College Physics I 1401 057 Spring On-Line Final 3(180 Minutes, 200 Points)Instructor: Dr. Ping LuStudent Name:_____________________________ Grade______________Notes: Please give me your all working on each question.For instance: (15 points)A grocery bagger pushes a 2.00 kg bag of potatoes in such a way that the net externalforce on itis 65.0 N. Calculate its acceleration.Given: m= 2.00 kg, F= 65.0 NFind: aSolve: From the second law F = ma, so we have a = F/m = 65.0/2.00 = 32.5 (m/s^2)Answer: Its acceleration is 32.5 m/s^2.All working on each question you should have four steps: given, find, solve, and answer.1. Multiple Choice: total 80 points 1. State how many significant figures are proper in the results of the following calculation: (106.7) (90) (46.210) (1.01) a. 1 b. 2 c. 3 d. 4A2. State how many significant figures are proper in the results of the following calculation: (5.9)^2 a. 1 b. 2 c. 3 d. 4 B3. What is the magnitude of the momentum of a 7000-kg truck whose speed is 20 m/s? Answer:_________Given: mass= 7000 kg velocity = 20 m/sFind: momentumSolve: Momentum = mass x velocity = (7000)(20) = 140000 kg m/sAnswer: 140000 kg m/s4. How much work does a supermarket checkout attendant do on a can of soup he pushes 0.310 m horizontally with a force of 5.20 N? Express your answer in joules and kilocalories. Answer:_________ Given: distance = 0.310 m Force = 5.20 NFind: workSolve: Work = Fd = (5.20N)(0.310m) =1.612 JAnswer: 1.612 J5. The density of brass is 8.6 g/cm3 . What is this value in kilograms per cubic meter? _______ kg/mGiven: D = 8.6g/cm3Find: D in kg/m3Solve: (8.6g/1cm3)(1 kg/1000g)(100cm/1m)3 = 8600 kg/m3Answer: 8600 kg/m3 6. Assume that an MX missile goes from rest to a suborbital velocity of 6.0 km/s in 40.0 s. What is its average acceleration in____ m/s^2? What is its average acceleration in multiples of g?______ Given: vo= 0 km/s vf=6.0 km/s = 6000 m/s t= 40.0s g=9.81 m/s2Find: average accelerationSolve: a= v/t = (6000/40) = 150 m/s2Answer: 150 m/s2Find: average acceleration in multiples of gSolve: 150/ 9.81 = 15.3 g m/s2Answer: 15.3 g m/s27. A stone is thrown horizontally from the top of a cliff. One second after it has left you hand its vertical distance below the cliff is ______. Given: t= 1 s a=g=9.81 m/s2Find: distanceSolve: Distance = (1/2) at2 =(1/2) 9.81 (1)2 =4.9 s =5 mAnswer: 5m8. Compared to the mass of a certain object on earth, the mass of the same object on the moon is a. more. b. less. c. the same C9. When the net force that acts on a hockey puck is 12 N, the puck accelerates at a rate of 48 m/s/s. Determine the mass of the puck.______Given: Fnet= 12 N a= 48 m/s/sFind: massSolve: F=m/a m=F/a m= 12/48 = 0.25 kgAnswer: 0.25 kg2. Questions: total 120 points 1. (10 points) A grocery bagger pushes a 2.00 kg bag of potatoes in such a way that the net external force on it is 20.0 N. Calculate its acceleration. Given: m= 2.00 kg Fnet= 20.0 NFind: accelerationSolve: F=m/a m=F/a m= 20/2 = 10.0 m/s2 Answer: 10.0 m/s22. (15 points) An ordinary workshop grindstone has a radius of 9.50 cm and rotates at 7000.0 rpm. Calculate the centripetal acceleration at its edge in m/s/s and convert it to multiples of g.Given: r= 0.095 