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Studentka2010 [4]
3 years ago
13

Can you cool a kitchen by leaving the refrigerator door open while the refrigerator is operating? a. No. Since the refrigerator

is operating, the “heat pump” transfers thermal energy from the inside of the refrigerator to the outside of the refrigerator which is presumably in the same room. Since this process is not 100% efficient due to the second law, the average temperature of the room will increase over time. b. No. Since the refrigerator is operating, the “heat pump” transfers thermal energy from the inside of the refrigerator to the outside of the refrigerator which is presumably in the same room. Due to the second law, there will be no change in temperature due to the perfect efficiency of the heat pump.
Physics
2 answers:
slava [35]3 years ago
7 0

Answer:

no

Explanation:

because will the cool air will be leaving the fridge the motor in the bottom will be running and eventually the mixture of the cool and hot air will balance out

posledela3 years ago
7 0

'a' is the correct statement.

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0.2^{2} – 20.8333^{2} = 2 * (-30 * 9.8) * distance

note:

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3 years ago
A 5.7 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4.5 + 13.7 x − 1.5 x 2 , where
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Answer:

The work done by the force on the particle is 29.85 J.

Explanation:

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W = ^{x_{2}}_{x_{1}}\int F_{x} dx

Where:

x₁: is the lower limit = 0 m    

x₂: is the upper limit = 1.9 m

Fₓ: is the force in the horizontal direction =  (4.5 + 13.7x - 1.5x²)N

W = ^{1.9}_{0}\int (4.5 + 13.7x - 1.5x^{2}) dx  

W = 4.5x|^{1.9}_{0} + \frac{13.7}{2}x^{2}|^{1.9}_{0} - \frac{1.5}{3}x^{3}|^{1.9}_{0}  

W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}    

W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}      

W = 29.85 J

Therefore, the work done by the force on the particle is 29.85 J.

I hope it helps you!                                

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