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HW1S - Nguyen Phan Homework 1 Due 4:00 am Inst V...

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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 ρ = m V Solution: Since the masses are the same, ρ Al V Al = ρ Fe V Fe ρ Al 4 3 π r 3 Al = ρ Fe 4 3 π r 3 Fe r Al r Fe 3 = ρ Fe ρ Al r Al = ρ Fe ρ 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 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. a = g + kv t + v 2 s 0 2. t = v a + x v 3. s = s 0 + 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 0 [ a ] = LT - 2 g + kv t + v 2 s 0 = 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
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Nguyen, Phan – Homework 1 – Due: Jan 27 2004, 4:00 am – Inst: V Kaplunovsky 2 is also consistent. (5): s = s 0 + vt + v 2 a [ s ] = L s 0 + 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 πrηv. F is a force, r the radius and v the velocity. The parameter η has the dimension of 1. [ η ] = M T L correct 2. [ η ] = T 2 L 2 M 3. [ η ] = M T L 2 4. [ η ] = T M 5. [ η ] = T 2 L M 6. [ η ] = M T 7. [ η ] = M T 2 L 2 8. [ η ] = M T 2 L 9. [ η ] = T L M 10. [ η ] = T L 2 M Explanation: Recall that the dimension of force is [ F ] = [ m a ] = M L T 2 . The number 6 π is dimensionless. The dimen- sions of the radius and velocity are [ r ] = L ; [ v ] = L T so [ F ] = [6 π r η v ] M L T 2 = L L T [ η ] Solving for [ η ] yields [ η ] = M LT . 004 (part 2 of 2) 10 points Consider a simple pendulum which consists of a string with a bob attached to its end. Its period, ( i.e. , the time interval taken for the bob to complete one cycle of motion), may be written in the form T = k m x g y b z , where k is a dimensionless constant, g the magnitude of the gravitational acceleration, m the mass of the bob, and b the length of the string. The appropriate x , y , and z values are given respectively by 1. ( x, y, z ) = 1 , 1 2 , - 1 2 2. ( x, y, z ) = 0 , - 1 2 , - 1 2 3. ( x, y, z ) = 1 , - 1 2 , 1 2 4. ( x, y, z ) = 0 , - 1 2 , 1 2 correct 5. ( x, y, z ) = 0 , 1 2 , - 1 2 6. ( x, y, z ) = 1 , - 1 2 , - 1 2 7. ( x, y, z ) = (1 , - 1 , 1) 8. ( x, y, z ) = 0 , 1 2 , 1 2 9. ( x, y, z ) = (0 , - 1 , 1)
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Nguyen, Phan – Homework 1 – Due: Jan 27 2004, 4:00 am – Inst: V Kaplunovsky 3 10. ( x, y, z ) = 1 , 1 2 , 1 2 Explanation: [ T ] = [ k m x g y b z ] T = M x L y T 2 y L z T = M x
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