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aleksandr82 [10.1K]
3 years ago
5

20 J of work is done per cycle on a refrigerator with a coefficient of performance of 4.0.

Physics
1 answer:
larisa [96]3 years ago
5 0

Answer:

A)Qa=80 J

B)Qr= 100 J

Explanation:

Given that

W= 20 J

COP = 4

Heat rejected from cold reservoir = Qa

Heat exhausted to hot reservoir = Qr

The COP of refrigerator is given as

COP=\dfrac{Qa}{W}

4=\dfrac{Qa}{20}

Qa= 4 x 20 J

Qa=80 J

By using first law of refrigerator

Qr= Qa + W

Qr= 80 + 20 J

Qr= 100 J

A)Qa=80 J

B)Qr= 100 J

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Light travels in a straight line at a constant speed of 300 000 km/s what is the lights acceleration
exis [7]

Answer:

it's acceleration is 0

Explanation:

since it is travelling at a constant speed it is not accelerating so its acceleration is 0

3 0
3 years ago
What the electron and outermost energy level of periods 2and3,4,5 and 6
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A certain parallel-plate capacitor is filled with a dielectric for which Κ = 5.5 .The area of each plate is 0.034 m2 , and the p
Nesterboy [21]

Answer:

The maximum energy that can be stored in the capacitor is  6.62 x 10⁻⁵ J

Explanation:

Given that,

dielectric constant k = 5.5

the area of each plate, A = 0.034 m²

separating distance, d =  2.0 mm = 2 x 10⁻³ m

magnitude of the electric field =  200 kN/C

Capacitance of the capacitor is calculated as follows;

C = \frac{k \epsilon A}{d} = \frac{5.5*8.85*10^{-12}*0.034}{2*10^{-3}} = 8.275 *10^{-10} \ F

Maximum potential difference:

V = E x d

V = 200000 x 2 x 10⁻³ = 400 V

Maximum energy that can be stored in the capacitor:

E = ¹/₂CV²

E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²

E = 6.62 x 10⁻⁵ J

Therefore, the maximum energy that can be stored in the capacitor is  6.62 x 10⁻⁵ J

4 0
3 years ago
A wheel, starting from rest, rotates with a constant angular acceleration of 1.80 rad/s^2. During a certain 7.00 s interval, it
lora16 [44]

Answer:

a) 1.3 rad/s

b) 0.722 s

Explanation:

Given

Initial velocity, ω = 0 rad/s

Angular acceleration of the wheel, α = 1.8 rad/s²

using equations of angular motion, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

where

θ2 - θ1 = 53.2 rad

t2 - t1 = 7s

substituting these in the equation, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²

53.2 = 7.ω(0) + 1/2 * 1.8 * 49

53.2 = 7.ω(0) + 44.1

7.ω(0) = 53.2 - 44.1

ω(0) = 9.1 / 7

ω(0) = 1.3 rad/s

Using another of the equations of angular motion, we have

ω(0) = ω(i) + α*t1

1.3 = 0 + 1.8 * t1

1.3 = 1.8 * t1

t1 = 1.3/1.8

t1 = 0.722 s

5 0
3 years ago
Star A and Star B have measured stellar parallax of 1.0 arc second and 0.75 arc second, respectively. Which star is closer? How
zhuklara [117]

Answer:

Star A is closer than Star B

Explanation:

As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle

Here we can say

angle = \frac{1 Parsec}{distance}

so we have

distance = \frac{1 Parsec}{angle}

so here we have

angle subtended by Star A = 1 arc sec

angle subtended by star B = 0.75 arc sec

now we have

distance for star A is given as

d_a = \frac{1 Parsec}{1} = 1 Parsec

distance of star B is given as

d_b = \frac{1 Parsec}{0.75} = 1.33 Parsec

So star A is closer than star B

7 0
3 years ago
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