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3 years ago

8

If the force of static friction on a crate is 67 N and the weight of the create is 289 N, what is the coefficient of static fric

tion?
a
19363
b
0.23
c
4.31
d
222

1 answer:

RUDIKE [14]3 years ago

8 0

Answer:

B) μ = 0.23

Explanation:

The coefficient of friction is equal to μ = F / N where μ (mu) is the coefficient of friction, F is the friction force, and N is the normal force (the force of an object being applied onto the earth by gravity).

F, the friction force, is given as 67 N

Since the weight of the crate is 289 N, that means the normal force is 289 N.

This means that the coefficient of friction is μ = F / N = 67 N / 289 N = 0.23183391 = 0.23

Therefore, B is the correct answer

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3 years ago

a small table has a mass of 4kg, stands on four legs, each leg having an area of 0.001 m2. what is the pressure exerted by the t

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Answer:

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Explanation:

Given that,

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2 years ago

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Black_prince [1.1K]

Ais the correct answer

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2 years ago

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In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

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Our values are:

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That way during each section the equations should be modified depending on the previous one, let's start:

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$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

$V_{f1}=0.8511m/s$

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$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

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$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

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3 years ago

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Answer:

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2 years ago