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In a golf ball game, a person hits the golf ball with a club. The club is in contact with the ball, which is initially at rest, for 2.45 ms. The ball has a mass of 0.0550 kg and leaves the club with a speed of 1.80 X 102 m/s. What is the average force exerted on the ball by the club

Respuesta :

Answer:

F=4040.81 N

Explanation:

Given that

Time ,t= 2.45 ms

Mass ,m= 0.055 kg

v= 1.8 x 10² m/s = 180 m/s

We know that rate of change in the linear momentum is known as force.

Momentum P = m v

[tex]F=\dfrac{dP}{dt}[/tex]

Therefore force F

[tex]F=\dfrac{\Delta P}{\Delta t}[/tex]

[tex]F=\dfrac{0.055\times 180}{2.45\times 10^{-3}}\ N[/tex]

F=4040.81 N

Therefore force on the ball will be 4040.81 N

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