What is the moment of inertia of a pendulum with a mass of 7 kg that is 2 m from the pivot?

$28 \setminus k g \setminus {m}^{2}$
Our moment of inertia can be defined as $I = m {r}^{2}$.
$I = \left(7 \setminus k g\right) \cdot {\left(2 \setminus m\right)}^{2} = 28 \setminus k g \setminus {m}^{2}$