Новое решение

Arman правка от 22 апр. 2026 г., 07:01

правка #17943 предыдущая #11962 ← раньше позже →

en/4.5.17.md +29 −0

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		+	### Statement
		+
		+	$4.5.17.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	The pressure inside the droplet is the sum of the Laplace pressure and the pressure of the surrounding liquid. Let $x$ be the distance from the liquid surface to the point inside the droplet,then the pressure inside the droplet:
		+
		+	\begin{equation}
		+	P(x)=\rho gx+\frac{2\sigma}{R}
		+	\end{equation}
		+
		+
		+	Now we see that the resulting function is linear,so we get that the pressure reaches the minimum value when $x=h-R$ and maximum value when $x=h+R$,so:
		+
		+
		+	\begin{equation}
		+	P_{max}=\rho g(h+R)+\frac{2\sigma}{R}
		+	\end{equation}
		+
		+	\begin{equation}
		+	P_{min}=\rho g(h-R)+\frac{2\sigma}{R}
		+	\end{equation}
		+
		+
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,29 @@
			### Statement

			$4.5.17.$ [Insert the problem statement]

			### Solution

			The pressure inside the droplet is the sum of the Laplace pressure and the pressure of the surrounding liquid. Let $x$ be the distance from the liquid surface to the point inside the droplet,then the pressure inside the droplet:

			\begin{equation}
			P(x)=\rho gx+\frac{2\sigma}{R}
			\end{equation}


			Now we see that the resulting function is linear,so we get that the pressure reaches the minimum value when $x=h-R$ and maximum value when $x=h+R$,so:


			\begin{equation}
			P_{max}=\rho g(h+R)+\frac{2\sigma}{R}
			\end{equation}

			\begin{equation}
			P_{min}=\rho g(h-R)+\frac{2\sigma}{R}
			\end{equation}



			#### Answer

			[Insert a concise answer or boxed result]