ssm_ch14 - Chapter 14 Student Solutions Manual 1. The...

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Chapter 14 – Student Solutions Manual 1. The pressure increase is the applied force divided by the area: Δ p = F / A = F / π r 2 , where r is the radius of the piston. Thus Δ p = (42 N)/ π (0.011 m) 2 = 1.1 × 10 5 Pa. This is equivalent to 1.1 atm. 3. 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 . Since 1 atm = 1.013 × 10 5 Pa, 54 (1.0 atm 0.96 atm)(1.013 10 Pa/atm)(3.4 m)(2.1 m) = 2.9 10 N. F =− × × 19. When the levels are the same the height of the liquid is h = ( h 1 + h 2 )/2, where h 1 and h 2 are the original heights. Suppose h 1 is greater than h 2 . The final situation can then be achieved by taking liquid with volume A ( h 1 h ) and mass ρ A ( h 1 h ), in the first vessel, and lowering it a distance h h 2 . The work done by the force of gravity is W = A ( h 1 h ) g ( h h 2 ). We substitute h = ( h 1 + h 2 )/2 to obtain () 2 33 2 4 2 12 11 (1.30 10 kg/m )(9.80 m/s )(4.00 10 m )(1.56 m 0.854 m) 44 0.635 J Wg A h h = × × = 2 . 27. (a) We use the expression for the variation of pressure with height in an incompressible fluid: p 2 = p 1 g ( y 2 y 1 ). We take y 1 to be at the surface of Earth, where the pressure is p 1 = 1.01 × 10 5 Pa, and y 2 to be at the top of the atmosphere, where the pressure is p 2 = 0. For this calculation, we take the density to be uniformly 1.3 kg/m 3 . Then, 5 3 1 21 32 1.01 10 Pa 7.9 10 m = 7.9 km. (1.3 kg/m )(9.8m/s ) p yy g × −= = (b) Let h be the height of the atmosphere. Now, since the density varies with altitude, we integrate 0 . h p pg d y Assuming = 0 (1 - y / h ), where 0 is the density at Earth’s surface and g = 9.8 m/s 2 for 0 y h , the integral becomes
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ssm_ch14 - Chapter 14 Student Solutions Manual 1. The...

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