What is the volume of the secret cavity in this submerged sculpture?

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The discussion focuses on calculating the volume of a secret cavity within a submerged brass sculpture using Archimedes' principle. The sculpture weighs 15.76N in air and 13.86N when submerged, indicating a buoyant force due to displaced water. The volume of the cavity is derived from the difference in weight, applying the formula Volume = (Weight in air - Weight in water) / Density of brass. Substituting the values yields a cavity volume of 0.000224 cubic meters. This calculation highlights the relationship between buoyancy and density in determining submerged object characteristics.
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A small sculpture made of brass (density = 8470 kg/m cubed) is believed to have a secret central cavity. The weight of the sculpture in air is 15.76N. When it is submerged in water, the weight is 13.86 N. What is the volume of the secret cavity?
 
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what have you done?

you must be familiar with archimedes principle, i suppose.

Well you know the buoyant force is equal to the mass of the fluid displaced times gravity. When the object is submerged, its weight is less due to the buoyant force.

Use the relationship \rho = \frac{m}{v} in the problem, and remember the volume of the fluid displaced is equal to the volume of what is submerged (in this case the entire object).
 


To calculate the volume of the secret cavity, we can use the formula:
Volume = (Weight in air - Weight in water) / Density of brass

Substituting the given values, we get:
Volume = (15.76N - 13.86N) / 8470 kg/m^3
= 1.9N / 8470 kg/m^3
= 0.000224 m^3

Therefore, the volume of the secret cavity in this submerged sculpture is 0.000224 cubic meters.
 
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