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(a) Calculate the power per square meter reaching Earth’s

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli ISBN: 9780130606204 3

Solution for problem 43PE Chapter 7

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 43PE

(a) Calculate the power per square meter reaching Earth’s upper atmosphere from the Sun. (Take the power output of the Sun to be 4.00×1026 W.) (b) Part of this is absorbed and reflected by the atmosphere, so that a maximum of 1.30 kW/m2 reaches Earth’s surface. Calculate the area in km2 of solar energy collectors needed to replace an electric power plant that generates 750 MW if the collectors convert an average of 2.00% of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States’ energy needs (1.05×1020 J)? Australia’s energy needs (5.4×1018 J)? China’s energy needs (6.3×1019 J)? (These energy consumption values are from 2006.)

Step-by-Step Solution:

Step-by-step solution Step 1 of 6 Power per meter square is called intensity of the light, Here is the power emitted by the sun and is the power per square meter.

Step 2 of 6

Chapter 7, Problem 43PE is Solved
Step 3 of 6

Textbook: Physics: Principles with Applications
Edition: 6
Author: Douglas C. Giancoli
ISBN: 9780130606204

The answer to “(a) Calculate the power per square meter reaching Earth’s upper atmosphere from the Sun. (Take the power output of the Sun to be 4.00×1026 W.) (b) Part of this is absorbed and reflected by the atmosphere, so that a maximum of 1.30 kW/m2 reaches Earth’s surface. Calculate the area in km2 of solar energy collectors needed to replace an electric power plant that generates 750 MW if the collectors convert an average of 2.00% of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States’ energy needs (1.05×1020 J)? Australia’s energy needs (5.4×1018 J)? China’s energy needs (6.3×1019 J)? (These energy consumption values are from 2006.)” is broken down into a number of easy to follow steps, and 136 words. The full step-by-step solution to problem: 43PE from chapter: 7 was answered by , our top Physics solution expert on 03/03/17, 03:53PM. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6. Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204. This full solution covers the following key subjects: Energy, power, sun, needs, needed. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. Since the solution to 43PE from 7 chapter was answered, more than 692 students have viewed the full step-by-step answer.

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(a) Calculate the power per square meter reaching Earth’s