HW1 - Pickering, Tracey Homework 1 Due: Jan 23 2006, noon...

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Unformatted text preview: Pickering, Tracey Homework 1 Due: Jan 23 2006, noon Inst: Drummond 1 This print-out should have 9 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 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 = 11 m 2 and density = 1950 kg / m 3 . The linear density of the rope , is defined to be the mass per unit length, in the form = x A y . Based on dimensional analysis, find the powers x and y . 1. x = 1 , y =- 1 2. x =- 1 , y = 1 3. x =- 2 , y = 1 4. x =- 1 , y =- 1 5. x =- 2 , y =- 1 6. x = 1 , y = 1 correct 7. x =- 1 , y = 2 8. x = 1 , y = 2 9. x =- 2 , 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 [ ] = M L = M L- 1 and the right hand side has [ 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). 002 (part 1 of 1) 10 points The modern standard of length is 1 m and the speed of light is approximately 2 . 99792 10 8 m / s . Find the time t for light to cover 1 m at the given speed. 1. t 3 10- 6 s 2. t 3 . 3 10- 8 s 3. t 3 10- 9 s 4. t 3 . 3 10- 10 s 5. t 3 . 3 10- 7 s 6. t 3 . 3 10- 6 s 7. t 3 10- 8 s 8. t 3 10- 10 s 9. t 3 10- 7 s 10. t 3 . 3 10- 9 s correct Explanation: Since distance is speed times velocity, we...
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HW1 - Pickering, Tracey Homework 1 Due: Jan 23 2006, noon...

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