Because $\triangle RST$ is a right triangle, $\sin R = \frac{ST}{RT}$. So $\sin R = \frac{2}{5} = \frac{ST}{5}$. Then $ST=2$.

We know that $\sin T = \frac{RS}{RT}$. By the Pythagorean Theorem, $RS = \sqrt{RT^2 - ST^2} = \sqrt{25-4} = \sqrt{21}$. Then $\sin T = \boxed{\frac{\sqrt{21}}{5}}$.
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
In $\triangle{RST}$, shown, $\sin{R}=\frac{2}{5}$.  What is $\sin{T}$?

[asy]
pair R,S,T;
T = (0,0);
S = (2,0);
R = (2,sqrt(21));
draw(R--S--T--R);
draw(rightanglemark(R,S,T,10));
label("$T$",T,SW);
label("$S$",S,SE);
label("$R$",R,NE);
label("$5$",(R+T)/2,NW);
[/asy]
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
\frac{\sqrt{21}}{5}
In $\triangle{RST}$, shown, $\sin{R}=\frac{2}{5}$.  What is $\sin{T}$?

[asy]
pair R,S,T;
T = (0,0);
S = (2,0);
R = (2,sqrt(21));
draw(R--S--T--R);
draw(rightanglemark(R,S,T,10));
label("$T$",T,SW);
label("$S$",S,SE);
label("$R$",R,NE);
label("$5$",(R+T)/2,NW);
[/asy]
test