How do you do this question?

I = 2/5 M (R₂⁵ − R₁⁵) / (R₂³ − R₁³)

Explanation:

We can look at this as the difference of two spheres, a large one and a small one.  If the density is ρ, then the mass of the large sphere is:

M₂ = ρ 4/3 π R₂³

The mass of the small sphere is:

M₁ = ρ 4/3 π R₁³

The difference is:

M = M₂ − M₁

M = ρ 4/3 π R₂³ − ρ 4/3 π R₁³

M = ρ 4/3 π (R₂³ − R₁³)

Solving for ρ:

ρ = 3M / (4π (R₂³ − R₁³))

The moment of inertia of the large sphere is:

I₂ = 2/5 M₂R₂²

I₂ = 2/5 (ρ 4/3 π R₂³) R₂²

I₂ = ρ 8/15 π R₂⁵

The moment of inertia of the small sphere is:

I₁ = 2/5 M₁R₁²

I₁ = 2/5 (ρ 4/3 π R₁³) R₁²

I₁ = ρ 8/15 π R₁⁵

The difference is:

I = I₂ − I₁

I = ρ 8/15 π R₂⁵ − ρ 8/15 π R₁⁵

I = ρ 8/15 π (R₂⁵ − R₁⁵)

Substituting:

I = 3M / (4π (R₂³ − R₁³)) 8/15 π (R₂⁵ − R₁⁵)

I = 2/5 M (R₂⁵ − R₁⁵) / (R₂³ − R₁³)

15.3 cos 65 degrees trig

luster is not a good characteristic to use by itself because luster is the property of minerals that indicates how much the surface of a mineral reflects light. the luster of a mineral is affected by the brilliance of the light used to observe the mineral surface.