Homework 1 solutions
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Homework 1 solutions

Course Number: PHYSICS Physics 7C, Winter 2010

College/University: UC Irvine

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1. What is the total mass of the Earth's atmosphere? (The radius of the Earth is 6.37 106 m, and atmospheric pressure at the surface is 1.013 105 N/m2). Solution: The Earth's surface area is 4 R 2 . The force pushing inward over this area amounts to F = P0 A = P0 4 R 2 ( ). ) 6 2 This force is the weight of the air: Fg = m g = P0 4 R 2 ( so the mass of the air is m = P0 4 R 2 g ( ) = (1.013 10 5 N m...

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What 1. is the total mass of the Earth's atmosphere? (The radius of the Earth is 6.37 106 m, and atmospheric pressure at the surface is 1.013 105 N/m2). Solution: The Earth's surface area is 4 R 2 . The force pushing inward over this area amounts to F = P0 A = P0 4 R 2 ( ). ) 6 2 This force is the weight of the air: Fg = m g = P0 4 R 2 ( so the mass of the air is m = P0 4 R 2 g ( ) = (1.013 10 5 N m 2 ) 4 ( 6.37 10 m ) 2 9. m s 80 = 5. 1018 kg 27 . 2. (a) Calculate the absolute pressure at an ocean depth of 1 000 m. Assume the density of seawater is 1 024 kg/m3 and that the air above exerts a pressure of 101.3 kPa. (b) At this depth, what force must the frame around a circular submarine porthole having a diameter of 30.0 cm exert to counterbalance the force exerted by the water? Solution: (a) P = P0 + gh = 1. 105 Pa + 1024 kg m 013 ( 3 )(9.80 m s ) (1000 m ) 2 P = 1. 107 Pa 01 (b) The gauge pressure is the difference in pressure between the water outside and the air inside the submarine, which we suppose is at 1.00 atmosphere. Pgauge = P - P0 = gh = 1. 107 Pa 00 The resultant inward force on the porthole is then F = PgaugeA = 1. 107 Pa ( 0. m 00 150 )2 = 7. 105 N . 09 3. The small piston of a hydraulic lift has a cross-sectional area of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2 (Figure 14.4). What force must be applied to the small piston for the lift to raise a load of 15.0 kN? (In service stations, this force is usually exerted by compressed air.) Solution: Since the pressure is the same on both sides, In this case, 15 000 F = 2 200 3. 00 F1 F = 2 A1 A 2 or F2 = 225 N 4. Figure on the right shows Superman attempting to drink water through a very long straw. With his great strength he achieves maximum possible suction. The walls of the tubular straw do not collapse. (a) Find the maximum height through which he can lift the water. (b) What If? Still thirsty, the Man of Steel repeats his attempt on the Moon, which has no atmosphere. Find the difference between the water levels inside and outside the straw. Solution: (a) We imagine the superhero to produce a perfect vacuum in the straw. Take point 1 at the water surface in the basin and point 2 at the water surface in the straw: P1 + gy1 = P2 + gy2 1. 105 N m 2 + 0 = 0 + 1000 kg m 013 y2 = 10. m 3 (b) No atmosphere can lift the water in the straw through zer height o difference. ( 3 )( 9.80 m s ) y 2 2 5. Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 20.0 mm from the normal height. What is the atmospheric pressure? (The density of mercury is 13.59 g/cm3). Solution: P0 = gh = -2. 103 Pa : 66 P = P0 + P0 = ( 1. - 0. 6) 105 Pa = 0. 