Updated spacing between @ latex expressions

astrosander edited
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revision #10406 parent #9721 GitHub c702ebe ← older newer →
en/1.4.9.md +10 −10
@@ -71,7 +71,7 @@
<h3>Solution</h3>
<p>
<p>$a)$ Since the collision is elastic, then according to the Law of Conservation of Momentum:</p>
−$$v \sin \alpha = u \sin \alpha $$
+$$v \sin\alpha = u \sin\alpha$$
$$\fbox{$v = u $}$$
@@ -88,18 +88,18 @@
</figure>
</center>
<p>Let's find the projections of the vector $\vec{u}$ on the horizontal and vertical axes:</p>
−$$u_y = v \sin \alpha $$
+$$u_y = v \sin\alpha$$
−$$u_x = v \cos \alpha + 2w$$
+$$u_x = v \cos\alpha + 2w$$
<p>Using the Pythagorean theorem, we find the modulus of the vector $\vec{u}$</p>
$$u = \sqrt{u_x^2+u_y^2}$$
−$$u = \sqrt{(v \sin \alpha)^2 + (v \cos \alpha + 2w)^2}$$
+$$u = \sqrt{(v \sin\alpha )^2 + (v \cos\alpha + 2w)^2}$$
−$$\fbox{$u=\sqrt{v^{2}+4vw\cos\alpha+4w^{2}}$}$$
+$$\fbox{$u=\sqrt{v^{2}+4vw\cos\alpha +4w^{2}}$}$$
<p>$в)$ Similar to the previous subparagraph</p>
$$\vec{u} = \vec{v} - 2\vec{w}$$
@@ -115,18 +115,18 @@
</figure>
</center>
<p>We will find the projections of the vector $\vec{u}$ on the horizontal and vertical axis:</p>
−$$u_y = v \sin \alpha - 2w \sin \beta$$
− $$u_x = v \cos \alpha + 2w \cos \beta$$
+$$u_y = v \sin\alpha - 2w \sin\beta$$
+ $$u_x = v \cos\alpha + 2w \cos\beta$$
<p>Using the Pythagorean equation, we find the modulus of the vector $\vec{u}$
$$u = \sqrt{u_x^2+u_y^2}$$
− $$u = \sqrt{(v \sin \alpha - 2w \sin \beta)^2 + (v \cos \alpha + 2w \cos \beta)^2}$$
− $$\fbox{$u=\sqrt{v^{2}+4vw\cos\alpha\cos\beta+4w^{2}\cos^{2}\beta}$}$$
+ $$u = \sqrt{(v \sin\alpha - 2w \sin\beta )^2 + (v \cos\alpha + 2w \cos\beta )^2}$$
+ $$\fbox{$u=\sqrt{v^{2}+4vw\cos\alpha\cos\beta +4w^{2}\cos^{2}\beta}$}$$
</p>
<h4>Answer</h4>
<p>
− $\text{a) } u=v.\quad\text{b) } u=\sqrt{v^{2}+4vw\cos\alpha+4w^{2}}.\quad\text{c) } u=\sqrt{v^{2}+4vw\cos\alpha\cos\beta+4w^{2}\cos^{2}\beta}.$
+ $\text{a) } u=v.\quad\text{b) } u=\sqrt{v^{2}+4vw\cos\alpha +4w^{2}}.\quad\text{c) } u=\sqrt{v^{2}+4vw\cos\alpha\cos\beta +4w^{2}\cos^{2}\beta}.$
</p>
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