'The moment of inertia for a 2200 kg solid disc is 11800 kg-m Find the radius of the disc: 3.275 m 3.932 m 2.951 m 3.893 m 2.687 m 3.810 m'
Added by Kristin B.
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\[ 11800 = \frac{1}{2} \times 2200 \times r^2 \] Show more…
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