# hw19 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 19....

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PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 19. 17.96, 33.10, 33.13, 33.24; Q36.03, Q36.06 17.96. = - = 1 2 1 2 log dB) 10 ( I I β a) 7 . 3 dB 10 dB 37 dB 10 1 2 1 2 1 2 10 10 10 and , dB 10 log so , log dB) 10 ( = = = = = I I I I I I 7 . 0 3 7 . 0 3 7 . 3 1 2 10 10 10 10 = = = + I I 5000 10 0 . 5 3 1 2 = × = I I b) I p m ², so 71 5000 1 2 1 2 = = = I I p p m m c) s m p m , so 71 1 2 1 2 = = m m m m p p s s 33.10. As shown in equation 33-27, on page 8998 of the textbook, 2 15 6 2 )] m/ly 10 ly)(9.46 3 . 4 [( 4 W 10 0 . 1 4 × × = = π r P I s I = 4.8 ×10 -29 W/m² 33.13. a) 2 0 2 1 m E c I μ = , so ) W/m 10 40 . 1 )( H/m 10 4 )( m/s 10 00 . 3 ( 2 2 2 3 7 8 0 × × × = = - I c E m E m = 1.03 ×10 3 V/m b) m/s 10 3.00 V/m 10 03 . 1 8 3 × × = = c E B m m B m = 3.43 ×10 -6 T

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) s / m 8 . 9 (where 320 s / m 10 14 . 3 kg 10 31 . 5 N 10 67 . 1 Then, kg 10 31 . 5 ) m / kg 10 00 . 5 ( m) 10 633 ( 3 4 3 4 and , d) N 10 67 . 1 s / m 10 0 . 3 W 10 00 . 5 c) ) m / (N Pa 2 . 13 s / m 10 0 . 3 m / W 10 97 . 3 , absorption For total b) m / W 10 97 . 3 ) 10 633 ( W 10 00 . 5 a) 33.24 2 2 3 15 11 15 2 3 3 9 3 11 8 3 2 8 2 9 2 9 2 9 3 2 = = × = × × = × = × × = = = × = × × = = = = = = × × = = = × = × × = = = - - - - - - - - g g a a r m m F a F c P A Ac P A c I A r P F r P c I r P pressure I r P Area Power I π ρ Q35.04. a) The lower ray reflects off the front surface. The upper ray reflects off the rear
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## This note was uploaded on 09/17/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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hw19 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 19....

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