From 412f230a7359e74ab419d8205c1471740bf4b2ca Mon Sep 17 00:00:00 2001 From: guyotv Date: Mon, 26 Sep 2022 19:22:09 +0200 Subject: [PATCH] Correction de quelques petites erreurs dans les exos. --- Annexe-Exercices/Annexe-Exercices.tex | 19 +++++++++--------- Annexe-Exercices/Annexe-Exercices.tex.bak | 16 ++++++++------- Annexe-Incertitudes/Annexe-Incertitudes.tex | 2 +- .../Annexe-Incertitudes.tex.bak | 2 +- CoursMecaniqueOSDF.pdf | Bin 12861768 -> 12861885 bytes SolutionsOS.tex | 19 +++++++++--------- 6 files changed, 31 insertions(+), 27 deletions(-) diff --git a/Annexe-Exercices/Annexe-Exercices.tex b/Annexe-Exercices/Annexe-Exercices.tex index 65af5d4..969a697 100644 --- a/Annexe-Exercices/Annexe-Exercices.tex +++ b/Annexe-Exercices/Annexe-Exercices.tex @@ -2593,7 +2593,7 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r Un gaz parfait comprenant trois moles de particules est à la pression de \SI{4}{\kilo\pascal} et à la température de \SI{20}{\celsius}. Calculez son volume. Puis recalculez-le si on augmente sa température à \SI{30}{\celsius} et diminue sa pression de 15\%. Réponses~: \SI{2,45}{\metre\cubed} et \SI{2,22}{\metre\cubed}. \begin{solos} De la loi des gaz parfaits, on tire~: - \[V=n\cdot R\cdot \frac{T}{p}=3\cdot 8,31\cdot \frac{293,15}{4\cdot 10^3}=\SI{2,45}{\metre\cubed}\] + \[V=n\cdot R\cdot \frac{T}{p}=3\cdot 8,31\cdot \frac{293,15}{4\cdot 10^3}=\SI{1,83}{\metre\cubed}\] Puis, de la même manière~: \[V=3\cdot 8,31\cdot \frac{303,15}{0,85\cdot 4\cdot 10^3}=\SI{2,22}{\metre\cubed}\] \end{solos} @@ -2805,16 +2805,17 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r I(F)&=I(G\cdot M\cdot m/d^2)\\ &=G\cdot M\cdot m/d^2\cdot i(G\cdot M\cdot m/d^2)\\ &=G\cdot \frac{M\cdot m}{d^2}\cdot (i(G)+i(M)+i(m)+2\cdot i(d))\\ - &=G\cdot \frac{M\cdot m}{d^2}\cdot (\frac{I(G)}{G}+\frac{I(M)}{M}+\frac{I(m)}{m}+2\cdot \frac{I(d)}{d}) + &=G\cdot \frac{M\cdot m}{d^2}\cdot (\frac{I(G)}{G}+\frac{I(M)}{M}\,+\\ + &\;\;\;\;\frac{I(m)}{m}+2\cdot \frac{I(d)}{d}) \end{align*} Pour la position d'un MRUA, on a~: \begin{align*} I(x)&=I(1/2\cdot a\cdot t^2 +v_0\cdot t+ x_0)\\ &=I(1/2\cdot a\cdot t^2)+I(v_0\cdot t)+I(x_0)\\ - &=\frac{1}{2}\cdot a\cdot t^2\cdot (i(1/2)+i(a)+2\cdot i(t))\\ - &+v_0\cdot t\cdot (i(v_0)+i(t))+I(x_0)\\ - &=\frac{1}{2}\cdot a\cdot t^2\cdot (0+\frac{I(a)}{a}+2\cdot \frac{I(t)}{t})\\ - &+v_0\cdot t\cdot (\frac{I(v_0}{v_0}+\frac{I(t)}{t})+I(x_0)) + &=\frac{1}{2}\cdot a\cdot t^2\cdot (i(1/2)+i(a)+2\cdot i(t))\,+\\ + &\;\;\;\;v_0\cdot t\cdot (i(v_0)+i(t))+I(x_0)\\ + &=\frac{1}{2}\cdot a\cdot t^2\cdot (0+\frac{I(a)}{a}+2\cdot \frac{I(t)}{t})\,+\\ + &\;\;\;\;v_0\cdot t\cdot (\frac{I(v_0}{v_0}+\frac{I(t)}{t})+I(x_0) \end{align*} Pour la vitesse, on a~: \begin{align*} @@ -2823,14 +2824,14 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r &=\frac{x-x_0}{t}\cdot (\frac{I(x)+I(x_0)}{x-x_0}+\frac{I(t)}{t})\\ &=\frac{I(x)+I(x_0)}{t}+(x-x_0)\cdot \frac{I(t)}{t^2} \end{align*} - Dans le dernier cas, on peut aussi procéder différemment~: + On aurait pu s'arrêter à l'avant dernière égalité, mais on est allé plus loin, car dans le dernier cas, on peut aussi procéder différemment~: \begin{align*} I(v)&=I((x-x_0)/t)=I(x/t-x_0/t)\\ &=I(x/t)+I(x_0/t)\\ &=\frac{x}{t}\cdot i(x/t)+\frac{x_0}{t}\cdot i(x_0/t)\\ &=\frac{x}{t}\cdot (i(x)+i(t))+\frac{x_0}{t}\cdot (i(x_0)+i(t))\\ - &=\frac{x}{t}\cdot (\frac{I(x)}{x}+\frac{I(t)}{t})\\ - &+\frac{x_0}{t}\cdot (\frac{I(x_0)}{x_0}+\frac{I(t)}{t})\\ + &=\frac{x}{t}\cdot (\frac{I(x)}{x}+\frac{I(t)}{t})\,+\\ + &\;\;\;\;\frac{x_0}{t}\cdot (\frac{I(x_0)}{x_0}+\frac{I(t)}{t})\\ &=\frac{I(x)}{t}+\frac{x\cdot I(t)}{t^2}+\frac{I(x_0)}{t}+\frac{x_0\cdot I(t)}{t^2}\\ &=\frac{I(x)+I(x_0)}{t}+(x+x_0)\cdot \frac{I(t)}{t^2} \end{align*} diff --git a/Annexe-Exercices/Annexe-Exercices.tex.bak b/Annexe-Exercices/Annexe-Exercices.tex.bak index ed5f4a2..f55d536 100644 --- a/Annexe-Exercices/Annexe-Exercices.tex.bak +++ b/Annexe-Exercices/Annexe-Exercices.tex.bak @@ -2593,7 +2593,7 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r Un gaz parfait comprenant trois moles de particules est à la pression de \SI{4}{\kilo\pascal} et à la température de \SI{20}{\celsius}. Calculez son volume. Puis recalculez-le si on augmente sa température à \SI{30}{\celsius} et diminue sa pression de 15\%. Réponses~: \SI{2,45}{\metre\cubed} et \SI{2,22}{\metre\cubed}. \begin{solos} De la loi des gaz parfaits, on tire~: - \[V=n\cdot R\cdot \frac{T}{p}=3\cdot 8,31\cdot \frac{293,15}{4\cdot 10^3}=\SI{2,45}{\metre\cubed}\] + \[V=n\cdot R\cdot \frac{T}{p}=3\cdot 8,31\cdot \frac{293,15}{4\cdot 10^3}=\SI{1,83}{\metre\cubed}\] Puis, de la même manière~: \[V=3\cdot 8,31\cdot \frac{303,15}{0,85\cdot 4\cdot 10^3}=\SI{2,22}{\metre\cubed}\] \end{solos} @@ -2765,7 +2765,7 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r \smallskip L'équation pour la transformation à pression constante et correcte puisqu'elle dit~: \[\frac{2\cdot V_1}{T_2}=\frac{V_1}{T_0}\;\Rightarrow\;\frac{2\cdot V_1}{586,3}=\frac{V_1}{293,15}\] - Mais ne permet de calculer aucun des volumes initial et final. + mais ne permet de calculer aucun des volumes initial et final. \end{solos} \end{exos} @@ -2805,16 +2805,17 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r