HW1S - Nguyen, Phan Homework 1 Due: Jan 27 2004, 4:00 am...

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Unformatted text preview: Nguyen, Phan Homework 1 Due: Jan 27 2004, 4:00 am Inst: V Kaplunovsky 1 This print-out should have 24 questions. Multiple-choice 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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HW1S - Nguyen, Phan Homework 1 Due: Jan 27 2004, 4:00 am...

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