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Exam1-solutions

# Exam1-solutions - Version 180 – Exam1 – Chiu –(60180...

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Unformatted text preview: Version 180 – 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 μ . Assume v = F x μ 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 4. 0 = x + y, 1 = x − y, − 1 = − 2 x correct 5. 1 = 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. 0 = 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 , [ μ ] = M L hence L T = [ F x μ 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. a = g + k v t + v 2 s 2. s = s + v t + v 2 a 3. t = k radicalbigg s g + a v correct 4. t = v a + x v 5. v 2 = 2 as + k sv t 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 180 – 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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Exam1-solutions - Version 180 – Exam1 – Chiu –(60180...

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