I(F)&=I(G\cdot M\cdot m/d^2)\\ &=G\cdot M\cdot m/d^2\cdot i(G\cdot M\cdot m/d^2)\\ &=G\cdot \frac{M\cdot m}{d^2}\cdot (i(G)+i(M)+i(m)+2\cdot i(d))\\ - &=G\cdot \frac{M\cdot m}{d^2}\cdot (\frac{I(G)}{G}+\frac{I(M)}{M}+\frac{I(m)}{m}+2\cdot \frac{I(d)}{d}) + &=G\cdot \frac{M\cdot m}{d^2}\cdot (\frac{I(G)}{G}+\frac{I(M)}{M}\,+\\ + &\;\;\;\;\frac{I(m)}{m}+2\cdot \frac{I(d)}{d}) \end{align*} Pour la position d'un MRUA, on a~: \begin{align*} I(x)&=I(1/2\cdot a\cdot t^2 +v_0\cdot t+ x_0)\\ &=I(1/2\cdot a\cdot t^2)+I(v_0\cdot t)+I(x_0)\\ - &=\frac{1}{2}\cdot a\cdot t^2\cdot (i(1/2)+i(a)+2\cdot i(t))\\ - &+v_0\cdot t\cdot (i(v_0)+i(t))+I(x_0)\\ - &=\frac{1}{2}\cdot a\cdot t^2\cdot (0+\frac{I(a)}{a}+2\cdot \frac{I(t)}{t})\\ - &+v_0\cdot t\cdot (\frac{I(v_0}{v_0}+\frac{I(t)}{t})+I(x_0)) + &=\frac{1}{2}\cdot a\cdot t^2\cdot (i(1/2)+i(a)+2\cdot i(t))\,+\\ + &\;\;\;\;v_0\cdot t\cdot (i(v_0)+i(t))+I(x_0)\\ + &=\frac{1}{2}\cdot a\cdot t^2\cdot (0+\frac{I(a)}{a}+2\cdot \frac{I(t)}{t})\,+\\ + &\;\;\;\;v_0\cdot t\cdot (\frac{I(v_0}{v_0}+\frac{I(t)}{t})+I(x_0) \end{align*} Pour la vitesse, on a~: \begin{align*} @@ -2823,6 +2824,7 @@ Les valeurs des c\oe fficients de dilatation thermique sont celles du tableau \r &=\frac{x-x_0}{t}\cdot (\frac{I(x)+I(x_0)}{x-x_0}+\frac{I(t)}{t})\\ &=\frac{I(x)+I(x_0)}{t}+(x-x_0)\cdot \frac{I(t)}{t^2} \end{align*} + On aurait pu s'arrêter à l'avant dernière égalité, mais on est allé plus loin, car Dans le dernier cas, on peut aussi procéder différemment~: \begin{align*} I(v)&=I((x-x_0)/t)=I(x/t-x_0/t)\\ diff --git a/Annexe-Incertitudes/Annexe-Incertitudes.tex b/Annexe-Incertitudes/Annexe-Incertitudes.tex index a0de073..79c021b 100644 --- a/Annexe-Incertitudes/Annexe-Incertitudes.tex +++ b/Annexe-Incertitudes/Annexe-Incertitudes.tex @@ -361,7 +361,7 @@ E_{pot}&=m\cdot g\cdot h\\ I(E_{pot})&=E_{pot}\cdot i(E_{pot})\\ &=m\cdot g\cdot h\cdot i(m\cdot g\cdot h)\\ &=m\cdot g\cdot h\cdot (i(m)+i(g)+i(h))\\ -&=m\cdot g\cdot h\cdot (\frac{i(m)}{m}+\frac{i(g)}{g}+\frac{i(h)}{h})\\ +&=m\cdot g\cdot h\cdot (\frac{I(m)}{m}+\frac{I(g)}{g}+\frac{I(h)}{h})\\ \end{align*} C'est l'exemple le plus simple mettant en jeu des incertitudes relatives. \item Pour obtenir l'incertitude sur l'énergie cinétique\index{incertitude@incertitude!