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Svetllana [295]
3 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]3 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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Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject
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Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}

where

\mu_{sun} is gravitational parameter of sun =  1.32712 x 10^20 m^3 s^2.t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}

t = 12,105.96 sec

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Cables and conductors installed on the side of a pole or building shall be protected up to ___feet above finished grade.
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Answer:

10 feet.

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A diver named Jacques observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the
Studentka2010 [4]

Answer:

A) the ratio of volumes of the bubble is Vs/Vb= 3.74

B)  would not be safe since Vs/Vb== 3.5 and there is no lung capacity to store such amount of volume, and thus would generate an unsafe pressure over the lungs. would be better to release air accordingly to maintain safe conditions.

Explanation:

assuming the gas of the bubble behaves as ideal gas

P * V = n * R * T

where P= absolute pressure, V= volume occupied by the gas, n = number of moles , R = ideal gas constant , T = absolute temperature

if we assume that the mass of the bubble remains constant ( that is, it does not capture other bubbles during ascension of disaggregate into smaller ones and there is no mass transfer into the bubble due to diffusion)

inicial state)  Pb * Vb = n * R * Tb

final  state)  Ps * Vs = n * R * Ts

dividing both equations

(Ps/Pb)(Vs/Vb) = Ts/Tb

therefore

Vs/Vb= (Ts/Tb) (Pb/Ps)

since Tb = 4°C = 277 K and Ts= 23°C = 296 K

Vs/Vb= (Ts/Tb) (Pb/Ps) = (296K/277K)*(3.5 atm/1 atm) = 3.74

B) if the T remains constant Ts=Tb and thus

Vs/Vb= (Ts/Tb) (Pb/Ps)= 1* (Pb/Ps) = 3.5 atm/1 atm = 3.5

it would not be safe since there is no lung capacity to store such amount of volume, and thus would generate an unsafe pressure over the lungs. would be better to release air accordingly to maintain safe conditions.

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