Correction d'une erreur dans la thermodynamique OS et passage au nouveau package siunitx.
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Guyot Vincent
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@ -93,7 +93,7 @@
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D&=2\cdot R_{Lune}=d\cdot \alpha\;\Rightarrow\;\alpha=\frac{2\cdot R_{Lune}}{d_{Terre-Lune}}\\
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\alpha&=\frac{2\cdot 0,2725\cdot 6,371\cdot 10^6}{3,84\cdot 10^8}=\SI{0,009}{\radian}
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\end{align*}
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Comme \(\SI{180}{\degree}=\SI{\pi}{\radian}\), l'angle considéré est~:
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Comme \(\SI{180}{\degree}=\pi\unit{\radian}\), l'angle considéré est~:
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\[\alpha=0,009\cdot \frac{180}{\pi}=\SI{0,5157}{\degree}\]
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Comme un degré vaut soixante minutes d'arc, \(\SI{1}{\degree}=\SI{60}{\arcminute}\), on a~:
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\[\alpha=\SI{0,5157}{\degree}=0,5157\cdot 60=\SI{31}{\arcminute}\]
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@ -111,7 +111,7 @@ Cette vitesse est incroyable. Nous ne la ressentons à nouveau pas ou peu toujou
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Notons que cette vitesse est la même pour toutes les étoiles proches du Soleil qui participent au mouvement de rotation autour du centre de la galaxie. Mais le Soleil a aussi un mouvement propre, c'est-à-dire qu'une partie de sa vitesse ne correspond pas à sa vitesse de rotation autour du centre de la galaxie. Cette composante vaut environ \SI{20}{\kilo\metre\per\second}.\endnote{Voir le site \url=http://www.dil.univ-mrs.fr/~gispert/enseignement/astronomie/5eme_partie/voieLactee.php=}
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Relevons enfin une règle bien pratique pour la transformation d'unité entre les \si{\metre\per\second} et les \si{\kilo\metre\per\hour}. On a en effet~:
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\[\SI{1}{\kilo\metre\per\hour}=\frac{1\,km}{1\,h}=\frac{1000\,m}{3600\,s}=\SI{1 / 3,6}{\metre\per\second}\]
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\[\SI{1}{\kilo\metre\per\hour}=\frac{1\,km}{1\,h}=\frac{1000\,m}{3600\,s}=1 / 3,6\unit{\metre\per\second}\]
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Ainsi, pour transformer des \si{\kilo\metre\per\hour} en \si{\metre\per\second}, il faut diviser le nombre correspondant aux \si{\kilo\metre\per\hour} par le facteur 3,6. Inversement, pour passer de \si{\metre\per\second} en \si{\kilo\metre\per\hour}, il faut multiplier les \si{\metre\per\second} par 3,6.
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\section{Vitesse et référentiel}
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@ -170,7 +170,7 @@ Les unités\index{unité} de la table \ref{grandeurs} ne font pas partie du Syst
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\medskip{}
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L'unité astronomique correspond à la longueur du demi-grand axe de l'orbite terrestre.
