Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 9 - Work and Kinetic Energy - Exercises and Problems - Page 229: 39


The area of the solar collector is $41.7~m^2$

Work Step by Step

Let $A$ be the area of the solar collector. The total energy collected by the solar collector is the intensity multiplied by the area multiplied by the time. Therefore; $I~A~t = E$ $A = \frac{E}{I~t}$ $A = \frac{150\times 10^6~J}{(1000~W/m^2)(3600~s)}$ $A = 41.7~m^2$ The area of the solar collector is $41.7~m^2$.
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