= Solution
To obtain a <power-law river kinematic wave>, let $P_b$ be the wetted perimeter and $I$ the small bed slope. A uniform-flow <force> balance per unit river <length> is $\rho gAI=\tau_bP_b$. With a <Chézy friction law> $\tau_b=\rho f u^2$, it gives $u=(gI/f)^{1/2}(A/P_b)^{1/2}$. Thus the <hydraulic radius> determines the <power> of <area>.
For a channel family with $P_b=P_0A^q$, where $q<1$, we have $u=C A^{(1-q)/2}$, with $C=(gI/(fP_0))^{1/2}$. Set $m=(1-q)/2>0$ and $c=(m+1)C$. Then
$$
\boxed{Q=Au=\frac c{m+1}A^{m+1}.}
$$
For example a broad fixed-width rectangular channel has approximately constant wetted perimeter, giving $m=1/2$; a geometrically similar channel family has $P_b\propto A^{1/2}$, giving $m=1/4$. The exponent is therefore a geometry-dependent consequence of the <friction> balance, not a universal arbitrary value fixed by Chezy's law alone.
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