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Physics Homework 1 - Solutions

# Physics Homework 1 - Solutions - homework 01 FRENNEA KYLE...

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homework 01 – FRENNEA, KYLE – Due: Jan 25 2007, 4:00 am 1

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homework 01 – FRENNEA, KYLE – Due: Jan 25 2007, 4:00 am 2 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. 37 cm 4 . 7 cm 2 . 8 cm The density is 7 . 6 g / cm 3 . What is the mass of this pipe? Correct answer: 12 . 5887 kg (tolerance ± 1 %). Explanation: Let : r 1 = 4 . 7 cm , r 2 = 2 . 8 cm , = 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 ) = π [ r 2 1 - r 2 2 ] = π [(4 . 7 cm) 2 - (2 . 8 cm) 2 ] (37 cm) = 1656 . 4 cm 3 . Thus the density is ρ = m V so m = ρ V = ρ π [ r 2 1 - r 2 2 ] = (7 . 6 g / cm 3 ) π [(4 . 7 cm) 2 - (2 . 8 cm) 2 ] (37 cm) = 12588 . 7 g = 12 . 5887 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 4 . 54 cm. Correct answer: 6 . 48249 cm (tolerance ± 1 %). Explanation: Let : m Al = 2700 kg , m Fe = 7860 kg , and r Fe = 4 . 54 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 = (4 . 54 cm) parenleftbigg 7860 kg 2700 kg parenrightbigg 1 3 = 6 . 48249 cm . Question 3 part 1 of 1 10 points A cylinder, 18 cm long and 5 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 . 2 g / cm 3 .
homework 01 – FRENNEA, KYLE – Due: Jan 25 2007, 4:00 am 3 18 cm 5 cm What length of the lighter metal is needed if the total mass is 7928 g? Correct answer: 6 . 66098 cm (tolerance ± 1 %). Explanation: Let : = 18 cm , r = 5 cm , ρ 1 = 4 . 6 g / cm 3 , ρ 2 = 6 . 2 g / cm 3 , and m = 7928 g . Volume of a bar of radius r and length is V = π r 2 and its density is ρ = m V = m π r 2 so that m = ρ π 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 = ρ 1 π r 2 x + ρ 2 π r 2 ( - x ) = ρ 1 π r 2 x + ρ 2 π r 2 - ρ 2 π r 2 x . Therefore m - ρ 2 π r 2 = ρ 1 π r 2 x - ρ 2 π r 2 x and x π r 2 ( ρ 1 - ρ 2 ) = m - ρ 2 π r 2 ℓ .

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Physics Homework 1 - Solutions - homework 01 FRENNEA KYLE...

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