Which of the following is the correct performance equation?

Question

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

$\big[\frac{\lambda}{u}+τ\big]+\frac{1}{2}\big[u^2+\frac{1}{u^2}\big]$=Emax$\big\{sin\gamma_2-\frac{V}{g}\big\}$

TuteeHUB · Answer

$\big[\frac{\lambda}{u}+τ\big]-\frac{1}{2}\big[u^2+\frac{1}{u^2}\big]$=Emax$\big\{sin\gamma_2-\frac{V}{g}\big\}$

TuteeHUB · Answer

$\big[\frac{\lambda}{u}+τ\big]+\frac{1}{2}\big[u^2+\frac{1}{u^2}\big]$=Emax$\big\{sin\gamma_2+\frac{V}{g}\big\}$

TuteeHUB · Answer

$\big[\frac{\lambda}{u}+τ\big]-\frac{1}{2}\big[u^2+\frac{1}{u^2}\big]$=Emax$\big\{sin\gamma_2+\frac{V}{g}\big\}$

1.	Which of the following is the correct performance equation?
A.	\(\big[\frac{\lambda}{u}+τ\big]+\frac{1}{2}\big[u^2+\frac{1}{u^2}\big]\)=Emax\(\big\{sin\gamma_2-\frac{V}{g}\big\}\)
B.	\(\big[\frac{\lambda}{u}+τ\big]-\frac{1}{2}\big[u^2+\frac{1}{u^2}\big]\)=Emax\(\big\{sin\gamma_2-\frac{V}{g}\big\}\)
C.	\(\big[\frac{\lambda}{u}+τ\big]+\frac{1}{2}\big[u^2+\frac{1}{u^2}\big]\)=Emax\(\big\{sin\gamma_2+\frac{V}{g}\big\}\)
D.	\(\big[\frac{\lambda}{u}+τ\big]-\frac{1}{2}\big[u^2+\frac{1}{u^2}\big]\)=Emax\(\big\{sin\gamma_2+\frac{V}{g}\big\}\)
Answer» E.

Discussion