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Contact [7]
4 years ago
7

Consider steady heat transfer between two large parallel plates at constant temperatures of T1 = 210 K and T2 = 150 K that are L

= 2 cm apart. Assume that the surfaces are black (emissivity ε = 1). Determine the rate of heat transfer between the plates per unit surface area assuming the gap between the plates is filled with atmospheric air.
Physics
1 answer:
snow_lady [41]4 years ago
8 0

Answer:

Q=81.56\ W/m^2

Explanation:

Given that

T_1= 210 K

T_2= 150 K

Emissivity of surfaces(∈) = 1

We know that heat transfer between two surfaces due to radiation ,when both surfaces are black bodies

Q=\sigma (T_1^4-T_2^4)\ W/m^2

So now by putting the values

Q=\sigma (T_1^4-T_2^4)\ W/m^2

Q=5.67\times 10^{-8}(210^4-150^4)\ W/m^2

Q=81.56\ W/m^2

So rate of heat transfer per unit area

Q=81.56\ W/m^2

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A wind generator, such as the one shown , uses the power of wind to generate electricity What do you think is happening inside t
sesenic [268]

Answer:

A wind turbine captures the wind, which then produces a renewable energy source. The wind makes the rotor spin; as the rotor spins, the movement of the blades drives a generator that creates energy. The motion of the blades turning is kinetic energy. It is this power that we convert into electricity.

8 0
3 years ago
Read 2 more answers
A circular test track for cars has a circumference of 3.5 kmkm . A car travels around the track from the southernmost point to t
elixir [45]

Answer:

(A) Distance will be equal to 1.75 km

(B) Displacement will be equal to 1.114 km

Explanation:

We have given circumference of the circular track = 3.5 km

Circumference is given by 2\pi r=3.5

r = 0.557 km

(a) It is given that car travels from southernmost point to the northernmost point.

For this car have to travel the distance equal to semi perimeter of the circular track

So distance will be equal to =\frac{3.5}{2}=1.75km

(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track

So displacement will be equal to d = 2×0.557 = 1.114 m

8 0
3 years ago
In a series lrc circuit, the frequency at which the circuit is at resonance is f0. If you double the resistance, the inductance,
taurus [48]

When you double capacitance and inductance, the new resonance frequency becomes f/2.

  • Resonance frequency:

The resonance frequency of RLC series circuit, is the frequency at which the capacity reactance is equal to inductive reactance.

It can also be defined as the natural frequency of an object where it tends to vibrate at a higher amplitude.

Xc = Xl

which gives the value for resonance frequency:

f = \frac{1}{2\pi \sqrt{LC} }

where;

f is the resonance frequency

L is the inductance

C is the capacitance

When you double capacitance and inductance, the new resonance frequency becomes;

f' = \frac{1}{2\pi \sqrt{2L2C} }

f' = \frac{1}{2\pi \sqrt{4LC} }

f' = \frac{1}{\pi \sqrt{LC} }\frac{1}{2}

f' = \frac{1}{2} f

Thus from above,

When you double capacitance and inductance, the new resonance frequency becomes f/2.

Learn more about resonance frequency here:

<u>brainly.com/question/13040523</u>

#SPJ4

6 0
2 years ago
Please send me solution of the question pls​
ELEN [110]

Answer:

20m

Explanation:

P.E=mgh

2000=10×10×h

2000=100h

Divide both side by 100

2000/100=20

4 0
3 years ago
The breaking car had 10,000 J of kinetic energy before breaking after breaking it had 2000 J of kinetic energy. How much thermal
WINSTONCH [101]

Answer:

8000J

Explanation:

The kinetic energy of the car lost during breaking are converted to thermal energy and are gained by the brakes.

Kinetic energy loss by car = thermal energy gained by brakes.

∆K.E = ∆T.E ....1

The Kinetic energy loss by car can be expressed as;

∆K.E = K.E1 - K.E2

Initial K.E = K.E1 = 10000J

Final K.E = K.E2 = 2000J

∆K.E= 10000J - 2000J = 8000J

From equation 1,

∆K.E = ∆T.E

∆T.E = 8,000J

thermal energy gain by brakes = 8,000J

8 0
3 years ago
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