G A flotation device is in the shape of a right cylinder, with a height of 0.301 m and a face area of 4.96 m2 on top and bottom, and its density is 0.565 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  =  1917.66584  J

Explanation:

When the cylinder is floating in the fluid, the buoyant force is given by,

F  =  ρ *V *g

Where,

F = Buoyant force

g = acceleration due to gravity = 9.81 m/s²

V = Displaced volume of the fluid

ρ  = Density = Density of fluid - density of object

Given:

Height of the cylinder, h = 0.301 m

Area of the face, A = 4.96  m ²

Volume of the cylinder, V = (4.96× 0.301)  m³  =  1.49296  m³

Density ,  ρ  =  ρ w  −  ρ₀

Calculation:

the buoyant force is given by,

F  =  ρ *V *g

⇒ F  =  ρ *V *g =  [ 1000  −  ( 1000 ×0.565 ) ]  Kg/m³× 1.49296 m³ × 9.8 1 m/s²

⇒  F = 6370.98286 N

The work done , W = F*h

W  =  F * h  =  6370.98286  N*0.301 m  =  1917.66584  J

The work done by the buoyant force is 1917.66584 J.

you would expect a sound wave to speed up. - b)

a displacement distance is size of displ