The tomato is dropped. what is the velocity, v, of the tomato when it hits the ground? assume 85.6 % of the work done in part a is transferred to kinetic energy, e, by the time the tomato hits the ground.

Ok, we know that you drop a tomato of mass M from a height h, because you drop it, it has not initial velocity.

Now, the kinetic energy is:

K = (M/2)*v^2

because at the begginig there is not velocity, the kinetic energy is zero.

the potential energy is:

U = M*g*h

where g = 9.8m/s^2

We know that when the tomato hits the ground, the 85.6% of these potential energy is converted in kinetic; so we have:

K = 0.856*M*g*h = (1/2)*M*v^2

M can be canceled in both sides:

v^2 = 2*0.856*g*h

v = √(1.712*g*h)

is the velocity of the tomato when it hits the ground.

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

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there are no atoms