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Question Number 8695 by vuckintv last updated on 22/Oct/16
solving for B? U_b = U_s − (2/(n+1))(((ρ×g×sinα)/B))^n H^(n+1)
$${solving}\:{for}\:{B}? \\ $$$${U}_{{b}} \:=\:{U}_{{s}} \:−\:\frac{\mathrm{2}}{{n}+\mathrm{1}}\left(\frac{\rho×{g}×{sin}\alpha}{{B}}\right)^{{n}} {H}^{{n}+\mathrm{1}} \\ $$
Answered by FilupSmith last updated on 22/Oct/16
U_b = U_s − (2/(n+1))(((ρ×g×sinα)/B))^n H^(n+1) U_b −U_s = − (2/(n+1))(((ρ×g×sinα)/B))^n H^(n+1) ((U_b −U_s )/H^(n+1) ) = − (2/(n+1))(((ρ×g×sinα)/B))^n −(((U_b −U_s )(n+1))/(2H^(n+1) )) = (((ρ×g×sinα)/B))^n (−(((U_b −U_s )(n+1))/(2H^(n+1) )))^(1/n) =((ρ×g×sinα)/B) (−(((U_b −U_s )(n+1))/(2H^(n+1) )))^n ρ×g×sin(α)=B
$${U}_{{b}} \:=\:{U}_{{s}} \:−\:\frac{\mathrm{2}}{{n}+\mathrm{1}}\left(\frac{\rho×{g}×{sin}\alpha}{{B}}\right)^{{n}} {H}^{{n}+\mathrm{1}} \\ $$$${U}_{{b}} −{U}_{{s}} \:=\:\:−\:\frac{\mathrm{2}}{{n}+\mathrm{1}}\left(\frac{\rho×{g}×{sin}\alpha}{{B}}\right)^{{n}} {H}^{{n}+\mathrm{1}} \\ $$$$\frac{{U}_{{b}} −{U}_{{s}} }{{H}^{{n}+\mathrm{1}} }\:=\:\:−\:\frac{\mathrm{2}}{{n}+\mathrm{1}}\left(\frac{\rho×{g}×{sin}\alpha}{{B}}\right)^{{n}} \\ $$$$−\frac{\left({U}_{{b}} −{U}_{{s}} \right)\left({n}+\mathrm{1}\right)}{\mathrm{2}{H}^{{n}+\mathrm{1}} }\:=\:\left(\frac{\rho×{g}×{sin}\alpha}{{B}}\right)^{{n}} \\ $$$$\left(−\frac{\left({U}_{{b}} −{U}_{{s}} \right)\left({n}+\mathrm{1}\right)}{\mathrm{2}{H}^{{n}+\mathrm{1}} }\right)^{\frac{\mathrm{1}}{{n}}} =\frac{\rho×{g}×{sin}\alpha}{{B}} \\ $$$$\left(−\frac{\left({U}_{{b}} −{U}_{{s}} \right)\left({n}+\mathrm{1}\right)}{\mathrm{2}{H}^{{n}+\mathrm{1}} }\right)^{{n}} \rho×{g}×\mathrm{sin}\left(\alpha\right)={B} \\ $$
Commented by Rasheed Soomro last updated on 22/Oct/16
Didn′t understand last two steps.
$$\mathrm{Didn}'\mathrm{t}\:\mathrm{understand}\:\mathrm{last}\:\mathrm{two}\:\mathrm{steps}. \\ $$
Answered by Rasheed Soomro last updated on 22/Oct/16
U_b = U_s − (2/(n+1))(((ρ×g×sinα)/B)8)^n H^(n+1) (2/(n+1))(((ρ×g×sinα)/B))^n H^(n+1) = U_s −U_b (((ρ×g×sinα)/B))^n = (((U_s −U_b )(n+1))/(2H^(n+1) )) (((ρ×g×sinα)^n )/B^n )= (((U_s −U_b )(n+1))/(2H^(n+1) )) (B^n /((ρ×g×sinα)^n ))=((2H^(n+1) )/((U_s −U_b )(n+1))) B^n =((2H^(n+1) (ρ×g×sinα)^n )/((U_s −U_b )(n+1))) B=((2^(1/n) H^((n+1)/n) (ρ×g×sinα))/((U_s −U_b )^(1/n) (n+1)^(1/n) ))
$${U}_{{b}} \:=\:{U}_{{s}} \:−\:\frac{\mathrm{2}}{{n}+\mathrm{1}}\left(\frac{\rho×{g}×{sin}\alpha}{{B}}\mathrm{8}\right)^{{n}} {H}^{{n}+\mathrm{1}} \\ $$$$\:\frac{\mathrm{2}}{{n}+\mathrm{1}}\left(\frac{\rho×{g}×{sin}\alpha}{{B}}\right)^{{n}} {H}^{{n}+\mathrm{1}} =\:{U}_{{s}} \:−{U}_{{b}} \\ $$$$\:\left(\frac{\rho×{g}×{sin}\alpha}{{B}}\right)^{{n}} =\:\frac{\left({U}_{{s}} \:−{U}_{{b}} \right)\left({n}+\mathrm{1}\right)}{\mathrm{2}{H}^{{n}+\mathrm{1}} } \\ $$$$\:\frac{\left(\rho×{g}×{sin}\alpha\right)^{{n}} }{{B}^{{n}} }=\:\frac{\left({U}_{{s}} \:−{U}_{{b}} \right)\left({n}+\mathrm{1}\right)}{\mathrm{2}{H}^{{n}+\mathrm{1}} } \\ $$$$\frac{{B}^{{n}} }{\left(\rho×{g}×{sin}\alpha\right)^{{n}} }=\frac{\mathrm{2}{H}^{{n}+\mathrm{1}} }{\left({U}_{{s}} \:−{U}_{{b}} \right)\left({n}+\mathrm{1}\right)} \\ $$$${B}^{{n}} =\frac{\mathrm{2}{H}^{{n}+\mathrm{1}} \left(\rho×{g}×{sin}\alpha\right)^{{n}} }{\left({U}_{{s}} \:−{U}_{{b}} \right)\left({n}+\mathrm{1}\right)} \\ $$$${B}=\frac{\mathrm{2}^{\frac{\mathrm{1}}{{n}}} {H}^{\frac{{n}+\mathrm{1}}{{n}}} \left(\rho×{g}×{sin}\alpha\right)}{\left({U}_{{s}} \:−{U}_{{b}} \right)^{\frac{\mathrm{1}}{{n}}} \left({n}+\mathrm{1}\right)^{\frac{\mathrm{1}}{{n}}} } \\ $$$$ \\ $$
Commented by vuckintv last updated on 22/Oct/16
Thank you
$${Thank}\:{you} \\ $$

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