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Physics 110 Homework Solutions Week #9 - Wednesday Friday, May
24, 2013 Chapter 8 Questions 8.10 From the continuity equation Q =
Av = constant, as the A decreases the v increases 8.11 Try it!
Un-intuitively, the papers come together.
Multiple-Choice 8.10 B 8.13 A 8.17 A 8.18 B 8.22 E Problems
8.11120 mmHg corresponds to a pressure difference of
. This corresponds to a depth of
.
8.16 A fountain of water a. By conservation of energy we
calculate the height of the column of water
assuming it to be almost completely vertical. We have
b. By continuity we have
where the velocity of the water at a height of 10 m is
calculated using
.
Thus the diameter if the water at 10m is 11.7cm. 8.19 Fluids in
the human heart
a. The velocity is calculated from the flow rate and we have
where the flow rate is
calculated in part b.
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b. From the information in the problem we calculate the flow
rate to be
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Q = flow rate = Aaortavaorta = πraorta2 vaorta = π 0.0125m(
)
2× 0.3 ms =1.47 ×10
−4 m 3s .
c. By the equation of continuity we have
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flow rate = Acapvcap → Acap =1.47 ×10−4 m 3s
3×10−4 ms= 0.49m2 .
d. Assuming that the capillaries all have the same cross
sectional area we have from
the equation of continuity .
e. For each of the velocities we calculated we can calculate the
volume kinetic energy of the blood in the aorta, the arteries and
the capillaries. Thus we find
f. If the blood flow is constant we find the time from the
distance traveled and the
speed. The time is .
8.20 The human heart as a pump a. To calculate the velocity we
use the equation of continuity and we find
.
b. The volume kinetic energy is given as
c. The power is the rate at which work is done and work is the
force that is applied to the blood. We relate the force applied to
the pressure and we have
d. To calculate the velocity of the blood in the blockage we use
Bernoulli’s equation.
Thus we have
8.26 Spinal Fluid
a. The pressure at a point 75cm lower than the base of the brain
is given as
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and this corresponds to a height of a column of water of
height
b. If the person is lying down then there is no difference in
height of the fluid. There
fore we have which corresponds to a height of a
column of water
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hwater =P
ρwaterg=
1333 Nm 2
1000 kgm 3× 9.8 m
s2= 0.136m =13.6cm .
Monday, May 27, 2013 Chapter 8 Questions - none Multiple-Choice
- none Problems - none Tuesday, May 28, 2013 Chapter 8 Questions -
none Multiple-Choice - none Problems 8.21 The speed of the exiting
water is calculated from Bernoulli’s equation where we
assume that the pressure on the top and sides of the tank is due
to air and is the same in both places. Further we assume that the
area of the tank’s top is much larger than the hole in the tanks
side so that we can, to a reasonable approximation ignore the speed
of the falling fluid. Thus we have from Bernoulli’s equation
where we
took the zero of the gravitational potential energy (per unit
volume) to be zero at the hole.
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8.22 The gauge pressure is given by and the
initial efflux velocity is .
8.27 The total mass of the airplanes is . The
weight of the airplanes is equal to the weight of a slice of
water 22cm thick with cross sectional area that is bounded by the
waterline of the ship. Thus we have
8.31 Power is the rate at which energy is transferred or and
Bernoulli’s
equation represents the energy per unit volume in a fluid, so
multiplying and dividing our expression for power by a volume we
find
. The
speed given converts to and the power becomes
8.33 Using Bernoulli’s equation we have
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P + 12 ρv2 = constant , we have that
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ΔP =Flift2Awing
=mplaneg2Awing
= 12 ρ vtop2 − vbottom
2( ) = 12 ρ 1.25( )2−1( )vbottom2
vbottom =mplaneg
0.563ρairAwing=
10,000kg × 9.8 ms2
0.563×1.3 kgm 3× 70m2
= 43.8 ms = 98 mihr
8.34 A Boeing 777
a. If the airplane is not accelerating up or down then the
difference in pressures above and below the wing gives rise to the
lifting force when multiplied by the wing area. Therefore we have
for the pressure on the upper wing surface
.
b. The upward acceleration is calculated from the unbalanced
force that acts on the wing (assuming that the pressure on the
upper surface remains constant) using Newton’s 2nd law. Thus we
have
