Respuesta :
Answer:
The magnitude of the force exerted by the club on the ball is 3015 N.
Explanation:
Given that,
Mass of the ball, m = 45 g = 0.045 kg
Speed acquired by the ball, v = 67 m/s
It is assumed that the club is in contact with the ball for about 0.0010 s.
To find,
The magnitude of the force exerted by the club on the ball.
Solution,
Initial speed of the ball is 0, as it is at rest.
The force acting on the ball is given by second law of motion as :
[tex]F=m\dfrac{v-u}{t}[/tex]
[tex]F=m\dfrac{v}{t}[/tex]
[tex]F=0.045\times \dfrac{67}{0.001}[/tex]
F = 3015 N
So, the magnitude of the force exerted by the club on the ball is 3015 N.
Answer:
3015 N
Explanation:
m = 45 g = 0.045 kg
u = 67 m/s
t = 0.001 s
v = 0
Force = rate of change of momentum
F = 0.045 x 67 / 0.001
F = 3015 N
