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mote1985 [20]
2 years ago
14

A 50-kg block is at rest on a 15o slope. A force of 250 N is acting on the block up the slope parallel to it. If the block does

not slide up the slope, what is the minimum value of the coefficient of static friction between the block and the slope
Physics
1 answer:
Iteru [2.4K]2 years ago
6 0

The minimum value of the coefficient of static friction between the block and the slope is 0.53.

<h3>Minimum coefficient of static friction</h3>

Apply Newton's second law of motion;

F - μFs = 0

μFs = F

where;

  • μ is coefficient of static friction
  • Fs is frictional force
  • F is applied force

μ = F/Fs

μ = F/(mgcosθ)

μ = (250)/(50 x 9.8 x cos15)

μ = 0.53

Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.

Learn more about coefficient of friction here: brainly.com/question/20241845

#SPJ1

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W=-21,870,000\ J

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We have

\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}

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\boxed{W=-21,870,000\ J}

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Leno4ka [110]

Answer:

- After throwing the snow, velocity of the thrower is 2.33 m/s

- the velocity of the receiver is 0.026 m/s

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(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s

Now, When he throws it at 31.5 m/s, these constitutes a momentum of;

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To get his velocity, we say;

161.94035 = mv

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Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s

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the receiver will gain momentum of 1.49625 kg.m/s

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1.49625 kg.m/s = mv

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3 0
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Answer:

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This means the answer is B) x^2+6x+8

             

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