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Homework 1 - homework 01 BAUTISTA ALDO Due 4:00 am Version...

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homework 01 – BAUTISTA, ALDO – Due: Jan 23 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 1. 10 points. A newly discovered giant planet has an av- erage radius 18 times that of the Earth and a mass 792 times that of the Earth. Calculate the ratio of the new planet’s den- sity to the Earth’s density. Correct answer: 0 . 135802 (tolerance ± 1 %). Explanation: Let : R n = 18 R E and m n = 792 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 = 792 (18) 3 = 0 . 135802 . Question 2 Part 1 of 1. 10 points. A sphere of metal has a radius of 6 cm and a density of 9 . 11 g / cm 3 . What is the mass of the sphere? Correct answer: 8242 . 53 g (tolerance ± 1 %). Explanation: Let : r = 6 cm and ρ = 9 . 11 g / cm 3 . Density is mass per unit volume, so ρ = m V m = ρ V = ρ 4 3 π r 3 = ( 9 . 11 g / cm 3 ) 4 3 π (6 cm) 3 = 8242 . 53 g . Question 3 Part 1 of 1. 10 points. A cylinder, 16 cm long and 3 cm in radius, is made of two different metals bonded end- to-end to make a single bar. The densities are 4 . 7 g / cm 3 and 6 . 4 g / cm 3 . 16 cm 3 cm What length of the lighter metal is needed if the total mass is 2587 g? Correct answer: 6 . 41388 cm (tolerance ± 1 %). Explanation: Let : = 16 cm , r = 3 cm , ρ 1 = 4 . 7 g / cm 3 , ρ 2 = 6 . 4 g / cm 3 , and m = 2587 g .
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homework 01 – BAUTISTA, ALDO – Due: Jan 23 2006, 4:00 am 2 Volume of a bar of radius r and length is V = π r 2 and its density is ρ = m V = m π r 2 so that m = ρ π r 2 x - x r Let x be the length of the lighter metal; then - x is the length of the heavier metal.
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