Homework 1 Solutions - homework 01 ALIBHAI, ZAHID Due: Jan...

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Unformatted text preview: homework 01 ALIBHAI, ZAHID Due: Jan 26 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 7 c m 4 . 6 cm 2 . 7 cm b The density is 7 . 6 g / cm 3 . What is the mass of this pipe? Correct answer: 12 . 253 kg (tolerance 1 %). Explanation: Let : r 1 = 4 . 6 cm , r 2 = 2 . 7 cm , = 37 cm , and = 7 . 6 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 . 6 cm) 2- (2 . 7 cm) 2 ] (37 cm) = 1612 . 23 cm 3 . Thus the density is = m V so m = V = [ r 2 1- r 2 2 ] = (7 . 6 g / cm 3 ) [(4 . 6 cm) 2- (2 . 7 cm) 2 ] (37 cm) = 12253 g = 12 . 253 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 2 . 68 cm. Correct answer: 3 . 82667 cm (tolerance 1 %). Explanation: Let : m Al = 2700 kg , m Fe = 7860 kg , and r Fe = 2 . 68 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 = (2 . 68 cm) parenleftbigg 7860 kg 2700 kg parenrightbigg 1 3 = 3 . 82667 cm . Question 3 part 1 of 1 10 points A cylinder, 16 cm long and 8 cm in radius, is made of two different metals bonded end- to-end to make a single bar. The densities are 4 . 6 g / cm 3 and 6 . 1 g / cm 3 . 1 6 c m 8 cm homework 01 ALIBHAI, ZAHID Due: Jan 26 2007, 4:00 am 2 What length of the lighter metal is needed if the total mass is 17741 g? Correct answer: 6 . 24234 cm (tolerance 1 %). Explanation: Let : = 16 cm , r = 8 cm , 1 = 4 . 6 g / cm 3 , 2 = 6 . 1 g / cm 3 , and m = 17741 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. Thus, m = m 1 + m 2 = 1 r 2 x + 2 r 2 ( - x ) = 1 r 2 x + 2 r 2 - 2 r 2 x. Therefore m- 2 r 2 = 1 r 2 x- 2 r 2 x and x r 2 ( 1- 2 ) = m- 2 r 2 . Consequently, x = m- 2 r 2 r 2 ( 1- 2 ) = (17741 g)- (6 . 1 g / cm 3 ) (8 cm) 2 (16 cm) (8 cm) 2 (4 . 6 g / cm 3- 6 . 1 g / cm 3 ) = 6 . 24234 cm . Question 4 part 1 of 1 10 points Convert 46 . 2 to radians....
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Homework 1 Solutions - homework 01 ALIBHAI, ZAHID Due: Jan...

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