Updated greek laters @ latex compiling

astrosander правка от
Посмотреть решение История
правка #10071 предыдущая #9373 GitHub dbb31ba ← раньше позже →
en/2.1.38.md +7 −7
@@ -71,15 +71,15 @@
<p>
Let's write the equation after a long period of time:
−$$ mg=γrv\;(1) $$
+$$ mg=\gamma rv\quad(1) $$
Let's find $m$ through the volume $V$:
−$$ m=ρV=\frac{4}{3} ρ \pi r^3 $$
+$$ m=\rho V=\frac{4}{3} \rho \pi r^3 $$
And we substitute into $(1)$:
−$$ \frac{4}{3} ρ \pi r^3 g=γrv $$
+$$ \frac{4}{3} \rho \pi r^3 g=\gamma rv $$
From here:
−$$ v = \frac{4}{3} \frac{ρ \pi g}{γ} \cdot r^2 =\alpha r^2\;(2) $$
+$$ v = \frac{4}{3} \frac{\rho \pi g}{\gamma } \cdot r^2 =\alpha r^2\quad(2) $$
−$$ \alpha = \frac{4}{3} \frac{ρ \pi g}{γ} =\frac{v}{r^2}=10^8 \,\frac{1}{\text{m}\cdot\text{s}} $$
+$$ \alpha = \frac{4}{3} \frac{\rho \pi g}{\gamma } =\frac{v}{r^2}=10^8 \,\frac{1}{\text{m}\cdot\text{s}} $$
We substitute and find the answer
$$ v(\frac{r}{2}) = \alpha \frac{r^2}{4}=0.25~\text{m/s} $$
@@ -89,8 +89,8 @@
<h4>Answer</h4>
<p>
− $$v_1 ≈ 0.25 ~\text{m/s}$$
− $$v_2 ≈ 0.01 ~\text{m/s}$$
+ $$v_1 \approx 0.25 ~\text{m/s}$$
+ $$v_2 \approx 0.01 ~\text{m/s}$$
</p>
ещё строк без изменений 12