ࡱ>  |dq abbjbjt+t+ ) AA[Wn]PPPPd+d+d+x+x+x+x+8+D,x+uL. 7"888ZZZuuuuuuu$vxxud+Z]D*ZZZuqPP88I L.qqqZPPb8d+8ux+x+PPPPZuqqu d+u8.DW\x+x+\uChapter 10. Uniform Circular Motion Centripetal Acceleration 10-1. A ball is attached to the end of a 1.5 m string and it swings in a circle with a constant speed of 8 m/s. What is the centripetal acceleration?  EMBED Equation.DSMT4  ac = 42.7 m/s2 10-2. What are the period and frequency of rotation for the ball in Problem 10-1?  EMBED Equation.DSMT4 ; T = 1.18 s  EMBED Equation.DSMT4 ; f = 0.849 rev/s 10-3. A drive pulley 6-cm in diameter is set to rotate at 9 rev/s. What is the centripetal acceleration of a point on the edge of the pulley? What would be the linear speed of a belt around the pulley? [ R = (0.06 m/2) = 0.03 m ]  EMBED Equation.DSMT4 ; ac = 95.9 m/s2  EMBED Equation.DSMT4 ; v = 1.70 m/s  10-4. An object revolves in a circle of diameter 3 m at a frequency of 6 rev/s. What is the period of revolution, the linear speed, and the centripetal acceleration? [ R = (3 m/2) = 1.5 m ]  EMBED Equation.DSMT4 ; T = 0.167 s ;  EMBED Equation.DSMT4 ; v = 56.5 m/s  EMBED Equation.DSMT4 ; ac = 2130 m/s2 10-5. A car moves around a curve 50 m in radius and undergoes a centripetal acceleration of 2 m/s2. What is its constant speed?  EMBED Equation.DSMT4 ; v = 10.0 m/s 10-6. A 1500-kg car moves at a constant speed of 22 m/s along a circular track. The centripetal acceleration is 6 m/s2. What is the radius of the track and the centripetal force on the car?  EMBED Equation.DSMT4 ; R = 80.7 m  EMBED Equation.DSMT4 ; Fc = 9000 N 10-7. An airplane dives along a curved path of radius R and velocity v. The centripetal acceleration is 20 m/s2. If both the velocity and the radius are doubled, what will be the new acceleration?  EMBED Equation.DSMT4 ; a2 = 2a1 = 2(20 m/s2; a = 40 m/s2 Centripetal Force 10-8. A 20-kg child riding a loop-the-loop at the Fair moves at 16 m/s through a track of radius 16 m. What is the resultant force on the child?  EMBED Equation.DSMT4 ; Fc = 320 N 10-9. A 3-kg rock, attached to a 2-m cord, swings in a horizontal circle so that it makes one revolution in 0.3 s. What is the centripetal force on the rock? Is there an outward force on the rock?  EMBED Equation.DSMT4 ; Fc = 2630 N, No 10-10. An 8-lb object swings in horizontal circle with a speed of 95 ft/s. What is the radius of the path, if the centripetal force is 2000 lb?  EMBED Equation.DSMT4   EMBED Equation.DSMT4 ; R = 1.13 ft *10-11. Two 8-kg masses are attached to the end of a tin rod 400 mm long. The rod is supported in the middle and whirled in a circle. The rod can support a maximum tension of only 800 N. What is the maximum frequency of revolution? [ R = (400 mm/2) = 200 mm ]  EMBED Equation.DSMT4 ; v = 4.47 m/s  EMBED Equation.DSMT4 ; f = 3.56 rev/s *10-12. A 500-g damp shirt rotates against the wall of a washer at 300 rpm. The diameter of the rotating drum is 70 cm. What is the magnitude and direction of the resultant force on the shirt? [ R = (70 cm/2) = 35 cm; f = 300 rpm(60 s/min) = 1800 rev/s ]  EMBED Equation.DSMT4 ; Fc = 2.24 x 107 N, toward the center *10-13. A 70-kg runner rounds a track of radius 25 m at a speed of 8.8 m/s. What is the central force causing the runner to turn and what exerts the force?  