What is the density of a 3.25 in. diameter sphere of steel that weighs 3.21 lbs? Round to two decimal places. DO NOT INCLUDE UNITS.

The density of the steel sphere is approximately 9.59

Explanation:

The parameters of the steel sphere are

The diameter of the sphere = 3.25 in.

The weight of the sphere, m = 3.21 lbs = 0.09976985 slug

The density of an object, ρ = (The mass of the object, m)/(The volume of the object, V)

The volume of a sphere, V = (4/3) × π·r³

Where;

V = The volume of the sphere

r = The radius of the sphere = (The diameter of the sphere)/2 = d/2

The radius of the given sphere, r = (3.25 in.)/2 = 1.625 in.

∴ The volume of the given sphere, V = (4/3) × π × 1.625³ = (2197/384)·π ≈ 17.974 in.³

The volume of the given sphere, V ≈ 17.974 in.³ ≈ 0.010401792195 ft.³

Therefore, the density of the steel sphere, ρ = m/V = (0.09976985  slug)/(0.010401792195 ft.³) ≈ 9.59 slugs/ft.³

The density of the steel sphere, ρ ≈ 9.59 slugs/ft.³

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explanation: