ANSWER
661.61 kg/m³
EXPLANATION
Given:
• The mass of the mine, m = 190 kg
,
• The volume of the mine, V = 1.150 m³
,
• The tension in the chain, T = 5.6x10³ N = 5600 N
Find:
• The density of the fluid, ρ
Let's draw a free-body diagram of this situation first,
If the forces are in equilibrium,
[tex]F_b-T-F_g=0[/tex]
The buoyant force is given by the equation,
[tex]F_b=\rho gV[/tex]
Where ρ is the density of the fluid, and V is the submerged volume - in this case, the volume of the mine. We know the magnitude of the tension in the chain and the weight of the mine is,
[tex]F_g=mg[/tex]
Replace in the first equation,
[tex]\rho gV-T-mg=0[/tex]
Solving for ρ,
[tex]\rho=\frac{T+mg}{gV}[/tex]
Replace with the known values and use g = 9.81 m/s²,
[tex]\rho=\frac{5600N+190kg\cdot9.81m/s^2}{9.81m/s^2\cdot1.150m^3}\approx661.61kg/m^3[/tex]
Hence, the density of the fluid the mine is in is 661.61 kg/m³.
