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Llana [10]
2 years ago
11

A rope is wrapped three and a half times around a cylinder. Determine the range of force T exerted on the free end of the rope f

or maintaining equilibrium that is required to just support a 5 kN weight. The coefficient of friction between the rope and the cylinder is 0.2
Physics
1 answer:
Taya2010 [7]2 years ago
8 0

The range of force exerted at the end of the rope is 285.7 N to 1,000 N.

<h3>Net horizontal force of the cylinder</h3>

The net horizontal force of the cylinder when it is at equilibrium position is determined by applying Newton's second law of motion.

∑F = 0

F - μFn = 0

F - 0.2(5,000) = 0

F - 1,000 = 0

F = 1,000 N

The strength of the applied force increases as the number of turns of the rope increases.

minimum force = total force/number of turns of rope

minimum force = 1,000/3.5

minimum force = 285.7 N

Thus, the range of force exerted at the end of the rope is 285.7 N to 1,000 N.

Learn more about Newton's second law of motion here: brainly.com/question/3999427

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Answer:

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Which poison was used in the Jonestown massacre?
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That would be Cyanide. 

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3 years ago
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Jonathan has six strings that are the same thickness and are all the same material. He cuts the strings to different lengths and
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7 0
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Pls help
xenn [34]

Answer:

A. It must be zero

Explanation:

A spacecraft leaves the solar system at a velocity of 1,500 m/s. The net force on this spacecraft is zero. What can we say about the spacecraft's acceleration?

According to Newton's second law

Force = Mass × acceleration

If the net force is zero

0 = mass × acceleration

0 = ma

a = 0/m

a = 0m/s²

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4 0
2 years ago
A 65-kg person stands on a scale in a moving elevator while holding a 5.0 kg mass suspended from a massless spring with spring c
disa [49]

Answer:

The question is incomplete. However, I believe, it is asking for the acceleration of the elevator. This is 3.16 m/s².

Explanation:

By Hooke's law, F = ke

F is the force on a spring, k is the spring constant and e is the extension or compression.

From the question,

F = (1.08\text{ kN/m}) \times (6.0 \times 10^{-2}\text{ m}) = 64.8 \text{ N}

This is the force on the mass suspended on the spring. Its acceleration, a, is given by

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a = \dfrac{64.8 \text{ N}}{5\text{ kg}} = 12.96\text{ m/s}^2

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