Answer:
ρ = 20.4 kg / m³
Explanation:
For this exercise we use Archimedes' principle, the thrust on a submerged body is equal to the weight of the released liquid
B = ρ g V_liquid (1)
let's write the equilibrium equation
F -W + B = 0
B = W -F
let's calculate
B = 0.900 9.8 - 2.94
B = 5.88 N
From equation 1
V_{liquid} = [tex]\frac{B}{\rho_{Hg} g}[/tex]
V_{liquid} = [tex]\frac{5.88}{13.6 \ 10^3 \ 9.8}[/tex]
V_{liquid} = 4.41176 10⁻⁵ m³
in this case it is indicated that the body is completely submerged
V_body = V_liquid = V
the definition of density is
ρ = m / V
ρ = [tex]\frac{0.900 \ 10^{-3}}{ 4.41176 \ 10^{-5} }[/tex]
ρ = 2.04 10¹ kg / m³
ρ = 20.4 kg / m³
