The system starts at rest. m1 = 2 kg m2 = 6 kg the angle is 20 degrees and uk=0.15. %3D What is the velocity of the system after m2 moves a distance of 10 meters?
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- The motion of a human body through space can be modeled as the motion of a particle at the bodys center of mass as we will study in Chapter 9. The components of the displacement of an athletes center of mass from the beginning to the end of a certain jump are described by the equations xj = 0 + (11.2 m/s) (cos 18.5)t 0.360 m = 0.840 m + (11.2 m/s)(sin 18.5)t - 12(9.80 m/s2)t2 where t is in seconds and is the time at which the athlete ends the jump. Identify (a) the athletes position and (b) his vector velocity at the takeoff point. (c) How far did he jump?Each Voyager spacecraft was accelerated toward escape speed from the Sun by the gravitational force exerted by Jupiter on the spacecraft. (a) Is the gravitational force a conservative or a nonconservative force? (b) Does the interaction of the spacecraft with Jupiter meet the definition of an elastic collision? (c) How could the space-craft be moving faster after the collision?What is the final horizontal velocity of the 83-kg block if it is initially at rest? 12 N 40% 15 N 30 N 30° 18 N A. 8.7 m/s B. 0 m/s C. 38 m/s D.76 m/s E. None of the above
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