Acue stick hits a cue ball with an average force of 22 n for a duration of 0.029 s. if the mass of the ball is 0.15 kg, how fast is it moving after being struck?

4.25 m/s

Explanation:

Force, F = 22 N

Time, t = 0.029 s

mass, m = 0.15 kg

initial velocity of the cue ball, u = 0

Let v be the final velocity of the cue ball.

Use newton's second law

Force = rate of change on momentum

F = m (v - u) / t

22 = 0.15 ( v - 0) / 0.029

v = 4.25 m/s

Thus, the velocity of cue ball after being struck is 4.25 m/s.

4.25 m/s

Explanation:

An impulse is defined as a change in momentum, and can be found my multiplying an applied force by its duration. In this case,

that's 22×0.029 = 0.638 kg-m/s. Momentum is also mass times velocity, and your velocity is unknown.

Momentum P= mv

0.15 v= 0.638

v= 4.25 m/s

a=d/t

a="triangle"v/t

explanation:

a = acceleration

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