HW9_Sol_F10

# HW9_Sol_F10 - Oct 26 2010 PHYS 2101 HW#9 WileyPlus Problem...

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Oct. 26, 2010 PHYS 2101 HW#9 WileyPlus Problem Solutions 1 HW #9 Solutions – or at least most of them… (USING BOOK VALUES 9 th Edition in problems; problem # indicated by []) 1) [14-5] 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 F = ( p i p o ) A = (1.0 atm 0.96 atm)(1.013 × 10 5 Pa/atm)(3.4 m)(2.1 m) = 2.9 × 10 4 N. 2) [14-17] The pressure p at the depth d of the hatch cover is p 0 + ρ gd , where ρ is the density of ocean water and p 0 is atmospheric pressure. Thus, the gauge pressure is , and the minimum force that must be applied by the crew to open the hatch has magnitude , where A is the area of the hatch. Substituting the values given, we find the force to be F = p gauge A = ( ρ gd ) A = (1024 kg/m 3 )(9.8 m/s 2 )(100 m)(1.2 m)(0.60 m) = 7.2 × 10 5 N. 3) [14-20] . (a) The force on face A of area A A due to the water pressure alone is F A = p A A A = ρ w gh A A A = ρ w g (2 d ) d 2 = 2 1.0 × 10 3 kg m 3 ( ) 9.8m s 2 ( ) 5.0m ( ) 3 = 2.5 × 10 6 N.

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