A tennis ball with a speed of 11.5 m/s is moving perpendicular to a wall. After striking the wall, the ball rebounds in the opposite direction with a speed of 5.842 m/s. If the ball is in contact with the wall for 0.0129 s, what is the average acceleration of the ball while it is in contact with the wall? Take "toward the wall" to be the positive direction. Answer in units of m/s 2

1344.34 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 11.5 m/s

Rebound velocity i.e final velocity (v) = 5.842 m/s

Time (t) = 0.0129 s

Acceleration (a) =.?

The acceleration of the ball while it is in contact with the wall can be obtained as follow:

a = (v + u) / t

a = (5.842 + 11.5) / 0.0129

a = 17.342 / 0.0129

a = 1344.34 m/s

Therefore, the average acceleration of the ball while it is in contact with the wall is 1344.34 m/s

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2