Physics 07 HOMEWORK SSIGNMENT #9b Cutnell & Johnson, 7 th edition Chapter : Problems 5, 58, 66, 67, 00 5 Concept Simulation. reiews the concept that plays the central role in this problem. (a) The olume flow rate in an artery supplying the brain is. If the radius of the artery is 5. mm, determine the aerage blood speed. (b) Find the aerage blood speed at a constriction in the artery if the constriction reduces the radius by a factor of 3. ssume that the olume flow rate is the same as that in part (a). 58 See Multiple-Concept Example 5 to reiew the concepts that are pertinent to this problem. The blood speed in a normal segment of a horizontal artery is 0. m/s. n abnormal segment of the artery is narrowed down by an arteriosclerotic plaque to one-fourth the normal cross-sectional area. What is the difference in blood pressures between the normal and constricted segments of the artery? *66 The concepts that play roles in this problem are similar to those in Multiple-Concept Example 5, except that the fluid here moes upward rather than remaining horizontal. liquid is flowing through a horizontal pipe whose radius is 0.000 m. The pipe bends straight upward through a height of 0.0 m and joins another horizontal pipe whose radius is 0.0400 m. What olume flow rate will keep the pressures in the two horizontal pipes the same? *67 n airplane has an effectie wing surface area of 6 m that is generating the lift force. In leel flight the air speed oer the top of the wings is 6.0 m/s, while the air speed beneath the wings is 54.0 m/s. What is the weight of the plane? 00 Concept uestions Water flows straight down from an open faucet. The effects of air resistance and iscosity can be ignored. (a) fter the water has fallen a bit below the faucet, is its speed less than, greater than, or the same as it was on leaing the faucet? (b) Is the olume flow rate in cubic meters per second less than, greater than, or the same as it was when the water left the faucet? (c) Is the cross-sectional area of the water stream less than, greater than, or the same as it was when the water left the faucet? Gie your reasoning.
5. RESONING a. ccording to Equation.0, the olume flow rate is equal to the product of the crosssectional area of the artery and the speed of the blood, =. Since and are known, we can determine. b. Since the olume flow rate through the constriction is the same as the olume flow rate in the normal part of the artery, =. We can use this relation to find the blood speed in the constricted region. SOLUTION a. Since the artery is assumed to hae a circular cross-section, its cross-sectional area is = π, where r is the radius. Thus, the speed of the blood is r 6 3 3.6 0 m / s π r 3 ( 5. 0 m) = = = = 4. 0 m/s π (.0) b. The olume flow rate is the same in the normal and constricted parts of the artery, so =. Since =, the blood speed is = / = /. We are gien that the radius of the constricted part of the artery is one-third that of the normal artery, so r = r. Thus, the speed of the blood at the constriction is 6 3 3.6 0 m / s π r π 3 ( r ) 3 π 3 ( 5. 0 m) = = = = = 0.38 m/s 3 58. RESONING We assume that region contains the constriction and region is the normal region. The difference in blood pressures between the two points in the horizontal artery is gien by Bernoulli s equation (Equation.) as P P = ρ ρ, where and are the speeds at the two points. Since the olume flow rate is the same at the two points, the speed at is related to the speed at by Equation.9, the equation of continuity: =, where and are the cross-sectional areas of the artery. By combining these two relations, we will be able to determine the pressure difference. SOLUTION Soling the equation of continuity for the blood speed in region gies = /. Substituting this result into Bernoulli s equation yields = ρ ρ = ρ ρ P P Since =, the pressure difference is 4
ρ ρ ρ 4 3 ( )( ) ( ) ( ) P P = = 6 = 060 kg/m 0. m/s 5 = 96 Pa We hae taken the density ρ of blood from Table.. 66. RESONING ND SOLUTION s seen in the figure, the lower pipe is at the leel of zero potential energy. h If the flow rate is uniform in both pipes, we hae (/)ρ = (/)ρ gh (since P = P ) and =. We can sole for, i.e., = ( / ), and plug into the preious expression to find, so that The olume flow rate is then gien by ( )( ) gh 9.80 m/s 0.0 m = = = 3.6 m/s π ( 0.0400 m) π ( 0.000 m) = = π r = π (0.0400 m) (3.6 m/s) = 3.8 0 m /s
67. RESONING In leel flight the lift force must balance the plane s weight W, so its magnitude is also W. The lift force arises because the pressure P B beneath the wings is greater than the pressure P T on top of the wings. The lift force, then, is the pressure difference times the effectie wing surface area, so that W = (P B P T ). The area is gien, and we can determine the pressure difference by using Bernoulli s equation. SOLUTION ccording to Bernoulli s equation, we hae B B B = T T T P gy P gy Since the flight is leel, the height is constant and y B = y T, where we assume that the wing thickness may be ignored. Then, Bernoulli s equation simplifies and may be rearranged as follows: P = P or P P = ρ ρ B B T T B T T B Recognizing that W = (P B P T ), we can substitute for the pressure difference from Bernoulli s equation to show that ( B T ) ρ ( T B ) W = P P = ( 3 ) ( ) ( ) ( ) =.9 kg/m 6.0 m/s 54.0 m/s 6 m = 9600 N We hae used a alue of.9 kg/m 3 from Table. for the density of air. This is an approximation, since the density of air decreases with increasing altitude aboe sea leel. 00. Concept uestions a. Since the effects of air resistance and iscosity are being ignored, the water can be treated as a freely-falling object, as Chapter discusses. It accelerates with the acceleration due to graity. Therefore, it has a greater speed at the lower point than it did upon leaing the faucet. b. The olume flow rate in cubic meters per second is the same as it was when the water left the faucet. This is because no water is added to or taken out of the stream after the water leaes the faucet. c. The cross-sectional area of the water stream is less than it was when the water left the faucet. With the olume flow rate unchanging, the equation of continuity applies in the form of Equation.9. The olume flow rate is the cross-sectional area of the stream times the speed of the water. When the speed increases, as it does when the water falls, the crosssectional area decreases.
SOLUTION Using the equation of continuity as stated in Equation.9, we hae { = { or = Below faucet t faucet To find the cross-sectional area, we must find the speed. To do this, we use Equation.9 from the equations of kinematics: = + ay or = + ay In using this result, we choose downward as the positie direction. Substituting into the equation of continuity gies 4 (.8 0 m )( 0.85 m/s) = = = 9.3 0 m 0.85 m/s 9.80 m/s 0.0 m + ay + ( ) ( )( ) 5