Answer:
The K.E of the bowling ball right before it hits the ground, K.E = 2450 J
Explanation:
Given data,
The mass of the bowling ball, m = 10 kg
The height of the building, h = 25 m
The total mechanical energy of the body is given by,
E = P.E + K.E
At height 'h' the P.E is maximum and the K.E is zero,
According to the law of conservation of energy, the K.E at the ground before hitting the ground is equal to the P.E at 'h'
Therefore, P.E at 'h'
P.E = mgh
= 10 x 9.8 x 25
= 2450 J
Hence, the K.E of the bowling ball right before it hits the ground, K.E = 2450 J
So directly the cell density value in a rectangle is:
In this exercise, we must first calculate the area of the rectangle, remembering that you need to convert from nm to cm is:
Next, let's calculate the density from the calculated area:
In this way, the calculation represents that there is:
Learn more: brainly.com/question/18320053
Answer:
12N
Explanation:
given- mass, acelation
Fnet=ma= .16kg*75m/s2
Fnet=12 N only force no friction given.
