1) What is the escape velocity (in m/s) for an object orbitingthe Earth at a hei
ID: 1675640 • Letter: 1
Question
1) What is the escape velocity (in m/s) for an object orbitingthe Earth at a height above the ground of 6.9E6 m? 2) A pitcher throws a 0.145-kg baseball to a batter at 36 m/s.The batter hits the ball straight back at the pitcher at 49 m/s.The amount of time that the bat and ball are in contact is 9.0E-3s. What is the force (in N, assumed to be constant) between theball and bat? 1) What is the escape velocity (in m/s) for an object orbitingthe Earth at a height above the ground of 6.9E6 m? 2) A pitcher throws a 0.145-kg baseball to a batter at 36 m/s.The batter hits the ball straight back at the pitcher at 49 m/s.The amount of time that the bat and ball are in contact is 9.0E-3s. What is the force (in N, assumed to be constant) between theball and bat?Explanation / Answer
(a) Height abovethe ground, h = 6.9 x 106 m = 6900 km Radius ofthe earth, R = 6400 km Accelerationdue to gravity at the given height, g' = [ R / (R + h) ] 2 * g = [ 6400 / 13300 ]2 * 9.8 = 2.27m/s2 Escapevelocity, Ve = ( 2 g' r ) = [ 2 g' ( R + h ) ] = [ 2 * 2.27 * 13300 x 103 ] = 7769.3 m/s = 7.8 km/s (b) Mass of theball, m = 0.145 kg Initialvelocity, Vi = - 36 m/s Finalvelocity, Vf = 49 m/s Time ofcontact, t = 9 x 10-3 s Vi and Vf carry opposite signs because theyare in opposite directions Force, F = ma = m [ ( Vf - Vi ) / t] = ( 0.145 * 85 ) / ( 9 X 10-3 ) = 1369.4 N
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.