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Nata [24]
4 years ago
7

A boxed 12.0 computer monitor is dragged by friction 7.50 up along the moving surface of a conveyor belt inclined at an angle of

35.9 above the horizontal. If the monitor's speed is a constant 2.40 , how much work is done on the monitor by friction, gravity, and the normal force of the conveyor belt?
Physics
1 answer:
sasho [114]4 years ago
8 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

Below is the solution:

W done by Normal = 0. (make the incline flat, Normal force goes directly up: no work done) 
<span>W done by gravity = w*displacement = (11kg*9.8) * 7.5sin(35) = -463J </span>
<span>W done by friction is the opposite of the work done by weight because the object is not moving. Therefore W done by friction = 463J</span>
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Answer:

T_f=5854.76 °C

Explanation:

Given:

mass of hiker, m= 63 kg

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We find the energy required for climbing 828 m height:

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W=63\times 9.8\times 828

W= 511207.2 J

∵Hike eats 2 energy bars= 2\times 1.1\times 10^{6} J

Energy produced= 2.2\times 10^{6} J

Now, according to her efficiency:

Total energy required for producing the work of W= 511207.2 J which is required to climb the given height will be (say, E):

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&

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Q=2044828.8-511207.2\\Q=1533621.6J

As we know that:

heat, Q=m.c. (T_f-T_i).................(1)

where:

T_f is the final temperature

Putting respective values in the eq. (1)

1533621.6= 63\times 4.184\times (T_f-36.6)

(T_f-36.6)\approx 5818.16

T_f\approx 5854.76 °C

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