# OLDHW1 - oldhomewk 01 PAPAGEORGE MATT Due 4:00 am...

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oldhomewk 01 – PAPAGEORGE, MATT – Due: Jan 17 2008, 4:00 am 1 Question 1, chap 1, sect 5. part 1 of 1 10 points A newly discovered giant planet has an av- erage radius 16 times that of the Earth and a mass 483 times that of the Earth. Calculate the ratio of the new planet’s den- sity to the Earth’s density. Correct answer: 0 . 11792 (tolerance ± 1 %). Explanation: Let : R n = 16 R E and m n = 483 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 = parenleftbigg m 1 m 2 parenrightbigg parenleftbigg R 1 R 2 parenrightbigg 3 = 483 (16) 3 = 0 . 11792 . Question 2, chap 1, sect 5. part 1 of 1 10 points A sphere of metal has a radius of 5 . 4 cm and a density of 7 . 95 g / cm 3 . What is the mass of the sphere? Correct answer: 5243 . 69 g (tolerance ± 1 %). Explanation: Let : r = 5 . 4 cm and ρ = 7 . 95 g / cm 3 . Density is mass per unit volume, so ρ = m V m = ρ V = ρ 4 3 π r 3 = ( 7 . 95 g / cm 3 ) 4 3 π (5 . 4 cm) 3 = 5243 . 69 g . Question 3, chap 1, sect 99. 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 . 1 g / cm 3 and 6 . 2 g / cm 3 . 16 cm 3 cm What length of the lighter metal is needed if the total mass is 2386 g? Correct answer: 7 . 05358 cm (tolerance ± 1 %). Explanation: Let : = 16 cm , r = 3 cm , ρ 1 = 4 . 1 g / cm 3 , ρ 2 = 6 . 2 g / cm 3 , and m = 2386 g .

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oldhomewk 01 – PAPAGEORGE, MATT – Due: Jan 17 2008, 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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