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Amiraneli [1.4K]
3 years ago
11

A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

e 0.8 and 0.4, respectively. A horizontal string is attached to the block and a constant tension is maintained in the string. What is the force of friction acting on the block if the tension is 24 N? Answer in units of N.
Physics
1 answer:
Ne4ueva [31]3 years ago
6 0

Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

Coefficients of static = 0.8

Kinetic friction = 0.4

Tension = 24 N

We need to calculate the maximum friction force

Using formula of friction force

f=\mu mg

Put the value into the formula

f=0.8\times37

f=29.6\ N

So, the tension must exceeds 29.6 N for the block to move

We need to calculate the frictional force acting on the block

Using formula of frictional force

f = \mu N

Put the value in to the formula

f=0.4\times37

f=14.8\ N

Hence, The frictional force acting on the block is 14.8 N.

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Answer:

393.6m/s

Explanation:

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Answer:

6.3\cdot 10^{-4} V

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According to Faraday-Newmann-Lenz, the induced emf in the loop is given by:

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We know that the magnetic flux through the loop is given by

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