# f04x2aan - 10 30 . ( 2 2 4 1 4 6 3--× × = = = LC T t t =...

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NAME PHYS-1200 PHYSICS II QUIZ 2 NOVEMBER 3, 2004 SOLUTIONS PART A. 1. A 7. D 2. A 8. D 3. E 4. D 5. C 6. A PART B . 1. c = d > b > a 2. It helps to remember that the graph of x vs. t will look like the negative of the graph of a vs. t which is shown, since a = – ϖ ² x . a) x m b) negative The graph of x vs t will have a negative slope here. c) positive 3. a) NEGATIVE x DIRECTION The direction of B E × . b) From the graph, λ = 2.0 m. Then, m 0 . 2 m/s 10 0 . 3 8 × = = c f f = 1.5 ×10 8 Hz

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NAME PART C . 1. a) F) 10 5 . 1 ( 2 C) 10 0 . 4 ( 2 6 2 5 2 - - × × = = C q U E U E = 5.3 ×10 -4 J b) F) 10 5 . 1 H)( 10 30 . 0 ( 1 2 1 1 2 1 6 3 - - × × = = π LC f f = 7.5 ×10 3 Hz = 7500 Hz = 7.5 kHz c) F) 10 5 . 1 H)( 10 30 . 0 ( C 10 0 . 4 6 3 5 - - - × × × = = = LC Q Q I ϖ I = 1.9 A d) The time required is one quarter of a period, so F) 10 5 . 1 H)(
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Unformatted text preview: 10 30 . ( 2 2 4 1 4 6 3--× × = = = LC T t t = 3.3 ×10-5 s = 33 μ s 2. a) Hz 500 m/s 343 = = f v λ = 0.69 m b) In this case, the source is stationary, and the detector is moving toward the source. m/s 343 m/s) . 3 m/s 343 ( Hz) 500 ( ' + = + = v v v f f D f’ = 504 Hz c) Now the car is moving away from the wall that reflects the wave. m/s 343 m/s) . 3 m/s 343 ( Hz) 500 ( "-=-= v v v f f D f” = 496 Hz d) YES Beats could be heard. The beat frequency would be f beat = f’ – f” = 504 Hz – 496 Hz f beat = 8 Hz...
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## This test prep was uploaded on 04/18/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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f04x2aan - 10 30 . ( 2 2 4 1 4 6 3--× × = = = LC T t t =...

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