An object with height h, mass M, and a uniform cross-sectional area A floats upright in a liquid with density ρ.

1 answer:

7 0

** Missing information: The vertical distance from surface of liquid to bottom of the object is sought in this question, with the condition that the object is at equilibrium **

Ans: The vertical distance = y = M/(ρA)

Explanation:

Support the vertical distance = y

Object's density = M/(A*h) (since A*h = volume)

By applying the condition,

(M/(Ah))/ρ = y/h

M/(ρAh) = y/h

y = M/(ρA)

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According to the question

$\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y$

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