Answer: the sphere's inner radius is 4.21 cm so the Inner diameter is 8.42 cm
Explanation:
Given that;
mass m = 746 g
diameter d = 10.4 cm
radius r = 5.2 cm
density of aluminum = 2.7 g/cm³
we know that
density × volume = mass
P × 4/3π( r₀³ - ri³) = m
so we substitute
2.7 × 4/3π( 5.2³ - ri³) = 746
11.3097( 5.2³ - ri³) = 746
( 5.2³ - ri³) = 746/11.3097
140.608 - ri³ = 65.961
ri³ = 140.608 - 65.961
ri³ = 74.647
ri = ∛74.647
ri = 4.21 cm
the sphere's inner radius is 4.21 cm so the Inner diameter is 8.42 cm
The inner diameter of the aluminum sphere is 8.42 cm.
The given parameters;
The volume of the aluminum sphere is calculated as follows;
[tex]volume = \frac{mass}{density} \\\\volume = \frac{746}{2.7} \\\\volume = 276.3 \ cm^3[/tex]
The volume of the sphere from the given outer and inner diameter is calculated as;
[tex]V = \frac{4}{3}\pi r^3\\\\V = \frac{4}{3}\pi \times \frac{d^3}{8} \\\\V = \frac{1}{6} \pi d^3\\\\V = \frac{1}{6} \pi (d_o^3 - d_i^3)\\\\d_o^3 - d_i^3 = \frac{6V}{\pi} \\\\d_i^3 = d_0^3 - \frac{6V}{\pi} \\\\d_i^3 = (10.4)^3 - \frac{6(276.3)}{\pi} \\\\d_i^3 = 597.23\\\\d_i = \sqrt[3]{597.23} \\\\d_i = 8.42 \ cm[/tex]
Thus, the inner diameter of the aluminum sphere is 8.42 cm.
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