Download Homework 10 - NMSU Physics

Solution to HW#10
CJ5 11.CQ.004.
REASONING AND SOLUTION As you climb a mountain, your
ears "pop" because of the change in atmospheric pressure.
a. As you climb up, the outside pressure becomes lower than the pressure in your
inner ear. The outward force per unit area on your eardrum is greater than the
inward force per unit area; therefore, your eardrum moves outward.
b. As you climb down, the outside pressure becomes greater than the pressure in
your inner ear. The outward force per unit area on your eardrum is less than the
inward force per unit area on your eardrum; therefore, your eardrum moves inward.
CJ5 11.CQ.012.
REASONING AND SOLUTION Archimedes' principle states
that any fluid applies a buoyant force to an object that is partially or completely
immersed in it; the magnitude of the buoyant force equals the weight of the fluid that
the object displaces.
A glass beaker, filled to the brim with water, is resting on a scale. A block is
placed in the water, causing some of it to spill over. The water that spills is wiped
away, and the beaker is still filled to the brim.
a. According to Table 11.1, the density of wood is 550 kg/m3,
while the density of water at 4 °C is 1.000 × 103 kg/m3 . Since the
Buoyant force
on the block
density of the wood is less than the density of water, the block will
float with part of its volume above the surface of the water.
According to Archimedes' principle, the buoyant force exerted on
Weight of
the bottom surface of the block must be equal in magnitude to the
the block
weight of water that is displaced by the block. However, since the
block floats on the surface of the water, the buoyant force must be
equal in magnitude and opposite in direction to the weight of the
block, as suggested in the free body diagram at the right. Hence,
the weight of the block must be equal to the weight of the water displaced by the
block. The weight of the contents of the beaker remains the same. Therefore, the
initial and final scale readings are the same if the block is made of wood.
b. According to Table 11.1, the density of iron is 7860 kg/m3, while the density of
water at 4 °C is 1.000 × 103 kg/m3 . Since the density of iron is greater than the
density of water, the iron block will sink to the bottom of the beaker. Let V represent
the volume of water that spills over. According to the definition of mass density
(Equation 11.1) the mass of water that spills over is equal to mwater = ρ water V . The
beaker is still filled to the brim, but the block now occupies the volume V where the
water used to be. According to Equation 11.1, the mass that now occupies the
volume V is given by mblock = ρiron V . Since ρiron > ρwater, the mass of the block is
greater than the mass of the water that spilled over. The weight of the contents of the
beaker increases. Therefore, the final reading on the scale will be greater than the
initial reading.
CJ5 11.CQ.015 .
REASONING AND SOLUTION As a person dives toward the
bottom of a swimming pool, the pressure increases noticeably. The buoyant force,
however, remains the same. According to Archimedes' principle, the water applies a
buoyant force to the swimmer, the magnitude of which is equal to the weight of the water
displaced by the swimmer. In other words, the buoyant force experienced by the
swimmer is given by ( ρ water Vswimmer ) g , where ρ water is the density of the water and
V swimmer is the volume of water displaced by the swimmer. If we neglect any change in
water density with depth, then the quantity ρ water Vswimmer remains the same as the
swimmer approaches the bottom of the pool.
CJ5 11.P.003. REASONING AND SOLUTION
The weight of the concrete in the
highway is
W = mg
(1)
where the mass m of the concrete can be found from the density of concrete and the
volume of the roadbed. By definition, density is the mass per unit volume:
ρ = m/V , where the volume is V = wd. Therefore,
m = ρ( wd)
(2)
Combining Equations (1) and (2) and obtaining the density of concrete from Table
11.1, we find that
W = ρ( wd)g = (2200 kg/m3)(150 × 103 m)(7.0 m)(0.30 m)(9.80
9
m/s2)= 6.8 × 10 N
CJ5 11.P.015. REASONING AND SOLUTION The least number of bricks are used
when the surface area is the smallest. The area of the brick in contact with the
ground has a surface area of
A = (0.0570 m)(0.0890 m) = 5.07 × 10–3 m2
The pressure exerted by a single brick is
P = F/A = (17.8 N)/(5.07 × 10–3 m2) = 3.51 × 103 N/m2
In order to equal or exceed atmospheric pressure
Number of bricks = (1.013 × 105 N/m2)/(3.51 × 103 N/m2) = 29 bricks
CJ5 11.P.023. 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 second-floor
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 first-floor 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 )
CJ5 11.P.026. REASONING AND SOLUTION The difference in pressure between the
ground and the roof is ∆P = ρgh, so
133 Pa
13.0 mm Hg
1 mm Hg
h=
= 137 m
1.29 kg / m 3 9.80 m / s 2
gFGH
hc
b
c
IJ
K
h
Note that we have used the fact that 133 Pa = 1 mm Hg as a conversion factor.
CJ5 11.P.037. REASONING AND SOLUTION From Pascal's principle, the pressure in
the brake fluid at the master cylinder is equal to the pressure in the brake fluid at the
plungers: PC = PP, or
FC
AC
=
FP
AP
or
F I
GH JK
πr2
r
= FC P2 = FC P
FP = FC
AC
π rC
rC
AP
2
The torque on the pedal is equal to the torque that is applied to the master cylinder so
that
F = FC
C
or
FC = F
C
Combining the expression for FC with the expression for FP above, we have
2
 rp 
0.150 m   1.90 × 10 –2 m 

