Explain how changing this characteristic on these two different waves affectstheirsound

What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a

viktelen [127]

Complete Question

A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.

What is the refrigerator's coefficient of performance? COP

(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP

Answer:

a

$COP = 8.49$

b

$COP_1 = 9.49$

Explanation:

From the question we are told that

The lower operation temperature of refrigerator is $T_1 = -8.00^oC = 265 \ K$

The upper operation temperature of the refrigerator is $T_2 = 23.2 ^oC = 296.2 \ K$

Generally the refrigerators coefficient of performance is mathematically represented as

$COP = \frac{T_1}{T_2 - T_1 }$

=> $COP = \frac{265}{296.2 - 265 }$

=> $COP = 8.49$

Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as

$COP_1 = \frac{T_2}{ T_2 - T_1}$

=> $COP_1 = \frac{296.2}{ 296.2 - 265 }$

=> $COP_1 = 9.49$

8 0

3 years ago

When 1 kg of water and 1 kg of wood absorb the same amount of heat, the change in temperature of the wood is greater than the ch

vampirchik [111]

Your Answer Will Be C

8 0

4 years ago

Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzz

amid [387]

Answer:

(C) The frequency decrease and intensity decrease

Explanation:

The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.

if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.

The frequency is the inverse from the wavelength, so the frequency heard will increase.

The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.

7 0

4 years ago

Please help! Will mark best answer!!

MA_775_DIABLO [31]

The answer is (A) hope it helps

7 0

3 years ago

Read 2 more answers

15 points. give me the method.

AveGali [126]

Answer:

$\boxed{{160 \: m(s)}^{ - 1} }$

Explanation:

$if \:the \: frequencies \: are \to \\ f_{1} = 640Hz \\ and \\f_{2} = 480Hz \: \\ but \: \boxed{v = f \gamma }: f = \frac{v}{ \gamma } \\ if \: \gamma_{1} - \gamma _{2} = 1 = \gamma \\ f_{1} - f_{2} = 640 - 480 = \boxed{ 160Hz} = f \\ v = f \gamma = 160 \times 1 = \boxed{{160 \: m(s)}^{ - 1} }$

5 0

3 years ago