# hw 1 - shah (rps587) – HW 1 – Kleinman – (58225) 1...

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Unformatted text preview: shah (rps587) – HW 1 – Kleinman – (58225) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An acceleration of 2 . 1 mi / h / s is equal to Correct answer: 0 . 938583 m / s 2 . Explanation: Let : a = 2 . 1 mi / h / s . a = 2 . 1 mi h · s · 1 . 609 km 1 mi · 1000 m 1 km · 1 h 3600 s = . 938583 m / s 2 . 002 10.0 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 5 g / cm 3 and 6 . 3 g / cm 3 . 1 8 c m 5 cm , radius What length of the lighter-density part of the bar is needed if the total mass is 7785 g? Correct answer: 10 . 9833 cm. Explanation: Let : ℓ = 18 cm , r = 5 cm , ρ 1 = 5 g / cm 3 , ρ 2 = 6 . 3 g / cm 3 , and m = 7785 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 ) = (7785 g)- (6 . 3 g / cm 3 ) π (5 cm) 2 (18 cm) π (5 cm) 2 (5 g / cm 3- 6 . 3 g / cm 3 ) = 10 . 9833 cm . 003 10.0 points A piece of pipe has an outer radius, an in- ner radius, and length as shown in the figure below. shah (rps587) – HW 1 – Kleinman – (58225) 2 3 8 c m 4 . 6 cm 2 .....
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## This note was uploaded on 04/16/2009 for the course PHY 58225 taught by Professor Klienman during the Spring '09 term at University of Texas at Austin.

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hw 1 - shah (rps587) – HW 1 – Kleinman – (58225) 1...

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