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goldfiish [28.3K]
3 years ago
9

If you use a horizontal force of 33.0 N to slide a 11.0 kg wooden crate across a floor at a constant velocity, what is the coeff

icient of kinetic friction between the crate and the floor?
Physics
1 answer:
alukav5142 [94]3 years ago
3 0

Answer:

μ= 0.3

Explanation:

Given that

F= 33 N

m = 11 kg

Given that crate is moving with constant velocity is means that acceleration of the crate is zero.If acceleration of the system is zero then the total force on the system will be balance.

Therefore

F= Friction force

F= μ m g

μ=Coefficient of friction

33 = μ x 10 x 11                       ( take g= 10 m/s²)    

3 = 10 μ

μ= 0.3

Therefore coefficient of friction will be 0.3 .

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Answer:

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A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wir
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Answer:

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Explanation:

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solution

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so

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and w = \frac{2*\pi}{T}  

w = \frac{d\theta }{dt}

w = 1.80 × 10^{6} × \frac{2*\pi}{1}

w = 11304000 rad/s

so here from equation 2

- 9 × 10^{9}  × \frac{2*linear\ charge\ density}{0.012} 1.80 × 10^{6} =  1.672 × 10^{-27} × 11304000² × 0.0120  

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