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Instructor: Dr. Scott A. Yost Office: 307 Nielsen Physics Building Hours: Monday, Wednesday and Friday, 1-2 PM or by appointment Textbook: L.A. Bloomfield, How Things Work, 2nd ed. |
Phone: 974-7852 E-Mail: syost@utk.edu Web Site: http://homework.phys.utk.edu/phys101 |
These are the answers to the even-numbered exercises and problems, and all of the cases assigned in Chapter 4. The answers to the others appear in the back of the textbook.You must have the symbol fonts installed to view Greek letters properly. You can download the file SYMBOL.TTF to your WINDOWS/FONTS folder if you do not have it already.
Exercises
2. The log also displaces more water. It is the density of the object which determines whether it will float. Objects with less density than water will float.6. The air in the car will continue to move forward, displacing the lighter helium balloon, which will move toward the rear of the car.
10. Fish must actively adjust the pressure on an air bladder by squeezing it or letting it relax so that the average density of the fish matches that of water. Without this effort, they could not float motionlessly.
12. The air pressure is higher at lower elevations, due to the extra weight of the air above. This extra weight compresses the bottle.
16. -300 degrees C is below absolute zero, so there is no temperature that low.
20. The tea flows faster when there is more pressure on the tea. The pressure decreases as the bottle empties, reducing the speed of the flow.
28. The air in your lungs must be less than the surrounding air when you breathe in.
30. Higher pressure allows for a higher nozzle velocity when the water leaves the hose. This allows the water to reach further before it falls to the ground.
Problems
4. If the temperatures are the same, the pressure must be proportional to the particle density. So the pressure inside the tank is 30 times as great as outside.
6. The net force on the log is the upward boyant force minus the weight of the log. The water displaced is 2 kg heavier than the log, so the net force upward is 2 kg x 9.8 m/s = 19.6 N.
8. If the density of gold is nineteen times that of water, the crown displaces 1/19 of its weight in water. If the crown weighs 30 N, the upward bouyant force is then 30N / 19 = 1.6 N.
10. Problem 9 showed that you must exert 28.4 N upward to keep the crown from accelerating. In other words, the crown weighs 30 N - 1.6 N = 28.4 N under water. If the crown were made of copper, the bouyant force would be 30N / 9 = 3.3 N, since copper is 9 times as heavy the water it displaces. Then the crown would weigh 30 N - 3.3 N = 26.7 N under water. The copper crown would weigh less under water than the gold crown because of its lower density, even though both weigh as much above water.
12. The top of the tank must be 25 meters above the ground. The water from the fountain will rise to the same height, by conservation of energy (which is equivalent to Bernoulli's Principle).
Cases
1a. The net force on the ship is zero.
1b. If the ship is displacing less fluid, the fluid must be denser, since the weight of the ship hasn't changed.
1c. The ship would displace less water, so the bouyant force would decrease.
1d. The ship would displace more water, so the bouyant force would increase.
1e. The last two answers show that the ship is in stable equilibrium. Either displacement results in a restoring force pushing the boat toward the equilibrium position where it is displacing exactly its weight in water.
1f. The ship is heavier with another passenger, so it must sink a little to displace more water.
1g. The ship sinks further the more cargo is on board. With experience, a captain could recognize how far the ship sinks into the water with different loads, and use this knowledge to estimate the load.
1h. Yes, but the ship is denser than air, so it doesn't float in air. Its weight is greater than the bouyant force of air.3a. Her average density should match the density of the water.
3b. She would feel the atmospheric pressure plus
rgh = 1000 kg/m3 x 9.8 m/s2 x 20 m = 2.0 x 105 Pa. The net pressure, including the atmosphere, would be 1.0 Atm + 2.0 Atm = 3.0 Atm.
3c. The pressure in her mouth must be greater than in her lungs.
3d. The pressure in her mouth would be less than in her lungs, so she would not be able to breathe.
3e. If the temperature of the air remains constant, then the pressure is proportional to the particle density. This means she is breathing more air per breath when it is under higher pressure, and depletes her tanks more quickly.
5a. The pressure is dropping from the high pressure in the hose to atmospheric pressure. Conservation of energy then makes the speed increase. This is an example of Bernoulli's Principle.
5b. The water slows down when it hits the bucket. The pressure must increase to compensate, by Bernoulli's Principle.
5c. Energy conservation says that the kinetic energy of the water when it falls back down is the same as when it first went up. So the speed is the same as when it left the nozzle. (Actually, it is less if you take into account air resistance.)
5d. The sponge's density must be greater than water if it sinks.
7a. The highest level h the water can reach is given by P = rgh where the pressure is P = 5 x 105 Pa. Since the density is r = 103 kg/m3 for water, the height is
h = P/(rg) = 51 m.
7b. Bernoulli's equation says that the pressure in the hose is equal to rv2/2 outside the hose. Then
v2 = 2P/ r = 1000 m2/s2. Taking the square root gives a water velocity of v = 32 m/s.
7c. The pressure just before the water leaves the hose will be equal to rgh, where h is the height the water can rise to. This gives 51 m, as in part a.
7d. If the pressure is 3 times as great, the water can rise 3 times as high. The water can then reach 153 m above ground.
Department of Physics University of Tennessee