Unformatted text preview: 10 –4 m) 85. SSM REASONING Since the faucet is closed, the water in the pipe may be treated as a
static fluid. The gauge pressure P2 at the faucet on the first floor is related to the gauge
pressure P1 at the faucet on the second floor by Equation 11.4, P2 = P1 + ρ gh .
SOLUTION
a. Solving Equation 11.4 for P1 , we find the gauge pressure at the secondfloor faucet is
P1 = P2 − ρ gh = 1.90 × 10 5 Pa – (1.00 × 10 3 kg / m 3 )( 9.80 m / s 2 )( 6.50 m ) = 1.26 × 10 5 Pa b. If the second faucet were placed at a height h above the firstfloor faucet so that the
gauge pressure P1 at the second faucet were zero, then no water would flow from the
second faucet, even if it were open. Solving Equation 11.4 for h when P1 equals zero, we
obtain
h= P2 − P1 ρg = 1.90 × 10 5 Pa − 0
= 19.4 m
(1.00 × 10 3 kg / m 3 )(9.80 m / s 2 ) Chapter 11 Problems 615 86. REASONING AND SOLUTION Using Bernoulli's equation
2 2 3 2 2 ∆P = P1 − P2 = (1/2)ρv2 − (1/2)ρv1 = (1/2)(1.29 kg/m )[(8.5 m/s) − (1.1 m/s) ]
∆P = 46 Pa
The air flows from high pressure to low pressure (from lower to higher velocity), so it
enters at B and exits at A . 87. SSM REASONING According to Archimedes principle, the buoyant force that acts on
the block is equal to the weight of the water that is displaced by the block. The block
displaces an amount of water V, where V is the volume of the block. Therefore, the weight
of the water displaced by the block is W = mg = ( ρ waterV ) g .
SOLUTION The buoyant force that acts on the block is, therefore, F = ρ waterV g = (1.00 × 103 kg/m3 )(0.10 m × 0.20 m × 0.30 m)(9.80 m/s 2 ) = 59 N 88. REASONING The pressure P2 at a lower point in a static fluid is related to the pressure P1 at a higher point by Equation 11.4, P2 = P1 + ρ gh, where ρ is the density of the fluid, g is
the magnitude of the acceleration due to gravity, and h is the difference in heights between
the two points. This relation can be used directly to find the pressure in the artery in the
brain. SOLUTION Solving Equation 11.4 for pressure P1 in the brain (the higher point), gives ( )( ) P = P2 − ρ gh = 1.6 ×104 Pa − 1060 kg/m3 9.80 m/s 2 ( 0.45 m ) = 1.1×104 Pa
1 89. REASONING According to Equation 4.5, the pillar’s weight is W = mg . Equation 11.1
can be solved for the mass m to show that the pillar’s mass is m = ρ V . The volume V of the
cylindrical pillar is its height times its circular crosssectional area.
SOLUTION Expressing the weight as W = mg (Equation 4.5) and substituting m = ρ V
(Equation 11.1) for the mass give W = mg = ( ρ V ) g (1) 616 FLUIDS The volume of the pillar is V = h π r 2 , where h is the height and r is the radius of the pillar.
Substituting this expression for the volume into Equation (1), we find that the weight is ( ) ( ) ( ) 2
W = ( ρ V ) g = ρ h π r 2 g = 2.2 ×103 kg/m3 ( 2.2 m ) π ( 0.50 m ) 9.80 m/s 2 = 3.7 ×104 N Converting newtons (N) to pounds (lb) gives ( ) 0.2248 lb 3
W = 3.7 ×104 N = 8.3 ×10 lb
1N 90. REASONING The flow rate Q of water in the pipe is given by Poiseuille’s law
π R 4 ( P2 − P )
1
Q=
(Equation 11.14), where R is the radius of the pipe, P2 − P is the
1
8η L
difference in pressure between the ends of the pipe, η is the viscosity of water, and L is the
length of the pipe. As the radius R of the pipe gets smaller, its length L does not change, nor
does the viscosity η of the water. We are told that the pressure difference P2 − P also
1
remains constant. Moving all of the constant quantities to one side of Equation 11.14, we
obtain
π ( P2 − P ) Q
1
=4
(1)
8η L
R
14 3
24
Each quantity
is constant We will use Equation (1) to find the final flow rate Qf in terms of the initial flow rate Q0, the
initial radius R0 and the final radius Rf of the pipe. Q
of the flow rate to the fourth power of
R4
the pipe’s inner radius remains constant as the inner radius decreases to Rf from R0.
Therefore, we have that
SOLUTION Equation (1) shows that the ratio Qf
Rf4 =
4 Q0
4
R0 or Qf = Q0 Rf4
4
R0 R = Q0 f R 0 4 (2) R Note that the ratio f multiplying the initial flow rate Q0 in Equation (2) is unitless, so
R 0
that Qf will have the same units as Q0. Thus, to obtain the final flow rate in gallons per Chapter 11 Problems 617 minute, there is no need to convert Q0 to SI units (m3/s). Substituting the given values into
Equation (2) yields the final flow rate:
4 0.19 m Qf = ( 740 gal/min ) = 290 gal/min 0.24 m 91. REASONING The density of the brass ball is given by Equation 11.1: ρ = m / V . Since
the ball is spherical, its volume is given by the expression V = ( 4 / 3) πr 3 , so that the density
may be written as
m
3m
ρ= =
V 4 πr 3 This expression can be solved for the radius r; but first, we must eliminate the mass m from
the equation since its value is not specified explicitly in the problem. We can determine an
expression for the mass m of the brass ball by analyzing the forces on the ball.
The only two forces that act on the ball are the upward tension T in the wire and the
downward weight mg of the ball....
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 Spring '13
 CHASTAIN
 Physics, The Lottery

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