Posted by Rosa on Thursday, January 10, 2013 at 9:50pm.
(Force)*(Contact time) = (Change in momentum) = 0.05*80 = 4.0 kg*m/s
Contact time
= (4.0 kg*m/s)/(3000 kg*m/s^2)
= 1.33*10^-3 s
Related Questions
physics - please help - A golf ball of mass 0.045 kg is hit off the tee at a ...
calculus - A golf ball of mass 0.05 kg is struck by a golf club of mass m kg ...
College Physics - A golf ball (mass 0.045 kg) is hit with a club from a tee. The...
physics - The physics of a golf ball and kinetic energy. When a 0.045 kg golf ...
physics - When a 0.045 kg golf ball takes off after being hit, its speed is 35 m...
Physics - When a golf club strikes a golf ball, the club is in contact with the ...
physics - A 10.67 kg bowling ball has speed 4.35 m/s before hitting the floor. A...
physics - A 10.67 kg bowling ball has speed 4.35 m/s before hitting the floor. A...
physics - After a golf ball is hit it takes off with an initial speed of 22.0 m/...
Physics - The head of a golf club exerts a force of 350N on a golf ball for 0.3m...
For Further Reading