Sspiked so that its incoming velocity of +3.27 m/s is changed to an outgoing velocity of -23.9 m/s. the mass of the volleyball is 0.350 kg. what is the magnitude of the impulse that the player applies to the ball?

Impulse, J = 9.50 kg-m/s

Explanation:

Mass of the volleyball, m = 0.35 kg

Incoming velocity, u = +3.27 m/s

Outgoing velocity, v = -23.9 m/s

We need to find the magnitude of the impulse that the player applies to the ball. It is equal to the change in momentum. It is given by :

J = -9.50 kg-m/s

So, the impulse that the player applies to the ball is 9.5 m/s but in opposite direction.

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j