A fast ball thrown with a velocity of 40 m/s (approximately 90 MPH) is struck by a baseball bat, and a line drive comes back toward the pitcher with a velocity of 60 m/s. The ball is in contact with the bat for a time of just 0.04 s. The baseball has a mass of 120 g (0.120 kg). What is the change in momentum of the baseball during this process

12 kg•m/s

Explanation:

Let "toward the pitcher" be the positive direction

Δp = mΔv = m(vf - vi)

Δp = 0.120(60 - (-40)) = 12 kg•m/s

Not asked for, but the average force during impact is

Δp = FΔt   (Here the units are N•s which are identical to kg•m/s)

F = 12/0.04 = 300 N

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

ok

explanation: