Answer:

Explanation:
We use the kinematics equation to solve this question:

because the ball is dropped
the acceleration is the gravity, negative because it points downwards
initial height
final height
So:


Answer:

Explanation:
The gravitational potential energy gets transformed into translational and rotational kinetic energy, so we can write
. Since
(the ball rolls without slipping) and for a solid sphere
, we have:

So our translational speed will be:

<span>A spring is water coming from under the ground to the surface of the earth and a stream is water that is running along the ground through a trench like place on earth down a hill or steep a area.</span>
Momentum of the wagon increases by (200 x 3)
= 600 newton-sec
= 600 kg-m/sec