132_ch15 - PHYS 132, WEEK # 6, CHAPTER 15:3, 5, 6, 9, 10,...

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Unformatted text preview: PHYS 132, WEEK # 6, CHAPTER 15:3, 5, 6, 9, 10, 30, 31, 32, 35, 39 15.3. Model: The density of water is 1000 kg/m3. Visualize: Solve: Volume of water in the swimming pool is V = 6 m 12 m 3 m 1 2 ( 6 m 12 m 2 m ) = 144 m 3 3 The mass of water in the swimming pool is m = rV = 1000 kg m ( )(144 m ) = 1.44 10 3 5 kg 15.5. Model: The density of sea water is 1030 kg/m3. Solve: The pressure below sea level can be found from Equation 15.6 as follows: p = p 0 + rgd = 1. 013 10 Pa + 1030 kg m 5 8 5 5 ( 3 )(9.80 m s )(1.1 10 2 8 4 m ) = 1.013 10 Pa + 1.1103 10 Pa = 1.1113 10 Pa = 1097 atm where we have used the conversion 1 atm = 1.013 10 Pa . Assess: The pressure deep in the ocean is very large. 15.6. Model: The density of water is 1,000 kg/m3 and the density of ethyl alcohol is 790 kg/m3. Solve: (a) The volume of water that has the same mass as 8.0 m3 of ethyl alcohol is Vwater = m water m alcohol r water = r water = r alcohol Valcohol r water 790 kg m 3 ^ 3 3 = ~ 8.0 m = 6.32 m 1000 kg m 3 ( ) (b) The pressure at the bottom of the cubic tank is p = p 0 + rwater gd : p = 1.013 10 Pa + 1000 kg m 5 ( 3 )(9.80 m s )(6.32 ) 2 13 = 1.194 10 Pa 5 where we have used the relation d = (Vwater ) 13 . 15.9. Model: The density of seawater rseawater = 1030 kg m 2 . Visualize: Solve: The pressure outside the submarine's window is p out = p0 + rseawater gd, where d is the maximum safe depth for the window to withstand a force F. This force is F A = p out - p in , where A is the area of the window. With pin = p0, we simplify the pressure equation to p out - p0 = F A = rseawater gd fi d = F Ar seawater g d= 1.0 10 6 N p ( 0.10 m ) 1030 kg m 2 9.8 m s 2 2 ( )( ) = 3153 m Assess: A force of 1.0 106 N corresponds to a pressure of r= F A = 1.0 10 6 N p ( 0.10 m ) 2 = 314 atm A depth of 3 km is therefore reasonable. 15.10. Visualize: We assume that the seal is at a radius of 5 cm. Outside the seal, atmospheric pressure presses on both sides of the cover and the forces cancel. Thus, only the 10 cm diameter opening inside the seal is relevant, not the 20 cm diameter of the cover. Solve: Within the 10 cm diameter area where the pressures differ, Fto left = p atmos A Fto right = p gas A 2 -3 2 where A = pr = 7.85 10 m is the area of the opening. The difference between the forces is Fto left - Fto right = p atmos - p gas A = (101, 300 Pa - 20, 000 Pa ) 7.85 10 ( ) ( -3 m 2 ) = 639 N Normally, the rubber seal exerts a 639 N force to the right to balance the air pressure force. To pull the cover off, an external force must pull to the right with a force 639 N. 15.30. Model: We assume that there is a perfect vacuum inside the cylinders with p = 0 Pa. We also assume that the atmospheric pressure in the room is 1 atm. Visualize: Please refer to Figure P15.30. 2 2 2 Solve: (a) The flat end of each cylinder has an area A = p r = p (0.30 m ) = 0.283 m . The force on each end is thus Fatm = p 0 A = 1.013 10 Pa 0.283 m ( 5 )( 2 ) = 2.86 10 4 N (b) The net vertical force on the lower cylinder when it is on the verge of being pulled apart is F y = Fatm - wplayers = 0 N fi wplayers = Fatm = 2.86 10 4 N 2.86 10 4 N fi number of players = (100 kg )(9.8 m s2 ) = 29.2 That is, 30 players are needed to pull the two cylinders apart. 15.31. Model: Assume that the oil is incompressible and its density is 900 kg/m3. Visualize: Please refer to Figure P15.31. Solve: (a) The pressure at depth d in a fluid is p = p 0 + rgd . Here, pressure p0 at the top of the fluid is due both to the atmosphere and to the weight of the piston. That is, p0 = patm + wp/A. At point A, w p A = patm + P + rg (1.00 m - 0.30 m ) A = 1.013 10 Pa 5 (10 kg )(9.8 m s2 2 p (0.02 m ) )+ (900 2 kg m 3 )( 9.8 m s )(0.70 m ) = 185, 460 Pa 2 fi FA = p A A = (185, 460 Pa )p (0.10 m ) = 5830 N (b) In the same way, p B = p atm + wP A + rg (1.30 m ) = 190, 752 Pa fi FB = 5990 N Assess: FB is larger than FA, because pB is larger than pA. 15.32. Model: The tire flattens until the pressure force against the ground balances the upward normal force of the ground on the tire. Solve: The area of the tire in contact with the road is A = (0.15 m )(0.13 m ) = 0.0195 m . The normal force on each tire is n= w 4 = 2 (1500 kg )(9.8 4 m s2 ) = 3675 N 14.7 psi 1 atm = 27.3 psi Thus, the pressure inside each tire is p inside = n A = 3675 N 0.0195 m 2 = 188, 500 Pa = 1.86 atm 15.35. Solve: The fact that atmospheric pressure at sea level is 101.3 kPa = 101,300 N/m2 means that the weight of the atmosphere over each square meter of surface is 101,300 N. Thus the mass of air over each square meter is m = (101,300 N)/g = (101,300 N)/(9.80 m/s2) = 10,340 kg per m2. Multiplying by the earth's surface area will give the total mass. Using Re = 6.27 106 m for the earth's radius, the total mass of the atmosphere is M air = Aearth m = ( 4p Re )m = 4 p (6.37 10 m) (10 ,340 kg / m ) = 5.27 10 2 6 2 2 18 kg 15.39. Model: Assume that oil is incompressible and its density is 900 kg/m3. Visualize: Please refer to Figure P15.39. Solve: (a) The hydraulic lift is in equilibrium and the pistons on the left and the right are at the same level. Equation 15.11, therefore, simplifies to Fleft piston Aleft piston fi rstudent = = Fright piston Aright piston fi w student p ( rstudent ) 2 = w elephant p relephant ( ) 2 w student ^ ~ relephant = w elephant ( ) (70 kg )g ( 1.0 m ) = 0.2415 m (1200 kg )g The diameter of the piston the student is standing on is therefore 2 0.2415 m = 0.483 m. (b) From Equation 15.13, we see that an additional force DF is required to increase the elephant's elevation through a distance d2. That is, DF = rg Aleft piston + Aright piston d 2 fi (70 kg ) 9.8 m s ( ) ( 2 ) = (900 kg m 3 )(9.8 m s p (0.2415 m ) + (1.0 m ) d 2 2 )[ 2 2 fi d 2 = 0.0234 m = 2.34 cm 5 ...
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This note was uploaded on 05/05/2008 for the course PHYS 132 taught by Professor Sharpe during the Spring '08 term at Cal Poly.

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