Homework 01

# Homework 01 - Bautista Aldo Homework 1 Due Sep 8 2005 4:00...

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Bautista, Aldo – Homework 1 – Due: Sep 8 2005, 4:00 am – Inst: Maxim Tsoi 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 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 3 . 45 cm. Correct answer: 4 . 92612 cm. Explanation: Let : m Al = 2700 kg , m Fe = 7860 kg , and r Fe = 3 . 45 cm . Density is ρ = m V . 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 = r Fe ρ Fe ρ Al 1 3 = (3 . 45 cm) 7860 kg 2700 kg 1 3 = 4 . 92612 cm . 002 (part 1 of 2) 10 points This problem shows how dimensional analysis helps us check our work and sometimes even help us find a formula. A rope has a cross section A = 12 m 2 and density ρ = 2310 kg / m 3 . The “linear” density of the rope μ , is defined to be the mass per unit length, in the form μ = ρ x A y . Based on dimensional analysis, find the powers x and y . 1. x = - 2 , y = 2 2. x = - 1 , y = 1 3. x = - 2 , y = 1 4. x = 1 , y = - 1 5. x = - 2 , y = - 1 6. x = 1 , y = 2 7. x = - 1 , y = - 1 8. x = 1 , y = 1 correct 9. x = - 1 , y = 2 Explanation: Kilogram (kg): a unit of mass ( M ). Meter (m): a unit of length ( L ). [ x ] means ”the units of x ”. The units of both sides of any equation must be the same for the equation to make sense. The units of the left hand side (LHS) are given as [ μ ] = M L = ML - 1 and the right hand side has [ ρ x A y ] = M L 3 x × ( L 2 ) y = M x L - 3 x L 2 y = M x L 2 y - 3 x The powers on the units of mass and length need to be the same as for the LHS above, so x = 1 2 y - 3 x = - 1 2 y = - 1 + 3 = 2 y = 1 Thus the answer is ( x, y ) = (1 , 1). 003 (part 2 of 2) 10 points A simple pendulum is made out of a string with length L and a mass m attached to one

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Bautista, Aldo – Homework 1 – Due: Sep 8 2005, 4:00 am – Inst: Maxim Tsoi 2 end of the string. Its period of oscillation T may depend on the gravitational acceleration g , and also depend on L and m . Based on dimensional analysis, which of the following expressions is dimensionally accept- able? k is a dimensionless constant. 1. T = k m g L 2. T = k g L 3. T = k r g L 4. T = k m L g 5. T = k L g 6. T = k r m g L 7. T = k s L m g 8. T = k s L g correct Explanation: Here we proceed in the same way: a pe- riod is a measure of time, thus the correct expression must have units of time. " k s L g # = 1 · s L L/T 2 = T This the correct one. As for the others, h k m g L i = 1 · ML/T 2 L = MT - 2 k m L g = 1 · ML L/T 2 = MT 2 k r m g L = 1 · r ML/T 2 L = M 1 / 2 T - 1 " k s L m g # = 1 · s L ML/T 2 = M - 1 / 2 T k L g = 1 · L L/T 2 = T 2 h k g L i = 1 · L/T 2 L = T - 2 k r g L = 1 · r L/T 2 L = T - 1 So these are all incorrect. 004 (part 1 of 3) 10 points Stokes law says F = 6 πrηv.
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