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You are watching: A race car has a mass of 710 kg

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1. A race automobile has a fixed of 710 kg. The starts native rest and travels 40.0m in 3.0s. The car is uniformly sped up duringthe entire time. How large is the net force acting ~ above the car? do a quantitative pressure diagram. Create a network forceequation for the axis along which pressures are no balanced.vi = 0.0 m/s vf = ___ x = 40 mt = 3.0 sa = ___Fnet = ____datatime velocity position0.0 0.00.01.0 _______2.0 ____ ____3.0 26.6 _40_FNx 40  0mv  13 .3 m st3sv vv  ns f  v f  2v v i2v f  2 13 .3 m s  26 .6 m sFf,car,roadmv 26 .6  0 ms 8.9 st3ssFa  network  Fnet  mamFroad  710 k g 8.9 m 2  6320 Ns aFg2. Mean that a 1000 kg car istraveling in ~ 25 m/s (55 mph). That is brakes can apply a pressure of 5000 N. What is theminimum distance compelled for the auto to stop? make a quantitative pressure diagram. Create a net force equation forthe axis along which pressures are no balanced.vi = 55 m/svf = 0.0 m/sa = ___Fbrakes = 5000 NFbrakesaFnetmaa5000 N1000 kga  5.0ms2FfmFNFbrakeor  5.0mseach second3. A 65 kg human being dives right into the water indigenous the 10 m platform.a.What is her speed as she start the water?vi = 0.0 m/s vf = ____y = -10 mv f2  v i2  2ayv f2  0  2( 10 sm2 )( 10m)v f2  200 multiple sclerosis 2  v f   200  14 ms2a = -10 m/s/sFgb.She comes to a avoid 4.0 m below the surface of the water. Find the force on the swimmer by the water.vi = 14.14 m/svf = 0.0 m/s y = - 4.0 ma = ____v f2  v i2  2ayaaav f2 v i22 y0 ( 14.14 ms )22( 4 m )20(200 m2s)8 mma  25 sm2  25 s sFnetmFwater Fgm Fwater  Fg  maFwaterFw  ma  FgNFwater  65kg (25 sm2 )  65kg (10 kg)  1625N  650N  2275N  2300NFnetFg4. During a head-on collision, a passenger in the front chair of a car speeds up from 13.3 m/s (30 miles/hour) to rest in0.10 s.a.Calculate the acceleration that the passenger.vi = 13.3 m/s vf = 0.0 m/st = 0.10 sa = ____assume passenger in left car and to ideal is positiveavtam0  13.3 ms 133 sm2  133 ss.10sb.The driver the the automobile holds the end his eight to store his 25 kg child (who is not wearing a seat belt) from wrecking intothe dashboard. Exactly how much pressure must that exert on the child?vi = 13.3 m/saFnetmFNmvf = 0.0 m/st = 0.10 s FN  maa = - 133 m/s/s (from part a.)m = 25 kgFnetFN  25kg ( 133 sm2 )  3325NFarmc.What is the load of the child?NFg  mg  25kg (10 kg)  250Nd.Convert the pressures in parts b and c indigenous Newtons to pounds. (1 lb = 4.45N). What room the possibilities the driver willbe able to protect against the child?1lb)  747 pounds4.45N1lb250N()  56 pounds4.45N3325N(The driver has actually no possibility of protecting against the child.6.The following questions describe the movement of a baseball.a.While gift thrown, a net force of 132 N acts on a baseball (mass = 140 g) because that a period of 4.5 x 10-2 sec.What is the magnitude of the readjust in inert of the ball?F t  mv  F t  132N (0.045s )  5.9N s  5.9 kgs mb.If the initial speed of the baseball is v = 0.0 m/s, what will certainly its speed be as soon as it leaves the pitcher"s hand?mv  5.9 kgs m  v 5.9 kgs m5.9 kgs m 42 ms1kg140g 10000.140kggc.When the batter hits the ball, a net pressure of 1150 N, opposite to the direction that the ball"s initial motion, actson the round for 9.0 x 10-3 s during the hit. What is the last velocity that the ball?F t  mv  F t  1150N(0.0090s )  (0.140kg )v  v  74 msv  v f  v i  v f  v  v i  v f  74 multiple sclerosis  42 ms  32 msd.viHow huge is the pressure the ball exerts top top the bat? Explain.The ball exerts a force on the bat of 1150N. Round on bat and bat on ball are a third Law pair.7.A rocket, weighing 4.36 x 104N, has actually an engine that offers an upward force of 1.2 x 105N. The reaches amaximum speed of 860 m/s.Fenginea.Draw a pressure diagram because that the rocket.Fgb. For how much time have to the engine burn throughout the launch in bespeak to with thisspeed?4.3510 NFg(860 ms )N10 kgmvg vFnet t  mv  t  49sFnet1.2  105 N  4.35  104 N7.65  104 N4