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Hunter-Best [27]
3 years ago
10

Air conditioners operate on the same principle as refrigerators. Consider an air conditioner that has 7.00 kg of refrigerant flo

wing through its circuit each cycle. The refrigerant enters the evaporator coils in phase equilibrium, with 54.0 % of its mass as liquid and the rest as vapor. It flows through the evaporator at a constant pressure and when it reaches the compressor 95% of its mass is vapor. In each cycle, how much heat Qc is absorbed by the refrigerant while it is in the evaporator? The heat of vaporization of the refrigerant is 1.50×105 J/kg
Physics
1 answer:
kozerog [31]3 years ago
4 0

Answer:

5.15J

Explanation:

First. 54% of the 7kg refrigerant is liquid

So we find mass of vapour at inlet generator

M1 = ( 1-0.54)*7= 3.2kg

At compressor mass of vapour will be

M2= 0.95*7= 6.7kg

So the Mass of vapour at exit generator is

M2-M1= 3.5kg

So to find heat absorbed by refrigerant in evaporation

Its using

Q= mh

°= 3.5x 1.50×10^5 J/kg

=5.15J

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Anestetic [448]
Need and answer choice if you have one
4 0
3 years ago
A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 3.7 V/m and points eastwar
UNO [17]

Answer:

Magnetic field, B = 1.24\times 10^{-8}\ T (eastwards)

Explanation:

Given that,

The magnitude of electric field, E = 3.7 V/m

A plane electromagnetic wave travels northward. We need to find the magnitude and direction of the magnetic field at this instant.

The relation between the magnetic field and the electric field of the plane electromagnetic wave is given by :

B=\dfrac{E}{c}

c is the speed of light

B=\dfrac{3.7}{3\times10^8}\\\\B=1.24\times 10^{-8}\ T

So, the magnetic field is 1.24\times 10^{-8}\ T. The direction of magnetic field is perpendicular to the electric field and the direction of wave propagation. Hence, the direction of magnetic field is eastward.

3 0
2 years ago
A mass of 5 kg stretches a spring 20 cm. The mass is acted on by an external force of 10 sin t 6 N (newtons) and moves in a medi
nordsb [41]

Answer:

u" + 40u' + 49u = 2 sin(t/6)        

upp + 40up + 49u = 2 sin(t/6)

Explanation:

Step 1: Data given

mass = 5 kg

L = 20 cm = 0.2 m

F = 10 sin(t/6)N

Fd(t) = - 6 N

u(0) = 0.03 m/s

u(0) = 0

u'(0) = 3 cm/s

Step 2:

ω =kL

k = ω/L = m*g /L = (5*9.8)/0.2 = 245 kg/s²

Since Fd(t) = -γu'(t)  we know:

γ =- Fd(t) / u'(t) = 6N/ 0.03 m/s = 200 Ns/m

The initial value problem which describes the motion of the mass is given by

5u" + 200u' + 245u = 10 sin(t/6)   u(0) = 0  ;  u'(0) = 0.03

This is equivalent to:

u" + 40u' + 49u = 2 sin(t/6)        u(0) = 0  ;  u'(0) = 0.03

upp + 40up + 49u = 2 sin(t/6)

With u in m and t in s

5 0
3 years ago
A sailboat travels a distance of 600 m in 40 seconds. What speed is it going?
Gekata [30.6K]

Answer:

15 miles /seconds

Explanation:

Distance = 600m

Time = 40 seconds

Speed=?

speed =  \frac{distance}{time}  \\ speed =  \frac{600}{40}

Simplify

\frac{600}{40}  =  \frac{60}{4}  \\  = 15

7 0
3 years ago
A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3
den301095 [7]

Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m

h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m

m = 148g*\frac{1kg}{1000g} = 0.148kg

g = 9.8 \frac{m}{s^2}

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J

The increase in potential energy of the ball is 115.82 J

5 0
2 years ago
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