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

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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 = m a 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 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 , [ 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 0 , 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 0 + v t + v 2 a 2. t = k radicalbigg s g + a v correct 3. t = v a + x v 4. v 2 = 2 a s + k s v t 5. a = g + k v t + v 2 s 0 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 0 [ a ] = LT 2 bracketleftbigg g + kv t + v 2 s 0 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
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Version 012 – Exam1 – Chiu – (60180) 2 This is not dimensionally consistent. (4) v 2 = 2 a s + k s v t [ v 2 ] = L 2 T 2 bracketleftbigg 2 a s + k s v t bracketrightbigg = LT 2 L + LLT 1 T 1 = L 2 T 2 This is also consistent. (5) s = s 0 + v t + v 2 a [ s ] = L bracketleftbigg s 0 + vt + v 2 a bracketrightbigg = L + LT 1 T + L 2 T 2 LT 2 = L This is also consistent. So only t = k radicalbigg s g + a v is dimensionally incorrect . 003 10.0 points A newly discovered giant planet has an aver- age radius 19 . 999 times that of the Earth and a mass 985 times that of the Earth. Calculate the ratio of the new planet’s den- sity to the Earth’s density. 1. 0.301749 2. 0.123143 3. 0.0644409 4. 0.196148 5. 0.106891 6. 0.806 7. 0.0354938 8. 0.0632716 9. 0.0268675 10. 0.251707 Correct answer: 0 . 123143.
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