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Likurg_2 [28]
3 years ago
14

Trevor is pulling his younger brother in a wagon. He pulls the wagon 50 m to the corner with a force that is a parallel with the

ground. He then turns the wagon around and pulls it 50 m back to the starting point with the same force at an angle of 30 degrees with respect to the ground. How does the amount of work Trevor performs going to the corner compare with the amount of work he performs coming back?
A) they are the same because he uses the same amount of force
B) they are the same because he travels the same distance
C) he does less work coming back because the force moving the wagon is only 50cos30
D)he does more work coming back because the force moving the wagon is only 50cos60
Physics
2 answers:
Nady [450]3 years ago
6 0

The answer is C I just took the quiz & got 100%

lidiya [134]3 years ago
4 0

Work has the greatest value when the force done is parallel to the direction of motion. In this case, the work done in pulling the wagon to the corner is greater than the amount of work done in pulling the wagon back to the starting point. This is because an angle of 30 degrees is applied when pulling it back. The formula for work when angle is applied is:

Wx = W cos θ

Where in this case, θ = 30

Therefore, the answer to this is letter:

<span>C) he does less work coming back because the force moving the wagon is only 50cos30</span>

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Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.
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To solve the problem it is necessary to take into account the concepts related to energy efficiency in the engines, the work done, the heat input in the systems, the exchange and loss of heat in the soupy the radius between the work done the lost heat ( efficiency).

By definition the efficiency of the heat engine is

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Where,

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The work done is defined as,

dW = \epsilon(dQ_h)

Where Q_h represents the input heat and at the same time is defined as

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Where,

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The change at the work would be defined then as

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On the other hand we have that the heat lost by the soup is equal to

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The ratio between both would be,

\frac{W}{Q_h} = \frac{c_v (T_h-T_c)-c_v T_c ln(\frac{T_h}{T_c})}{c_v (T_h-T_c)}

\frac{W}{Q_h} = \frac{1+ln(\frac{T_h}{T_c})}{1-\frac{T_h}{T_c}}

Replacing with our values we have,

\frac{W}{Q_h} = 1+\frac{ln(\frac{340K}{300K})}{1-\frac{340K}{300K}}

\frac{W}{Q_h} = 0.0613

Therefore the fraction of heat lost by the soup that can be turned into useable work by the engine is 0.0613.

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