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Unformatted text preview: he same everywhere. Since the volume flow rate is equal to the speed
of the water times the crosssectional area of the pipe (see Equation 11.10), the speed is
greatest where the crosssectional area is smallest.
15. (b) The volume flow rate Q of the blood is the same everywhere. Since Q = Av, we have
that A2v2 = A1v1, where the subscript 2 denotes the unblocked region of the artery and 1 the
partially blocked region. Since the area of a circle is A = π r 2 , the speed v1 of the blood in ( 2.0 mm )2 r22 the partially blocked region is v1 = v2 2 = ( 0.40 m/s ) = 1.6 m/s.
2
r (1.0 mm ) 1 16. (a) Since blood is an incompressible fluid, the volume flow rate Q is the same everywhere
within the artery. Q is equal to the crosssectional area of the artery times the speed of the
blood, so the blood slows down, or decelerates, as it moves from the narrow region into the
wider region. Because the hemoglobin molecule decelerates, the direction of the net force
acting on it must be opposite to its velocity. Therefore, the pressure ahead of the molecule is
greater than that behind it, so the pressure in the wider region is greater than that in the
narrow region.
17. P1 − P2 = 207 Pa Chapter 11 Answers to Focus on Concepts Questions 563 18. (b) The pressure at C is greater than that at B. These two points are at the same elevation,
but the fluid is moving slower at C since it has a greater crosssectional area. Since the fluid
is moving slower at C, its pressure is greater. The pressure at B is greater than that at A. The
speed of the fluid is the same at both points, since the pipe has the same crosssectional area.
However, B is at the lower elevation and, consequently, has more water above it than A.
The greater the height of fluid above a given point, the greater is the pressure at that point,
provided the crosssectional area does not change.
19. PB − PA = 12 000 Pa
20. (d) A longer pipe offers a greater resistance to the flow of a viscous fluid than a shorter pipe
does. The volume flow rate depends inversely on the length of the pipe. For a given pipe
radius and pressure difference between the ends of the pipe, the volume flow rate is less in
longer pipes. In this case the longer pipe is twice as long, so its volume flow rate QB is
onehalf that of QA.
21. QB = 1.62 × 104 m3/s 564 FLUIDS CHAPTER 11 FLUIDS PROBLEMS
1. SSM REASONING The weight W of the water bed is equal to the mass m of water times
the acceleration g due to gravity; W = mg (Equation 4.5). The mass, on the other hand, is
equal to the density ρ of the water times its volume V, or m = ρ V (Equation 11.1). SOLUTION Substituting m = ρ V into the relation W = mg gives
W = mg = ( ρ V ) g ( ) ( ) = 1.00 × 103 kg/m 3 (1.83 m × 2.13 m × 0.229 m ) 9.80 m/s 2 = 8750 N We have taken the density of water from Table 11.1. Since the weight of the water bed is
greater than the additional weight that the floor can tolerate, the bed should not be
purchased. 2. m
(Equation 11.1), where m is
V
the mass of the solvent and V is its volume. The solvent occupies a cylindrical tank of radius
r and height h. Its volume V, therefore, is the product of the circular crosssectional area
π r 2 of the tank and the height h of the solvent:
REASONING The density ρ of the solvent is given by ρ = V = π r 2h
SOLUTION Substituting Equation (1) into ρ = (1) m
(Equation 11.1), we obtain the density
V of the solvent: ρ= 3. m
m
14 300 kg
= 2=
= 824 kg/m3
V π r h π (1.22 m )2 ( 3.71 m ) SSM REASONING Equation 11.1 can be used to find the volume occupied by 1.00 kg
of silver. Once the volume is known, the area of a sheet of silver of thickness d can be found
from the fact that the volume is equal to the area of the sheet times its thickness. Chapter 11 Problems 565 SOLUTION Solving Equation 11.1 for V and using a value of ρ = 10 500 kg/m3 for the
density of silver (see Table 11.1), we find that the volume of 1.00 kg of silver is
V= m ρ = 1.00 kg
= 9.52 ×10−5 m3
3
10 500 kg/m The area of the silver, is, therefore, A= 4. V 9.52 ×10 –5 m3
=
= 317 m 2
−7
d 3.00 × 10 m REASONING AND SOLUTION
a. We will treat the neutron star as spherical in shape, so that its volume is given by the
familiar formula, V = 4 π r 3 . Then, according to Equation 11.1, the density of the neutron
3 star described in the problem statement is ρ= 3m
3(2.7 × 1028 kg)
m
m
=
=
=
= 3.7 × 1018 kg/m3
3
3
3
4 π r3
V3
4π r
4π (1.2 × 10 m) b. If a dime of volume 2.0 × 10 –7 m3 were made of this material, it would weigh
W = mg = ρVg = (3.7 × 1018 kg/m3 )(2.0 × 10 –7 m3 )(9.80 m/s 2 )= 7.3 ×1012 N
This weight corresponds to 1 lb 12
7.3 × 1012 N = 1.6 × 10 lb 4.448 N 5. REASONING AND SOLUTION 14.0 karat gold is (14.0)/(24.0) gold or 58.3%. The weight
of the gold in the necklace is then (1.27 N)(0.583) = 0.740 N. This corresponds to a volume
given by V = M/ρ = W/(ρg) . Thus,
V= 6. 0.740 N (19 300 kg/m )(9.80 m/s )
3 REASONING AND SOLUTION 2 = 3.91 × 10−6 m3 If the concrete were solid, it would have a...
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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