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Unformatted text preview: homework 01 – KIM, JI – 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 9 c m 4 . 5 cm 2 . 8 cm b The density is 7 . 8 g / cm 3 . What is the mass of this pipe? Correct answer: 11 . 8599 kg (tolerance ± 1 %). Explanation: Let : r 1 = 4 . 5 cm , r 2 = 2 . 8 cm , ℓ = 39 cm , and ρ = 7 . 8 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 . 5 cm) 2 (2 . 8 cm) 2 ] (39 cm) = 1520 . 5 cm 3 . Thus the density is ρ = m V so m = ρ V = ρ π [ r 2 1 r 2 2 ] ℓ = (7 . 8 g / cm 3 ) π [(4 . 5 cm) 2 (2 . 8 cm) 2 ] (39 cm) = 11859 . 9 g = 11 . 8599 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 2 . 77 cm. Correct answer: 3 . 95518 cm (tolerance ± 1 %). Explanation: Let : m Al = 2700 kg , m Fe = 7860 kg , and r Fe = 2 . 77 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 . 77 cm) parenleftbigg 7860 kg 2700 kg parenrightbigg 1 3 = 3 . 95518 cm . Question 3, chap 1, sect 99. part 1 of 1 10 points A cylinder, 16 cm long and 2 cm in radius, is made of two different metals bonded end toend to make a single bar. The densities are 4 . 9 g / cm 3 and 6 g / cm 3 . homework 01 – KIM, JI – Due: Sep 5 2007, 4:00 am 2 1 6 c m 2 cm What length of the lighter metal is needed if the total mass is 1120 g? Correct answer: 6 . 24839 cm (tolerance ± 1 %). Explanation: Let : ℓ = 16 cm , r = 2 cm , ρ 1 = 4 . 9 g / cm 3 , ρ 2 = 6 g / cm 3 , and m = 1120 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....
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This note was uploaded on 05/05/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Work

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