Physics 110 Homework Solutions Week #9 Friday, May 22, 2009 Chapter 8 Questions 8.6 For the vertical walled container, the pressure = mg/A with mg = total weight and A = surface area at bottom. For the other two containers the sidewalls exert vertical forces (either up for the container on the left, or down for that on the right) since they are slanted and their normals have a vertical component. These additional forces account for the same pressure. Multiple-Choice 8.3 D Problems 8.1 The radius is found from the density. We have

.

8.2 We find the mass of each from their respective densities and volumes. Thus the total

mass is m = ρwVw + ρiceVice = (103)(πd2Hw/4) + (0.917 x 103)(πd2Hice/4) = 3.31 x 104

kg 8.6 The pressure is given by

. Since the pressure is proportional to the inverse of the volume and if the diver were to rise the pressure would decrease by approximately 2x105 times then the volume would increases by the same fraction.

Monday, May 25, 2009 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 Problems 8.7 The height is given from the difference in pressure

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8.9 P = Patm + ρgH = 1.01105 N/m2 + (1.025 x 103kg/m3)(9.8m/s2)(35,800 ft x 0.305 m/ft) = 1.1 x 108 Pa, assuming ρ = constant 8.12 This maximum force represents the gauge pressure and is given by

. From this we can calculate the mass

of water in the tube and knowing the density we can calculate the height (through the volume of liquid.) Solving the equation for the mass we find

. This corresponds to a weight

of water of , or about 25,600 times the force from the water. From the density we can now calculate the height of the column of water. We find

.

Tuesay, May 26, 2009 Chapter 8 Questions

- None

Multiple-Choice 8.22 E Problems 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 . c. By the equation of continuity we have

.

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 .

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

Wednesday, May 27, 2009 Chapter 8

Questions - None

Multiple-Choice - None Problems 8.11120 mmHg corresponds to a pressure difference of

. This corresponds to a depth of

.

8.33. Using Bernoulli’s equation we have P + ½ ρv2 = constant, we have that ΔP = ½

ρΔv2 and ΔF = AΔP + ½ AρΔv2. Given the wing area, density of air, weight of the plane, and the fractional difference in air speed above/below the wing, we can write that ΔF = ½ (2 x 70)(1.3) v2[1.252 - 1] = Mg = 10,000(9.8); then solving for v we have v = 43.8 m/s = 98 mph.

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