Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 22 - Electromagnetic Waves - Search and Learn - Page 643: 1

Answer

a. $2.5\times10^{-4}m^2$ b. $7.5m^2$ c. $149m^2$ d. On the item, on the roof, and neither (respectively)

Work Step by Step

The area of the solar cell is the power required, divided by $20\%$ of the received intensity. a. $A=\frac{P}{I_{effective}}=\frac{50\times10^{-3}W}{200W/m^2}=2.5\times10^{-4}m^2$ b. $A=\frac{P}{I_{effective}}=\frac{1500W}{200W/m^2}=7.5m^2$ c. $A=\frac{P}{I_{effective}}=\frac{40hp(746W/hp)}{200W/m^2}=149m^2$ d. The calculator’s solar panel is small enough to be mounted directly on it. The hair dryer’s solar panel is small enough to be mounted on a house roof. The car’s solar panel is NOT small enough to be mounted directly on it.
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