EMBED Equation.DSMT4 ; Fc = 217 N, friction  *10-14. In Olympic bobsled competition, a team takes a turn of radius 24 ft at a speed of 60 mi/h. What is the acceleration? How many gs do passengers experience? (60 mi/h = 88 ft/s)  EMBED Equation.DSMT4 ; ac = 323 ft/s2 or 10.1 gs Flat Curves and Banked Curves 10-15. On a rainy day the coefficient of static friction between tires and the roadway is only 0.4. What is the maximum speed at which a car can negotiate a turn of radius 80 m?  EMBED Equation.DSMT4 ; vc = 17.7 m/s or 63.8 km/h  10-16. A bus negotiates a turn of radius 400 ft while traveling at a speed of 60 mi/h. If slipping just begins at this speed, what is the coefficient of static friction between the tires and the road? (60 mi/h = 88 ft/s)  EMBED Equation.DSMT4 ; (s = 0.605 10-17. Find the coefficient of static friction necessary to sustain motion at 20 m/s around a turn of radius 84 m.  EMBED Equation.DSMT4 ; (s = 0.486 *10-18. A 20-kg child sits 3 m from the center of a rotating platform. If (s = 0.4, what is the maximum number of revolutions per minute that can be achieved without slipping? (Slipping occurs when the centripetal force equals the maximum force of static friction.)  EMBED Equation.DSMT4 ; f = 0.182 rev/s (60 s/min); f = 10.9 rpm  *10-19. A platform rotates freely at 100 rpm. If the coefficient of static friction is 0.5, how far from the center of the platform can a bolt be placed to without slipping? f = 100 rev/min (1 min/60 s) = 1.67 rev/s; (s = 0.5; R = ?  EMBED Equation.DSMT4 ; R = 21.1 cm  10-20. Find the required banking angle to negotiate the curve of Prob.10-15 without slipping.  EMBED Equation.DSMT4 ; ( = 21.80 10-21. Find the required banking angle for Problem 10-16 to prevent slipping?  EMBED Equation.DSMT4 ; ( = 31.20 10-22. The optimum banking angle for a curve of radius 20 m is found to be 280. For what speed was this angle designed?  EMBED Equation.DSMT4 ; v = 10.3 m/s *10-23. A curve in a road 9 m wide has a radius of 96 m. How much higher than the inside edge should the outside edge be for an automobile to travel at the optimum speed of 40 km/h? v = 40 km/h = 11.1 m/s; h = (9 m) sin (;  EMBED Equation.DSMT4   EMBED Equation.DSMT4 ; ( = 7.480; h = (9 m) sin 7.480; h = 1.17 m The Conical Pendulum 10-24. A conical pendulum swings in a horizontal circle of radius 30 cm. What angle does the supporting cord make with the vertical when the liner speed of the mass is 12 m/s?  EMBED Equation.DSMT4 ; ( = 88.80  10-25. What is the linear speed of the flyweights in Fig. 10-16 if L = 20 cm and ( = 600? What is the frequency of revolution? L = 20 cm = 0.20 m; R = L sin ( = (0.2 m) sin 600; R = 0.173 m   EMBED Equation.DSMT4 ; v = 1.71 m/s  EMBED Equation.DSMT4 ; f = 1.58 rev/s 10-26. If the length of the L in Fig. 10-16 is 60 cm. What velocity is required to cause the flyweights to move to an angle of 300 with the vertical? R = L sin ( = (60 cm) sin 300; R = 30 cm = 0.30 m  EMBED Equation.DSMT4 ; v = 1.30 m/s 10-27. Each of the flyweights in Fig. 10-16 has a mass of 2 kg. The length L = 40 cm and the shaft rotates at 80 rpm. What is the tension in each arm? What is the angle (? What is the height h? 80 rpm = 1.33 rev/s; T sin( = Fc = mv2/R   EMBED Equation.DSMT4 ;  EMBED Equation.DSMT4  T = (0.4 m)(4 (2)(1.33 rev/s)2(2 kg); T = 56.1 N  T cos ( = mg;  EMBED Equation.DSMT4 ; ( = 69.60 h = L cos ( = (0.4 m) cos 69.60 ; h = 0.14 m or h = 14.0 cm  10-28. In Fig. 10-16, assume that L = 6 in., each flyweight is 1.5 lb, and the shaft is rotating at 100 rpm. What is the tension in each arm? What