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Alex
2 years ago
11

A refrigerator has, to transfer an average of 263 J of heat per second from temperature – 10°C to 25°C. Calculate the average po

wer consumed assuming ideal reversible cycle and no other losses.
thankyou ~​
Physics
1 answer:
Zolol [24]2 years ago
6 0

\huge \mathfrak \red{Answer} \\  \\  \:   \:  \implies \: \sf 35 \: watt \\  \\  \\  \bf \: Given \:  :  \\  \\  \:  \:  \tt Q_{1} = 263j \\    \:   \tt \:  T_{1} =  - 10    {}^{ \circ} C \:  =  - 10 + 273 = 263 \: k \\  \:  \tt \:  T_{2}  = 25^{ \circ} C = 25 + 273 = 298 \:k \\  \\  \\  \bf \: To \: Find \:   :   \: \:  P ?\:  \\  \\ \:  \sf \green{Solution :} \\  \\  \:  \:   \large \: \bf \: P = Q _{2} -  Q_{1} \\  \\   \:  \:  \: \tt \: \rightarrow \frac{Q_{2}}{Q_{1}}  =  \frac{T_{2}}{T_{1}}  \\  \\  \:\rightarrow  \tt \:  \: \:   Q_{2} =  \frac{T_{2}}{T_{1}}  \times Q_{1} \\  \\  \:  \tt \:  \: \rightarrow \:  Q_{2}=  \frac{298}{ \cancel 263}  \times \cancel 263  \\   \\ \:   \: \tt \:\rightarrow \: Q_{2} =298\: \: J / s\\  \\   \:  \:  \:  \sf \:  \bf \: \: P= Q_{2} -  Q_{1}  \\   \tt \: \:  \:  \:  \:  \:  \:  \:  \:  \:=298 - 263  \\ \\  \:  \:  \:  \tt \:  \:  \:  \:  \:  \:  \rightarrow \:  \fbox{ \blue{35 \: watt}}

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Answer:

x = 6.94 m

Explanation:

For this exercise we can find the speed at the bottom of the ramp using energy conservation

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            Em₀ = K + U = ½ m v₀² + m g h

Final point. Lower

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            Em₀ = Em_{f}

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            v² = v₀² + 2 g h

             

Let's calculate

             v = √(1.23² + 2 9.8 1.69)

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In the horizontal part we can use the relationship between work and the variation of kinetic energy

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Newton's second law

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The equation for the friction is

               fr = μ N

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We replace

             μ m g x = ½ m v²

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Let's calculate

            x = 5.89² / (2 0.255 9.8)

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6 0
3 years ago
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2 years ago
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vampirchik [111]

Answer:

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F = ?

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3 years ago
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Answer:

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v=u-gt\\\\v=39.54-9.8\times 5.4\\\\v=-13.38\ m/s

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