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3 years ago

9

When a golf club hits a 0.0459 kg ball at rest, it exerts a 2380 N force for 0.00100 s. What is the speed of the ball afterwards

? (Unit = m/s)

2 answers:

aksik [14]3 years ago

8 0

Answer:

51.85m/s

Explanation:

Given parameters:

Mass of ball = 0.0459kg

Force = 2380N

Time taken = 0.001s

Unknown:

Speed of the ball afterwards = ?

Solution:

To solve this problem, we use Newton's second law of motion:

F = m x $\frac{v - u}{t}$

F is the force

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

2380 = 0.0459 x $\frac{v- 0}{0.001}$

0.0459v = 2.38

v = 51.85m/s

stepladder [879]3 years ago

5 0

Answer:

2.38

Explanation:

F = dp/dt --> dp = F*dt ---> p = F*t if F is constant over time so

p = 2380N* 0.001 s = 2.38 kg-m/s

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Part a)

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