hw11-12_sol_s11

hw11-12_sol_s11 - Phys 2101 Homework 11&12 Solution...

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Phys 2101 Homework 11&12 Solution Spring `11 These solutions use the parameter values from the problems printed in the book, not those that appear in your personal homework assignment. The logic used to get to the answer is the same, however. 1. The magnitude F of the force required to pull the lid off is F = ( p o p i ) A , where p o is the pressure outside the box, p i is the pressure inside, and A is the area of the lid. Recalling that 1N/m 2 = 1 Pa, we obtain 54 42 480 N 1.0 10 Pa 3.8 10 Pa. 77 10 m io F pp A 2. The pressure difference between two sides of the window results in a net force acting on the window. The air inside pushes outward with a force given by p i A , where p i is the pressure inside the room and A is the area of the window. Similarly, the air on the outside pushes inward with a force given by p o A , where p o is the pressure outside. The magnitude of the net force is F = ( p i p o ) A . With 1 atm = 1.013 10 5 Pa, the net force is 5 4 ( ) (1.0 atm 0.96 atm)(1.013 10 Pa/atm)(3.4 m)(2.1 m) 2.9 10 N. F p p A  3. The pressure p at the depth d of the hatch cover is p 0 + gd , where is the density of ocean water and p 0 is atmospheric pressure. Thus, the gauge pressure is gauge p gd , and the minimum force that must be applied by the crew to open the hatch has magnitude () F p A gd A  , where A is the area of the hatch. Substituting the values given, we find the force to be 32 5 ( ) (1024 kg/m )(9.8 m/s )(100 m)(1.2 m)(0.60 m) 7.2 10 N. F p A gd A 4. (a) The force on face A of area A A due to the water pressure alone is       3 2 3 3 2 6 (2 ) 2 1.0 10 kg m 9.8m s 5.0m 2.5 10 N. A A A w A A w F p A gh A g d d  Adding the contribution from the atmospheric pressure, F 0 = (1.0 10 5 Pa)(5.0 m) 2 = 2.5 10 6 N, we have 6 6 6 0 2.5 10 N 2.5 10 N 5.0 10 N. AA F F F   (b) The force on face B due to water pressure alone is
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(b) The force on face B due to water pressure alone is       3 2 3 3 3 2 avg 6 5 5 5 1.0 10 kg m 9.8m s 5.0m 2 2 2 3.1 10 N. B B B w d F p A g d gd      Adding the contribution from the atmospheric pressure, F 0 = (1.0 10 5 Pa)(5.0 m) 2 = 2.5 10 6 N, we obtain 6 6 6 0 2.5 10 N 3.1 10 N 5.6 10 N. BB F F F   5. Taking “down” as the positive direction, then using Eq. 14 - 16 in Newton’s second law, we have (5.00 kg) g (3.00 kg) g = 5 a. This gives a = 2 5 g = 3.92 m/s 2 , where g = 9.8 m/s 2 . Then (see Eq. 2- 15) 1 2 at 2 = 0.0784 m (in the downward direction).
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hw11-12_sol_s11 - Phys 2101 Homework 11&12 Solution...

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