What will happen to the sky diver once the air drag equals their weight?  (Newton's 2nd Law)?
969296
Their acceleration becomes zero\nThey will maintain constant velocity from then on\nThey will no longer accelerate
830124
Nothin would happen.\nIf you consider two bodies the earth (name it E) and a large mass (m). \ninitially M rests on E. and you lift mass M. the force used to lift it will also act on E.\nAlso if you consider the mass M and E. as a system force exerted on dropping M would be internal and since internal forces do not cause acc. the Earth will not get unbalanced.\n\nThats good for us isn't it...
1152217
Yes since it would reach it's terminal velocity after a few feet. &  DUE TO THE DRAG IT WOULD NOT ACHIEVE THE SPEED OTHERS are THINKING IT  would ACHIEVE.Given that air is present .
989278
The effect which causes weight is actually an acceleration, and in the case of on the surface of Earth it is due to gravity.  Weight may also be caused by acceleration due to motion in a straight line, such as with astronauts during liftoff having their weight "increased".\n\nSo to answer your question depends upon if any acceleration or deceleration is happening during the journey.  If the astronauts are moving toward another planet at a constant speed, then they are completely weightless.  (Well, they will of course be pulled a very tiny amount toward any gravitational sources, but unless they are massive and nearby, the effect will be practically insignificant.)  If they are moving under acceleration, then they will feel "weight" based upon the amount of acceleration.
225537
At sea level, the pressure the air exerts downward is at 1 atmosphere (normally).  If you decend 32 feet below water, the water pressure acting on you is 1 atmosphere as well, even greater as you go deeper.\n\nWhen you try to suck air down underwater, the water is trying to push it back up.  At some point, you just wont have enough "sucking power" to over come the weight of the water.
400397
I guess the only opposite reaction Archimedes got was from the townsfolk when he did his infamous streak.\nAll Archimedes described was displacement for a given volume.If it weighs less than the same volume of water it'll float.( Same applies to Zeppelins in air)\n\nAn equal opposite reaction can be observed when you drop a life raft from a helicopter. It'll bounce up out of the water, when that energys spent it'll settle back down niceley conforming to Archimedes.
152879
No, it will not.  The other answers to this question are on the right track, in that they point out that an object's density is the key thing in most situations, since if it is less dense than the fluid it is in, it will displace fluid equal to its weight before it is completely submerged, and hence float.\n\nBut in *this* problem, we are told from the outset that there is not enough water to be displaced.  Therefore, the object will displace all the water, all 1 lb. of it, and still not have displaced enough water to equal its weight.  Therefore, it will rest on the bottom, i.e. not float.
868820
the answer above is wrong to a degree - in your example i would have to imagine falling through a vacuum therefore if he didn't touch the sides his momentum would carry him to the other side and then back again over and over\n\nthe answer above ( 1st ) assumes air resistance would slow him down but he would go back and forth for a long time before stopping\n\nthe gravity would decrease on the way down and increase on the way up but not in the way you think what would happen is that the gravity would be pulling from the side more and more the deeper you got ( so the effect is only virtual) you would always be in a 1g field it is just the direction(s) of the pull that would change\nthe direction of the most mass at any one time would appear to be down and at the center all pulls ( 360 degrees ) would seem to cancel one another ( but they would still be there )
679199
What you are experiencing in the roller coaster, or an airplane in turbulence, for that matter, is Newton's laws of motion, i.e.,"a body in motion will continue that motion until an outside force is applied".\n\nSo, in the case of the roller coaster, when it reaches the top of the course and starts down (that's when you feel it), your body continues its up-ward motion and direction according to Newton until a force is applied to it by way of a seat belt, or other restraint, thereby changing your body's direction.  This particular force is sometimes called a "negative g force" by astronauts and pilots, because it acts counter to gravity (g).  \n\n\nAnother way to look at it is that when the roller coaster starts down, it takes your seat with it, but your body continues going up! So it seems to you that the seat has dropped out from under you (same deal in the turbulent airplane!)\n\nNow, when the roller coaster reaches the bottom of the track, and starts back up, or the airplane climbs out of a dive, a positive g force (acting the same direction as gravity)is created. Then, you are jammed down in your seat, or, as Newton would explain, "the seat you are in is applying an upward force on your body, which wants to continue down in the direction it was going!\n\nAin't physics fun?
