A solid 0.6750 kg 0.6750 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m R=0.6750 m . What minimum translational speed v min vmin must the ball have when it is a height H = 1.039 m H=1.039 m above the bottom of the loop in order to complete the loop without falling off the track? Assume that the radius of the ball itself is much smaller than the loop radius R R . Use g = 9.810 m/s 2 g=9.810 m/s2 for the acceleration due to gravity.

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.

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