holding these items at the same level then let them go what will hit the floor first
210275
The paper is out, because of wind resistance, leaving only the small and big rocks.  Refering to the 11th century-actually, Newton's law was supplanted some years ago, and we know that the acceleration due to gravity is determined by the combined gravitational forces of both objects.  Assuming an exactly equal shape and an exactly equal density of the two rocks, the larger would fall more rapidly, because of the involvement of it's own gravity.  If we assume, for example, that the small rock is the size of a baseball, and the big rock is the size of, oh...the moon, then the big rock will fall quite a bit faster (the combined gravity will be greater, because of greater mass).  As the large rock gets smaller, the gravity will decrease.  However, no matter how close in size the two rocks, there will be some difference.  It would not be easily measurable at anything less that a few million tons, but it would still be there.\nNewton was likely the greatest genius that ever lived...but we have moved beyond him in a lot of areas, including our understanding of gravity-
282659
Bang em on a table or other hard surface.  The shock may disrupt the magnetism.
16479
You can find the floor in a maximum of 14 attempts.\n\nDrop a bulb from the floors in this order:14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100\n\nIf the bulb breaks, stop and test starting from just above the last floor that it didn't break on, and go up one floor at a time. If the bulb breaks on the first attempt, resume the test from the first floor.\n\nExamples:\n14 (breaks), 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\n14, 27, 39 (breaks), 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38\n14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99 (breaks), 96, 97, 98\n14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100 (breaks)
391153
Trick question... all things regardless of weight fall at the same speed. You need to know how big the item is first... this is the only way to calculate resistance.
147950
to get to the other slide
431206
8 total shoes were dropped on the floor.\n\nEmily grabs 1. We are now down to 7 shoes on the floor. 1 of them is the match with Emily and 6 are not.\n\n6/7 is the chance its not the matching shoe.\n\nGood Luck!
386276
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m
450067
according to the conservation of momentum the second ball has the same momentum, though it depends on the angle at which the first ball collides with the second\n\nalso if you put English on the cue ball\n\nif it is a solid shot, the cue ball can stop and transfer all momentum to the second ball\n\nhope i could help
5070
im going to assume this ia an elastic collision (when the objects bounce off of each other).\nyou just need the formula for conservation of momentum. p=mv (momentum = mass times velocity)  conservation of momentum says that (assuming no friction) the total momentum before the collision would equal the total momentum after the collision.  in other words mv=mv.  now, there are two balls so your formula looks like this:\n mv+mv = mv+mv.  the reason it has 2 "mv" 's  on each side, is because there are two balls.  now the second mv on the first side can be taken out since the velocity is 0, so mass times velocity would be 0. that means mv + 0 = mv + mv  so the mass times velocity of the first ball before the collision equals the mass times velocity of the first ball after plus the mass times velocity of the second ball after.  this would look like this:\nmv= mv + mv\n(1kg)(10m/s)=(1kg)v1 + (2kg)v2 (where the 1 and 2 just represent the velocities of the first and second balls respectively.)\nso basically:\n10=v1 + 2v2\nyou would need one of the velocities in order to solve for the other. there has to be more to the problem to either give you one of these velocities, or enough information to solve for it somehow... the balls would move in opposite directions.\n\nhowever, we can solve this problem if it was an inelastic collision(meaning the balls stick together).  if this were the case, then (if this is a problem for class) it probably would have said it was inelastic and it probably isnt because balls generally bounce off of each other unless they're sticky or balls of clay or something, but lets say it was inelastic just in case so i can solve it this way for you.  if it was inelastic then mv + mv=mv so the mv of the first ball plus the mv of the second ball(which is zero like before because the ball is at rest so v is zero) equals the combined masses of the balls plus the new velocity. so mv + 0 = mv, because we have only one object(because they move as one object if they stick).  the combined masses of the balls is 1kg + 2kg = 3kg so now:\nmv=mv\n(1kg)(10m/s) = (3kg)v\nsolving for v we get 10/3 or 3.3333.  they would be moving in the same direction as the initial ball.
575581
use an inanimate object like a tree pnce u master basic drills find somthing that movess like a dog or other
129242
Top of the grip to the ground.
125490
Pretend not to notice and get off at the very next floor, whether it's the right one or not. Taking the stairs is good exercise and stairwells are usually empty, in case I need to let off a few more.
526671
The range of the marble as it hits the floor, should be 5.5m.\nHere's how I figured it out :\nFirst, you would use the Dy = ViT + 1/2 gT^2 formula, to find your time. you should end up with 1.5 = 4.9T^2\nThen, divide the 1.5 by the 4.9, and take the square root of that number. You should end up with something like 0.55328.....\nSo, your time = 0.5s.\n\nThen, use the Dx = VxT formula to find your distance when it hits the floor.\nSo, Dx = 10m/sx0.5s\nand, Dx should equal 5.5m\nTherefore, it took the marble 0.5s traveling at a velocity of 10m/s to hit the floor 5.5m away after being launched horizontally off a table that's 1.5m high.\n\nHope this helps!
8023
This question sounds like your teacher/ta/whoever wants you to make 3 assumptions (or they already did and you just didnt mention it):\nThey have the same mass (otherwise he would have to specify the masses of each)\nAir resistance is negligible (pretty common)\nThey all roll without slipping (otherwise they all get to the bottom together)\n\nIf you make these assumptions, the order is based on their respective moments of inertia about the center.  They each move downward at a speed based on their rotation, so the order they reach the bottom will be smallest rotational inertia first. Sphere, disk, ring.
429614
2 marbles.  It will be the one that was sitting alone, as well as the last of the 5.  In these perfectly elastic collisions, which we idealize for this thought experiment, the momentum is transferred without loss down the line, ending up only in the last recipient.  Similarly, if you shot 2 marbles at a single marble sitting alone, only the second of the 2 marbles would stop.  The first would keep moving, now with the one that had been sitting alone, so the quantity of moving marbles would be 2, the same as what was moving in the beginning, even though it is different individual marbles.
662458
yes .. in vacuum, all items will fall at same rate... it is the rate at which the gravity acceleration attacts everything to the center of teh Earth. But you might have a queation, that if the acceleration  is same then why different things weigh different... \n\nweight =mass X acceleration of gravity (9.8meter/second^2) and mass is the one which is a measure of matter in a body makes the difference... so if you let a 5kg weight and a 5,000kg weight fall in a vaccum, they will fall at the same rate and will reach the ground at the same time.
138242
put the sharp needles or stones on the floor