# s01x2ans - U . At the time in question, the same energy is...

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% PHYS-1200 PHYSICS II QUIZ 2 MARCH 7, 2001 SOLUTIONS PART A. 1. D 7. B 2. A 8. C 3. B 4. B 5. C 6. E PART B . 1. a) TO THE RIGHT . The displacement current flows in the same direction as the charge current in the external circuit. b) OUT OF THE PAGE . Use the right hand rule. 2. a) From either graph, y m = 5.0 cm b) From the graph on the left, λ = 3.0 m c) From the graph on the right, T = 0.20 s d) v = / T = (3.0 m)/(0.20 s) v = 15 m/s e) Since the graph on the left is at t = 0, it is a graph of y = y m sin( kx + φ ). It starts at y = 0, and becomes negative. That means that at x = 0, the phase is ± π (or 180°). Either sign will work. = ±

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PART C. 1. a) H) C) s , and -1 ϖ π = = × × = × = - - 1 1 4 0 10 10 10 50 10 2 4 6 4 LC f ( . ( . . . f = 8.0 ×10 3 Hz b) c) d) e) H) A J J B B U Li U = = × = × = - - 1 2 2 1 2 4 2 5 4 0 10 050 50 10 50 ( . ( . ) . μ f) At the start, all of the energy was stored in the capacitor; call that energy
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Unformatted text preview: U . At the time in question, the same energy is divided between the capacitor and the inductor. Then, U = U E + U B , so U E = U – U B , and J 10 . 2 F) 10 . 1 ( 2 C) 10 . 2 ( 2 4 6 2 5 2---× = × × = = C Q U = 200 J. U B is known from the answer to part e) above, so U E = U – U B = 2.0 ×10-4 J – 5.0 ×10-5 J U E = 1.5 ×10-4 J = 150 J Hz)] 400 ( 2 [ m / N 6400 ) 2 ( and , ) 2 ( 1 Then, . 2 1 a) 2. 2 2 2 2 = = = = f k m m k f m k f m = 1.0 × 10-3 kg = 1.0 g ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ b) λ = v / f = (343 m/s)/(400 Hz) = 0.86 m = 86 cm ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯...
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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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s01x2ans - U . At the time in question, the same energy is...

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