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Dmitriy789 [7]
3 years ago
14

A 51-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37.0° above the hor

izontal. If the tension in the rope is 115 N, how much work is done on the crate to move it 7.0 m?
Physics
1 answer:
fiasKO [112]3 years ago
3 0

work is done by the pulling force which is same as the tension force in the rope. the net work done is zero for the crate since crate moves at constant velocity. but there is work done by the tension force which is equal in magnitude to the work done by the frictional force.

T = tension force in the rope = 115 N

d = displacement of the crate = 7.0 m

θ = angle between the direction of tension force and displacement = 37 deg

work done on the crate is given as

W = F d Cosθ

inserting the values given above

W = (115) (7.0) Cos37

W = 643 J

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Answer:

v = 1.15*10^{7} m/s

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given data:

charge/ unit area= \sigma = 1.99*10^{-7} C/m^2

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we know that

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Work done by charge q id\Delta X =\frac{ q\sigma \Delta x}{\epsilon}

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solving for v

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Use the given data to calculate the total mass of hydrogen available for fusion over the lifetime of the sun.
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Answer:

The new kinetic energy would be 16 times greater than before.

Explanation:

Kinetic energy is found using this formula:

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We can see that kinetic energy is directly proportional to the square of the velocity, meaning that if the speed was increased by 4 times, then the kinetic energy would get increased by a factor of 16.

The velocity just before the ball hits the ground can be found by the equation:

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Let's substitute h = 10 m and h = 40 m into this formula.

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We can see that the velocity increases by a factor of 4 (10 m → 40 m).

Therefore, this means that the kinetic energy would also be increased by a factor of (4)² = 16. Thus, the answer is D) The new kinetic energy would be 16 times greater than before.

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