## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 22 - Problems - Page 864: 35

#### Answer

It would take 2.4 seconds to absorb the same amount of energy.

#### Work Step by Step

$U = I~A~t$ $U$ is energy $I$ is intensity $A$ is the area $t$ is the time Let $P$ be the power of the source. Let $I_1$ be the intensity when the panel is a distance of 1.00 AU from the source. We can find an expression for $I_1$: $I_1 = \frac{P}{4\pi (1.00~AU)^2}$ $I_1 = 0.07958~P$ We can find an expression for $I_2$: $I_2 = \frac{P}{4\pi (1.55~AU)^2}$ $I_2 = 0.03312~P$ We can write an expression for the energy collected by the first panel: $U = I_1~A~t_1$ We can write an expression for the energy collected by the second panel: $U = I_2~A~t_2$ Since the energy should be equal, we can equate the two expressions to find $t_2$: $I_2~A~t_2 = I_1~A~t_1$ $t_2 = \frac{I_1~t_1}{I_2}$ $t_2 = \frac{(0.07958~P)(1~s)}{0.03312~P}$ $t_2 = 2.4~s$ It would take 2.4 seconds to absorb the same amount of energy.

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