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Sagot :
Answer:
[tex]\frac{h_{liquid} }{ h_{body} }[/tex] = 5/9
Explanation:
This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.
B = ρ_liquid g V_liquid
let's write the translational equilibrium condition
B - W = 0
let's use the definition of density
ρ_body = m / V_body
m = ρ_body V_body
W = ρ_body V_body g
we substitute
ρ_liquid g V_liquid = ρ_body g V_body
[tex]\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }[/tex]
In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar
V = A h_bogy
Thus
[tex]\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }[/tex]
we substitute
5/9 = [tex]\frac{h_{liquid} }{ h_{body} }[/tex]
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