Updated greek laters @ latex compiling

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правка #10175 предыдущая #9167 GitHub dbb31ba ← раньше
en/5.5.3.md +7 −7
@@ -70,21 +70,21 @@
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Initially, considering a massless piston
− $$P_0 = P_a \;(1)$$
+ $$P_0 = P_a \quad(1)$$
When load of mass $m$ is putted over
− $$PS=mg+P_aS \;(2)$$
+ $$PS=mg+P_aS \quad(2)$$
After the addition of the another load,
− $$P'S=(m+\Delta m)g+P_aS \;(3)$$
+ $$P'S=(m+\Delta m)g+P_aS \quad(3)$$
From Boyle-Mariotte Law
$$P_0V_0 = P\frac{V_0}{n}$$
− $$P_0 = \frac{P}{n} \;(4)$$
+ $$P_0 = \frac{P}{n} \quad(4)$$
Substituting $(2)$ into $(3)$, according to $(1)$ and separating $P_a$
− $$P_a = \frac{mg}{S(n-1)} \;(5)$$
+ $$P_a = \frac{mg}{S(n-1)} \quad(5)$$
Applying Boyle-Mariotte Law again
$$P\frac{V_0}{n}=P'\frac{V_0}{nk}$$
− $$P=\frac{P'}{k} \;(6)$$
+ $$P=\frac{P'}{k} \quad(6)$$
Substituting $(2)$ and $(3)$ into $(6)$ and developing algebraically
− $$\left(\frac{mg}{S}+P_a\right)(k-1)=\frac{\Delta mg}{S} \;(7)$$
+ $$\left(\frac{mg}{S}+P_a\right)(k-1)=\frac{\Delta mg}{S} \quad(7)$$
Putting $(5)$ into $(7)$ and separating $\Delta m$
$$\boxed{\Delta m=m\frac{(k-1)n}{(n-1)}}$$
This problem can also be solved considering the mass of piston.
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