105 Pa 013 026 986 6. (a) A light balloon is filled with 400 m3 of helium. At 0C, the balloon can lift a payload of what mass? (b) What If? In Table 14.1, observe that the density of hydrogen is nearly one-half the density of helium. What load can the balloon lift if filled with hydrogen? Solution: (a) The balloon is nearly in equilibrium: Fy = m ay B - ( Fg )helium or This reduces to - Fg ( ) payl oad =0 ai gV - helum gV - m payl g = 0 r i oad m payl = ( ai - helum ) V = 1. kg m 3 - 0. 29 179 kg m oad r i m payl = 444 kg oad ( 3 )( 400 m ) 3 3 (b) Similarly, 29 089 m payl = ai - hydrogen V = 1. kg m 3 - 0. 9 kg m oad r m payl = 480 kg oad ( ) ( )( 400 m ) 3 The air does the lifting, nearly the same for the two balloons. 7. A spherical aluminum ball of mass 1.26 kg contains an empty spherical cavity that is concentric with the ball. The ball just barely floats in water. Calculate (a) the outer radius of the ball and (b) the radius of the cavity. Solution: (a) The weight of the ball must be equal to the buoyant force of the water: 4 1. kgg = w ater r3 erg 26 out 3 3 1. kg 26 r er = out 3 4 1000 kg m (b) 13 = 6. cm 70 The mass of the ball is determined by the density of aluminum: 4 4 m = A lV = A l r3 - r3 3 0 3 i 4 1. kg = 2 700 kg m 3 ( 0. m 26 067 3 ( ) 3 - r3 ) i 1. 10-4 m 11 3 = 3. 10-4 m 3 - r3 01 i 3 13 r = 1. 10-4 m 89 i ( ) = 5. cm 74 8. How many cubic meters of helium are required to lift a balloon with a 400-kg payload to a height of 8 000 m? (Take He = 0.180 kg/m3.) Assume that the balloon maintains a constant volume and that the density of air decreases with the altitude z according to the expression air = 0ez/8 000, where z is in meters and 0 = 1.25 kg/m3 is the density of air at sea level. Solution: The balloon stops rising when ( air - H e ) gV = M g ( air - H e ) V = M , Therefore, V = M = and air - H e 400 1. e-1 - 0. 25 180 V = 1 430 m 3 9. A legendary Dutch boy saved Holland by plugging a hole in a dike with his finger, 1.20 cm in diameter. If the hole was 2.00 m below the of surface the North Sea (density 1 030 kg/m3), (a) what was the force on his finger? (b) If he pulled his finger out of the hole, how long would it take the released water to fill 1 acre of land to a depth of 1 foot, assuming the hole remained constant in size? (A typical U.S. family of four uses 1 acre-foot of water, 1 234 m3, in 1 year). Solution: (a) Between sea surface and clogged hole: P1 + 1 2 1 2 v1 + gy1 = P2 + v2 + gy2 2 2 1 at + 0 + 1030 kg m m ( 3 )( 9.8 m s ) ( 2 m ) = P + 0 + 0 2 2 P2 = 1 at + 20. kPa m 2 The air on the back of his hand pushes opposite the water, so the net force on his hand is F = PA = 20. 103 N m 2 (b) ( 2 ) (1.2 10 4 -2 m ) 3 2 F = 2. N 28 Now, Bernoulli's theorem is 1 at + 0 + 20. kPa = 1 at + m 2 m The volume rate of flow is 1 1030 kg m 2 ( )v -4 2 2 +0 v2 = 6. m s 26 A 2v2 = (1.2 10 4 -2 m ) (6.26 m s) = 7.08 10 2 m 3 s 3 One acrefoot is 4 047 m 2 0. 8 m = 1234 m 304 Requiring 1234 m 3 3 7. 10-4 m 08 s = 1. 106 s = 20. days 74 2 10. A siphon is used to drain water from a tank, as illustrated in figure below. The siphon has a uniform diameter. Assume steady flow without friction. (a) If the distance h = 1.00 m, find the speed of outflow at the end of the siphon. (b) What If? What is the limitation on the height of the top of the siphon above the water surface? (For the flow of the liquid to be continuous, the pressure must not drop below the vapor pressure of the liquid.) Solution: (a) P0 + gh + 0 = P0 + 0 + 00 If h = 1. m , 1 2 v3 2 v3 = 2gh v3 = 4. m s 43 (b) P + gy + 1 2 1 2 v2 = P0 + 0 + v3 2 2 P = P0 - gy Since v2 = v3 , Since P 0 y P0 1. 