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elena-14-01-66 [18.8K]
3 years ago
13

In the pulley system shown below, a 360 N weight is slowly lifted. Assuming the system is 100% efficient and each pulley is weig

htless, what is the minimum input force needed to lift the weight? Rope Pile L= 16.74m
A. 61 N
B. 121 N
C. 181 N
D. 361 N
Physics
2 answers:
STALIN [3.7K]3 years ago
8 0
Your answer is D. 361 N
Sveta_85 [38]3 years ago
7 0

Answer:

A. 61 N

Explanation:

In this system, there are three moveable pulleys. In this case, the rope will have to move six times as far as the weight will rise. In other words, the input distance is 6 times the output distance  ( d in =6⋅ d out ) . Because the efficiency is 100%, the work input is equal to the work output, so we have:

work input  = work output

f( in)  ⋅  d( in) =   f( out)  ⋅  d( out)

 f( in)  ⋅ (6 ⋅  d( out))  =  f( out)  ⋅  d( out)

 f( in)  ⋅ 6 =  f( out)

  f( in)  ⋅ 6 = 360 N

  f( in)  = 60 N  

Another way to think of this is that since the rope has to move 6 times as far as the height of the weight, the mechanical advantage will be 6. Therefore, the input force needed is  360 ÷ 6 = 60 N . Of course, 60 N will balance the object, so a force slightly greater than 60 N (like 61 N) will be required to lift it.

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Answer plz also give reason
user100 [1]

See this suggested solution.

1. Let a force F' is the vector sum of the forces P and Q, then it is shown on the attached picture and marked with red color.

2. according to the condition the force F holds the object, then F should have the same length as the force F' and the opposite direction.

3. using the conditions described in 2. the answer is C.

4 0
3 years ago
Determine the net work a hiker must do on a 3.35-kg backpack to carry it up a
Sphinxa [80]

Answer:

410.4J

Explanation:

Step one:

given

mass= 3.35kg

weight= 3.35*9.81= 32.86N

h=12.49m

Required

The net work done

Step two:

the work done is given  as

WD= force* distance

WD= 32.86*12.49

WD= 410.4J

8 0
3 years ago
70 POINTS AND BRAINLIEST PLEASE HELP!!!The Moon orbits Earth. This orbit causes the Moon to look different over the course of ab
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Answer:

It always acurs after a 1st quarter

do you have photo?

Explanation:

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3 years ago
The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an
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Answer:

-4.941*10^8J.

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

\Delta U = \Delta_{perogee}-\Delta_{Apogee}

Where,

U= \frac{-GmM_e}{r}

Replacing

\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}

\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})

Our values are given by,

m = 85.5Kg

M_e = 5.97*10^{24}Kg

r_A = 7330Km

r_p = 6610Km

G = 6.67*10^{-11}Nm^2/Kg^2

Replacing at the equation,

\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})

\Delta U = -4.941*10^8J

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was -4.941*10^8J.

4 0
3 years ago
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Explanation:

Given that,

Distance between two long wires, d = 16 cm = 0.16 m

Current in one wire, I_1=2.9\ A

Current in wire 2, I_2=5.7\ A    

The magnetic force per unit length of one wire on the other is given by the following expression as :

\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi d}

\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 2.9\times 5.7}{2\pi \times 0.16}

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The current is flowing in opposite direction, the magnetic force acting on it is repulsive. Hence, this is the required solution.

5 0
4 years ago
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