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Snowcat [4.5K]
3 years ago
11

An input for of 80 N is used to lift an object weighing 240 N with a system of pulleys. How far must the rope around the pulleys

be pulled in order to lift the object a distance of 1.4 m?
Physics
1 answer:
sammy [17]3 years ago
7 0

Answer:

4.2 m

Explanation:

Note: If energy is conserved, i.e no work is done against friction

Work input = work output.

Work output = Force output × distance,

Work input = force input × distance moved moved.

Therefore,

input force×distance moved = output force × distance moved........................Equation 1

Given: input force = 80 N, output force = 240 N, output distance = 1.4 m

Let input distance = d

Substitute into equation 1

80×d = 240×1.4

80d = 336

d = 336/80

d = 4.2 m.

Thus the rope around the pulley must be pulled 4.2 m

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What is the principal of moment​
antiseptic1488 [7]

Answer:

hope it helps...

Explanation:

The Principle of Moments states that when a body is balanced, the total clockwise moment about a point equals the total anticlockwise moment about the same point.

4 0
3 years ago
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Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal t
MrRa [10]

Answer:

(C). The line integral of the magnetic field around a closed loop

Explanation:

Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux.

This can be written mathematically as;

EMF = -\frac{\delta \phi _B}{\delta t}

(\frac{\delta \phi _B}{\delta t} ) is the rate of change of the magnetic flux through a surface bounded by the loop.

ΔФ = BA

where;

ΔФ is change in flux

B is the magnetic field

A is the area of the loop

Thus, according to Faraday's law of electric generators

∫BdL = \frac{\delta \phi _B}{\delta t} = EMF

Therefore, the line integral of the magnetic field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop.

The correct option is "C"

(C). The line integral of the magnetic field around a closed loop

8 0
3 years ago
Can someone check if what i wrote so far makes sense and if i made a mistake.
Serjik [45]
I don't really know what it's about but everything looks okay to me. There might be some mistakes on the last sentence but i'm not completely sure.
7 0
3 years ago
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A force of 10. N toward the right is exerted on a wooden crate initially moving to the right on a horizontal wooden floor. the c
Artyom0805 [142]

1) 5 N

The crate is initially moving, so we must calculate the force of kinetic friction, which is given by:

F_f = \mu_k mg

where

\mu_k=0.2 is the coefficient of friction between the crate (made of wood) and the floor (made of wood). The coefficient of kinetic friction between wood and wood is about 0.2.

mg=25 N is the weight of the crate

Substituting the numbers into the formula, we find

F_f=(0.2)(25 N)=5 N


2) 5 N

There are two forces acting on the crate along the horizontal direction:

- The force that pushes the crate toward the right, of magnitude F=10 N

- The force of friction, which acts in the opposite direction (so, towards the left), of magnitude F_f = 5 N

Since the two forces are in opposite directions, the net force is given by their difference:

F_{net}=F-F_f = 10 N-5 N=5 N


3) Yes

The crate is accelerating. In fact, according to Second Newton's Law:

F_{net}=ma (1)

where Fnet is the net force on the crate, m is its mass, a is its acceleration. We can immediately see that since Fnet is not zero, the acceleration is also non-zero, so the crate is accelerating.

We can even calculate the magnitude of the acceleration. In fact, the mass of the crate is given by:

m=\frac{Weight}{g}=\frac{25 N}{9.8 m/s^2}=2.55 kg

And by using (1) we find

a=\frac{F_{net}}{m}=\frac{5 N}{2.55 kg}=1.96 m/s^2


3 0
3 years ago
Answer this question thanks
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