Exam1-solutions - Version 012 Exam1 Chiu (60180) 1 This...

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Unformatted text preview: Version 012 Exam1 Chiu (60180) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The velocity of a transverse wave traveling along a string depends on the tension F = ma of the string and its mass per unit length . Assume v = F x y . The powers of x and y may be determined based on dimensional analysis. By equating the powers of mass, length, and time, one arrives correspondingly at a set of three equations. Choose the correct expressions for x and y . 1. 0 = x + y, 1 = x + y, 1 = 2 x 2. 0 = x + y, 1 = x y, 1 = 2 x 3. 0 = x + y, 1 = x y, 1 = 2 x correct 4. 0 = x y, 1 = x y, 1 = 2 x 5. 0 = x y, 1 = x y, 1 = 2 x 6. 0 = x y, 1 = x + y, 1 = 2 x 7. 0 = x + y, 1 = x + y, 1 = 2 x 8. 0 = x + y, 0 = x y, 1 = 2 x 9. 1 = x + y, 1 = x y, 1 = 2 x 10. 0 = x y, 1 = x + y, 1 = 2 x Explanation: Recall that the dimensions of velocity, ten- sion, and mass per unit length are [ v ] = L T , [ F ] = ML T 2 , [ ] = M L hence L T = [ F x y ] = parenleftbigg ML T 2 parenrightbigg x parenleftbigg M L parenrightbigg y = M x + y L x y T 2 x Equating the powers of M yields 0 = x + y . Equating the powers of L yields 1 = x y . Equating the powers of T yields 1 = 2 x . 002 10.0 points Consider the following set of equations, where s , s , x and r have units of length, t has units of time, v has units of velocity, g and a have units of acceleration, and k is dimensionless. Which one is dimensionally incorrect ? 1. s = s + v t + v 2 a 2. t = k radicalbigg s g + a v correct 3. t = v a + x v 4. v 2 = 2 as + k sv t 5. a = g + k v t + v 2 s Explanation: For an equation to be dimensionally cor- rect, all its terms must have the same units. (1) t = v a + x v [ t ] = T bracketleftBig v a bracketrightBig + bracketleftBig x v bracketrightBig = LT 1 LT 2 + L LT 1 = T + T = T It is consistent. (2) a = g + k v t + v 2 s [ a ] = LT 2 bracketleftbigg g + kv t + v 2 s bracketrightbigg = LT 2 + LT 1 T + L 2 T 2 L = LT 2 It is also consistent. (3) t = k radicalbigg s g + a v [ t ] = T bracketleftbigg k radicalbigg s g + a v bracketrightbigg = radicalbigg L LT 2 + LT 2 LT 1 = T + T 1 Version 012 Exam1 Chiu (60180) 2 This is not dimensionally consistent. (4) v 2 = 2 as + k sv t [ v 2 ] = L 2 T 2 bracketleftbigg 2 as + k sv t bracketrightbigg = LT 2 L + LLT 1 T 1 = L 2 T 2 This is also consistent. (5) s = s + v t + v 2 a [ s ] = L bracketleftbigg s + vt + v 2 a bracketrightbigg = L + LT 1 T + L 2 T 2 LT 2 = L This is also consistent....
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This note was uploaded on 03/11/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Exam1-solutions - Version 012 Exam1 Chiu (60180) 1 This...

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