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Zolol [24]
3 years ago
5

During each cycle, a refrigerator ejects 610 kJ of energy to a high-temperature reservoir, and takes in 505 kJ of energy from a

low-temperature reservoir. (a) Determine the work done on the refrigerant in each cycle. kJ (b) Determine the coefficient of performance of the refrigerator.
Physics
1 answer:
jenyasd209 [6]3 years ago
8 0

Answer

A. the work done on the refrigerant in each cycle is 105kJ

B the coefficient of performance of the refrigerator is 4.8

Explanation

Given data

Work done at high temperature T2 Qh=610kJ

Work done at low temperature T1 Ql=505kJ

We know that the net work done by the refrigerator is expressed as

Wnet= Qh-Ql

=610-505

=105kJ

Also we know that the coefficient of performance is expressed as

COP= Ql/Wnet

COP= 505/105

= 4.8

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A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k
Luba_88 [7]

Answer:

The frictional force  F_{fri} = 6.446 N

The acceleration of the block a = 6.04 \frac{m}{s^{2} }

Explanation:

Mass of the block = 3.9 kg

\theta = 40°

\mu = 0.22

(a). The frictional force is given by

F_{fri} = \mu R_{N}

R_{N} = mg \cos \theta

R_{N} = 3.9 × 9.81 × \cos 40

R_{N} = 29.3 N

Therefore the frictional force

F_{fri} = 0.22 × 29.3

F_{fri} = 6.446 N

(b). Block acceleration is given by

F_{net} = F - F_{fri}

F = 30 N

F_{fri} = 6.446 N

F_{net} = 30 - 6.446

F_{net} = 23.554 N

The net force acting on the block is given by

F_{net}  = ma

23.554 = 3.9 × a

a = 6.04 \frac{m}{s^{2} }

This is the acceleration of the block.

8 0
3 years ago
You are pushing on a heavy desk with a force of 65 N the desk does not slide the force of friction between the desk and the floo
astra-53 [7]

When an object does not move even on pushing , static frictional force acts on in opposite direction of the applied force to stop the object from moving. static frictional force is a self adjusting force and it adjust its value according to the applied force if the applied force is smaller than the maximum value of static frictional force. The object starts moving once the applied force on it becomes greater than the maximum static frictional force.  hence the statement is true.

7 0
4 years ago
A car has uniform velocity of 108km/hour. How far does it travel in 1 1/2 minutes
Nostrana [21]

Answer:

2.7km

Explanation:

Two methods: Convert km/hour to km/minutes or convert 3/2 minutes to hours.

Then multiply time to get the distance of the car traveled.

3 0
3 years ago
If the speed of a wave doubles while the frequency remains the same, it means
sattari [20]

Answer:

Explanation:

Velocity of a wave is describe as

velocity =Frequency × Wavelength

Mathematically

v = fλ

Hence, Frequency, F = v / λ

Wavelength λ = v/f

So, if the frequency is kept constant, wavelength of the wave becomes directly proportional to velocity of the wave.

And this implies that, as the speed double, the wavelength is double.

5 0
3 years ago
-g A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The amplitude of the
Nina [5.8K]

Answer:

a= 92. 13 m/s²

Explanation:

Given that

Amplitude ,A= 0.165 m

The maximum speed ,V(max) = 3.9 m/s

We know that maximum velocity in the SHM  given as

V(max)  = ω A

ω=Angular speed

A=Amplitude

\omega =\dfrac{3.9}{0.165}\ rad/s

ω=23.63 rad/s

The maximum acceleration given as

a = ω² A

a= (23.63)² x 0.165 m/s²

a= 92. 13 m/s²

Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².

5 0
3 years ago
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