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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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Answer:

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8 0
2 years ago
What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56×104v/m and a magnetic fiel
sashaice [31]

1) 3.38\cdot 10^6 m/s

When both the electric field and the magnetic field are acting on the electron normal to the beam and normal to each other, the electric force and the magnetic force on the electron have opposite directions: in order to produce no deflection on the electron beam, the two forces must be equal in magnitude

F_E = F_B\\qE = qvB

where

q is the electron charge

E is the magnitude of the electric field

v is the electron speed

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Solving the formula for v, we find

v=\frac{E}{B}=\frac{1.56\cdot 10^4 V/m}{4.62\cdot 10^{-3} T}=3.38\cdot 10^6 m/s

2) 4.1 mm

When the electric field is removed, only the magnetic force acts on the electron, providing the centripetal force that keeps the electron in a circular path:

qvB=m\frac{v^2}{r}

where m is the mass of the electron and r is the radius of the trajectory. Solving the formula for r, we find

r=\frac{mv}{qB}=\frac{(9.1 \cdot 10^{-31} kg)(3.38\cdot 10^6 m/s)}{(1.6\cdot 10^{-19} C)(4.62\cdot 10^{-3}T)}=4.2\cdot 10^{-3} m=4.1 mm

3) 7.6\cdot 10^{-9}s

The speed of the electron in the circular trajectory is equal to the ratio between the circumference of the orbit, 2 \pi r, and the period, T:

v=\frac{2\pi r}{T}

Solving the equation for T and using the results found in 1) and 2), we find the period of the orbit:

T=\frac{2\pi r}{v}=\frac{2\pi (4.1\cdot 10^{-3} m)}{3.38\cdot 10^6 m/s}=7.6\cdot 10^{-9}s

7 0
2 years ago
If a force of 65 N is exerted on a 45 kg sofa and the sofa is moved 6.0 meters, how much work is done in moving the sofa? 17,550
abruzzese [7]

Answer:

The answer to your question is:   W = 390 J

Explanation:

Work is the transfer of energy when a body is moved from one place to another.

Data

Force = 65 N

mass = 45 kg

distance = 6 meters

work = ? J

Formula

W = F x d

Process

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2 years ago
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2 years ago
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A quarterback throws a football at 40km/hr to a receiver 50yd away. How much time does it take the ball to reach the receiver
Akimi4 [234]

Given:

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Time = 4.12 second

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