Finalchiu - Nguyen Don Final 1 Due Dec 8 2004 5:00 pm Inst...

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Nguyen, Don – Final 1 – Due: Dec 8 2004, 5:00 pm – Inst: Charles Chiu 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The velocity, v , of a sound wave traveling in the air depends on B , the bulk modulus, and ρ , the density of the air. The bulk mod- ulus is defined by the variation of pressure Δ P = - B Δ V/V , where Δ V/V is the frac- tional change of the volume. Assume v = B x ρ y . The powers of x and y may be determined based on a dimensional analysis. Through equating the powers of M, of L and of T, one arrives at correspondingly a set of three equations. Choose the correct set. 1. 0 = x + y , 1 = x + 3 y and 1 = - 2 x 2. 0 = x + y , 1 = - x + 3 y and - 1 = - 2 x 3. 0 = x - y , 1 = x - 3 y and 1 = - 2 x 4. 1 = x + y , 0 = x - 3 y and 1 = - 2 x 5. 0 = x + y , 1 = - x - 3 y and - 1 = - 2 x correct 6. 0 = x - y , - 1 = x - 3 y and - 1 = - 2 x 7. 0 = x - y , 1 = x + 3 y and 1 = - 2 x 8. 0 = x - y , 2 = x + 3 y and - 1 = - 2 x 9. 0 = x + y , 1 = x - 3 y and 1 = - 2 x 10. 0 = x + y , 0 = x - 3 y and 1 = - 2 x Explanation: Denote the dimension of a quantity by a square bracket. [ v ]=L/T, [ B ]=[ F/A ], [ ρ ]=M/L 3 . The last two equations lead to [ B x ρ y ] =(ML/(TL) 2 ) x (M/L 3 ) y . Now we write [ v ]=[ B x ρ y ]. Equating the power of M gives 0 = x + y . Equating power of L gives 1 = - x - 3 y , and powers of T gives - 1 = - 2 x . 002 (part 1 of 1) 10 points The graph shows position as a function of time for two trains running on parallel tracks. At time t =0 (origin) the position of both trains is 0. Which is true: position time t B B A 1. Both trains speed up all the time 2. In the time interval from t =0 to t = t B , train B covers more distance than train A 3. At time t B , both trains have the same velocity 4. Somewhere before time t B , both trains have the same acceleration 5. Both trains have the same velocity at some time before t B correct Explanation: The slope of the curve B is parallel to line A at some point t < t B . 003 (part 1 of 1) 10 points Consider a man standing on a scale which is placed in an elevator. When the elevator is stationary, the scale reading is S s .
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Nguyen, Don – Final 1 – Due: Dec 8 2004, 5:00 pm – Inst: Charles Chiu 2 Scale Find S up , the scale reading when the el- evator is moving upward with acceleration ~a = 1 6 g ˆ , in terms of S s . 1. S up = 8 7 S s 2. S up = 4 3 S s 3. S up = 6 5 S s 4. S up = 6 7 S s 5. S up = 5 7 S s 6. S up = 0 m/s 2 7. S up = 7 5 S s 8. S up = 2 3 S s 9. S up = 7 6 S s correct 10. S up = 5 6 S s Explanation: Basic Concepts: Newton’s 2nd law X ~ F = m~a . (1) Solution We consider the forces acting on the man . Taking up (ˆ ) as positive, we know that m g acts on the man in the downward ( - ˆ ) di- rection. The only other force acting on the man is the normal force ~ S s from the scale. By the law of action and reaction, the force on the scale exerted by the man ( i.e. , the scale reading) is equal in magnitude but opposite in direction to the ~ S s vector. Initially, the el- evator is moving upward with constant speed (no acceleration) so S s - m g = 0 , or S s = m g as we would expect.
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