A flotation device is in the shape of a right cylinder, with a height of 0.620 m and a face area of 2.24 m2 on top and bottom, and its density is 0.415 times that of fresh water. It is initially held fully submerged in fresh water, with its top face at the water surface. Then it is allowed to ascend gradually until it begins to float. How much work does the buoyant force do on the device during the ascent?

W = 4941.47 J

The workdone by the buoyant force on the device during the ascent is 4941.47J

Explanation:

Given;

Height h = 0.620m

Face area A= 2.24m^2

density p = 0.415g/cm^3 = 415kg/m^3

Buoyancy force is the force exerted on the device to push it to the surface.

F = pVg =phAg

Where;

F = buoyancy force

p = density of fluid - density of device

V = volume

g = acceleration due to gravity

h = height

A = face area

And work done by buoyanct force is;

W = Fl = Fh = phAgh = pAgh^2

Substituting the given values;

W = (1000-415)×2.24×9.81×0.620^2

W = 4941.47 J

The workdone by the buoyant force on the device during the ascent is 4941.47J

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j