Updated spacing between @ latex expressions

astrosander правка от
Посмотреть решение История
правка #10482 предыдущая #10116 GitHub c702ebe ← раньше
en/2.4.16.md +3 −3
@@ -72,20 +72,20 @@
$$ v = \sqrt{\frac{k}{2m}}\left(l-l_0\right)$$
The greatest compression will be when it has nowhere to compress $(v_y=0)$
Alternative for horizontal projection
−$$v_x=v\cos\alpha;v_x=\text{const}$$
+$$v_x=v\cos\alpha ;v_x=\text{const}$$
After substituting
$$ \boxed{v_x = \sqrt{\frac{k}{2m}}\left(l-l_0\right)\cos\alpha} $$
Conservation of energy with $x_\text{max}$ deformation
$$\frac{k\left(l-l_0\right)^2}{2} = 2\frac{mv_x^2}{2} + \frac{kx_\text{max}^2}{2}$$
From where
$$x_\text{max}^2=\left(l-l_0\right)^2\left(1-\cos^2 \alpha\right)$$
After using pythagorean trigonometric identity:
−$$\boxed{x_\text{max} = \left(l-l_0\right) \sin \alpha}$$
+$$\boxed{x_\text{max} = \left(l-l_0\right) \sin\alpha}$$
</p>
<h4>Answer</h4>
<p>
− $$\begin{aligned}v_x=(l-l_0)\sqrt{k/(2m)} \cos\alpha; x=(l-l_0)\sin\alpha.\end{aligned}$$
+ $$\begin{aligned}v_x=(l-l_0)\sqrt{k/(2m)} \cos\alpha ; x=(l-l_0)\sin\alpha .\end{aligned}$$
</p>
<p style="text-align: right; font-style: italic; font-size: 14;">
Almaskhan Arsen<br>
ещё строк без изменений 14