Apparent weight of a submerged rock.

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SUMMARY

The apparent weight of a submerged rock can be calculated using the formula: Apparent Weight = mg - ρ_water * V * g, where mg is the weight of the rock in air, ρ_water is the density of water, V is the volume of the rock, and g is the acceleration due to gravity. For a rock weighing 58 N with a volume of 1.8 * 10^-3 m^3 submerged in water, the buoyant force is equal to the weight of the displaced water. Additionally, for a rock weighing 57 N with a volume of 2.3 * 10^-3 m^3 submerged in a liquid with double the density of water, the apparent weight can be recalculated using the same principles, adjusting for the increased buoyancy from the denser liquid.

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Homework Statement


What is the apparent weight of a rock submerged in water if the rock weighs 58 N in air and has a volume of 1.8 * 10^-3 m^3?

And then as a continuation:

A rock weighing 57 N with a volume of 2.3 10-3 m3 is submerged in a liquid with a density exactly twice that of water, what will be its new apparent weight reading in the liquid?


Homework Equations





The Attempt at a Solution


I did m*(a+g) but that was incorrect unfortuneately.
 
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When the rock is submerged completely then there will be the force of buoyancy. This force is equal to the weight of the displaced fluid. If all the rock is submerged then the force of buoyancy will be \rho_{water}Vg where \rho_{water} is the density of the water and V is the volume of the rock. So there will be force of gravity ,mg, downwards and the buoyancy force upwards. So the net force downwards is the apparent weight.

\mbox{Apparent Weight}\quad =mg- \rho_{water}Vg
 
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