Updated spacing between @ latex expressions

astrosander правка от
правка #10299 предыдущая #10132 GitHub c702ebe ← раньше позже →
@@ -77,7 +77,7 @@
<p>
From the drawing
−$$\tan \alpha = \frac{v_y}{v_x}$$
+$$\tan\alpha = \frac{v_y}{v_x}$$
Law of conservation of energy
$$\frac{mv^2_x}{2}=eU$$
From where
@@ -96,16 +96,16 @@
Find the angle of velocity to the horizontal
$$\tan\theta = \frac{v_y}{v_x}=\frac{eU_0}{m\omega d}\sqrt{\frac{m}{2eU}} \left( 1-\cos\omega l \sqrt{\frac{m}{2eU}}\right)$$
Whence the angle $\angle \theta$
−$$\boxed{\theta=\operatorname{\arctan } \left( \frac{U_0}{\omega d}\sqrt{\frac{e}{2mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] \right)}$$
+$$\boxed{\theta =\operatorname{\arctan } \left( \frac{U_0}{\omega d}\sqrt{\frac{e}{2mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] \right)}$$
A little extra. Given that $d \ll l$ we can use the approximation $\tan x \approx x$
−$${\theta= \frac{U_0}{\omega d}\sqrt{\frac{e}{2mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] }$$
+$${\theta = \frac{U_0}{\omega d}\sqrt{\frac{e}{2mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] }$$
Find the required scattering angle
−$${\Delta \alpha=2\theta= \frac{U_0}{\omega d}\sqrt{\frac{2e}{mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] }$$
+$${\Delta \alpha =2\theta = \frac{U_0}{\omega d}\sqrt{\frac{2e}{mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] }$$
</p>
<h4>Answer</h4>
<p>
− $$\Delta\alpha=\pm\operatorname{\arctan }\bigg\{\frac{V_{0}}{d\omega}\sqrt{\frac{2e}{m_{e}V}}\bigg[1-\operatorname{cos}\bigg(\omega l\sqrt{\frac{m_{e}}{2eV}}\bigg)\bigg]\bigg\}$$
+ $$\Delta\alpha =\pm\operatorname{\arctan }\bigg\{\frac{V_{0}}{d\omega}\sqrt{\frac{2e}{m_{e}V}}\bigg[1-\operatorname{cos}\bigg(\omega l\sqrt{\frac{m_{e}}{2eV}}\bigg)\bigg]\bigg\}$$
</p>
<p style="text-align: right; font-style: italic; font-size: 14;">
Lutfulloyev Shukurullo<br>
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