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Solving-for-A-U-z-U-b-2A-h-1-g-sin-n-H-n-1-H-Z-n-1-




Question Number 8704 by vuckintv last updated on 22/Oct/16
Solving for A.  U(z) = U_b +((2A)/(h+1))(ρ×g×sin(α))^n [H^(n+1) −(H−Z)^(n+1) ]
SolvingforA.U(z)=Ub+2Ah+1(ρ×g×sin(α))n[Hn+1(HZ)n+1]
Answered by Rasheed Soomro last updated on 22/Oct/16
U(z) = U_b +((2A)/(h+1))(ρ×g×sin(α))^n [H^(n+1) −(H−Z)^(n+1) ]  ((2A)/(h+1))(ρ×g×sin(α))^n [H^(n+1) −(H−Z)^(n+1) ]=U(z) − U_b   ((2A)/(h+1))=((U(z) − U_b )/((ρ×g×sin(α))^n [H^(n+1) −(H−Z)^(n+1) ]))  A=(((U(z) − U_b )(h+1))/(2(ρ×g×sin(α))^n [H^(n+1) −(H−Z)^(n+1) ]))
U(z)=Ub+2Ah+1(ρ×g×sin(α))n[Hn+1(HZ)n+1]2Ah+1(ρ×g×sin(α))n[Hn+1(HZ)n+1]=U(z)Ub2Ah+1=U(z)Ub(ρ×g×sin(α))n[Hn+1(HZ)n+1]A=(U(z)Ub)(h+1)2(ρ×g×sin(α))n[Hn+1(HZ)n+1]

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