# Chapter 22 - Problems - Page 866: 65

It takes $19.4~seconds$ for the detector to measure 420 kJ of energy.

#### Work Step by Step

We can find the power absorbed by the detector: $P = (I~cos~\theta)~A$ $P = (1000~W/m^2)~(cos~30.0^{\circ})~(5.00~m)^2$ $P = 21,650.6~W$ We can find the time it takes to measure 420 kJ of energy: $t = \frac{Energy}{Power} = \frac{420,000~J}{21,650.6~W} = 19.4~s$ It takes $19.4~seconds$ for the detector to measure 420 kJ of energy.

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