m angular velocity= 7000.0 rpmFind: centripetal accelerationSolve: a = rw2 = (0.095m)(7000.0 (2πrad/60s)2= (50999.04 m/s2)/ 9.81 m/s2 = 5200 g m/s2 Answer: 5200 g m/s23. Suppose a yo-yo has a center shaft 0.300 cm in radius and that string is being pulled from it. (10 points) (a) If the string is stationary and the yo-yo accelerates away from it at a rate of 1.70 m/s2 , what is its angular acceleration? (5 points)Given: r= 0.003m a= 1.70 m/s2 Find: angular accelerationSolve: angular acceleration = a/r = (1.70 m/s2)/ 0.003 m = 566.7 rad/s2Answer: 566.7 rad/s2 (b) What is the angular velocity after 0.750 s if it starts from rest? (5 points) Given: angular acceleration = 566.7 rad/s2 Time = 0.750sFind: angular velocitySolve: w = at = (566.7 rad/s2 )(0.750s) = 425 rad/sAnswer: 425 rad/s(c) The outside radius of the yo-yo is 3.50 cm. In the reference fram of the yo-yo's mass center, what is the tangential acceleration of a point on its edge? Given: a=566.7 rad/s2 r= 0.035 mFind: tangential accelerationSolve: ta=ar = (566.7)(0.035) = 19.83 m/s2Answer: 19.83 m/s2 4. Suppose a 57.0-kg gymnast climbs a rope. (a) (8 points) What is the tension in the rope if he climbs at a constant speed? Given: m= 57.0 kg a=g=9.80 m/s2Find: tensionSolve: T=ma = (57.0)(9.80) =558.6 = 559 NAnswer: 559 N(b) (7 points) What is the tension in the rope if he accelerates upward at a rate of 1.80 m/s2 ? Given: m= 57.0 kg a=9.80 + 1.80 = 11.6 m/s2Find: tensionSolve: T=ma = (57.0)(11.6) = 661.2 = 661 NAnswer: 661 N5. (15 points) The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 10 ✕ 3 N with an effective perpendicular lever arm of 4.00 cm,producing an angular acceleration of the forearm of 130 rad/s2 . What is the moment of inertia of the boxer's forearm? Given: r= 0.04 m F= 2.00 x10 3 N l=130 rad/s2Find: Moment of inertiaSolve: angular acceleration = rF/l = ((.04)(2.00 x 103)/ 130 = 0.6 kg m2Answer: 0.6 kg m26. (8 points) (a) How much gravitational potential energy (relative to the ground on which it is built) is stored in an Egyptian pyramid, given its mass is about 6 10 ✕ 9 kg and its center of mass is 28.0 m above the surrounding ground? (7 points)Given: m= 6 x109 h= 28.0 m g= 9.80 m/s2Find: energySolve: Energy = mgh = (6 x109)(9.80)(28) = 1.65 e 12 JAnswer 1.65 e 12 J(b) What is the ratio of this energy to the daily food intake of a person (1.2 10 ✕ 7 J)? 1.65 e12/ 1.2 e7 =1.38 e5 JAnswer 1.38 e5 J7. (15 points) A 5.50 10 ✕ 5 kg subway train is brought to a stop from a speed of 0.600 m/s in 0.900 m by alarge spring bumper at the end of its track. What is the force constant k of the spring? Given: m=5.50 x 105 vo=0.600 m/s x=0.900 mFind: force constant kSolve: KE =(1/2)(mass)(velocity)2 = (1/2)( 5.50 x 105 )(.600)2 = 99000 JStrain energy = (1/2) Ke2 =99000jK=(2 x 99000)/(0.900 m)2= 244444 = 2.44 e5Answer: 2.44 e5 8. (8 points) (a) Calculate the force needed to bring a 1000 kg car to rest from a speed of 90.0 km/h in a distance of 116 m (a fairly typical distance for a nonpanic stop). Given: m=1000 kg vo= 90.0 km/h x=116m vf= 0Find: forceSolve: (90km/1hr)(1000m/1km)(1hr/60min)(1min/60s)= 25 m/sA= (v2-vo2)/ -2x =
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