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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 , ℓ = 37 cm , and ρ = 7 . 6 g / cm 3 . Basic Concepts: The volume of the pipe will be the cross sectional area times the length. Solution: V = ( π r 2 1 π r 2 2 ) ℓ = π [ r 2 1 r 2 2 ] ℓ = π [(4 . 6 cm) 2 (2 . 7 cm) 2 ] (37 cm) = 1612 . 23 cm 3 . Thus the density is ρ = m V so m = ρ V = ρ π [ r 2 1 r 2 2 ] ℓ = (7 . 6 g / cm 3 ) π [(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 ρ = m V . Since the masses are the same, ρ Al V Al = ρ Fe V Fe ρ Al parenleftbigg 4 3 π r 3 Al parenrightbigg = ρ Fe parenleftbigg 4 3 π r 3 Fe parenrightbigg parenleftbigg r Al r Fe parenrightbigg 3 = ρ Fe ρ Al r Al = r Fe parenleftbigg ρ Fe ρ 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 toend 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 : ℓ = 16 cm , r = 8 cm , ρ 1 = 4 . 6 g / cm 3 , ρ 2 = 6 . 1 g / cm 3 , and m = 17741 g . Volume of a bar of radius r and length ℓ is V = π r 2 ℓ and its density is ρ = m V = m π r 2 ℓ so that m = ρ π r 2 ℓ ℓ x ℓ x r Let x be the length of the lighter metal; then ℓ x is the length of the heavier metal. Thus, m = m 1 + m 2 = ρ 1 π r 2 x + ρ 2 π r 2 ( ℓ x ) = ρ 1 π r 2 x + ρ 2 π r 2 ℓ ρ 2 π r 2 x. Therefore m ρ 2 π r 2 ℓ = ρ 1 π r 2 x ρ 2 π r 2 x and xπ r 2 ( ρ 1 ρ 2 ) = m ρ 2 π r 2 ℓ . Consequently, x = m ρ 2 π r 2 ℓ π r 2 ( ρ 1 ρ 2 ) = (17741 g) (6 . 1 g / cm 3 ) π (8 cm) 2 (16 cm) π (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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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Mass, Work

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