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Unformatted text preview: homework 24 RAHMAN, TARIQUE Due: May 3 2008, 3:00 am 1 Question 1, chap 18, sect 2. part 1 of 2 10 points The aorta in an average adult has a cross sectional area of 2 . 0 cm 2 . a) Calculate the flow rate (in grams per second) of blood ( = 1 . 0 g / cm 3 ) in the aorta if the flow speed is 44 cm / s. Correct answer: 88 g / s (tolerance 1 %). Explanation: Basic Concept: flow rate = A v (in cm 3 / s) flow rate = A v (in g / s) Given: A = 2 . 0 cm 2 = 1 . 0 g / cm 3 v = 44 cm / s Solution: flow rate = (2 cm 2 )(44 cm / s)(1 g / cm 3 ) = 88 g / s Question 2, chap 18, sect 2. part 2 of 2 10 points Assume that the aorta branches to form a large number of capillaries with a combined crosssectional area of 3 . 10 3 cm 2 . b) What is the flow speed in the capillaries? Correct answer: 0 . 0293333 cm / s (tolerance 1 %). Explanation: Basic Concept: A 1 v 1 = A 2 v 2 Given: A 2 = 3 . 10 3 cm 2 Solution: v 2 = v 1 A 1 A 2 = (44 cm / s)(2 cm 2 ) 3000 cm 2 = 0 . 0293333 cm / s Question 3, chap 18, sect 3. part 1 of 1 10 points Why is the gas pressure inside an inflated balloons always greater than the air pressure outside? 1. Warmer air is inside the balloon. 2. The balloon is made of stretchy rubber that pushes inward on the gas. correct 3. The pressure is actually less; thats why the inflated balloons rise. The stretched rub ber supplies an inward force (and pressure). 4. Cooler air is inside the balloon. Explanation: The pressure inside is balanced by the pres sure of the outside air and the stretched rub ber. Question 4, chap 18, sect 3. part 1 of 2 10 points In a car lift, compressed air exerts a force on a piston with a radius of 2 . 75 cm. This pressure is transmitted to a second piston with a radius of 14 . 5 cm. a) How large a force must the compressed air exert to lift a 1 . 17 10 4 N car? Correct answer: 420 . 838 N (tolerance 1 %). Explanation: Basic Concepts: P = F A A = r 2 Given: r 1 = 2 . 75 cm r 2 = 14 . 5 cm F 2 = 1 . 17 10 4 N homework 24 RAHMAN, TARIQUE Due: May 3 2008, 3:00 am 2 Solution: P 1 = P 2 F 1 A 1 = F 2 A 2 F 1 = parenleftbigg F 2 A 2 parenrightbigg A 1 = parenleftbigg F 2 r 2 2 parenrightbigg ( r 2 1 ) = F 2 r 2 1 r 2 2 = (11700 N) (2 . 75 cm) 2 (14 . 5 cm) 2 = 420 . 838 N Question 5, chap 18, sect 3. part 2 of 2 10 points b) What pressure produces this force? Ne glect the weight of the pistons....
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This note was uploaded on 02/03/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics, Work

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