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Otrada [13]
3 years ago
10

A(n) 93 kg clock initially at rest on a horizontal floor requires a(n) 617 N horizontal force to set it in motion. After the clo

ck is in motion, a horizontal force of 528 N keeps it moving with a constant velocity. The acceleration of gravity is 9.81 m/s 2 . a) Find µs between the clock and the floor
Physics
1 answer:
Viefleur [7K]3 years ago
5 0

Answer:

\mu_s = 0.676

Explanation:

As we know that the force required to move the clock from rest position must be equal to the maximum limiting friction

So we will have

F = F_f

now we know that

F_f = \mu_s Mg

here we will have

F = 617 N

m = 93 kg

now from above formula we will have

617 = \mu_s (93)(9.81)

\mu_s = 0.676

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D

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The next four questions refer to the situation below.
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This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

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with the velocity constant we can use the relations of uniform motion

           v_{sg1} = D / t_{out}

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        v_{sg 2} =  v_{sr}  - v_{rg}

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           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

           t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

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3 years ago
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The capacitance of parallel plate capacitor in free space is given by,

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Where \epsilon_{o}  = permittivity of free space = 8.85 \times 10^{-12}

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