Aball whose mass is 0.4 kg hits the floor with a speed of 4 m/s and rebounds upward with a speed of 1 m/s. if the ball was in contact with the floor for 2.0 ms (2.0 × 10-3 s), what was the average magnitude of the force exerted on the ball by the floor? n calculate the magnitude of the gravitational force that the earth exerts on the ball: n

The average  magnitude of the force exerted on the ball by the floor is 600 N.

The gravitational force that Earth exerts on the ball is 3.92 N.

The given parameters;

The average  magnitude of the force exerted on the ball by the floor is calculated as follows;

The gravitational force that Earth exerts on the ball is calculated as follows;

F = mg

F = (0.4 x 9.8)

F = 3.92 N

Learn more here:link

Explanation:

It is given that,

Mass of the ball, m = 0.4 kg

Initial speed of the ball, u = 4 m/s

Final speed of the ball, v = 1 m/s

Time of contact of the ball and the floor,

(a) Let F is the force exerted on the ball by the floor. It can be calculated using the definition of impulse. It is given by :

F = -600 N

|F| = 600 N

(b) The magnitude of the gravitational force that the Earth exerts on the ball is equal to the weight of the ball. It is given by :

W = 3.92 kg

Hence, this is the required solution.

car a and b are moving in opposite directions

plz mark.me the brainliest i rlly need this kind of yours in exchange

colorectal cancer may also develop from areas of abnormal cells in the lining of the colon or rectum. this area of abnormal cells is called dysplasia and is more commonly seen in people with certain inflammatory diseases of the bowel such as crohn's disease or ulcerative colitis.