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Unformatted text preview: homework 01 ANDERSON, ZACH Due: Sep 5 2007, 4:00 am 1 Question 1, chap 1, sect 6. 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 . 5 cm 2 . 6 cm b The density is 8 . 7 g / cm 3 . What is the mass of this pipe? Correct answer: 13 . 6421 kg (tolerance 1 %). Explanation: Let : r 1 = 4 . 5 cm , r 2 = 2 . 6 cm , = 37 cm , and = 8 . 7 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 . 6 cm) 2 ] (37 cm) = 1568 . 06 cm 3 . Thus the density is = m V so m = V = [ r 2 1 r 2 2 ] = (8 . 7 g / cm 3 ) [(4 . 5 cm) 2 (2 . 6 cm) 2 ] (37 cm) = 13642 . 1 g = 13 . 6421 kg . Question 2, chap 1, sect 5. 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 4 . 13 cm. Correct answer: 5 . 89707 cm (tolerance 1 %). Explanation: Let : m Al = 2700 kg , m Fe = 7860 kg , and r Fe = 4 . 13 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 = (4 . 13 cm) parenleftbigg 7860 kg 2700 kg parenrightbigg 1 3 = 5 . 89707 cm . Question 3, chap 1, sect 99. part 1 of 1 10 points A cylinder, 18 cm long and 8 cm in radius, is made of two different metals bonded end toend to make a single bar. The densities are 4 . 2 g / cm 3 and 6 . 4 g / cm 3 . homework 01 ANDERSON, ZACH Due: Sep 5 2007, 4:00 am 2 1 8 c m 8 cm What length of the lighter metal is needed if the total mass is 18623 g? Correct answer: 10 . 2622 cm (tolerance 1 %). Explanation: Let : = 18 cm , r = 8 cm , 1 = 4 . 2 g / cm 3 , 2 = 6 . 4 g / cm 3 , and m = 18623 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 ) = (18623 g) (6 . 4 g / cm 3 ) (8 cm) 2 (18 cm) (8 cm) 2 (4 . 2 g / cm 3 6 . 4 g / cm 3 ) = 10 . 2622 cm ....
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This note was uploaded on 04/10/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
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