Verifying Solidity/Hollowness of a Copper Object

Thread starter jpfelgueiras
Start date Jul 6, 2010
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SUMMARY

The discussion focuses on determining the solidity of a copper object by analyzing its weight in air and water. The copper object has a density of 8.96 g/cm³ and weighs 2.8 N in air and 1.3 N when submerged in water with a density of 0.998 g/cm³. By applying Archimedes' principle, the buoyant force can be calculated, allowing for the determination of the object's volume and, subsequently, its density. If the calculated density matches the known density of copper, the object is solid; otherwise, it is hollow.

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Understanding of Archimedes' principle
Basic knowledge of density calculations
Familiarity with weight measurements in different mediums
Knowledge of unit conversions (g/cm³ to kg/m³)

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Jul 6, 2010

jpfelgueiras

I got a small conserning how to determine if a object is solid or hollow.
"Verify if a copper object with density of 8,96 g/m^3, which weights 2,8N on the air and when submersive on water (p=998 kg/m^3) weights 1,3N, is it solid or hollow?"

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Jul 6, 2010

Doc Al

Mentor

How could you solve for the density of the object? (Note that the density given for copper should be in g/cm^3, not g/m^3.)

Thread 'Magnitude of buoyant force in fluids of different densities'

The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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