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Answer 1.1: a) The flour is leaking out at a rate of .04 lb/s Answer 1.2: You would do 16.6 J more work Answer 1.3: a) Luisa does 532 kJ of work, Miguel does 684 kJ Answer 1.4: a) Sam does 2144 J of work b) Cory is moving at 9.24 m/s at the bottom Answer 1.5: Because the direction of the force changes with the distance the roller travels, we need to take the integral of the force dotted with the distance over the distance and end up with 40.2 J Answer 2.1: a) 449 J b) It is moving at 8.47 m/s Answer 2.2: a) 1.41 x 10^{11 }J b) 2.87 x 10^{11} J; since her distance is almost 10% of the earth’s radius, the gravitational field strength is noticeably reduce Answer 2.3: a) P=E/t and E=mgh, so water goes over the dam at a rate of1.93 x 10^{6} kg/s b) 46.7 m/s Answer 2.4: a) .859 m/s b) 388 m/s Answer 2.5: a) 81.75 N/m b) 501 J Answer 2.6: a) 450 J b) His maximum speed is when the net force on him is equal, when the cord has stopped accelerating him and friction is not yet slowing him down, and at that point his speed is 2.65 m/s c) 4.1 m d) 450 J Answer 3.1: a) 816 J b) 4.2 m/s c) 7.8 m/s Answer 3.2: a) .0566 J b) .00579^{o} Celsius Answer 3.3: a) 4.85 m/s b) .295 m c) If the monkeys drop the bananas at the top of their swing, it won’t affect anything about their swing back down. However, the bananas will probably be stolen by someone else before they get back down. Answer 3.4: a) 2 m vertically above the lowest point b) 6.26 m/s c) 462 J Answer 4.1: 1.6 x 10^{4} m^{3} Answer 4.2: a) 78.2 g b) 210 o F c) No, mass is so small Answer 4.3: a) 24 photons at 690 nm is 1.38 x 10^{17} J of energy b) 38.6 % Answer 5.1: a) 15 s b) Using pulleys means that Joe won’t have to pull as hard because the ropes divide the load. Unfortunately, it also means that he has to walk four times as far to raise the water from the well. So the power (rate at which he does work) is smaller, but he still has to do the same amount of work. Answer 5.2: a) 3.17 x 10^{13} kg of coal, 1.16 x 10^{14} kg of CO 2 b) 9000 kg of enriched uranium Answer 5.3: a) 40 cells b) About 1 yr, 10 mos Answer 5.4: Answer 5.5: Answer 5.6: where r is the density of the air. The kinetic power of the wind depends on the cube of the wind velocity, so a small increase in the wind speed will slow the ball down significantly. Answer 5.7: The power in the wind is Answer 5.8: We assume that the wind is steady in direction and magnitude, that the density of the air is constant, and that the flow is uniform and incompressible. That means that we can assume the air flows through the turbine in a streamline, as shown above, which means that the flow of mass through the turbine is: These assumptions also mean that Bernoulli’s equation holds: Ï b + r v b 2 = Ï a + r v a 2 In the two equations above, r is the density of the air, A b, A t, and A a are the crosssectional area of the streamline before, at, and after the turbine, respectively; v b, v t, and v a are the speed of the air before, at, and after the turbine, respectively; and Ï b and Ï a are the air pressure before and after the turbine, respectively. First consider the force on the rotor, and conservation of momentum to derive: Further, consider force as a difference of pressure, and use Bernoulli’s equation to derive: F=A t ( Ï a Ï b) = r A t (v b 2v a 2) Setting these two equal, show that the speed of the wind at the turbine is an average of the speeds before and after: Now, the power extracted by the turbine is equal to the difference in the wind power before and after the turbine. Using the equation we derived in problem 1 for the kinetic power of the wind: we can find: For a given v b, we want to find the v a that maximizes P t. Show that this occurs for v a= v b/3. Substitute this back in to find: The quantity in parentheses is the power of the wind impinging on the turbine, so the turbine can extract 16/27 = 59% of the power of the wind. This is known as the Betz Limit. So Mariella is right, and Ignacio is unnecessarily pessimistic about the potential of wind power. Answer 5.9: a) The turbine/generator converts into electricity 45% of the power in the wind, so and A = 31.6 m 2. The area swept out is circular, so A = p r 2 and r = 3.2 m. b) We reuse the same power equation, substituting different windspeeds, and get: 7 W at v = 1 m/s 57 W at v = 2 m/s 7 kW at v = 10 m/s Note the very strong dependence on windspeed.Answer 5.10: Answer 5.11: Answer 5.12: Answer 5.13: 

Posted on 8/1/05 © 20002005 Physics For Everyone. All Rights Reserved. 