Edits to “Statement”, “Solution”

crico edited Dec 27, 2024, 8:47 PM

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revision #12596 newer →

en/5.4.17.md +33 −14

		@@ -1,39 +1,58 @@
			### Statement

		−	$5.4.17.$ [Insert problem description here]
		+	$5.4.16*.$ [Insert problem description here]

			__Example Statement__:
		−	$1.1.1.$ Det~~ermine the coordinate $x(t)$ of a body as a function of time $t$, given that its acceleration is defined as $a(t) = bt$~~, where ~~$b$ is a constant.~~
		+	$1.1.1.$ Between two flat parallel plates located at a distance δ from each other, there
		+	is a monatomic gas (the free path of atoms is much longer than δ). Estimate
		+	the heat flux density if the temperature of the plates is maintained at T and
		+	T + ∆T, respectively, and the unit volume of gas contains n' atoms; µ is the
		+	mass of the atom


			### Solution

			[Your solution should be placed here]

			__Example Solution__:
		−	~~The acceleration of the body~~ defined by
		+	Heat flow is defined by

		−	$$~~a(t) = bt~~$$
		+
		+	$$J =\frac{ΔE}{A*t} $$

		−	We know that ~~acceleration is the time derivative of velocity:~~
		+	We know that the change of the energy is given by

		−	$$~~a(t)~~ = \frac{~~d v(t)}{d t~~}$$
		+	$$ΔE = \frac{nRΔT}{\gamma-1}$$

		−	~~To find the velocity $v(t)$, we integrate $a(t)$ with respect to time:~~
		+	Where (n) is the number of moles

		−	$$v(t) = \int a(t) \, dt = \int b t \, dt$$
		+	We know that Heat flow then

		−	If the initial velocity is $v(0) = 0$, then the velocity becomes:
		+	$$J = \frac{nRΔT}{(\gamma-1)A*t}$$

		−	$$v(t) = \frac{b t^2}{2}$$
		+	We know that the time is given by because (the free path of atoms is much longer than δ) furthermore λ>>δ therefore the average collision between the molecules will be δ:

		−	Likewise, integrate $v(t)$ with respect to time:
		+	$$t = \frac{δ}{v}$$

		−	$$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$
		+	Likewise:

		−	From where the coordinate from time, considering the initial conditions:
		+	$$J = \frac{nRΔTv}{(\gamma-1)A*δ}$$

		−	$$~~\boxed{x(t)=\frac{bt^3}{6}~~}$$
		+	Now we have to:
		+
		+	$$A*δ=V$$
		+
		+	So from the previous equations we have to:
		+
		+	$$J = \frac{nRΔTv}{(\gamma-1)V}$$
		+
		+	Another way to write this equation is:
		+
		+	$$J = \frac{n'KΔTv}{(\gamma-1)}$$
		+
		+	Where (n') is the number of the atoms
		+
		+	$$v=\

			#### Answer

			[Insert a concise answer or boxed result, like this:]


			__Example Answer__:
			$$ x(t)=\frac{bt^3}{6} $$
		unchanged lines 5

	@@ -1,39 +1,58 @@
	### Statement		### Statement

	$5.4.17.$ [Insert problem description here]		$5.4.16*.$ [Insert problem description here]

	__Example Statement__:		__Example Statement__:
	$1.1.1.$ Det~~ermine the coordinate $x(t)$ of a body as a function of time $t$, given that its acceleration is defined as $a(t) = bt$~~, where ~~$b$ is a constant.~~		$1.1.1.$ Between two flat parallel plates located at a distance δ from each other, there
			is a monatomic gas (the free path of atoms is much longer than δ). Estimate
			the heat flux density if the temperature of the plates is maintained at T and
			T + ∆T, respectively, and the unit volume of gas contains n' atoms; µ is the
			mass of the atom


	### Solution		### Solution

	[Your solution should be placed here]		[Your solution should be placed here]

	__Example Solution__:		__Example Solution__:
	~~The acceleration of the body~~ defined by		Heat flow is defined by

	$$~~a(t) = bt~~$$
			$$J =\frac{ΔE}{A*t} $$

	We know that ~~acceleration is the time derivative of velocity:~~		We know that the change of the energy is given by

	$$~~a(t)~~ = \frac{~~d v(t)}{d t~~}$$		$$ΔE = \frac{nRΔT}{\gamma-1}$$

	~~To find the velocity $v(t)$, we integrate $a(t)$ with respect to time:~~		Where (n) is the number of moles

	$$v(t) = \int a(t) \, dt = \int b t \, dt$$		We know that Heat flow then

	If the initial velocity is $v(0) = 0$, then the velocity becomes:		$$J = \frac{nRΔT}{(\gamma-1)A*t}$$

	$$v(t) = \frac{b t^2}{2}$$		We know that the time is given by because (the free path of atoms is much longer than δ) furthermore λ>>δ therefore the average collision between the molecules will be δ:

	Likewise, integrate $v(t)$ with respect to time:		$$t = \frac{δ}{v}$$

	$$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$		Likewise:

	From where the coordinate from time, considering the initial conditions:		$$J = \frac{nRΔTv}{(\gamma-1)A*δ}$$

	$$~~\boxed{x(t)=\frac{bt^3}{6}~~}$$		Now we have to:

			$$A*δ=V$$

			So from the previous equations we have to:

			$$J = \frac{nRΔTv}{(\gamma-1)V}$$

			Another way to write this equation is:

			$$J = \frac{n'KΔTv}{(\gamma-1)}$$

			Where (n') is the number of the atoms

			$$v=\

	#### Answer		#### Answer

	[Insert a concise answer or boxed result, like this:]		[Insert a concise answer or boxed result, like this:]


	__Example Answer__:		__Example Answer__:
	$$ x(t)=\frac{bt^3}{6} $$		$$ x(t)=\frac{bt^3}{6} $$
	unchanged lines 5