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# hw1 - Overton Mays Homework 1 Due 4:00 am Inst Turner This...

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Overton, Mays – Homework 1 – Due: Jan 24 2005, 4:00 am – Inst: Turner 1 This print-out should have 11 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 A newly discovered giant planet has an aver- age radius 16 times that of the Earth and a mass 364 times that of the Earth. Calculate the ratio of the new planet’s den- sity to the Earth’s density. Correct answer: 0 . 0888672 . Explanation: Let : R n = 16 R E and m n = 364 m E . Density is the ratio of mass to volume, ρ = m V . A spherical planet of average radius R has volume 4 3 π R 3 and hence density ρ = m 4 3 π R 3 . For two planets of respective radii R 1 and R 2 and masses m 1 and m 2 we have ρ 1 ρ 2 = m 1 4 3 π R 3 1 m 2 4 3 π R 3 2 = m 1 m 2 R 1 R 2 3 = 364 (16) 3 = 0 . 0888672 . 002 (part 1 of 1) 10 points A sphere of metal has a radius of 5 . 5 cm and a density of 7 . 5 g / cm 3 . What is the mass of the sphere? Correct answer: 5226 . 82 g. Explanation: Let : r = 5 . 5 cm and ρ = 7 . 5 g / cm 3 . Density is mass per unit volume, so ρ = m V m = ρ V = ρ 4 3 π r 3 = ( 7 . 5 g / cm 3 ) 4 3 π (5 . 5 cm) 3 = 5226 . 82 g . 003 (part 1 of 1) 10 points A cylinder, 16 cm long and 2 cm in radius, is made of two different metals bonded end-to- end to make a single bar. The densities are 4 . 5 g / cm 3 and 6 . 4 g / cm 3 . 16 cm 2 cm What length of the lighter metal is needed if the total mass is 1076 g? Correct answer: 8 . 82876 cm. Explanation: Let : = 16 cm , r = 2 cm , ρ 1 = 4 . 5 g / cm 3 , ρ 2 = 6 . 4 g / cm 3 , and m = 1076 g . Volume of a bar of radius r and length is V = π r 2

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Overton, Mays – Homework 1 – Due: Jan 24 2005, 4:00 am – Inst: Turner 2 and its density is ρ = m V = m π r 2
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