physics

An irregularly shaped chunk of concrete has a hollow spherical cavity inside. The mass of the chunk is 32kg, and the volume enclosed by the outside surface of the chunk is 0.020m^3. What is the radius of the spherical cavity?

To solve this, you need to know the density of concrete without the hole. The internet gives a range of values, mostly from 2300 to 2500 kg/m^3. Call that density "rho".

If R is the radius of the spherical cavity inside,
[0.020 m^3 - (4/3)pi R^3]*(rho) = 32 kg
I suggest you use 2400 kg/m^3 for "rho" and solve for R.
48 - 32 = 16 = (4/3)pi*R^3*2400

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asked by Cavin

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