is the angle (? What is the distance h?  100 rpm = 1.67 rev/s; T sin( = Fc = mv2/R  EMBED Equation.DSMT4 ; L = 6 in. = 0.50 ft   EMBED Equation.DSMT4 ;  EMBED Equation.DSMT4  T = (0.5 ft)(4 (2)(1.67 rev/s)2(0.0469 slug); T = 2.75 lb  T cos ( = mg;  EMBED Equation.DSMT4 ; ( = 54.30 h = L cos ( = (0.5 ft) cos 54.30 ; h = 0.292 ft 10-29. Consider the rotating swings in Fig. 10-17. The length L = 10 m and the distance a = 3 m. What must be the linear velocity of the seat if the rope is to make an angle of 300 with the vertical? [ L = 10 m; a = 3 m, ( = 300 ] b = L sin ( = (10 m) sin 300 = 5 m; R = a + b = 7 m;  EMBED Equation.DSMT4    EMBED Equation.DSMT4  v = 6.73 m/s  10-30. What must be the frequency of revolution for the swing in Problem 10-17 if the angle ( is to be equal to 250? b = L sin 280 = (10 m) sin 280; b = 4.695 m; R = a + b = 7.695 m  EMBED Equation.DSMT4 ;  EMBED Equation.DSMT4 ; v = 6.33 m/s  EMBED Equation.DSMT4 ; f = 0.131 rev/s or 7.86 rpm Motion in a Vertical Circle 10-31. A rock rests on the bottom of a bucket moving in a vertical circle of radius 70 cm. What is the least speed the bucket must have as it rounds the top of the circle if it is to remain in the bucket?  [ Resultant force to center = mv2/R ]   EMBED Equation.DSMT4 ; Critical speed vc is when T = 0   EMBED Equation.DSMT4 ; vc = 6.86 m/s 10-32. A 1.2-kg rock is tied to the end of a 90-cm length of string. The rock is then whirled in a vertical circle at a constant speed. What is the critical velocity at the top of the path if the string is not to become slack?   EMBED Equation.DSMT4 ; Critical speed vc is when T = 0   EMBED Equation.DSMT4 ; vc = 2.97 m/s  *10-33. Assume that the rock of Problem 10-32 moves in a vertical circle at a constant speed of 8 m/s? What are the tensions in the rope at the top and bottom of the circle.  EMBED Equation.DSMT4  At Top:  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4    EMBED Equation.DSMT4 ; T = 73.6 N At Bottom:  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4    EMBED Equation.DSMT4 ; T = 97.1 N  *10-34. A test pilot in Fig. 10-18 goes into a dive at 620 ft/s and pulls out in a curve of radius 2800 ft. If the pilot weighs 160 lb, what acceleration will be experienced at the lowest point? What is the force exerted by the seat on the pilot? W = 160 lb; m =160 lb/32 ft/s2 = 5 slugs; v = 620 ft/s;  EMBED Equation.DSMT4 ; a = 137 ft/s2  EMBED Equation.DSMT4  *10-34. (Cont.)  EMBED Equation.DSMT4 ; N = 846 lb *10-35. If it is desired that the pilot in Problem 10-34 not experience an acceleration greater than 7 times gravity (7g), what is the maximum velocity for pulling out of a dive of radius 1 km?  EMBED Equation.DSMT4 ; v = 262 m/s or 943 km/h Note: The pilot actually feels a force that is eight times W: N mg = m(7g); N = 8mg *10-36. A 3-kg ball swings in a vertical circle at the end of an 8-m cord. When it reaches the top of its path, its velocity is 16 m/s. What is the tension in the cord? What is the critical speed at the top? [ R = 8 m; m = 3 kg; v = 16 m/s ] At Top:  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4    EMBED Equation.DSMT4 ; T = 66.6 N  When T = 0,  EMBED Equation.DSMT4 ; vc = 8.85 m/s *10-37. A 36-kg girl rides on the seat of a swing attached to two chains that are each 20 m long. If she is released from a position 8 m below the top of the swing, what force does the swing exert on the girl as she passes the lowest point?  EMBED Equation.DSMT4 ; h = 20 m 8 m = 12 m  EMBED Equation.DSMT4 ; v = 15.3 m/s *10-37. (Cont.) At Bottom:  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4   EMBED Equation.DSMT4 ; T = 776 N Gravitation 10-38. How far apart should a 2-ton weight be from a 3-ton weight if their mutual force of attraction is equal to 0.0004 lb? ( G = 3.44 x 10-8 lb ft2/slug2 )  EMBED Equation.DSMT4   EMBED Equation.DSMT4  R = 1.42 ft  10-39. A 4-kg mass is separated from a 2 kg mass by a distance of 8 cm. Compute the gravitational force of attraction between the two masses.  EMBED Equation.DSMT4 ; F = 8.34 x 10-8 N  *10-40. A 3-kg mass is located 10 cm away from a 6-kg mass. What is the resultant gravitational force on a 2-kg mass located at the midpoint of a line joining the first two masses?   EMBED Equation.DSMT4   EMBED Equation.DSMT4  F3 = -1.6 x 10-7 N, F6 = 3.20 x 10-7 N FR = -1.6 x 10-7 N + 3.60 x 10-7 N; FR = 1.60 x 10-7 N *10-41. On a distant planet, the acceleration due to gravity is 5.00 m/s2. and the radius of the planet is roughly 4560 m. Use the law of gravitation to estimate the mass of this planet.  EMBED Equation.DSMT4 ; mp = 1.56 x 1024 kg *10-42. The mass of the earth is about 81 times the mass of the moon. If the radius of the earth is 4 times that of the moon, what is the acceleration due to gravity on the moon? me = 81mm; Re = 4Rm ; Consider test mass m on moon and then on earth:   EMBED Equation.DSMT4 ;  EMBED Equation.DSMT4   EMBED Equation.DSMT4 ;  EMBED Equation.DSMT4 ; gm = 1.94 m/s2 *10-43. At 60-kg mass and a 20-kg mass are separated by 10 m. At what point on a line joining these charges will another mass experience zero resultant force? [ F2 = F6 ]   EMBED Equation.DSMT4  ;  EMBED Equation.DSMT4    EMBED Equation.DSMT4 ; x = 1.732(10 x); x = 17.32 1.732 x x = 6.34 m from the 60-kg mass. Keplers Laws and Satellites 10-44. What speed must a satellite have if it is to move in a circular orbit of 800 km above the surface of the earth? [ The central force Fc must equal the gravitational force Fg.] Note that : R = Re + h = (6.38 x 106 m) + (0.8 x 106 m) = 7.18 x 106 m 10-44. (Cont.) Fc = Fg ;  EMBED Equation.DSMT4 ;  EMBED Equation.DSMT4 ; v = 7450 m/s  10-45. The mass of the Jupiter is 1.90 x 1027 kg and its radius is 7.15 x 107 m. What speed must a spacecraft have to circle Jupiter at a height of 6.00 x 107 m above the surface of Jupiter? R = Rj + h = 7.15 x 107 m + 6 x 107 m; R = 1.315 x 108 m ;  EMBED Equation.DSMT4   EMBED Equation.DSMT4  v =31,400 m/s This represents a speed of approximately 69,800 mi/h.  10-46. What is the orbital speed of a satellite that moves in an orbit 1200 km above the earths surface? Note that : R = Re + h = (6.38 x 106 m) + (1.2 x 106 m) = 7.58 x 106 m Fc = Fg ;  EMBED Equation.DSMT4 ;  EMBED Equation.DSMT4 ; v = 7254 m/s 10-47. The radius of the moon is 1.74 x 106 m and the acceleration due to its gravity is 1.63 m/s2. Apply the law of universal gravitation to find the mass of the moon.  EMBED Equation.DSMT4 ; mm = 7.40 x 1022 kg *10-48.A satellite is located at a distance of 900 km above the earths surface. What is the period of the satellites motion? [ R = 6.38 x 106 m + 0.9 x 106 m = 7.28 x 106 m ]  EMBED Equation.DSMT4   EMBED Equation.DSMT4 ; T = 6180 s (about an hour and 43 minutes)  *10-49. How far above the earths surface must a satellite be located if it is to circle the earth in a time of 28 h? T = 28 h (3600 s/h) = 1.01 x 105 s; T2 = 1.02 x 1010 s2   EMBED Equation.DSMT4   EMBED Equation.DSMT4 ; R = 4.69 x 107 m; h = R Re = 4.05 x 107 m Challenge Problems 10-50. At what frequency should a 6-lb ball be revolved in a radius of 3 ft to produce a centripetal acceleration of 12 ft/s2? What is the tension in the cord?  