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solutions_to_practice_problems_for_midterm_i

# solutions_to_practice_problems_for_midterm_i - i =...

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Solutions to Practice Problems for Midterm I, Phy9B-B, Fall 2006 1) D 2) B 3) E 4) B 5) A 6) B 7) B 8) A 9) C 10) D 11) D 12) A 13) C 14) D 15) C 16) a) Hz, 25.0 m) 320 . 0 ( s) m 00 . 8 ( = = ! = v f T = 1 f = 1 (25.0 Hz) = 4.00 " 10 # 2 s, k = 2 \$ % = (2 \$ ) (0.320 m) = 19.6 rad m. b) y ( x , t ) = (0.0700 m) cos 2 " t (25.0 Hz) + x 0.320 m # \$ % & ( . c) (0.0700 m) cos 2 " ((0.150s)(25.0Hz) + (0.360m) (0.320 m)) [ ] = # 4.95 cm. d) The argument in the square brackets in the expression used in part (c) is 2 " (4.875), and the displacement will next be zero when the argument is 10 " ; the time is then s 0.1550 m)) (0.320 m) (0.360 Hz)(5 0 . 25 1 ( ) 5 ( = ! = " ! x T and the elapsed time is s. 02 . 0 2 e) s, 0050 . 0 = T 17) v 4 L = 6 " v 2 " 8.80 m , so L = 0.733 m . 18) The refraction angle at the first surface " 1 and the incident angle at the second surface " 2 are related by " 1 = " 2 # 45 o . The critical angle for total internal reflection at the second surface is sin " 2, crit = 1 1.21 , " 2, crit = 55.7 o . The use the Snell’s law at the first surface,

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Unformatted text preview: i = 1.21sin(55.7 o " 45 o ) = 0.225 , so i = 13.0 o . 19) a) From Eq. ( ) Hz. 375 , s m . 15 , with , 17 . 16 L S = ! " = = A f v v b) Hz. 371 , s m . 15 , s m 35.0 With L S = ! = = B f v v c) ) in figure extra an (keeping Hz 4 A B A f f f ! = ! " ! . The difference between the frequencies is known to only one figure. 20) For : cm 18 = s a) cm. . 63 cm . 18 1 cm . 14 1 1 1 1 1 = ! " # = ! " = ! + s s f s s b) . 50 . 3 . 18 . 63 ! = ! = " ! = s s m c) and d) From the magnification, we see that the image is real and inverted. For : cm . 7 = s a) cm. . 14 cm 00 . 7 1 cm . 14 1 1 1 1 1 ! = " # ! = " # = " + s s f s s b) . 00 . 2 00 . 7 . 14 = ! ! = " ! = s s m c) and d) From the magnification, we see that the image is virtual and erect....
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