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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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1. Frequency: 7.06\cdot 10^{14} Hz

The frequency of a light wave is given by:

f=\frac{c}{\lambda}

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f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz


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The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

The frequency of this light wave is 7.06\cdot 10^{14} Hz (found in the previous exercise), so the period is:

T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s


4 0
3 years ago
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Answer:

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