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515_Physics ProblemsTechnical Physics

515_Physics ProblemsTechnical Physics - Chapter 17 P17.57(a...

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Chapter 17 517 P17.57 (a) ′ = f f v v v diver b g so 1 = v v f f diver = F H G I K J v v f f diver 1 with v = 343 m s , ′ = f 1 800 Hz and f = 2 150 Hz we find v diver m s = F H G I K J = 343 1 1 800 2 150 55 8 . . (b) If the waves are reflected, and the skydiver is moving into them, we have ′′ = + ′′ = L N M M O Q P P + f f v v v f f v v v v v v diver diver diver b g b g b g so ′′ = + = f 1 800 343 55 8 343 55 8 2 500 . . a f a f Hz . P17.58 (a) ′ = f fv v u ′′ = − − f fv v u a f ′ − ′′ = + F H G I K J f f fv v u v u 1 1 f fv v u v u v u uvf v u v u v u v f = + + = = a f e j e j b g e j 2 2 2 2 2 2 2 2 1 2 1 (b) 130 36 1 km h m s = . = = f 2 36 1 400 340 1 36 1 340 85 9 2 2 . . . a fa f a f Hz P17.59 When observer is moving in front of and in the same direction as the source, ′ =
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