10. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with hypotenuse $AB$, $SA=SB=SC=2$, $AB=2$, and points $S, A, B, C$ all lie on a sphere with center $O$. Then the distance from point $O$ to the plane $ABC$ is $\qquad$.
$S A=S B=S C \Rightarrow S$ the projection of $S$ on the base $A B C$ is the circumcenter of $\triangle A B C$, which is the midpoint $D$ of the hypotenuse $A B$, thus the center of the sphere $O$ lies on the line $S D$.
$\triangle S A B$ is an equilateral triangle $\Rightarrow S D=\sqrt{3}$, let $O D=d$ $\Rightarrow(\sqrt{3}-d)^{2}=d^{2}+1 \Rightarrow d=\frac{\sqrt{3}}{3}$. Therefore, the distance from point $O$ to the plane $A B C$ is $\frac{\sqrt{3}}{3}$.