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pychu [463]
3 years ago
7

A gaseous refrigerant undergoes compression when 150 j of work is done on it. if the internal energy of the gas increases by 120

j, the amount of heat transfer is:
Physics
1 answer:
zhenek [66]3 years ago
3 0

The heat transferred is -30J

When the gas undergoes compression, work is done on the gas and its internal energy increases.

According to the first law of thermodynamics, the increase in internal energyΔU is the sum of the heat given to the gas ΔQ and the amount of work done on the gas ΔW

\Delta U=\Delta Q +\Delta W

The work done on the gas is 150 J due to compression and the internal energy of the gas increases by 120 J.

Therefore, the heat given to the gas is given by,

\Delta Q =\Delta U -\Delta W\\ =120 J -150 J\\  =-30 J

Thus, an amount of heat equal to 30 J flows out of the system.

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Answer:

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A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a forc
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Answer:

Part a)

a = 0.36 m/s^2

Part b)

a = -1.29 m/s^2

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Force applied by the student on the box is 80 N at an angle of 25 degree

so here two components of the force on the box is given as

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F_x = 80 cos25 = 72.5 N

F_y = Fsin25

F_y = 80 sin25 = 33.8 N

now in vertical direction we can use force balance for the box to find the normal force on it

F_n + F_y = mg

F_n = (25)(9.81) - 33.8

F_n = 211.45 N

now kinetic friction on the box opposite to applied force due to rough floor is given as

F_k = \mu F_n

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now the net force on the box in forward direction is given as

F_{net} = F_x - F_k

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now the acceleration of the box is given as

a = \frac{F_{net}}{m}

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Part b)

when box is pulled up along the inclined surface of angle 10 degree

now the two components of the force will be same along the inclined and perpendicular to inclined plane

F_x = 72.5 N

F_y = 33.8 N

now force balance perpendicular to inclined plane is given as

F_n + F_y = mgcos\theta

F_n = (25)(9.81)cos10 - 33.8 = 207.7 N

now the friction force opposite to the motion on the box is given as

f_k = \mu F_n

f_k = (0.300)(207.7) = 62.3 N

now the net pulling force along the inclined plane is given as

F_{net} = F_x - F_k - mgsin10

F_{net} = 72.5 - 62.3 - (25)(9.81)sin10

F_{net} = -32.38 N

now the box will decelerate and it is given as

a = \frac{F_{net}}{m}

a = \frac{-32.38}{25} = -1.29 m/s^2

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