Find the dimensions of a/b in the equation. `F=asqrtx+bt^(2)` where F is force, x is distance and t is time.

Explanation

`[asqrtx]=[F]`
`therefore [a]=([F])/([sqrtx])=(MLT^(-2))/(L^(1//2))=ML^(1//2)T^(-2)`
`[bt^(2)]=[F]`
`therefore [b]=([F])/([t^(2)])=(MLT^(-2))/(T^(2))`
`=MLT^(-4)`
`[a//b]=(ML^(1//2)T^(-2))/(MLT^(-4))=L^(-1//2)T^(2)`