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{ }\frac{0.842}{1.025}=\text{ }0.8215=\text{ 82.15\%} \end{gathered}$

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