Question

The specific gravity of an object is 0.842, whereasthat of seawater is 1.025.What percent of an object is above the surface of the water?

Answer 1

Answer:

82.15% of an object is above the surface of the water

Explanations:

Note that:

Mass = Density x Volume

That is, m = ρV

Force = Mass x acceleration due to gravity

That is, F = mg

Substitute m = ρV into F = mg

F = ρVg

Also note that:

The Buoyant Force = Gravitational force

Therefore:

`\begin{gathered} F_(Buoyant)=F_(gravitational) \\ \rho_(sw)V_(sw)g=\rho_(ob)V_(ob)g \\ \text{where:} \\ \rho_(sw)=specific\text{ gravity of sea water} \\ V_(sw)=\text{ Volume of sea water displaced} \\ \rho_(ob)=\text{ Specific gravity of the object} \\ V_(ob)=\text{ Volume of the object} \end{gathered}`
`\begin{gathered} \rho_(sw)V_(sw)g=\rho_(ob)V_(ob)g \\ \frac{V_(sw)}{V_{\text{obj}}}=\text{ }\frac{\rho_(ob)_{}}{\rho_(sw)}=\text{ }(0.842)/(1.025)=\text{ }0.8215=\text{ 82.15\%} \end{gathered}`

82.15% of an object is above the surface of the water