Updated spacing between @ latex expressions

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правка #10644 предыдущая #10147 GitHub c702ebe ← раньше позже →
en/3.1.1.md +3 −3
@@ -67,7 +67,7 @@
<h3>Solution</h3>
<p>
In addition to the force of gravity $mg$, the ball is subject to the force of tension of the string $F$, which has a vertical component $F'$:
−$$ F'=-2F\cos\alpha=-2F\frac{x}{\sqrt{l^2+x^2}}\quad(1) $$
+$$ F'=-2F\cos\alpha =-2F\frac{x}{\sqrt{l^2+x^2}}\quad(1) $$
Since the oscillations are small, we take into account that $x\ll l$
$$ \frac{x}{\sqrt{l^2+x^2}} = \frac{1}{\sqrt{1+\frac{l^2}{x^2}}} $$
Т.к. $x\ll l$, то $\frac{l^2}{x^2} \gg 1$:
@@ -79,13 +79,13 @@
Next, we integrate $dU$ from $0$ to $U$
$$ U=\int_{0}^{U}dU=\int_{0}^{x}2F\frac{x}{l}dx=\boxed{\frac{Fx^2}{l}} $$
From the law of conservation of energy, we obtain the desired speed at maximum displacement
−$$ \frac{Fx_0^2}{l}=\frac{m\upsilon^2}{2}\Rightarrow \boxed{\upsilon=x_0\sqrt{\frac{2F}{ml}}} $$
+$$ \frac{Fx_0^2}{l}=\frac{m\upsilon^2}{2}\Rightarrow \boxed{\upsilon =x_0\sqrt{\frac{2F}{ml}}} $$
</p>
<h4>Answer</h4>
<p>
− $$\upsilon=x_0\sqrt{\frac{2F}{ml}}$$
+ $$\upsilon =x_0\sqrt{\frac{2F}{ml}}$$
</p>
<p style="text-align: right; font-style: italic; font-size: 14;">
Dzikan Mikita<br>
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