Updated spacing between @ latex expressions

astrosander edited
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revision #10242 parent #9807 GitHub c702ebe ← older
en/2.2.44.md +6 −6
@@ -102,8 +102,8 @@
For the thread to be stretched at a constant speed $\vec{v}$, there must be a force on the first half that is modulo the frictional force $\vec{F}$:
−$$F'^2=F^2+F^2-2F^2\cos(\pi-\alpha)$$
−$$F'=F\sqrt{2(1+\cos\alpha)}$$
+$$F'^2=F^2+F^2-2F^2\cos(\pi -\alpha )$$
+$$F'=F\sqrt{2(1+\cos\alpha )}$$
$$F'=2F\cos\frac{\alpha}{2}$$
@@ -121,8 +121,8 @@
Similiarly,
−$$p'^2=p^2+p^2-2p^2\cos(\pi-\alpha)$$
−$$p'=p\sqrt{2(1+\cos\alpha)}$$
+$$p'^2=p^2+p^2-2p^2\cos(\pi -\alpha )$$
+$$p'=p\sqrt{2(1+\cos\alpha )}$$
$$p'=2p\cos\frac{\alpha}{2}$$
By definition of momentum
@@ -142,7 +142,7 @@
$$N=2\cos\frac{\alpha}{2}(\frac{-\Delta m v}{\Delta t}+F)$$
By definition of density
−$$\Delta m = \rho \Delta x$$
+$$\Delta m = \rho\Delta x$$
By definition of speed
$$ v = \frac{\Delta x}{\Delta t}$$
$$ N = 2F\cos\frac{\alpha}{2} - 2\rho v^2\cos\frac{\alpha}{2}$$
@@ -157,7 +157,7 @@
<h4>Answer</h4>
<p>
− $$N=2(F-\rho v^2)\cos\frac\alpha2;\text{ при }v\geqslant\sqrt{F/\rho}.$$
+ $$N=2(F-\rho v^2)\cos\frac\alpha 2;\text{ при }v\geqslant\sqrt{F/\rho}.$$
</p>
<p style="text-align: right; font-style: italic; font-size: 14;">
Almaskhan Arsen<br>
unchanged lines 15