a spherical ball of mass 5kg and specific gravity 0.8 is dropped from a height of 20m above the surface of a large vessel of water. The work done by the bouyant force toll the ball comes to rest
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here's ua answer :
Buoyancy Force = (dV)g
m = d’V
acceleration = net Force/m => dVg/d’V = > d*g/d’ = 10/0.8 = 12.5 m/s^2
net accln. = 12.5 – 10 = 2.5 m/s^2.
depth it gets sinked down:
=>
(2g*2) = 0^2 + 2*g*s
s = 8 meter.
《HOPE》《IT》《HELPS》《UH》
it's really nice to meet uh
here's ua answer :
Buoyancy Force = (dV)g
m = d’V
acceleration = net Force/m => dVg/d’V = > d*g/d’ = 10/0.8 = 12.5 m/s^2
net accln. = 12.5 – 10 = 2.5 m/s^2.
depth it gets sinked down:
=>
(2g*2) = 0^2 + 2*g*s
s = 8 meter.
《HOPE》《IT》《HELPS》《UH》
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