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Ludmilka [50]
3 years ago
15

A person is able to pull a rope with a force of 700 N. What is the minimum number of fixed and moveable pulleys that the person

will require to lift up a 27,800 N elephant? Assume the pulleys themselves are weightless and the system has 100% efficiency.
A. 10 fixed, 10 moveable
B. 15 fixed, 15 moveable
C. 20 fixed, 20 moveable
D. 30 fixed, 30 moveable
Physics
2 answers:
Sav [38]3 years ago
8 0

<u>Answer: </u>

C option is  correct

20 movable and 20 fixed pulleys required

<u>Explanation: </u>

To lift a 27,800 N elephant with a 700 N drive,  

27800/700=39.7  

the individual requires a mechanical favorable position of 39.7 which is equivalent to 40. A solitary movable pulley gives a mechanical favorable position of 2, two portable pulleys yield a mechanical preferred standpoint of 4, and a framework with three movable pulleys results in a mechanical favorable position of 6. In light of this example, 20 movable pulleys and 20 fixed are required to give a mechanical preferred standpoint of 40

Katen [24]3 years ago
4 0

Answer:

C. 20fixed, 20moveables

Explanation:

Efficiency of a machine = MA/VR × 100%

MA is the mechanical advantage

VR is the velocity ratio of the pulley

MA = Load/Effort

MA = 27,800N/700N

MA = 39.71

The velocity ratio formula depends on the type of machine. Since the machine in question is a block and tackle system, the velocity ratio will be the number of pulleys in the system.

Using the efficiency formula to get the VR

100 = 39.71/VR × 100

39.71/VR = 1

VR = 39.71

VR = 40(approximately)

This means that the minimum number of pulleys that the person will require to lift the load is 40pulleys i.e 20fixed, 20moveables

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where:

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\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

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\alpha = \frac{\omega_f-\omega_i}{t}

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\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

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t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

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\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

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\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

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