Answer
(a) $E_k$ = $7.4 \times 10^{-13}$ J
(b) $0.74$ pJ.
Work Step by Step
(a)
$E_k = \frac{1}{2}mv^2$
$E_k = \frac{1}{2}(6.6 \times 10^{-27}kg)(1.5 \times 10^7\frac{m}{s})^2$
$E_k = 7.4 \times 10^{-13}$ $\frac{kg*m^2}{s^2}$ = $7.4 \times 10^{-13}$ J
(b)
The appropriate prefix is pico (p), because it is the closest to $10^{-13}$ since it represents $10^{-12}$
$7.4 \times 10^{-13} $ J$ = 7.4 \times 10^{-13}\frac{1}{10^{-12}} $ pJ = $0.74$ pJ.
