Updated spacing between @ latex expressions

astrosander правка от
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правка #10565 предыдущая #10564 GitHub c702ebe ← раньше позже →
en/7.1.22.md +6 −6
@@ -92,7 +92,7 @@
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− $mg\sin{\theta}-\frac{kqQ}{\left[h+l(1-\cos\theta)\right]^2+l^2\sin^2 {\theta}}\sin{(\beta+\theta)} $$= m\frac{dv}{dt} \; (1)$
+ $mg\sin{\theta}-\frac{kqQ}{\left[h+l(1-\cos\theta )\right]^2+l^2\sin^2 {\theta}}\sin{(\beta +\theta )} $$= m\frac{dv}{dt} \; (1)$
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@@ -108,15 +108,15 @@
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− $\sin{(\beta+\theta)} = \sin{\beta}\cos{\theta}+\cos{\beta}\sin{\theta}$
+ $\sin{(\beta +\theta )} = \sin{\beta}\cos{\theta}+\cos{\beta}\sin{\theta}$
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as $\theta$ is small (small oscillations), $\sin{\theta}\approx\theta$ and $\cos{\theta}\approx1-\theta^2/2$, then
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− $\sin{(\beta+\theta)} = \sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta} \label{E3} \; (3)$
+ $\sin{(\beta +\theta )} = \sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta} \label{E3} \; (3)$
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@@ -132,15 +132,15 @@
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− $mg\theta-\frac{kqQ}{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}\left[\sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta}\right] = m\frac{dv}{dt} \label{NSLmod}$
+ $mg\theta -\frac{kqQ}{\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2}\left[\sin{\beta}\left(1-\frac{\theta^2}{2}\right)+\theta\cos{\beta}\right] = m\frac{dv}{dt} \label{NSLmod}$
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Since angular acceleration is related to lineal acceleration according $dv/dt = \ddot{\theta}l$, dividing all equation by $ml$ and take out $\cos{\beta}$ from second term of left side, it is obtained
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− $\frac{g}{l}\theta-\frac{kqQ\cos{\beta}}{ml\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]}\left[\left(1-\frac{\theta^2}{2}\right)\tan{\beta}+\theta\right] = \ddot{\theta} \label{NEq} \; (5)$
+ $\frac{g}{l}\theta -\frac{kqQ\cos{\beta}}{ml\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]}\left[\left(1-\frac{\theta^2}{2}\right)\tan{\beta}+\theta\right] = \ddot{\theta} \label{NEq} \; (5)$
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@@ -156,7 +156,7 @@
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− $\frac{g}{l}\theta-\frac{kqQ}{ml}\frac{h}{\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]^{\frac{3}{2}}}\left[\left(1+\frac{l}{h+\frac{l\theta^2}{2}}\right)\theta-\frac{l\theta^3}{2h+l\theta^2}\right] = \ddot{\theta}$
+ $\frac{g}{l}\theta -\frac{kqQ}{ml}\frac{h}{\left[\left(h+\frac{l\theta^2}{2}\right)^2+l^2\theta^2\right]^{\frac{3}{2}}}\left[\left(1+\frac{l}{h+\frac{l\theta^2}{2}}\right)\theta -\frac{l\theta^3}{2h+l\theta^2}\right] = \ddot{\theta}$
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