210848
Ignore Zelo's first parahgraph, he is pulling your leg.\nHis second paragraph is correct though. However, the difference in gravity between Denver and Orlando is barley measurable. I can provide calculations in anyone is interested in the details. \n\nOnto your main question. The air pressure inside is exactly the same indoors as it is outside. Houses and buildings are not perfectly sealed like a bicycle tire or light bulb is. Open doors/ windows and small leaks all over keep the pressure the same inside as it is outside, which is a good thing. If the  air pressure inside were much lower we couldn't breath or function normally. If it got to low too fast you would just plain explode! Don't worry though, the pressure on the ground simply can't get that low.\n\nAs far as pressure affecting your weight: It is true that the air pressure pushes down on your shoulders, but it also lifts up on your arm pits. Remember pressure is everywhere on you body in every direction. In the end all of those ups and downs cancel each other out, leaving only gravity to affect your weight. \n\nOddly enough, in  very very small way the air actually lifts you up a tiny tiny bit in the same way that a heavy object in a swimming pool is lighter when it is in the water. You are slightly, not an amount you could easily measure, lighter in the atmospher than you would be if there were no atmosphere at all.   \n\nThis and a few other reasons are why, in science, mass is used instead of weight. Mass is always the same while weight can be affected by outside things. \n\nThese are great questions, keep pondering and asking.
61784
That's an interesting qts. I assume that you wish to invlove pure physics in this qts. Forget all the heat at the core of the earth and also air resistance.\nConsider pure gravitation theory.\n\nIf say the fellow who fell in the hole has mass m\n\nThen force of gravity exerted by the earth is :\n\n  F = GMm/(r*r) ..... Where G - universal gravitation constant\n\n                            M - Mass of the earth\n                            r - distance between earth and body of mass m\n\n          acceleration due to gravity\n            = (1-h/R)g   .... where h is distance of body beneath the earth's surface and g is acceleration due to gravity at the surface of the earth.\n\nSo as the body comes closer to the core of the earth, h tends to R.\nSo the acceleration of the body goes on decreasing very much. So, body may not actually hover, since his velocity will be constant the instant it is at the core.
64850
A correction to the first answer. The formula terminal V = SQRT(2W/(Cd*R*A)) is correct, as it expresses the fact that W is balanced by the drag force V^2*Cd*R*A/2. However, for two same-sized smooth balls of different materials, only W changes; Cd stays the same. The air density R depends only on altitude.\nSince air density increases with decreasing altitude, the terminal speed decreases during a fall. If dropped from high enough, the actual fall speed, which initially = 0, increases until the (decreasing) terminal speed is reached, and then the sign of the net force (gravitational + drag) changes so the falling ball slows. For the remainder of the fall, the terminal and fall speeds both decrease, but the fall speed is slightly faster. This speed difference provides the net upward force to slow the ball.
616393
Any thing moving through the atmosphere experiences the force of gravity and a drag force due to the friction of the air as it moves through it. When the object is falling, the force of gravity is directed towards the ground and the drag force directed upwards. As the body moves faster, the drag force becomes larger and larger until it equals the force of gravity and the object then falls at a constant velocity. This velocity is called the terminal velocity. For a person jumping out of an airplane, in skydiving, it is roughly 120-150 miles per hour. The terminal velocity of a water drop is much smaller. A large raindrop (5 mm in diameter) will fall at a velocity of around 20 mph. If you get hit by a large raindrop you will definitely feel it. However its still not large enough to cause much pain. The smaller the raindrop the slower it falls and hence the less energy it has when it hits you. If you dropped a grain of sand the same size as a raindrop it wouldn't hurt either. Pebbles are much heavier than raindrops and hit your head with a lot more force. Force is directly proportional to the mass and velocity of an object.\n\nOf course hail, which is really just a bunch of frozen raindrops which have joined together, is actually less dense than water but will certainly hurt due to its large mass.
870489
E, G, J\n\nin this problem, it can be assumed that time of fall is of sec's only hence air resistance can be negligible. problem can be solved using hooke's law though i am not sure what principle bungee rope works. heant generation which will take place in rope will be much smaller then actual work done. There can't be a constant speed or accelaration coz both r changing constantly, for example to start of jump speed is zero, acc downwards, where as at bottom point speed is again zero but acc. is upwards. n in between speed is non zero. \nlast, u can treat him as a particle for simlification as long as u keep air resistance negligible. u must realize that neeta (coff. of air resistance is very small n play effective role at high speeds).\ncord applies different force at different moments becoz of hooke's law.\n\nI hope it helps......
777336
Everyone is weightless in space. \n\nWe float because of the lack of gravity. A persons weight, is basically a measure of gravitational force acting downward on them so no gravity= no weight. \n\nWeight is a persons mass (How much space they take up) , multiplied by the acceleration due to gravity at any point on earth ( the rate at which a falling object increases it's speed as it falls -- 9.8 meters per second) . \n\nIf you take away the force of gravity then no matter what the persons mass is, the weight is still zero because any number multiplied by zero is still zero. The person mass in space would still be the same. They would still be bigger, because their mass doesn't change, but yes they would float.
694162
nothing, the air is probably underneath the fluid and the position change will reverse that allowing the air to escape.
729735
both are the same if it is drop at the same height because the gravity speed is 10 ms-2 (9.81ms-2 more accurate)