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| rev | Admin | (Created page with "'''Answer: C''' <math display="block"> \begin{aligned} \ddot{a}_{35: 30}^{(2)} & \approx \ddot{a}_{35: \overline{30}}-\frac{(m-1)}{2 m}\left(1-v^{30}{ }_{30} p_{35}\right) \\ \ddot{a}_{35: \overline{30}} & =\frac{1-A_{35: 30}}{d}=\frac{1-A_{35: 30}^{1}-{ }_{30} E_{35}}{d} \\ & =\frac{1-\left(A_{35}-{ }_{30} E_{35} \times A_{65}\right)-{ }_{30} E_{35}}{d} \end{aligned} </math> Since <math>{ }_{30} E_{35}=v^{30}{ }_{30} p_{35}=0.2722</math>, then <math display="block...") | Jan 19'24 at 0:14 | +868 |