105 Pa 013 = = 10. m 3 3 g 10 kg m 3 9. m s2 8 ( )( ) 11. A U-tube open at both ends is partially filled with water (figure on the right). Oil having a density 750 kg/m3 is then poured into the right arm and forms a column L = 5.00 cm high (Fig. P 14.71b). (a) Determine the difference h in the heights of the two liquid surfaces. (b) The right arm is then shielded from any air motion while air is blown across the top of the left arm until the surfaces of the two liquids are at the same height (Fig. P14.71c). Determine the speed of the air being blown across the left arm. Take the density of air as 1.29 kg/m3. Solution: Note: Variation of atmospheric pressure with altitude is included in this solution. Because of the small distances involved, this effect is unimportant in the final answers. (a) Consider the pressure at points A and B in part (b) of the figure: Using the left tube: PA = Patm + agh + w g( L - h) where the second term is due to the variation of air pressure with altitude. Using the right tube: PB = Patm + 0 gL But Pascal's principle says that PA = PB . Therefore, or Patm + 0 gL = Patm + agh + w g( L - h) ( w - a ) h = ( w - 0 ) L , giving 1000 - 750 - 0 h= w 00 25 L = 1000 - 1. 5. cm = 1. cm 29 w - a (b) Consider part (c) of the diagram showing the situation when the air flow over the left tube equalizes the fluid levels in the two tubes. First, apply Bernoulli's equation to points A and B ( yA = yB , vA = v, and vB = 0) This gives: PA + 1 1 2 av2 + agyA = PB + a ( 0) + agyB 2 2 1 av2 2 (1) and since yA = yB , this reduces to: PB - PA = Now consider points C and D, both at the level of the oil water interface in the right tube. Using the variation of pressure with depth in static fluids, we have: PC = PA + agH + w gL and PD = PB + agH + 0 gL But Pascal's principle says that PC = PD . Equating these two gives: PB + agH + 0 gL = PA + agH + w gL or PB - PA = ( w - 0 ) gL (2) Substitute equation (1) for PB - PA into (2) to obtain 1 av2 = ( w - 0 ) gL 2 or v= 2gL ( w - 0 ) a 1000 - 750 = 2 9. m s2 ( 0. 0 m ) 80 050 1. 29 ( ) v = 13. m s 8 12. A piece of aluminum with mass 1.00 kg and density 2 700 kg/m3 is suspended from a string and then completely immersed in a container of water (figure on the right). Calculate the tension in the string (a) before and (b) after the metal is immersed. Solution: (a) Before the metal is immersed: Fy = T1 - M g = 0 or T1 = M g = (1. kg) 9. m s 00 80 = 9. N 80 ( 2 ) yyyyyyy ??????? yyyyyyy ??????? scale B T1 T2 (b) After the metal is immersed: Fy = T2 + B - M g = 0 or T2 = M g - B = M g - ( w V ) g V = M = 1. kg 00 2 700 kg m 3 Thus, 13. A horizontal pipe 10.0 cm in diameter has a smooth reduction to a pipe 5.00 cm in diameter. If the pressure of the water in the larger pipe is 8.00 104 Pa and the yyyyyyy ??????? Mg Mg a b pressure in the smaller pipe is 6.00 104 Pa, at what rate does water flow through the pipes? Solution: By Bernoulli's equation, 1 1 00 (1000) v2 = 6. 104 N m 2 + 2 (1000) 16v2 2 1 2. 104 N m 2 = (1000) 15v2 00 2 v = 1. m s 63 8. 104 N m 2 + 00 dm = A v = 1000 5. 10-2 00 dt ( ) (1.63 m s) = 2 12. kg s 8

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