Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 24, Problem 60P

(a)

To determine

The power received by the city from the space mirror reflecting light.

(a)

Expert Solution
Check Mark

Answer to Problem 60P

The power received by the city from the space mirror reflecting light is 32.1MW .

Explanation of Solution

Given info: The diameter of the space mirror is 200m and the intensity of the sunlight is 1370W/m2 . The percentage of light reflected from the space mirror to the city is 74.6% .

The formula to calculate the area of space mirror is,

A=πd24 (1)

Here,

d is the diameter of the space mirror.

Substitute 200m for d in equation (1) to find the value of A .

A=π(200m)24=31415m2

The formula to calculate the net power received by the mirror station is,

P=SA (2)

Here,

S is the intensity of radiations .

A is the area of space station.

Substitute 31415m2 for A and 1370W/m2 for S in the equation (2) to find the value of P .

P=(1370W/m2)(31415m2)=43.03×106W×106MW1W=43.03MW

The formula to calculate the power received by the city is,

Pcity=74.6%×P

Substitute 43.03MW for P in the above equation to find the value of Pcity .

Pcity=74.6%100%×(43.03MW)=32.1MW

Thus, the power received by the city from the space mirror reflecting light is 32.1MW .

Conclusion:

Therefore, the power received by the city from the space mirror reflecting light is 32.1MW .

(b)

To determine

The intensity of light received by the city.

(b)

Expert Solution
Check Mark

Answer to Problem 60P

The maximum intensity of light received by the city is 0.63W/m2 .

Explanation of Solution

Given info: The diameter of the space mirror is 200m and the intensity of the sunlight is 1370W/m2 . The diameter of the city is 8.00km .

The formula to calculate the area of space mirror is,

A=πd24 (1)

Here,

d is the diameter of the space mirror.

Substitute 8.00km for d in equation (1) to find the value of A .

A=π(8.00km×1000m1km)24=5.03×107m2

The formula to calculate the net power received by the city is,

P=SA

Here,

S is the intensity of radiations.

A is the area of the city.

Rewrite the above equation to find the value of S .

Scity=PcityA

Substitute 32.06MW for Pcity and 5.03×107m2 for A in the above equation to find the value of Scity .

S=(32.06MW×106W1MW)(5.03×107m2)=0.63W/m2

Thus, the maximum intensity of light received by the city is 0.63W/m2 .

Conclusion:

Therefore, the maximum intensity of light received by the city is 0.63W/m2 .

(c)

To determine

The percentage of vertical component of the intensity of sunlight at St. Petersburg, when the sun reaches at an angle of 7.0° above the horizon at noon.

(c)

Expert Solution
Check Mark

Answer to Problem 60P

The percentage of vertical component of the intensity of sunlight at St. Petersburg, when the sun reaches at an angle of 7.0° above the horizon at noon is 0.505% .

Explanation of Solution

Given info: The diameter of the space mirror is 200m and the intensity of the sunlight is 1370W/m2 . The diameter of the city is 8.00km .

The formula to calculate the intensity of light on earth is,

S=S×74.6%

Substitute 1370W/m2 for S in the above equation to find the value of S .

S=(1370W/m2)×74.6%100%=1022.02W/m2

The sunlight reaches the city St. Petersburg when the sun reaches at an angle of 7.0° above the horizon at noon so the vertical component of the intensity of sunlight will be,

SV=S×sin(7°) . (3)

Substitute 1022.02W/m2 for S in equation (3) to find SV .

SV=(1022.02W/m2)×sin(7°)=124.55W/m2

The formula to calculate the percentage of radiation is,

Spercentage=Scity×100%SV (4)

Substitute 124.55W/m2 for SV and 0.63W/m2 for Scity in equation (4) to find the value of Spercentage .

Spercentage=0.63W/m2×100%124.55W/m2=0.505%

Thus, the percentage of vertical component of the intensity of sunlight at St. Petersburg, when the sun reaches at an angle of 7.0° above the horizon at noon is 0.505% .

Conclusion:

Therefore, the percentage of vertical component of the intensity of sunlight at St. Petersburg, when the sun reaches at an angle of 7.0° above the horizon at noon is 0.505% .

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Chapter 24 Solutions

Principles of Physics: A Calculus-Based Text

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