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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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Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

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Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
labwork [276]

Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

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The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

\tau=NIAB\ \sin\theta

Let us assume that, A=0.0008\ m^2

\theta is the angle between normal and the magnetic field, \theta=90^{\circ}-30^{\circ}=60^{\circ}

Torque is given by :

\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

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3 years ago
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