Updated spacing between @ latex expressions

astrosander edited
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revision #10415 parent #6946 GitHub c702ebe ← older
en/2.2.39.md +5 −5
@@ -84,26 +84,26 @@
$$\Delta L=v\Delta t$$
Given a constant cross-sectional area $S=\pi r^2$, the expression from above can be rewritten as
$$\frac{m}{\rho_0S}-\frac{m}{\rho S}=v\Delta t$$
−$$\frac{m}{\Delta t}=Sv\frac{\rho\rho_0}{\rho-\rho_0}$$
+$$\frac{m}{\Delta t}=Sv\frac{\rho\rho_0}{\rho -\rho_0}$$
Because the mass remains constant ($m=\text{const}$):
$$m=\rho_0SL_0=\rho SL$$
From where
$$\frac{\rho}{\rho_0}=\frac{L_0}{L}$$
Representing the force by a small change in momentum
$$F_\text{average}=\frac{mv_\text{system}}{\Delta t}$$
Substituting the value of $\frac{m}{\Delta t}$:
−$$F_\text{average}=Sv\frac{\rho\rho_0}{\rho-\rho_0}v_\text{system}$$
+$$F_\text{average}=Sv\frac{\rho\rho_0}{\rho -\rho_0}v_\text{system}$$
$$v_\text{system}=\frac{0+v}{2}=\frac{v}{2}$$
$$F_\text{average}=\frac{0+F}{2}=\frac{F}{2}$$
Substitute into the formula for $F_\text{average}$:
−$$\frac{F}{2}=Sv\frac{\rho\rho_0}{\rho-\rho_0}\frac{v}{2}$$
+$$\frac{F}{2}=Sv\frac{\rho\rho_0}{\rho -\rho_0}\frac{v}{2}$$
From where
−$$\boxed{v=\sqrt{\frac{F(\rho-\rho_0)}{\pi r^2\rho\rho_0}}}$$
+$$\boxed{v=\sqrt{\frac{F(\rho -\rho_0)}{\pi r^2\rho\rho_0}}}$$
</p>
<h4>Answer</h4>
<p>
− $$v=\sqrt{F(\rho-\rho_0)/(\pi r^2\rho\rho_0)}$$
+ $$v=\sqrt{F(\rho -\rho_0)/(\pi r^2\rho\rho_0)}$$
</p>
<p style="text-align: right; font-style: italic; font-size: 16;">Almaskhan Arsen</p>
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