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choli [55]
3 years ago
15

Electronics and inhabitants of the International Space Station generate a significant amount of thermal energy that the station

must get rid of. The only way that the station can exhaust thermal energy is by radiation, which it does by using thin, 1.8-m-by-3.6-m panels that have a working temperature of about 6C. How much power is radiated from each panel? Assume that the panels are in the shade so that the absorbed radiation will be negligible. Assume that the emissivity of the panel is 1.0.
Physics
1 answer:
katrin2010 [14]3 years ago
5 0

Answer:

4462.0927 W

Explanation:

\epsilon = Emissivity of the panel = 1

\sigma = Stefan-Boltzmann constant = 5.67\times 10^{-8}\ W/m^2K^4

T = Temperature = (273.15+6)

Area of the panel is given by

A=2\times 1.8\times 3.6\\\Rightarrow A=12.96\ m^2

The power radiated is given by

P=\epsilon \sigma AT^4\\\Rightarrow P=1\times 5.67\times 10^{-8}\times 12.96\times (273.15+6)^4\\\Rightarrow P=4462.0927\ W

The power radiated from each panel is 4462.0927 W

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How does the direction of acceleration compare with the direction of the net force that produces it
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Explanation:

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3 years ago
Explain: What happens to the velocity of a stream as the size of the sediment increases?
artcher [175]

Answer:

Also, as stream depth increases, the hydraulic radius increases thereby making the stream more free flowing. Both of these factors lead to an increase in stream velocity. The increased velocity and the increased cross-sectional area mean that discharge increases.

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2 years ago
A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕
Slav-nsk [51]

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

6 0
3 years ago
A 70 kg student stands on top of a 5.0 m platform diving board . how much gravitational potential energy does the student have?
Marianna [84]

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

<u>Given the following data;</u>

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

<em>P.E = 3430J</em>

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

<em>Workdone = 3430Nm</em>

6 0
3 years ago
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