EMBED Equation.DSMT4 ; f = 0.318 rev/s  EMBED Equation.DSMT4 ; T = 2.25 lb 10-51. What centripetal acceleration is required to move a 2.6 kg mass in a horizontal circle of radius 300 mm if its linear speed is 15 m/s? What is the centripetal force?  EMBED Equation.DSMT4 ; a = 750 m/s2 Fc = mac = (2.6 kg)(750 m/s2); Fc = 1950 N 10-52. What must be the speed of a satellite located 1000 mi above the earths surface if it is to travel in a circular path? [ R = 4000 mi + 1000 mi = 5000 mi ; 5000 mi = 2.64 x 107 ft ]  EMBED Equation.DSMT4 ;  EMBED Equation.DSMT4   EMBED Equation.DSMT4 ; v = 7041 m/s  10-53. A 2-kg ball at swings in a vertical circle at the end of a cord 2 m in length. What is the critical velocity at the top if the orbit is to remain circular?   EMBED Equation.DSMT4 ; Critical speed vc is when T = 0   EMBED Equation.DSMT4 ; vc = 4.42 m/s  *10-54. A 4-kg rock swings at a constant speed of 10 m/s in a vertical circle at the end of a 1.4 m cord. What are the tensions in the cord at the top and bottom of the circular path?  At Top:  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4    EMBED Equation.DSMT4 ; T = 247 N At Bottom:  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4    EMBED Equation.DSMT4 ; T = 325 N *10-55. What frequency of revolution is required to raise the flyweights in Fig. 10-16 a vertical distance of 25 mm above their lowest position. Assume that L = 150 mm. h = 150 mm 25 mm = 124 mm; h = 0.124 m   EMBED Equation.DSMT4 ; f = 1.41 rev/s = 84.6 rpm *10-56. The combined mass of a motorcycle and driver is 210 kg. If the driver is to negotiate a loop-the-loop of radius 6 m, what is the critical speed at the top?  EMBED Equation.DSMT4 ; vc = 7.67 m/s  *10-57. If the speed at the top of the loop in Prob. 10-54 is 12 m/s, what is the normal force at the top of the loop? At Top:  EMBED Equation.DSMT4  and  EMBED Equation.DSMT4    EMBED Equation.DSMT4 ; N = 2980 N  10-58. The speed limit at a certain turn of radius 200 ft is 45 mi/h. What is the optimum banking angle for this situation. Are roads actually constructed at the optimum angles?  v = 45 mi/h = 66.0 ft/s;  EMBED Equation.DSMT4  ( = 34.20; NO *10-59. For the figure shown in Fig. 10-17, assume that a = 2 m and L = 4 m. Find the speed to cause the swing to move out to an angle of 200? b = L sin ( = (4 m) sin 200 = 1.37 m; R = a + b = 3.37 m;   EMBED Equation.DSMT4  v = 3.47 m/s Critical Thinking Questions *10-60. A coin rests on a rotating platform distance of 12 cm from the center of rotation. If the coefficient of static friction is 0.6, what is the maximum frequency of rotation such that he coin does not slip? Suppose the frequency is cut in half. Now how far from the center can the coin be placed? The maximum frequency occurs when Fc = Fs  EMBED Equation.DSMT4 ;  EMBED Equation.DSMT4   EMBED Equation.DSMT4 ; f = 1.11 rev/s  EMBED Equation.DSMT4   EMBED Equation.DSMT4   EMBED Equation.DSMT4 ; R = 0.480 m; R = 48.0 cm *10-61. The laboratory apparatus shown in Fig. 10-19 allows a rotating mass to stretch a spring so that the supporting cord is vertical at a particular frequency of rotation. Assume the mass of the bob is 400 g and the radius of revolution is 14 cm. With a stop watch the time for 50 revolutions is found to be 35 s. What is the magnitude and direction of the force acting on the bob? First find f in (rev/s).  