10. Given the line $x+y-k=0(k>0)$ intersects the circle $x^{2}+y^{2}=4$ at two distinct points $A$ and $B$, and $O$ is the origin, with $|\overrightarrow{O A}+\overrightarrow{O B}| \geqslant \frac{\sqrt{3}}{3}|\overrightarrow{A B}|$. Then the range of $k$ is ( ).
(A) $(\sqrt{3},+\infty)$
(B) $[\sqrt{2},+\infty)$
(C) $[\sqrt{2}, 2 \sqrt{2})$
(D) $[\sqrt{3}, 2 \sqrt{2})$
10. C.

Combining the equations of the line and the circle, and noting that there are two intersection points, we have $\Delta>0$. Solving this, we get $-2 \sqrt{2}<k<2 \sqrt{2}$.

As shown in Figure 7, construct rhombus $A O B C$, and let the diagonals $O C$ and $A B$ intersect at point $M$.

From the inequality in the problem, we have
$$
O M \geqslant \frac{\sqrt{3}}{3} M B \text {. }
$$

In the right triangle $\triangle O M B$, $O B=2$, so $O M \geqslant 1$.

Therefore, the distance from point $O(0,0)$ to the line $x+y-k=0$ is $O M \geqslant 1$.
Thus, $\frac{|k|}{\sqrt{1+1}} \geqslant 1 \Rightarrow k \geqslant \sqrt{2}$ or $k \leqslant-\sqrt{2}$.
In summary, $\sqrt{2} \leqslant k<2 \sqrt{2}$.