a rock is dropped from a tower and falls freely under the influence of gravity. which statement is true with regards to the rock as it falls. 1)gain an equal amount of momentum during each second 2)gain an equal amount of kinetic energy during each second 3)gain an equal amount of speed for each meter it falls 4)gain an equal amount of momentum for each meter is falls 5)the amount of momentum it gains will be proportional to the amount of potential energy it loses.
should be (1.) This question should be in the physics section lol
yea i know that =) but there is no one in the physics section thanks tho for your help
THe answer is 1, but do you know why it is one? It's important that you do. 1 says that it gains and EQUAL amount of momentum every second. What does that mean? It means that it is constant. The units of momentum include velocity right? Which is m/s, so for every second it falls a meter, every second. that is the unit is now m/s^2, which is the unit of acceleration. In this case, gravity is your acceleration.
yup and every second it falls it adds on another 9.8m/s^2 downwards
Well wait, gravity is constant, in free-fall an object is under the force of gravity and that is it, it doesn't add upon it's self right? What does the graph of acceleration vs. time look like for the acceleration of gravity on objects close to earth's surface?
under constant acceleration?
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9.81(m/s^2) is the slope, not the y value
You are correct, almost, razzputin. But remember, acceleration is a vector quantity. As cool as it would be to have the force of gravity going in the positive, or upwards direction, that might also be terrifying. Gravity pulls objects towards the center of the earth, not away from it. So in the case of free-fall, the acceleration of gravity is negative. Does that make sense? Things fall to the earth right? @shamuta is acceleration where the slope, you would be plotting velocity vs. time. not acceleration vs. time.
*if, were my bad... it's really late here.
yup this accelration is negative as it is pulling your center of mass towards the earths cebter of mass,
I meant in relation to the problem, since the original question's answer mentions momentum
Make sure you graph the acceleration correctly, it would be the same exact thing, but in the 4th quadrant right?
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