Updated greek laters @ latex compiling

astrosander правка от
Посмотреть решение История
правка #10178 предыдущая #9154 GitHub dbb31ba ← раньше позже →
en/5.6.22.md +11 −11
@@ -67,33 +67,33 @@
<h3>Solution</h3>
<p>
a) From Energy Conservation Law or Kinetic Energy - Work theorem
− $$\frac{Mv^2}{2} = Mg(H-y) + A \;(1)$$
+ $$\frac{Mv^2}{2} = Mg(H-y) + A \quad(1)$$
where $A$ is the work done over the piston, equals to work done by gas, in this case (isothermal process) defined by
− $$A = nRT\ln{\frac{Sy}{SH}} = -nRT\ln{\frac{H}{y}} \;(2)$$
+ $$A = nRT\ln{\frac{Sy}{SH}} = -nRT\ln{\frac{H}{y}} \quad(2)$$
Substituting $(2)$ into $(1)$
− $$\frac{Mv^2}{2} = Mg(H-y) - nRT\ln{\frac{H}{y}} \;(3)$$
+ $$\frac{Mv^2}{2} = Mg(H-y) - nRT\ln{\frac{H}{y}} \quad(3)$$
From Boyle-Mariotte Law
− $$PSH = P'Sy \;(4)$$
+ $$PSH = P'Sy \quad(4)$$
Applying state equation to initial state,
− $$n = \frac{PSH}{RT} \;(5)$$
+ $$n = \frac{PSH}{RT} \quad(5)$$
$$PSH = nRT$$
Applying Newton's Second Law
$$Mg - P(t)S = M\frac{dv}{dt}$$
At the moment of maximum velocity, $dv/dt = 0$ and $P(t) = P'$, hence
− $$Mg = P'S \;(6)$$
+ $$Mg = P'S \quad(6)$$
Putting $(6)$ into $(4)$ and separating $y$
− $$y = \frac{PSH}{Mg} \;(7)$$
+ $$y = \frac{PSH}{Mg} \quad(7)$$
Substituting $(5)$ and $(7)$ into $(3)$ and separating $v$
$$\boxed{v = \sqrt{2gH\left(1-\frac{PS}{Mg}+\frac{PS}{Mg}\ln{\frac{PS}{Mg}}\right)}}$$
b) From Poisson's Law
− $$P(SH)^\gamma = P'(Sy)^\gamma \;(8)$$
+ $$P(SH)^\gamma = P'(Sy)^\gamma \quad(8)$$
The work done by the gas in an adiabatic process is given by
− $$A = \frac{PSH-P'Sy}{\gamma-1} \;(9)$$
+ $$A = \frac{PSH-P'Sy}{\gamma-1} \quad(9)$$
According to $(6)$, we can modify $(8)$ and separating $y$, obtaining
− $$y = H\left(\frac{PS}{Mg}\right)^\frac{1}{\gamma} \;(10)$$
+ $$y = H\left(\frac{PS}{Mg}\right)^\frac{1}{\gamma} \quad(10)$$
Putting $(9)$ into $(1)$, and taking account $(6)$,
<div class="scroll-wrapper">
− $$\frac{Mv^2}{2} = H\left(Mg+\frac{PS}{\gamma-1}\right)-Mgy\left(1+\frac{1}{\gamma-1}\right) \;(11)$$
+ $$\frac{Mv^2}{2} = H\left(Mg+\frac{PS}{\gamma-1}\right)-Mgy\left(1+\frac{1}{\gamma-1}\right) \quad(11)$$
</div>
Substituting $(10)$ into $(11)$, since gas is monoatomic ($i = 3$ and $\gamma = 5/3$), and separating $v$
$$\boxed{v = \sqrt{2gH\left[1-\frac{5}{2}\left(\frac{PS}{Mg}\right)^\frac{3}{5}+\frac{3}{2}\frac{PS}{Mg}\right]}}$$
ещё строк без изменений 22