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solutions 1

# solutions 1 - Patel Kinal Homework 1 Due Sep 4 2007 7:00 pm...

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Patel, Kinal – Homework 1 – Due: Sep 4 2007, 7:00 pm – Inst: D Weathers 1 This print-out should have 10 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 The mass of the planet Saturn is 5 . 64 × 10 26 kg and its radius is 6 × 10 7 m. Calculate its density. Correct answer: 623 . 357 kg / m 3 . Explanation: Let : M = 5 . 64 × 10 26 kg and R = 6 × 10 7 m . Assuming Saturn to be a sphere (neglecting the rings), we can use the formula for the volume of a sphere of radius R : V = 4 3 π R 3 . The density of Saturn is ρ = M V = M 4 3 π R 3 = 5 . 64 × 10 26 kg 4 3 π (6 × 10 7 m) 3 = 623 . 357 kg / m 3 . keywords: 002 (part 1 of 2) 10 points This problem shows how dimensional analysis helps us check and sometimes even find a formula. A rope has a cross section A = 11 . 4 m 2 and density ρ = 2010 kg / m 3 . The “linear” density of the rope μ , defined to be the mass per unit length, can be written in the form μ = ρ x A y . Based on dimensional analysis, determine the powers x and y by choosing an expression below. 1. μ = ρ A 2. μ = A ρ 3. μ = 1 ρ A 2 4. μ = A ρ 2 5. μ = 1 ρ A 6. μ = A 2 ρ 2 7. μ = ρ A 2 8. μ = ρ A 2 9. μ = ρ A correct 10. μ = A 2 ρ 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 , thus M +1 L - 1 = M x L 2 y - 3 x . The powers of the units of mass and length need to be the same as for the LHS above, so x = 1 2 y - 3 x = - 1 .

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solutions 1 - Patel Kinal Homework 1 Due Sep 4 2007 7:00 pm...

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