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LiRa [457]
2 years ago
10

A weight lifter is trying to lift a 1500-N weight but can apply a force of only 1200 N on the weight. One of his friends helps h

im lift it all the way. What force was applied to the weight by the weight lifter's friend?
Physics
1 answer:
aliya0001 [1]2 years ago
4 0

Answer:

300 N

Explanation:

We have three forces acting on the object:

- W = 1500 N, downward: the weight of the object

- F1 = 1200 N, upward: the force exerted by the first man

- F2, upward: the force exerted by the friend

In order to lift the object, the sum of the two forces F1 and F2 must be at least equal to the weight of the object W. Therefore we can write

F_1 +F_2 = W

And from this equation, we can find the magnitude of F2:

F_2 = W-F_1 = 1500 N-1200 N=300 N

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A 12.0kg microwave oven is pushed 14.0m up the sloping surface of a loading ramp inclined at an angle 37 degrees above the horiz
Semenov [28]

Answer:

Answered

Explanation:

a) What is the work done on the oven by the force F?

W = F * x

W = 120 N * (14.0 cos(37))

<<<< (x component)

W = 1341.71

b) F_f=\mu_k N

F_f=0.25\times12\times9.8

= 29.4 N

W_f= F_f\times x

W_f= 29.0\times 14 cos(37)

W_f= 328.72 J = 329 J

c) increase in the internal energy

U_2 = mgh

= 12*9.81*14sin(37)

= 991 J

d) the increase in oven's kinetic energy

U_1 + K_1 + W_other = U_2 + K_2

0 + 0 + (W_F - W_f ) = U_2 + K_2

1341.71 J - 329 J - 991 J = K_2

K_2 = 21.71 J

e) F - F_f = ma

(120N - 29.4N ) / 12.0kg = a

a = 7.55m/s^2

vf^2 = v0^2 + 2ax

vf^2 = 2(7.55m/s)(14.0m)  

V_f = 14.5396m/s

K = 1/2(mv^2)

K = 1/2(12.0kg)(14.5396m/s)

K = 87.238J

4 0
2 years ago
The distance between two objects is increased by three times the oringinal distance. How will this change the force of attractio
tigry1 [53]
<span>The distance between two objects is increased by three times the oringinal distance.  Since they were already separated by one time the original distance,
the additional three times the oringinal distance now puts them four times the original distance apart.

Whether we're talking about the gravitational forces of attraction or
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So changing the distance to four times the original distance causes
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The forces become 1/16 of their original magnitude.<span>
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3 years ago
Amount of work done by a rotating object
Oduvanchick [21]
The work done by a rotating object can be calculated by the formula Work = Torque * angle.

This is analog to the work done by the linear motion where torque is analog to force and angle is analog to distance. This is Work = Force * distance.

An example will help you. Say that you want to calculate the work made by an engine that rotates a propeller with a torque of 1000 Newton*meter over 50 revolution.

The formula is Work = torque * angle.

Torque = 1000 N*m

Angle = [50 revolutions] *  [2π radians/revolution] = 100π radians

=> Work = [1000 N*m] * [100π radians] = 100000π Joules ≈ 314159 Joules of work.

 
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2 years ago
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Suppose that two objects attract each other with a gravitational force of 16 units. If the mass of both objects was doubled, and
Lady_Fox [76]

Answer:

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6 0
2 years ago
I need help with part D
natka813 [3]
1).  The little projectile is affected by friction all the way through the block.
Friction robs some kinetic energy.

2).  The block is affected by friction as it scrapes along the top of the post.
Friction robs some kinetic energy.

3).  The block is also affected by friction with the air (air resistance) as it
falls to the ground.  Friction robs some kinetic energy.
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3 years ago
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