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Le parsec\index{parsec} est la distance à laquelle \SI{1}{\astronomicalunit}\index{UA} est vue sous un angle de \SI{1}{\arcsecond} (une seconde\index{seconde!d'arc}) d'arc. Comme \SI{1}{\degree} est divisé en \SI{60}{\arcminute} (minutes\index{minute d'arc}) d'arc et \SI{1}{\arcminute} d'arc en \SI{60}{\arcsecond}, une seconde d'arc (\SI{1}{\arcsecond}) représente \SI{1/3600}{\degree}. Pour calculer ce que vaut \SI{1}{\parsec}\index{pc}, il faut une relation entre la distance réelle L de \SI{1}{\astronomicalunit} et l'angle \(\alpha\) (\SI{1}{\arcsecond}) sous lequel cette distance est vue. Cette relation est (voir figure \ref{arc}):
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Le parsec\index{parsec} est la distance à laquelle \SI{1}{\astronomicalunit}\index{UA} est vue sous un angle de \SI{1}{\arcsecond} (une seconde\index{seconde!d'arc}) d'arc. Comme \SI{1}{\degree} est divisé en \SI{60}{\arcminute} (minutes\index{minute d'arc}) d'arc et \SI{1}{\arcminute} d'arc en \SI{60}{\arcsecond}, une seconde d'arc (\SI{1}{\arcsecond}) représente 1/3600 \unit{\degree}. Pour calculer ce que vaut \SI{1}{\parsec}\index{pc}, il faut une relation entre la distance réelle L de \SI{1}{\astronomicalunit} et l'angle \(\alpha\) (\SI{1}{\arcsecond}) sous lequel cette distance est vue. Cette relation est (voir figure \ref{arc}):
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\begin{equation}\label{relationdarc}
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L=\alpha\cdot R
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@ -182,7 +182,7 @@ L=\alpha\cdot R
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\includegraphics{Arc.eps}
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\end{figure}
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où \(R\) est le rayon de l'arc de cercle de longueur \(L\) et d'angle au centre \(\alpha\). Mais, attention, \(\alpha\) doit être en radians. Or, comme \SI{180}{\degree}=\SI{\pi}{\radian}, \SI{1}{\degree}=\SI{\pi}{\radian}/\SI{180}{\radian} et \SI{1}{\arcsecond}=\SI{\pi}{\radian}/(\SI[output-product = \cdot]{180 x 3600}{\radian}). Ainsi on obtient la valeur du parsec~:
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où \(R\) est le rayon de l'arc de cercle de longueur \(L\) et d'angle au centre \(\alpha\). Mais, attention, \(\alpha\) doit être en radians. Or, comme \SI{180}{\degree}=\(\pi\)\unit{\radian}, \SI{1}{\degree}=\(\pi\)\unit{\radian}/\SI{180}{\radian} et \SI{1}{\arcsecond}=\(\pi\)\unit{\radian}/(180 x 3600)\unit{\radian}). Ainsi on obtient la valeur du parsec~:
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\begin{align*}
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R&=\frac{L}{\alpha}=\frac{L\cdot 180\cdot 3600}{\pi}\\
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&=\frac{1,496\cdot10^{11}\cdot 180\cdot 3600}{\pi}=\SI{3,0857e16}{\metre}
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@ -35,7 +35,7 @@
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D&=2\cdot R_{Lune}=d\cdot \alpha\;\Rightarrow\;\alpha=\frac{2\cdot R_{Lune}}{d_{Terre-Lune}}\\
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\alpha&=\frac{2\cdot 0,2725\cdot 6,371\cdot 10^6}{3,84\cdot 10^8}=\SI{0,009}{\radian}
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\end{align*}
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Comme \(\SI{180}{\degree}=\SI{\pi}{\radian}\), l'angle considéré est~:
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Comme \(\SI{180}{\degree}=\pi\unit{\radian}\), l'angle considéré est~:
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\[\alpha=0,009\cdot \frac{180}{\pi}=\SI{0,5157}{\degree}\]
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Comme un degré vaut soixante minutes d'arc, \(\SI{1}{\degree}=\SI{60}{\arcminute}\), on a~:
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\[\alpha=\SI{0,5157}{\degree}=0,5157\cdot 60=\SI{31}{\arcminute}\]
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0000001002 00000 n
|
||||
0000000228 00000 n
|
||||
0000000983 00000 n
|
||||
0000001239 00000 n
|
||||
trailer
|
||||
<< /Size 8 /Root 1 0 R /Info 2 0 R
|
||||
/ID [<9A77833D3271E3B40FE2E36EF3428221><9A77833D3271E3B40FE2E36EF3428221>]
|
||||
>>
|
||||
startxref
|
||||
2695
|
||||
%%EOF
|
||||
|
|
@ -164,8 +164,8 @@ On commence par calculer le volume initial~:
|
|||
Puis, on détermine la variation de volume~:
|
||||
|
||||
\begin{align*}
|
||||
\Delta V&=100\cdot 2\cdot 10^{-4}\cdot (25-17)\\
|
||||
&=\SI{0,16}{\metre\cubed}=\SI{160}{\deci\metre\cubed}=\SI{160}{\liter}
|
||||
\Delta V&=100\cdot 3\cdot 2\cdot 10^{-4}\cdot (25-17)\\
|
||||
&=\SI{0,48}{\metre\cubed}=\SI{480}{\deci\metre\cubed}=\SI{480}{\liter}
|
||||
\end{align*}
|
||||
|
||||
\section{Chaleurs spécifique et latente}
|
||||
|
|
|
|||
Loading…
Reference in New Issue