HW 29 - Morby, Grant – Homework 29 – Due: Apr 17 2006,...

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Unformatted text preview: Morby, Grant – Homework 29 – Due: Apr 17 2006, noon – Inst: Drummond 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Blood flows through a coronary artery that is partially blocked by deposits along the artery wall. blood flow Through which part of the artery is the flux (mass of blood per unit time) largest? 1. the flux is the same in both parts correct 2. the narrow part 3. the wide part Explanation: Because blood is incompressible, the flux for a same flow is always the same. 002 (part 1 of 2) 10 points A person rides up a lift to a mountain top, but the person’s ears fail to ”pop”; that is, the pressure of the inner ear does not equalize with the outside atmosphere. The radius of each eardrum is 0.40 cm. The pressure of the atmosphere drops from 1 . 008 × 10 5 Pa at the bottom of the lift to 0 . 983 × 10 5 Pa at the top. a) What is the net pressure on the inner ear at the top of the mountain? Correct answer: 2500 Pa. Explanation: Basic Concept: P net = P b- P t Given: P b = 1 . 008 × 10 5 Pa P t = 0 . 983 × 10 5 Pa Solution: P net = 100800 Pa- 98300 Pa = 2500 Pa 003 (part 2 of 2) 10 points b) What is the magnitude of the net force on each eardrum? Correct answer: 0 . 125664 N. Explanation: Basic Concepts: P = F A A = πr 2 Given: r = 0 . 40 cm Solution: F net = P net A = P net ( π r 2 ) = (2500 Pa) π (0 . 4 cm) 2 · µ 1 m 100 cm ¶ 2 = 0 . 125664 N 004 (part 1 of 1) 10 points In testing a new material for shielding space- craft, 138 small ball bearings, each moving at a supersonic speed of 393.9 m/s, collide head- on and elastically with the material during a 0.689 min interval. If the bearings each have a mass of 6.1 g and the area of the tested material is 0.77 m 2 , what pressure is exerted on the material? Correct answer: 20 . 8336 Pa. Explanation: Basic Concepts: P = F A Conservation of momentum: Δ p = F Δ t Elastic collision: v f =- v i Morby, Grant – Homework 29 – Due: Apr 17 2006, noon – Inst: Drummond 2 Given: Δ t = 0 . 689 min N = 138 m = 6 . 1 g v = 393 . 9 m / s A = 0 . 77 m 2 Solution: Δ p = m v f- m v i = m (- v i )- m v i =- 2 m v i and F = Δ p Δ t For one bearing, P = fl fl fl Δ p Δ t fl fl fl A = 2 m v A Δ t For all of the bearings, P = N µ 2 m v A Δ t ¶ = 2(138)(6 . 1 g)(393 . 9 m / s) (0 . 77 m 2 )(0 . 689 min) · µ 1 min 60 s ¶µ 1 kg 1000 g ¶ = 20 . 8336 Pa 005 (part 1 of 1) 10 points The hypodermic syringe shown in the figure contains a medicine with the same density as...
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This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas at Austin.

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HW 29 - Morby, Grant – Homework 29 – Due: Apr 17 2006,...

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