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geniusboy [140]
3 years ago
13

An ideal reversible refrigerator keeps its inside compartment at 6°C. What is the high temperature, Th, needed to give this refr

igerator a coefficient of performance of 2.7? An ideal reversible refrigerator keeps its inside compartment at 6°C. What is the high temperature, Th, needed to give this refrigerator a coefficient of performance of 2.7? a. 109°C b. 8°C c. 759°C d. 22°C
Physics
1 answer:
pishuonlain [190]3 years ago
8 0

Answer:

a. 109°C

Explanation:

The COP (coefficient of performance) of heating pump and refrigerator is the ratio of the output energy and input energy. These coefficients in terms of temperature is given as follows.

The COP of the heat pump = K = \frac{Th}{Th-Tc}

The COP of the refrigerator = K = \frac{Tc}{Th-Tc}

where Th and Tc are the hot and cold absolute temperatures.

Let's first convert Tc into kelvin.

Tc = 6°C = 6+273 = 279 K

K = \frac{Tc}{Th-Tc}

2.7 =\frac{279}{Th-279}

2.7(Th -279) = 279

2.7Th-753.3 = 279

2.7Th = 1032.3

Th = 382.33 K

Th = 382.33 - 273

Th = 109°C

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Answer:

W = 0.060 J

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Explanation:

solution:

for the spring:

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the work-energy theorem,

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3 years ago
a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres​
Scilla [17]

The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

       \text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}

Similarly,

      \text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}

We know the below formulas,

          \text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}

          \text {volume of wire}=\pi \times r^{2} \times l

When equating both the equations, we can find length of wire as below, where \pi=\frac{22}{7}

          \frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l

         \frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l

The \pi value gets cancelled as common on both sides, we get

           \frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l

The 10^{-6} value gets cancelled as common on both sides, we get

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Explanation:

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Así tenemos:

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Es el mismo resultado, pues recorrió la mitad de distancia en la mitad de tiempo.

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