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Svetllana [295]
2 years ago
10

Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope

rate and the maximum heat energy that can be removed per second QC,max/Δt from its interior are given.
Part A
Rank these refrigerators on the basis of their performance coefficient.
Rank from largest to smallest. To rank items as equivalent, overlap them.
Part B
Thsee six refrigerators are placed in six identical sealed rooms. Rank the refrigerators on the basis of the rate at which they raise the temperature of the room.
Rank from largest to smallest. To rank items as equivalent, overlap them.
1) P= 750 W, Qc,max/deltaT= 1500 J/s
2) P= 400 W, Qc,max/deltaT= 1200 J/s
3) P= 500 W, Qc,max/deltaT= 2000 J/s
4) P= 250 W, Qc,max/deltaT= 1000 J/s
5) P= 500 W, Qc,max/deltaT= 1500 J/s
6) P= 1000 W, Qc,max/deltaT= 3000 J/s
Physics
1 answer:
Mandarinka [93]2 years ago
5 0

Answer:

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s.

the rate at which they raise the temperature of the room.

2.1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

Explanation:

A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. .A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input

The performance coefficient  of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir  to the work  input to the refrigerator:

k=QC/W

power is defined as work per unit time

1.k=1500/750=2

2. 1200/400=3

3.2000/500=4

4.1000/250=4

5.1500/500=3

6.3000/1000=3

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s

2, Rate at which they raise the temperature of the room.

rate at which temperature rises in the inner chamber of the refrigerator is proportional to the rate of energy used to dispel heat from the refrigerator

1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

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Savatey [412]

Answer:

The friend on moon will be richer.

Explanation:

We must calculate the mass of gold won by each person, to tell who is richer. For that purpose we will use the following formula:

W = mg

m = W/g

where,

m = mass of gold

W = weight of gold

g = acceleration due to gravity on that planet

<u>FOR FRIEND ON MOON</u>:

W = 1 N

g = 1.625 m/s²

Therefore,

m = (1 N)/(1.625 m/s²)

m(moon) = 0.6 kg

<u>FOR ME ON EARTH</u>:

W = 1 N

g = 9.8 m/s²

Therefore,

m = (1 N)/(9.8 m/s²)

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Since, the mass of gold on moon is greater than the mass of moon on earth.

<u>Therefore, the friend on moon will be richer.</u>

7 0
2 years ago
When the balloon hits the ground, it rebounds slightly. What is the source of the energy for this rebound? A. When the balloon h
nevsk [136]

Answer:

The correct answer is c.    When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL

Explanation:

Let's analyze the situation of the globe

When it touches the ground, the part that is in contact decreases its velocity to zero, but the upper part of the ball continues to move, which creates that the molecules approach slightly, if we approximate the spring links, a repulsive force is created that after all the particles reach zero speed. The force of the springs moves the ball up until the force decreases to zero.

We can relate this force of Hooke with an elastic energy

This energy can be stored in the deformation of the system, as elastic potential energy, which is subsequently transformed into gravitational potential energy when the balloon is lifted.

The correct answer is c

7 0
2 years ago
A police siren of frequency fsiren is attached to a vibrating platform. The platform and siren oscillate up and down in simple h
babunello [35]

Answer:

he maximum frequency occurs when the denominator is minimum

 f’= f₀  \frac{343}{343 + v_s}

Explanation:

This is a doppler effect exercise, where the sound source is moving

           f = fo \frac{v}{v-v)s}      when the source moves towards the observer

           f ’=f_o  \frac{v}{v+v_{sy}}  Alexandrian source of the observer

the maximum frequency occurs when the denominator is minimum, for both it is the point of maximum approach of the two objects

          f’= f₀  \frac{343}{343 + v_s}

8 0
2 years ago
In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we
jekas [21]

Answer:

The weight of heart of a human is 0.93 lbs.

Explanation:

Given that,

Approximately weight of heart is 0.5 % of the total body weight.

Weight of human = 185 lbs

Let the the weight of total body is w and weight of heart is w_{h}.

We need to calculate the weight of heart of a human

Using given data

w_{h}=0.5\times w

Where, h = weight of heart

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w_{h}=\dfrac{0.5}{100}\times 185

w_{h}=0.93\ lbs

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8 0
2 years ago
A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equa
kozerog [31]

Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

given,                                  

weight of the person = 625 N

weight of the bike = 98 N        

Pressure on each Tyre = 7.60 x 10⁵ Pa

Area of contact on each Tyre = ?          

total weight of the system = 625 + 98

                                             = 723 N

Let F be the force on both the Tyre

F + F = W                                    

2 F  = 723                                    

F = 361.5 N                                

F = P A                                            

A = \dfrac{F}{P}                          

A = \dfrac{361.5}{7.60 \times 10^5}

A = 4.76 x 10⁻⁴ m²

7 0
2 years ago
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