This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: sousse (res2468) – HW 1 – Kleinman – (58225) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An acceleration of 3 mi / h / s is equal to Correct answer: 1 . 34083 m / s 2 . Explanation: Let : a = 3 mi / h / s . a = 3 mi h · s · 1 . 609 km 1 mi · 1000 m 1 km · 1 h 3600 s = 1 . 34083 m / s 2 . 002 10.0 points A cylinder, 18 cm long and 6 cm in radius, is made of two different metals bonded endto end to make a single bar. The densities are 4 . 9 g / cm 3 and 6 . 5 g / cm 3 . 1 8 c m 6 cm , radius What length of the lighterdensity part of the bar is needed if the total mass is 11212 g? Correct answer: 11 . 1651 cm. Explanation: Let : ℓ = 18 cm , r = 6 cm , ρ 1 = 4 . 9 g / cm 3 , ρ 2 = 6 . 5 g / cm 3 , and m = 11212 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 ) = (11212 g) (6 . 5 g / cm 3 ) π (6 cm) 2 (18 cm) π (6 cm) 2 (4 . 9 g / cm 3 6 . 5 g / cm 3 ) = 11 . 1651 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. sousse (res2468) – HW 1 – Kleinman – (58225) 2 3 9 c m 4 . 7 cm...
View
Full
Document
This note was uploaded on 04/08/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Acceleration

Click to edit the document details