# phys1hw1 - homework 01 – ANDERSON ZACH – Due Sep 5 2007...

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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 , ℓ = 37 cm , and ρ = 8 . 7 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 . 6 cm) 2 ] (37 cm) = 1568 . 06 cm 3 . Thus the density is ρ = m V so m = ρ V = ρ π [ r 2 1- r 2 2 ] ℓ = (8 . 7 g / cm 3 ) π [(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 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 . 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- to-end 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 : ℓ = 18 cm , r = 8 cm , ρ 1 = 4 . 2 g / cm 3 , ρ 2 = 6 . 4 g / cm 3 , and m = 18623 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 ) = (18623 g)- (6 . 4 g / cm 3 ) π (8 cm) 2 (18 cm) π (8 cm) 2 (4 . 2 g / cm 3- 6 . 4 g / cm 3 ) = 10 . 2622 cm ....
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phys1hw1 - homework 01 – ANDERSON ZACH – Due Sep 5 2007...

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