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4    EMBED Equation.DSMT4 ; Fc = 4.51 N, directed toward the center. The centripetal force is ON the bob. The outward force ON the spring is NOT the centripetal force. *10-62. In Problem 10-2 above, assume that a 100-g mass is added to the 400-g bob? The force required to stretch the spring should be the same as before, but the rotating mass has increased. What changes when the experiment is performed again so that the centripetal force is the same as before? On what does the centripetal force act in this experiment? Since the centripetal force must be the same, it is necessary that the velocity be reduced so that:  EMBED Equation.DSMT4  The product mv2 must be the same.  EMBED Equation.DSMT4  v1 = 2(f1R = 2((1.429 rev/s)(0.14 m) = 1.26 m/s  EMBED Equation.DSMT4 ; v2 = (0.894)(1.26 m/s); v2 = 1.13 m/s Thus, the object moves slower and the frequency of revolution decreases so that the centripetal force acting ON the bob does not change:  EMBED Equation.DSMT4   *10-63. A 10-in. diameter platform turns at 78 rpm. A bug rests on the platform 1 in. from the outside edge. If the bug weighs 0.02 lb, what force acts on it? What exerts this force? Where should the bug crawl in order to reduce this force by one-half? f = 78 rev/min = 1.30 rev/s; m = W/g = 0.02 lb/32 ft/s2; m = 0.000625 slugs R = 5 in. 1 in. = 4 in.; R = (4/12) ft = 0.333 ft; Fc = 4(2f2 mR  EMBED Equation.DSMT4 ; Fc = 0.0139 lb The central force ON the bug is exerted BY the platform (Static friction). Since Fc is proportional to R, halving the radius will also halve the force!  EMBED Equation.DSMT4 ; The bug should crawl to a point 2 cm from the center. *10-64. The diameter of Jupiter is 11 times that of the earth, and its mass is about 320 times that of earth. What is the acceleration due to gravity near the surface of Jupiter? mj = 11me = 320(5.98 x 1024 kg); mj = 1.914 x 1027 Rj = 11Re = 11(6.38 x 106 m); Rj = 7.018 x 107 m ;  EMBED Equation.DSMT4   EMBED Equation.DSMT4 ; gj = 25.9 m/s2 *10-65. Assume that L = 50 cm and m = 2 kg in Fig. 10-16. How many revolutions per second are needed to make the angle ( = 300? What is the tension in the supporting rod at that point? [ h = (50 cm) cos 300 = 0.433 m; m = 2.0 kg; ( = 300 ]  EMBED Equation.DSMT4   v = 1.19 m/s; v = 2(fR;  EMBED Equation.DSMT4  ; f = 0.757 rev/s  T cos ( = mg;  EMBED Equation.DSMT4 ; T = 22.6 N  *10-66. A 9-kg block rests on the bed of a truck as it turns a curve of radius 86 m. Assume that (k = 0.3 and that (s = 0.4. Does the friction force on the block act toward the center of the turn or away? What is the maximum speed with which the truck can make the turn without slipping? What if the truck makes the turn at a much greater speed, what would be the resultant force on the block? [ F = Fc ; m = 9 kg ]   EMBED Equation.DSMT4 ; vc = 18.4 m/s Fk = (kmg = (0.3)(9 kg)(9.8 m/s2); Fk = FR = 26.5 N Chapter 10. 