problem32_16

problem32_16 - m 10 00 3 m F 10 85 8 m W 800 8 12 2 rms =...

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32.16: a) The average power from the beam is W 10 4 . 2 ) m 10 0 . 3 ( ) m W 800 . 0 ( 4 2 4 2 - - × = × = = IA P b) We have, using Eq. 32.29, . 2 rms 0 2 max 0 2 1 cE cE I ε = = Thus, m V 4 . 17 s)
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Unformatted text preview: m 10 00 . 3 )( m F 10 85 . 8 ( m W 800 . 8 12 2 rms = × × = =-c I E...
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