9. Let $x=0.82^{0.5}, y=\sin 1, z=\log \sqrt[3]{7}$, then the size relationship of $x, y, z$ is $(\quad)$.
A. $x<y<z$
B. $y<z<x$
C. $z<x<y$
D. $z<y<x$
9. B. From $x=0.82^{0.5}>0.81^{0.5}=0.9, y=\sin 17^{5}$, we can get $9>5 \log _{3} 7$, which means $0.9>\log _{3} \sqrt{7}$, hence $x>z$. Also, from $3^{7}<7^{4}$, we can get $3^{\frac{7}{4}}<7, 3^{1.75}<7$, further leading to $3^{\sqrt{3}}<7, \sqrt{3}<\log _{3} 7, \frac{\sqrt{3}}{3}<\log _{3} \sqrt{7}$, which means $y<z$. In summary, $y<z<x$.