5. If $\omega=\cos 40^{\circ}+i \sin 40^{\circ}$, then $\mid \omega+2 \omega^{2}+3 \omega^{3}+\cdots$ $+\left.9 \omega^{9}\right|^{-1}$ equals ( ).
(A) $\frac{1}{18} \cos 20^{\circ}$
(B) $\frac{1}{9} \sin 40^{\circ}$
(C) $\frac{1}{9} \cos 40^{\circ}$
(D) $\frac{2}{9} \sin 20^{\circ}$
5. (D).

Let $s=\omega+2 \omega^{2}+3 \omega^{3}+\cdots+9 \omega^{9}$, where $\omega=e^{i \frac{2 \pi}{9}}$.
$$
\begin{array}{l}
\omega s=\omega^{2}+2 \omega^{3}+3 \omega^{4}+\cdots+9 \omega^{10}, \\
s(1-\omega)=\omega+\omega^{2}+\omega^{3}+\cdots+\omega^{9}-9 \omega^{10} . \\
\because \omega \neq 1, \\
\therefore s(1-\omega)=\frac{\omega^{0}-\omega}{\omega-1}-9 \omega^{10} .
\end{array}
$$

Notice that $\omega^{9}=\left(e^{i \frac{2 \pi}{9}}\right)^{9}=e^{i 2 \pi}=1 . \omega^{10}=\omega$,
$$
\therefore s(1-\omega)=-9 \omega \text {, }
$$

Thus $\frac{1}{2}=\frac{\omega-1}{9 \omega}, \frac{1}{|s|}=\frac{|\omega-\cdots|}{9|\omega|}=\frac{|\omega-1|}{9}$.
Since $\omega$ is a vertex of a regular nonagon inscribed in the unit circle, $\omega-1$ is the side length of the regular nonagon inscribed in the unit circle. Therefore, $|\omega-1|=2 \sin \frac{\pi}{9}$, which means $\frac{|\omega-1|}{9}=\frac{2}{9} \sin 20^{\circ}$.