I appreciate the help on this physics question

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

Mandarinka [93]

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

5 0

3 years ago

Filiberto va a 90m/s en su carri, de repente se distrae por un par de segundos y su velocdad pasa a 40 m/s ¿cual fue su acelerac

svet-max [94.6K]

Answer:

Filiberto experimenta una deceleración de $2\,\frac{m}{s^{2}}$ .

Explanation:

(The problem was written in Spanish. Hence, explanation will be held in Spanish).

Asúmase que Filiberto se distrae por 2 segundos, puesto que un par equivale a dos, y que el vehículo experimenta un aceleración constante. La deceleración experimentada por el vehículo se deriva de la siguiente fórmula:

$a = \frac{v-v_{o}}{\Delta t}$

$a = \frac{40\,\frac{m}{s}-90\,\frac{m}{s} }{2\,s}$

$a = -25\,\frac{m}{s^{2}}$

Filiberto experimenta una deceleración de $2\,\frac{m}{s^{2}}$ .

7 0

3 years ago

What is the mass of an object that has an acceleration of 2.63m/s2 when a unbalanced force of 112N is applied to it?

exis [7]

Answer:

42.58kg

Explanation:

By Newton's second law, F = ma.

F is the force being applied, in this case 112N. a is the acceleration, in this case 2.63 m/s^2.

Thus, with some simple algebraic manipulation, we get the mass to equal:

m = F/a = 112N / 2.63 m/s^2 = 42.58kg

4 0

3 years ago

NEED HELP ASAP

katrin [286]

Answer:

its In most solids, conduction takes place as particles vibrate in place.

Explanation:

i did it in a test

8 0

2 years ago

During a race, a runner runs at a speed of 6 m/s. 2 seconds later, she is running at a speed of 10 m/s. What is the runner’s acc

igor_vitrenko [27]

Remember what acceleration is? It's how fast the speed is changing over a period of time.

$a = \frac{dv}{dt}$

So in dt = 2 secs, she sped up from 6 to 10 m/s and so dv = (10-6) m/s.

Now that you know dv and dt, you can calculate the acceleration a!

5 0

3 years ago