Question

A geologist finds that a Moon rock whose mass is 9.50 kg has an apparent mass of 6.00 kg when submerged in water Part A What is the density of the rock? 0 AED O ? kg/m Submit Request Answer Provide Feedback

Answer 1

Buoyant Force = Density of the fluid * Volume of displaced water * g

Volume of displaced fluid = Volume of stone

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Apparent Weight = Weight of the stone - Buoyant Force

$W_{app}=W_{stone}-F_{buoyant}$

$m_{app}*g=m_{stone}*g-\rho _{fluid}*V_{displaced}*g$

$m_{app}=m_{stone}-\rho _{water}*V_{stone}$

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Density = Mass/Volume

Volume = Mass/Density

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$m_{app}=m_{stone}-\rho _{water}*\frac{m_{stone}}{\rho _{stone}}$

$6kg=9.50kg-1000kg/m^{3}*\frac{9.50kg}{\rho _{stone}}$

$6=9.5-1000*\frac{9.5}{\rho _{stone}}$

$6=9.5-\frac{9500}{\rho _{stone}}$

$\frac{9500}{\rho _{stone}}=9.5-6$

$\frac{9500}{\rho _{stone}}=3.5$

$\frac{9500}{3.5}=\rho _{stone}$

ANSWER: ${\color{Red} \rho _{stone}=2714.286kg/m^{3}}$

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