énergie cinétique}~: diff --git a/Annexe-Incertitudes/Annexe-Incertitudes.tex.bak b/Annexe-Incertitudes/Annexe-Incertitudes.tex.bak index a0de073..b4269b7 100644 --- a/Annexe-Incertitudes/Annexe-Incertitudes.tex.bak +++ b/Annexe-Incertitudes/Annexe-Incertitudes.tex.bak @@ -361,7 +361,7 @@ E_{pot}&=m\cdot g\cdot h\\ I(E_{pot})&=E_{pot}\cdot i(E_{pot})\\ &=m\cdot g\cdot h\cdot i(m\cdot g\cdot h)\\ &=m\cdot g\cdot h\cdot (i(m)+i(g)+i(h))\\ -&=m\cdot g\cdot h\cdot (\frac{i(m)}{m}+\frac{i(g)}{g}+\frac{i(h)}{h})\\ +&=m\cdot g\cdot h\cdot (\frac{I(m)}{m}+\frac{i(g)}{g}+\frac{i(h)}{h})\\ \end{align*} C'est l'exemple le plus simple mettant en jeu des incertitudes relatives. \item Pour obtenir 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\[V=3\cdot 8,31\cdot \frac{303,15}{0,85\cdot 4\cdot 10^3}=\SI{2,22}{\metre\cubed}\] @@ -1069,16 +1069,17 @@ I(F)&=I(G\cdot M\cdot m/d^2)\\ &=G\cdot M\cdot m/d^2\cdot i(G\cdot M\cdot m/d^2)\\ &=G\cdot \frac{M\cdot m}{d^2}\cdot (i(G)+i(M)+i(m)+2\cdot i(d))\\ - &=G\cdot \frac{M\cdot m}{d^2}\cdot (\frac{I(G)}{G}+\frac{I(M)}{M}+\frac{I(m)}{m}+2\cdot \frac{I(d)}{d}) + &=G\cdot \frac{M\cdot m}{d^2}\cdot (\frac{I(G)}{G}+\frac{I(M)}{M}\,+\\ + &\;\;\;\;\frac{I(m)}{m}+2\cdot \frac{I(d)}{d}) \end{align*} Pour la position d'un MRUA, on a~: \begin{align*} I(x)&=I(1/2\cdot a\cdot t^2 +v_0\cdot t+ x_0)\\ &=I(1/2\cdot a\cdot t^2)+I(v_0\cdot t)+I(x_0)\\ - &=\frac{1}{2}\cdot a\cdot t^2\cdot (i(1/2)+i(a)+2\cdot i(t))\\ - &+v_0\cdot t\cdot (i(v_0)+i(t))+I(x_0)\\ - &=\frac{1}{2}\cdot a\cdot t^2\cdot (0+\frac{I(a)}{a}+2\cdot \frac{I(t)}{t})\\ - &+v_0\cdot t\cdot (\frac{I(v_0}{v_0}+\frac{I(t)}{t})+I(x_0)) + &=\frac{1}{2}\cdot a\cdot t^2\cdot (i(1/2)+i(a)+2\cdot i(t))\,+\\ + &\;\;\;\;v_0\cdot t\cdot (i(v_0)+i(t))+I(x_0)\\ + &=\frac{1}{2}\cdot a\cdot t^2\cdot (0+\frac{I(a)}{a}+2\cdot \frac{I(t)}{t})\,+\\ + &\;\;\;\;v_0\cdot t\cdot (\frac{I(v_0}{v_0}+\frac{I(t)}{t})+I(x_0) \end{align*} Pour la vitesse, on a~: \begin{align*} @@ -1087,14 +1088,14 @@ &=\frac{x-x_0}{t}\cdot (\frac{I(x)+I(x_0)}{x-x_0}+\frac{I(t)}{t})\\ &=\frac{I(x)+I(x_0)}{t}+(x-x_0)\cdot \frac{I(t)}{t^2} \end{align*} - Dans le dernier cas, on peut aussi procéder différemment~: + On aurait pu s'arrêter à l'avant dernière égalité, mais on est allé plus loin, car dans le dernier cas, on peut aussi procéder différemment~: \begin{align*} I(v)&=I((x-x_0)/t)=I(x/t-x_0/t)\\ &=I(x/t)+I(x_0/t)\\ &=\frac{x}{t}\cdot i(x/t)+\frac{x_0}{t}\cdot i(x_0/t)\\ &=\frac{x}{t}\cdot (i(x)+i(t))+\frac{x_0}{t}\cdot (i(x_0)+i(t))\\ - &=\frac{x}{t}\cdot (\frac{I(x)}{x}+\frac{I(t)}{t})\\ - &+\frac{x_0}{t}\cdot (\frac{I(x_0)}{x_0}+\frac{I(t)}{t})\\ + &=\frac{x}{t}\cdot (\frac{I(x)}{x}+\frac{I(t)}{t})\,+\\ + &\;\;\;\;\frac{x_0}{t}\cdot (\frac{I(x_0)}{x_0}+\frac{I(t)}{t})\\ &=\frac{I(x)}{t}+\frac{x\cdot I(t)}{t^2}+\frac{I(x_0)}{t}+\frac{x_0\cdot I(t)}{t^2}\\ &=\frac{I(x)+I(x_0)}{t}+(x+x_0)\cdot \frac{I(t)}{t^2} \end{align*}