Question Number 8788 by vuckintv last updated on 27/Oct/16
![Solve for α U(z)=U_b +((2A)/(n+1))(ρ×g×α)^n [H−i(H−Z)^(n+1) ]](https://www.tinkutara.com/question/Q8788.png)
$${Solve}\:{for}\:\alpha \\ $$$$ \\ $$$${U}\left({z}\right)={U}_{{b}} +\frac{\mathrm{2}{A}}{{n}+\mathrm{1}}\left(\rho×{g}×\alpha\right)^{{n}} \left[{H}−{i}\left({H}−{Z}\right)^{{n}+\mathrm{1}} \right] \\ $$
Commented by ridwan balatif last updated on 28/Oct/16
![U(z)−U_b =((2A)/(n+1))(ρ×g)^n ×α^n [H−i(H−Z)^(n+1) ] (((U(z)−U_b )(n+1))/(2A(ρ×g)^n [H−i(H−Z)^(n+1) ]))=α^n ((1/(ρ×g)))(((((U(z)−U_b )(n+1))/(2A(H−i(H−Z)^(n+1) )))^(1/N) )=α](https://www.tinkutara.com/question/Q8797.png)
$$\mathrm{U}\left({z}\right)−\mathrm{U}_{\mathrm{b}} =\frac{\mathrm{2A}}{\mathrm{n}+\mathrm{1}}\left(\rho×\mathrm{g}\right)^{\mathrm{n}} ×\alpha^{\mathrm{n}} \left[{H}−{i}\left({H}−{Z}\right)^{{n}+\mathrm{1}} \right] \\ $$$$\frac{\left({U}\left({z}\right)−{U}_{{b}} \right)\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}{A}\left(\rho×{g}\right)^{{n}} \left[{H}−{i}\left({H}−{Z}\right)^{\mathrm{n}+\mathrm{1}} \right]}=\alpha^{\mathrm{n}} \:\: \\ $$$$\left(\frac{\mathrm{1}}{\rho×{g}}\right)\left(\sqrt[{\mathrm{N}}]{\frac{\left(\mathrm{U}\left({z}\right)−{U}_{{b}} \right)\left({n}+\mathrm{1}\right)}{\mathrm{2}{A}\left({H}−{i}\left({H}−{Z}\right)^{\mathrm{n}+\mathrm{1}} \right.}}\right)=\alpha \\ $$$$ \\ $$