# ch14 - 1 The air inside pushes outward with a force given...

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1. 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, 5 4 (1.0 atm 0.96 atm)(1.013 10 Pa/atm)(3.4 m)(2.1 m) = 2.9 10 N. F = × ×

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2. We note that the container is cylindrical, the important aspect of this being that it has a uniform cross-section (as viewed from above); this allows us to relate the pressure at the bottom simply to the total weight of the liquids. Using the fact that 1L = 1000 cm 3 , we find the weight of the first liquid to be 1 1 1 1 3 3 2 6 2 (2.6 g / cm )(0.50 L)(1000 cm / L)(980 cm/s ) 1.27 10 g cm/s 12.7 N. W m g V g ρ = = = = × = In the last step, we have converted grams to kilograms and centimeters to meters. Similarly, for the second and the third liquids, we have 3 3 2 2 2 2 2 (1.0 g/cm )(0.25 L)(1000 cm L)(980 cm s ) 2.5 N W m g V g ρ = = = = and 3 3 2 3 3 3 3 (0.80 g/cm )(0.40 L)(1000 cm / L)(980 cm/s ) 3.1 N. W m g V g ρ = = = = The total force on the bottom of the container is therefore F = W 1 + W 2 + W 3 = 18 N.
3. The pressure increase is the applied force divided by the area: p = F / A = F / π r 2 , where r is the radius of the piston. Thus p = (42 N)/ π (0.011 m) 2 = 1.1 × 10 5 Pa. This is equivalent to 1.1 atm.

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4. 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 5 4 4 2 480 N 1.0 10 Pa 3.8 10 Pa. 77 10 m i o F p p A = = × = × ×
5. Let the volume of the expanded air sacs be V a and that of the fish with its air sacs collapsed be V . Then 3 3 fish fish fish 1.08 g/cm and 1.00 g/cm w a m m V V V ρ ρ = = = = + where ρ w is the density of the water. This implies ρ fish V = ρ w ( V + V a ) or ( V + V a )/ V = 1.08/1.00, which gives V a /( V + V a ) = 7.4%.

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6. Knowing the standard air pressure value in several units allows us to set up a variety of conversion factors: (a) ( ) 5 2 2 1.01 10 Pa 28 lb/in. 190 kPa 14.7 lb/in P § · × = = ¨ ¸ © ¹ (b) 5 5 1.01 10 Pa 1.01 10 Pa (120 mmHg) 15.9 kPa, (80 mmHg) 10.6 kPa. 760 mmHg 760 mmHg § · § · × × = = ¨ ¸ ¨ ¸ © ¹ © ¹
7. (a) The pressure difference results in forces applied as shown in the figure. We consider a team of horses pulling to the right. To pull the sphere apart, the team must exert a force at least as great as the horizontal component of the total force determined by “summing” (actually, integrating) these force vectors. We consider a force vector at angle θ . Its leftward component is p cos θ dA , where dA is the area element for where the force is applied. We make use of the symmetry of the problem and let dA be that of a ring of constant θ on the surface. The radius of the ring is r = R sin θ , where R is the radius of the sphere. If the angular width of the ring is d θ , in radians, then its width is R d θ and its area is dA = 2 π R 2 sin θ d θ . Thus the net horizontal component of the force of the air is given by 2 2 2 2 / 2 2 0 0 2 sin cos sin .

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