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Unformatted text preview: Nguyen, Phan Homework 1 Due: Jan 27 2004, 4:00 am Inst: V Kaplunovsky 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points One cubic meter (1.0 m 3 ) of aluminum has a mass of 2700 kg, and a cubic meter of iron has a mass of 7860 kg. Find the radius of a solid aluminum sphere that has the same mass as a solid iron sphere of radius 2 . 2 cm. Correct answer: 3 . 1413 cm. Explanation: Let : m Al = 2700 kg , m Fe = 7860 kg , and r Fe = 2 . 2 cm . Basic Concept Density is = m V Solution: Since the masses are the same, Al V Al = Fe V Fe Al 4 3 r 3 Al = Fe 4 3 r 3 Fe r Al r Fe 3 = Fe Al r Al = Fe Al 1 3 ( r Fe ) = 7860 kg 2700 kg 1 3 (2 . 2 cm) = 3 . 1413 cm 002 (part 1 of 1) 10 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 + kv t + v 2 s 2. t = v a + x v 3. s = s + vt + v 2 a 4. t = k r s g + a v correct 5. v 2 = 2 as + ksv t Explanation: For an equation to be dimensionally correct, all its terms must have the same units. (1): t = v a + x v [ t ] = T h v a i + h x v i = LT 1 LT 2 + L LT 1 = T + T = T is consistent. (2): a = g + kv t + v 2 s [ a ] = LT 2 g + kv t + v 2 s = LT 2 + LT 1 T + L 2 T 2 L = LT 2 is also consistent. (3): t = k r s g + a v [ t ] = T k r s g + a v = r L LT 2 + LT 2 LT 1 = T + T 1 is not dimensionally consistent. (4): v 2 = 2 as + ksv t [ v 2 ] = L 2 T 2 2 as + ksv t = LT 2 L + LLT 1 T 1 = L 2 T 2 Nguyen, Phan Homework 1 Due: Jan 27 2004, 4:00 am Inst: V Kaplunovsky 2 is also consistent. (5): s = s + vt + v 2 a [ s ] = L s + vt + v 2 a = L + LT 1 T + L 2 T 2 LT 2 = L is also consistent. So only t = k r s g + a v is dimensionally in correct . 003 (part 1 of 2) 10 points Stokes law says F = 6 rv. F is a force, r the radius and v the velocity. The parameter has the dimension of 1. [ ] = M T L correct 2. [ ] = T 2 L 2 M 3. [ ] = M T L 2 4. [ ] = T M 5. [ ] = T 2 L M 6. [ ] = M T 7. [ ] = M T 2 L 2 8. [ ] = M T 2 L 9. [ ] = T L M 10. [ ] = T L 2 M Explanation: Recall that the dimension of force is [ F ] = [ m a ] = M L T 2 . The number 6 is dimensionless. The dimen sions of the radius and velocity are [ r ] = L ; [ v ] = L T so [ F ] = [6 r v ] M L T 2 = L L T [ ] Solving for [ ] yields [ ] = M LT ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
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