6. A 25-kg ball with a radius of 10 cm is submerged held halfway down a 1-m tall tank of water. Once released, will it float to the top or sink to the bottom? How long will it take to reach the surface/bottom?
Solution
volume of the sphere ,
V =4pi/3 * r^3
V = 4pi/3 * 0.10^3
V = 4.19 *10^-3 m^3
bouyant force , Fs = density of water * V * g
Fs = 4.19 *10^-3 *1000 * 9.8
Fs = 41 N
weight of the ball = 25 * 9.8 = 245 N
as the weight is larger than the buyoant force
the ball will sink to the bottom
for the acceleration of the ball
25 * a = 245 - 41
a = 8.16 m/s^2
let the time taken is t
1/2 = 0.50 * 8.16 * t^2
t = 0.35 s
the time taken to reach bottom is 0.35 s
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6- A 25-kg ball with a radius of 10 cm is submerged held halfway down.docx
1. 6. A 25-kg ball with a radius of 10 cm is submerged held halfway down a 1-m tall tank of water.
Once released, will it float to the top or sink to the bottom? How long will it take to reach the
surface/bottom?
Solution
volume of the sphere ,
V =4pi/3 * r^3
V = 4pi/3 * 0.10^3
V = 4.19 *10^-3 m^3
bouyant force , Fs = density of water * V * g
Fs = 4.19 *10^-3 *1000 * 9.8
Fs = 41 N
weight of the ball = 25 * 9.8 = 245 N
as the weight is larger than the buyoant force
the ball will sink to the bottom
for the acceleration of the ball
25 * a = 245 - 41
a = 8.16 m/s^2
let the time taken is t
1/2 = 0.50 * 8.16 * t^2
2. t = 0.35 s
the time taken to reach bottom is 0.35 s
