Answer
$$=2.6 \times 10^{4} \mathrm{g}$$
Work Step by Step
$v_{0}=0, \quad v=1.6 \mathrm{m} / \mathrm{s},$ and $\Delta x=5.0 \mu \mathrm{m}$ are Given,
$\therefore$ the acceleration of the spores during the launch is
$$a=\frac{v^{2}-v_{0}^{2}}{2 x}=\frac{(1.6 \mathrm{m} / \mathrm{s})^{2}}{2\left(5.0 \times 10^{-6} \mathrm{m}\right)}$$
$$=2.56 \times 10^{5} \mathrm{m} / \mathrm{s}^{2}$$
$$=2.6 \times 10^{4} \mathrm{g}$$