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   mm pp == ggRR 22 GG== ((55..0000 mm//ss 22 ))((44556600 mm)) 22 66..6677 xx 1100 --1111  NN mm 22 //kkgg 22 FMathType 4.0 Equation MathType EFEquation.DSMT49q%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APA_989938058[jbF@\@\Ole 2CompObjac3iObjInfod5Equation Native 6_989937946gF`g\`g\Ole >CompObjfh?iE%B_AC_A %!AHA_D_E_E_A  mmgg mm == GGmmmm mm RR mm22   aanndd   gg mm == GGmm mm RR mm22 FMathType 4.0 Equation MathType EFEquation.DSMT49q%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  mmgg ee == GGmmmm ee RR ee22   aanndd   ggObjInfoiAEquation Native B_989938111lF`g\`g\Ole J ee == GGmm ee RR ee22V FMathType 4.0 Equation MathType EFEquation.DSMT49q(%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APACompObjkmKiObjInfonMEquation Native ND_989938398`yqF`g\ \E%B_AC_A %!AHA_D_E_E_A   gg mm gg ee == mm mm RR ee22 mm ee RR mm22 == mm mm ((44RR mm )) 22 8811mm mm RR mm22Ole XCompObjprYiObjInfos[Equation Native \` FMathType 4.0 Equation MathType EFEquation.DSMT49qD%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A   gg mm 99..88 mm//ss 22 ==00..11997755 FMathType 4.0 Equation MathType EFEquation.DSMT49q|%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APA_989938922vF \ \Ole bCompObjuwciObjInfoxeEquation Native f_989939177t~{F@0 \ \Ole mCompObjz|niE%B_AC_A %!AHA_D_E_E_A   GGmm 22 mm''((1100"-xx)) 22 == GGmm 66 mm''xx 22 FMathType 4.0 Equation MathType EFEquation.DSMT49qObjInfo}pEquation Native qt_989939319F \ \Ole wX%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A   xx 22 ((1100"-xx)) 22 == mm 66 mm 22WCompObjxiObjInfozEquation Native {_989946703F̝\̝\ FMathType 4.0 Equation MathType EFEquation.DSMT49q%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A   xx  ((1100"-xx))==  mm 66 mm 22 ==  6600 kkgg2200 kkgg  ==11..773322 FMathType 4.0 Equation MathType EFEquation.DSMT49qܤ%MU2GxDSMT4WinANSITimes New RomOle CompObjiObjInfoEquation Native anSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A   mmvv 22 RR== GGmmmm ee RR 22 ;;    vv==  GGmm ee RR  FMathType 4.0 Equation MathTy_989947618F %\F\Ole CompObjiObjInfope EFEquation.DSMT49qP%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  vv==  ((66..6677 xx 1100 --1111  NN mm 22 //kkgEquation Native l_989947106QFF\F\Ole CompObjig 22 ))((55..9988 xx 1100 2244  kkgg))77..1188 xx 1100 66  mm 4 FMathType 4.0 Equation MathType EFEquation.DSMT49qObjInfoEquation Native \_989947206F{W\{W\Ole @%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A   mmvv 22 RR== GGmmmm ee RR 22 ;;  FMathType 4.0 Equation MathTyCompObjiObjInfoEquation Native d_989947689F_\_\pe EFEquation.DSMT49qH%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  vv==  GGmm ee RR++hh  ==  ((66..6677 xx 1100 --1111 ))((11..99 xx 1100 2277 kkgg))11..331155 xx 1100 88  mm  FMathType 4.0 Equation MathType EFEquation.DSMT49qOle CompObjiObjInfoEquation Native lP%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  vv==  ((66..6677 xx 1100 --1111  NN mm 22 //kkgg 22 ))((55..9988 xx 1100 2244  kkgg))77..5588 xx 1100 66  mm  FMathType 4.0 Equation MathType EFEquation.DSMT49q%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APA_989947841F_\_\Ole CompObjiObjInfoEquation Native _989949602Fo\`\Ole CompObjiE%B_AC_A %!AHA_D_E_E_A  mmgg== GGmmmm mm RR mm22 ;;    mm mm == ggRR mm22 GG== ((11..6633 mm//ss 22 ))((11..7744 xx 1100 66 mm)) 22 66..6677 xx 1100 --1111 NN mm 22 //kkgg 22 FMathType 4.0 Equation MathType EFEquation.DSMT49qܠ%MU2GxDSMT4WinANSITimes New RomObjInfoEquation Native _989949652F`\`\Ole anSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  TT 22 == 44p 22 GGmm ee ()RR 33 == 44p 22 ((77..2288 xx 