A: A ball is released at the bottom of a container of liquid. The ball rises upw

Question

A: A ball is released at the bottom of a container of liquid. The ball rises upward, pops out of the liquid and reaches a height of 20 cm, above the liquid, moving through a near vacuum. The experiment is carried out on Earth, and the time from the ball's release to the moment when it reaches its maximum height is 0.70 seconds. 1. Calculate the velocity of the ball when it leaves the liquid. 2. Determine the average acceleration in the liquid, assuming it to be constant. 3. Determine the depth of the liquid from the motion of the ball and the time.

Explanation / Answer

On coming out of liquid, energy =0.5 mv^2 =mgh

v=1.98 m/s

time in air = sqrt(2h/g) = 0.2 sec

time in liquid = 0.5 sec

accn = change in vel / time = 1.98 / 0.5 = 3.96 m/s^2

depth = 0.5accn x time^2 = 49.5 cm

B starting velocity = 12, ending velocity = 16, time = 240 sec

relative velocity = -4, separation = 55

55=-4(240)+0.5 accn 240^2

accn =0.0352 m/s^2

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