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Unformatted text preview: homework 01 HELD, ETHAN Due: Jan 23 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A piece of pipe has an outer radius, an inner radius, and length as shown in the figure below. 3 6 c m 4 . 5 cm 2 . 5 cm b The density is 7 . 8 g / cm 3 . What is the mass of this pipe? Correct answer: 12 . 3502 kg (tolerance 1 %). Explanation: Let : r 1 = 4 . 5 cm , r 2 = 2 . 5 cm , = 36 cm , and = 7 . 8 g / cm 3 . Basic Concepts: The volume of the pipe will be the cross sectional area times the length. Solution: V = ( r 2 1 r 2 2 ) = [ r 2 1 r 2 2 ] = [(4 . 5 cm) 2 (2 . 5 cm) 2 ] (36 cm) = 1583 . 36 cm 3 . Thus the density is = m V so m = V = [ r 2 1 r 2 2 ] = (7 . 8 g / cm 3 ) [(4 . 5 cm) 2 (2 . 5 cm) 2 ] (36 cm) = 12350 . 2 g = 12 . 3502 kg . Question 2 part 1 of 1 10 points One cubic meter (1.0 m 3 ) of aluminum has a mass of 2700 kg, and a cubic meter of iron has a mass of 7860 kg. Find the radius of a solid aluminum sphere that has the same mass as a solid iron sphere of radius 3 . 4 cm. Correct answer: 4 . 85473 cm (tolerance 1 %). Explanation: Let : m Al = 2700 kg , m Fe = 7860 kg , and r Fe = 3 . 4 cm . Density is = m V . Since the masses are the same, Al V Al = Fe V Fe Al parenleftbigg 4 3 r 3 Al parenrightbigg = Fe parenleftbigg 4 3 r 3 Fe parenrightbigg parenleftbigg r Al r Fe parenrightbigg 3 = Fe Al r Al = r Fe parenleftbigg Fe Al parenrightbigg 1 3 = (3 . 4 cm) parenleftbigg 7860 kg 2700 kg parenrightbigg 1 3 = 4 . 85473 cm . Question 3 part 1 of 1 10 points A cylinder, 17 cm long and 3 cm in radius, is made of two different metals bonded end toend to make a single bar. The densities are 4 . 5 g / cm 3 and 6 . 1 g / cm 3 . homework 01 HELD, ETHAN Due: Jan 23 2007, 4:00 am 2 1 7 c m 3 cm What length of the lighter metal is needed if the total mass is 2470 g? Correct answer: 10 . 2135 cm (tolerance 1 %). Explanation: Let : = 17 cm , r = 3 cm , 1 = 4 . 5 g / cm 3 , 2 = 6 . 1 g / cm 3 , and m = 2470 g . Volume of a bar of radius r and length is V = r 2 and its density is = m V = m r 2 so that m = 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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