1100 66 )) 33 ((66..6677 xx 1100 --1111 NN mm 22 //kkgg 22 ))((55..9988 xx 1100 2244 kkgg));;    TT 22 == 33..8822 xx 1100 77  ss 22 FMathType 4.0 Equation MathTyCompObjiObjInfoEquation Native P_989950281F ։\ ։\pe EFEquation.DSMT49q4%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  TT== 33..8822 xx 1100 77  ss 22Ole CompObjiObjInfoEquation Native  FMathType 4.0 Equation MathType EFEquation.DSMT49qܐ%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  TT 22 ==     "#$%&'*-./012589:;<=>ADEFGHIJKLORSTUVWZ]^_`abcfijklmnoruvwxyz{~     UV !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTXWZY\[]^`a~bcdefghijlkmnopqrstuvwyxz{|}44p 22 GGmm ee ()RR 33 ;; RR 33 == GGmm ee TT 22 44p 22 == ((66..6677 xx 1100 --1111 NN mm 22 //kkgg 22 ))((55..9988 xx 1100 2244 kkgg))((11..0022 xx 1100 1100 ss 22 ))44p 22 FMathType 4.0 Equation MathType EFEquation.DSMT49q<%MU2GxDSMT4WinANSITimes New Rom_989950304F\\Ole CompObjiObjInfoEquation Native X_989953631Fg\Ý\Ole  CompObjianSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  RR== 11..0033 xx 1100 2233 mm 3333a FMathType 4.0 Equation MathType EFEquation.DSMT49qObjInfoEquation Native _989953660FÝ\Ý\Ole ܰ%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  aa cc ==44p 22 ff 22 RR;;    ff 22 == aa cc 44p 22 RR== ((1122 fftt//ss 22 ))44p 22 ((33 fftt));;   ff== 00..11001133 ss --22 FMathType 4.0 Equation MathType EFEquation.DSMT49qCompObjiObjInfo Equation Native !_989953814F̝\̝\ܤ%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  TT==mmaa cc == 66 llbb3322 fftt//ss 22 ()((1122 fftt//ss 22 ))Ole (CompObj)iObjInfo+Equation Native , FMathType 4.0 Equation MathType EFEquation.DSMT49qܐ%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  aa cc == vv 22 RR== ((1155 mm//ss)) 22 ((00..330000 mm))`L FMathType 4.0 Equation MathType EFEquation.DSMT49q%MU2GxDSMT4WinANSITimes New Rom_989954319Fԝ\`\Ole 3CompObj4iObjInfo6Equation Native 7_989954440F,\,\Ole ?CompObj@ianSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  RR==22..6644 xx 1100 77 fftt 00..33004488 mm11 fftt()==88..004477 xx 1100 66 mm FMathType 4.0 Equation MathType EFEquation.DSMT49qT%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  vv==  ((66..6677 xx 1100 --111ObjInfoBEquation Native Cp_989954628F,\,\Ole M1  NN mm 22 //kkgg 22 ))((55..9988 xx 1100 2244  kkgg))88..004477 xx 1100 66  mm  FMathType 4.0 Equation MathType EFEquation.DSMT49qCompObjNiObjInfoPEquation Native Q_989954951F \P9\t%MU2GxDSMT4WinANSITimes New RomanSymbolCourier NewMT Extra!/'_!!/!"!APAE%B_AC_A %!AHA_D_E_E_A  vv cc == ggRR  == ((99..88 mm//ss 22 ))((22..00 mm)) mS61$jn-l0ׁØBǠE(_{G03y;a IL:'0q³œL>UCzu֭Bzbx?bJyӂT/DdlB D S A@? 2iX|+c'3E`!=X|+c'30hK xUKSQs̹uT2%JbT(H 7%p6vpɂ)"zC=]${w{>sC7  w 3fƛͦBB\*I|O]:aMiM {xKQz-?V+ӵ9ґ9'2y8u!^2Z :¾I[fmiGY?q޴3pV/K%kV1h6}*YY3[j\i- ;ړ'Uu+x4 C/dcj8.2(Pآ/ J2G03HCzP]+Oħ&:zz|i.*#t(k}]n34d[nWy^Cgٱkwwe8;ɥTʖ `pIdρ]߇ rOkQ$Awv7 D^'~G|Ok^Q7z?Yb $7hFŕFiWnu 3 *1Vu?Q{m2L{mJJwlD,lcnVϳVyi'j\h23F&mFjk#Lv̰x&P"'X U7fu| `R'5l,6 (v'+Im}!F& z!)bijc'fNe+?d$=5yf bUv'qK s_Q{\S~zoaQ^DS<#/q҉ShVOW'/--WO/bxWV*kB:h̴[kˈb :ۍbPk:3:C:` 0PxTKOSQ9PB- uc 1mKIE`,4u]p¥+?Ԯ49A)=;f曹 m7ތ7Meʒ$n[epM*dD6͇ MrO'#d!U3ʬ:R }|=HPgʆ< ؖUAM& YH5fe>s.'KBt3%Sp|8l4~]Y_\;1uҠOU=_ZSPCJ'ee|fomcЪCX\nNY1Kf[sYJ58mS̘&qssYxkI_q?T2.OK^yj~ŝ=;ýT1W [\䋷7vwrbe]:Jtus}_υB6d6 Ӎ9$Ql!2UotC2Z90Nz`ԓt} ,)ڼkyFZëP3!i<0`H]+|j/F'=O`{vu$zn&nQD]IU#pK"%qI.aEq ܬnIU7݆Q/$U>VViNj߳6f=,eNIah 66+&lpϤP? 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