Physics Solution Manual for 1100 and 2101

Physics Solution Manual for 1100 and 2101

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Unformatted text preview: 200 s 3.0 m / s bg b g 120 m 3 = 0.14 m 1200 s 5.0 m / s bg b g 58. REASONING a. According to Equation 11.10, the volume flow rate Q is equal to the product of the crosssectional area A of the artery and the speed v of the blood, Q = Av. Since Q and A are known, we can determine v. b. Since the volume flow rate Q2 through the constriction is the same as the volume flow rate Q1 in the normal part of the artery, Q2 = Q1. We can use this relation to find the blood speed in the constricted region. SOLUTION a. Since the artery is assumed to have a circular cross-section, its cross-sectional area is A1 = π r12 , where r1 is the radius. Thus, the speed of the blood is v1 = Q1 Q1 3.6 ×10−6 m3 / s = 2= = 4.2 ×10−2 m/s 2 A1 π r1 π 5.2 ×10−3 m ( ) (11.10) b. The volume flow rate is the same in the normal and constricted parts of the artery, so Q2 = Q1. Since Q2 = A2 v2 , the blood speed is v2 = Q2/A2 = Q1/A2. We are given that the radius of the constricted part of the artery is one-third that of the normal artery, so r2 = 1 r1. 3 Thus, the speed of the blood at the constriction is v2 = Q1 Q Q1 = 12 = A2 π r2 π 1 r 31 () 2 = 3.6 × 10−6 m3 / s ( ) π 1 5.2 ×10−3 m 3 2 = 0.38 m/s 598 FLUIDS 59. REASONING AND SOLUTION Let rh represent the inside radius of the hose, and rp the radius of the plug, as suggested by the figure below. hose plug rh rp 2 2 Then, from Equation 11.9, A1v1 = A2v2 , we have π rh v1 = (π rh2 − π rp )v2 , or v1 v2 ( rh2 − rp2 ) = 1 − rp 2 = or r h rh2 rp rh = 1− v1 v2 According to the problem statement, v2 = 3v1, or rp rh = 1− v1 3v1 = 2 = 0.816 3 60. REASONING The number N of capillaries can be obtained by dividing the total crosssectional area Acap of all the capillaries by the cross-sectional area acap of a single capillary. 2 We know the radius rcap of a single capillary, so acap can be calculated as acap = π rcap . To find Acap, we will use the equation of continuity. SOLUTION The number N of capillaries is N= Acap acap = Acap 2 π rcap (1) 2 where the cross-sectional area acap of a single capillary has been replaced by acap = π rcap . To obtain the total cross-sectional area Acap of all the capillaries, we use the equation of continuity for an incompressible fluid (see Equation 11.9). For present purposes, this equation is Aaorta vaorta (2) Aaorta vaorta = Acap vcap or Acap = vcap Chapter 11 Problems 599 2 The cross-sectional area of the aorta is Aaorta = π raorta , and with this substitution, Equation (2) becomes Acap = Aaorta vaorta vcap = 2 π raorta vaorta vcap Substituting this result into Equation (1), we find that N= = Acap 2 π rcap π r2 aorta vaorta vcap r2 v = = aorta aorta 2 2 rcap vcap π rcap (1.1 cm )2 ( 40 cm/s ) = 2 ( 6 ×10−4 cm ) ( 0.07 cm/s ) 2 ×109 61. SSM REASONING AND SOLUTION a. Using Equation 11.12, the form of Bernoulli's equation with y 1 = y 2 , we have ( ) 2 2 P − P2 = ρ v2 − v1 = 1 1 2 1.29 kg/m3 2 (15 m/s)2 − ( 0 m/s )2 = 150 Pa b. The pressure inside the roof is greater than the pressure on the outside. Therefore, there is a net outward force on the roof. If the wind speed is sufficiently high, some roofs are "blown outward." 62. REASONING We apply Bernoulli’s equation as follows: 2 2 PS + 1 ρ vS + ρ gyS = PO + 1 ρ vO + ρ gyO 24 24 144 2444 144 2444 3 3 At surface of vaccine in reservoir At opening SOLUTION The vaccine’s surface in the reservoir is stationary during the inoculation, so that vS = 0 m/s. The vertical height between the vaccine’s surface in reservoir and the opening can be ignored, so yS = yO. With these simplifications Bernoulli’s equation becomes 2 P = PO + 1 ρ vO S 2 Solving for the speed at the opening gives 600 FLUIDS vO = 2 ( P − PO ) S ρ = ( 2 4.1×106 Pa 1100 kg/m3 )= 86 m/s 63. SSM REASONING AND SOLUTION Let the speed of the air below and above the wing be given by v 1 and v 2 , respectively. According to Equation 11.12, the form of Bernoulli's equation with y 1 = y 2 , we have c P1 − P2 = ρ v − v 1 2 2 2 2 1 1.29 kg / m 3 = ( 251 m / s) 2 − ( 225 m / s) 2 = 7.98 × 10 3 Pa 2 h From Equation 11.3, the lifting force is, therefore, c h F = P1 − P2 A = ( 7 .98 × 10 3 Pa)(24.0 m 2 ) = 1.92 × 10 5 N 64. REASONING The absolute pressure in the pipe must be greater than atmospheric pressure. Our solution proceeds in two steps. We will begin with Bernoulli’s equation. Then we will incorporate the equation of continuity. SOLUTION According to Bernoulli’s equation, as given by Equation 11.11, we have 2 2 P + 1 ρ v1 + ρ gy1 = P2 + 1 ρ v2 + ρ gy2 12 24 1442443 144 2444 3 Pipe Nozzle The pipe and nozzle are horizontal, so that y1 = y2 and Bernoulli’s equation simplifies to 2 2 P + 1 ρ v1 = P2 + 1 ρ v2 12 2 or ( 2 2 P = P2 + 1 ρ v2 − v1 1 2 ) where P1 is the absolute pressure of the water in the pipe. We have values for the pressure P2 (atmospheric pressure) at the nozzle opening and the speed v1 in the pipe. However, to solve this expression we also need a value for the speed v2 at the nozzle opening. We obtain this value by us...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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