If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 55.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

14.79 kgm/s

Explanation:

Data provided in the question

Let us assume the mass of baseball =  m = 0.145 kg

The Initial velocity of pitched ball = = 47 m/s

Final velocity of batted ball in the opposite direction = = -55m/s

Based on the above information, the change in momentum is

= 14.79 kgm/s

Hence, the magnitude of the change in momentum of the ball is 14.79 kg m/s

i wanna say it’s between b and d, leaning more towards b.

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