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Andru [333]
3 years ago
12

A person applies a force of 200N to move an object initially at rest weighing 400N. What could be the coefficient of static fric

tion between the object and the surface?
Physics
1 answer:
SOVA2 [1]3 years ago
5 0

Answer:

\mu=0.5

Explanation:

Given that,

Applied force, F = 200 N

Weight of the object, or the normal force, N = 400 N

We need to find the coefficient of static friction between the object and the surface.

The force of friction has the same magnitude as that of applied force. It can be given by :

F=\mu mg

mg is weight of the object

\mu=\dfrac{F}{mg}\\\\\mu=\dfrac{200}{400}\\\\\mu=0.5

So, the coefficient of static friction between the object and the surface is 0.5.

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vichka [17]

Answer:

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Given;

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The angular speed of the club is calculated as follows;

\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega =  (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s

The average linear velocity (inches/second) of the golf club is calculated as;

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Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

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