 = (9.00 N) 


FP = F
 0.0500 m   9.50 × 10 –3 m 
C  rC 
2
=
108 N
CJ5 11.P.041. REASONING AND SOLUTION The upward buoyant force on the raft
must equal the weight of the raft plus the weight of the swimmers, FB = Wr + Ws.
Thus,
ρwgVr = ρpgVr + Msg
Ms = (ρw − ρp)Vr = (1.00 × 103 kg/m3 − 550 kg/m3)(4.0 m)(4.0 m)(0.30 m)
Ms = 2.2 × 103 kg
CJ5 11.P.053. REASONING AND SOLUTION
a. The volume flow rate is given by
Q = Av = (2.0 × 10–4 m2)(0.35 m/s) = 7.0 × 10 –5 m 3/s
b.
We know that A1v1 = A2v2 so that
v2 = v1(A1/A2) = (0.35 m/s)(2.0 × 10–4 m2)/(0.28 m2) = 2.5 × 10 –4 m/s
CJ5 11.P.058. REASONING AND SOLUTION
Using Equation 11.12,
1
1 2
1
2
2
2
P1 + ρv1 = P2 + ρv2 , we obtain P1 − P2 = ρ(v2 − v1 ). With v1 = 0 and v2 = 27
2
2
2
m/s, we have
1
P1 − P2 = 2 (1.29 kg/m 3 )(27 m/s) 2 =
4.7 × 10 2 Pa
CJ5 11.P.065. SSM
REASONING
Since the pressure difference is known,
Bernoulli's equation can be used to find the speed v 2 of the gas in the pipe.
Bernoulli's equation also contains the unknown speed v 1 of the gas in the Venturi
meter; therefore, we must first express v 1 in terms of v 2 . This can be done by using
Equation 11.9, the equation of continuity.
SOLUTION
a. From the equation of continuity (Equation 11.9) it follows that v1 = A2 / A1 v 2 .
c
Therefore,
v1 =
0.0700 m 2
v = ( 1.40 ) v 2
0.0500 m 2 2
h
Substituting this expression into Bernoulli's equation (Equation 11.12), we have
P1 + 21 ρ ( 1.40 v 2 ) 2 = P2 + 21 ρ v 22
Solving for v 2 , we obtain
v2 =
2 ( P2 − P1 )
ρ (1.40 ) − 1
2
=
2( 120 Pa)
(1.30 kg / m 3 ) (1.40) 2 − 1
= 14 m / s
b. According to Equation 11.10, the volume flow rate is
Q = A2 v 2 = (0.0700 m 2 )( 14 m / s) = 0.98 m 3 / s