Physics 110 Homework Solutions Week #9 – Tuesday 11/5 and Wednesday11/6 Tuesday, November 5, 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. 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 3

s . c. By the equation of continuity we have

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flow rate = Acapvcap → Acap =1.47 ×10−4 m 3

s

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

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 .

Wednesday, November 6, 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.

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

€

P + 12 ρv

2 = constant , we have that

€

ΔP =Flift2Awing

=mplaneg2Awing

= 12 ρ vtop

2 − vbottom2( ) = 1

2 ρ 1.25( )2 −1( )vbottom2

vbottom =mplaneg

0.563ρairAwing

=10,000kg × 9.8 m

s2

0.563×1.3 kgm 3 × 70m2 = 43.8 m

s = 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