation of the Earth. See Foucault's Pendulum. This would slow it down, causing it again to settle at the center of the Earth.\n\nThere you have it. No need to drill through the center of the Earth now.
547508
No. \nThe system is static and the sum of the forces equal zero, but there definately are forces flowing through the bar.  Replace the bar with a piece of plastic and then what happens.  The system is no longer in equilbruim, becomes dynamic and the plastic gets torn to shreads.  The same would be true if say you used a piece of wire.  Just because something is not moving does not mean there are no forces flowing through it.
618509
fill water balloons with different amounts of water and drop them at the same height to see if they fall differently.  =)