Changement de notation du module Siunitx avec qty et ajout de hyperref.
This commit is contained in:
Guyot Vincent
parent
e6602eadda
commit
1f7ac2d6cd
@ -21,9 +21,6 @@
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\global\let\hyper@last\relax
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\gdef\HyperFirstAtBeginDocument#1{#1}
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\providecommand*\HyPL@Entry[1]{}
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\@writefile{toc}{\boolfalse {citerequest}\boolfalse {citetracker}\boolfalse {pagetracker}\boolfalse {backtracker}\relax }
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\@writefile{lof}{\boolfalse {citerequest}\boolfalse {citetracker}\boolfalse {pagetracker}\boolfalse {backtracker}\relax }
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\@writefile{lot}{\boolfalse {citerequest}\boolfalse {citetracker}\boolfalse {pagetracker}\boolfalse {backtracker}\relax }
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\abx@aux@refcontext{nty/global//global/global}
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\HyPL@Entry{0<</S/D>>}
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\babel@aux{french}{}
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@ -38,9 +35,9 @@
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\@writefile{lof}{\addvspace {10\p@ }}
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@ -79,7 +76,7 @@
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\newlabel{LaFoudre}{{2.2.8}{21}{La foudre\index {foudre}}{subsection.2.2.8}{}}
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@ -88,9 +85,9 @@
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\newlabel{precurseur}{{2.8}{22}{Le précurseur}{figure.caption.23}{}}
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\newlabel{effetpointe}{{2.2.8}{22}{La foudre\index {foudre}}{figure.caption.23}{}}
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@ -100,13 +97,13 @@
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@ -125,7 +122,7 @@
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\newlabel{sub@casa}{{(a)}{27}{Subfigure 2 2.14(a)\relax }{subfigure.2.14.1}{}}
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@ -164,7 +161,7 @@
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\newlabel{tensiongeneral}{{2.19}{31}{tension charge ponctuelle}{figure.caption.36}{}}
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@ -180,15 +177,15 @@
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\newlabel{résistance}{{3.2}{35}{Résistance\index {resistance@résistance} et loi d'Ohm\index {loi!d'Ohm}\index {loi!d'Ohm}}{section.3.2}{}}
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@ -292,11 +289,11 @@
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\newlabel{aimantU}{{4.7}{54}{Aimant\index {aimant} en U\relax }{figure.caption.80}{}}
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@ -305,11 +302,11 @@
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@ -324,7 +321,7 @@
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@ -367,12 +364,12 @@
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@ -409,36 +406,36 @@
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@ -1265,14 +1265,14 @@
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@ -1282,14 +1282,14 @@
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<bcf:field fieldtype="field" datatype="datepart">origendyeardivision</bcf:field>
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@ -1299,14 +1299,14 @@
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@ -1339,10 +1339,10 @@
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<bcf:field>endhour</bcf:field>
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@ -1381,17 +1381,17 @@
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@ -1401,7 +1401,6 @@
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@ -1443,6 +1442,7 @@
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@ -1929,17 +1929,17 @@
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@ -1980,17 +1980,17 @@
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@ -2062,17 +2062,17 @@
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@ -2395,6 +2395,29 @@
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<bcf:citekey order="49">CB82</bcf:citekey>
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@ -1441,7 +1441,7 @@
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@ -1,5 +1,4 @@
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@ -87,7 +86,7 @@
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\contentsline {figure}{\numberline {E.2}{\ignorespaces Champ à \qty {33}{\centi \metre }\relax }}{103}{figure.caption.122}%
|
||||
\contentsline {figure}{\numberline {E.3}{\ignorespaces Force et champ\relax }}{103}{figure.caption.123}%
|
||||
\contentsline {figure}{\numberline {E.4}{\ignorespaces Champ entre deux charges\relax }}{103}{figure.caption.124}%
|
||||
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|
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|
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@ -1,123 +1,123 @@
|
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\BOOKMARK [0][-]{chapter.1}{Introduction}{}% 1
|
||||
\BOOKMARK [1][-]{section.1.1}{Introduction}{chapter.1}% 2
|
||||
\BOOKMARK [1][-]{section.1.2}{Rappels}{chapter.1}% 3
|
||||
\BOOKMARK [2][-]{subsection.1.2.1}{Les lois de Newton}{section.1.2}% 4
|
||||
\BOOKMARK [3][-]{section*.4}{Premi\350re loi \(loi d'inertie\)}{subsection.1.2.1}% 5
|
||||
\BOOKMARK [3][-]{section*.5}{Deuxi\350me loi \(principe fondamental\)}{subsection.1.2.1}% 6
|
||||
\BOOKMARK [3][-]{section*.6}{Troisi\350me loi \(action et r\351action\)}{subsection.1.2.1}% 7
|
||||
\BOOKMARK [2][-]{subsection.1.2.2}{La loi de la gravitation universelle}{section.1.2}% 8
|
||||
\BOOKMARK [2][-]{subsection.1.2.3}{Conclusion}{section.1.2}% 9
|
||||
\BOOKMARK [0][-]{chapter.2}{\311lectrostatique }{}% 10
|
||||
\BOOKMARK [1][-]{section.2.1}{Introduction}{chapter.2}% 11
|
||||
\BOOKMARK [1][-]{section.2.2}{La loi de Coulomb}{chapter.2}% 12
|
||||
\BOOKMARK [2][-]{subsection.2.2.1}{La charge \351lectrique}{section.2.2}% 13
|
||||
\BOOKMARK [2][-]{subsection.2.2.2}{Le pendule \351lectrique}{section.2.2}% 14
|
||||
\BOOKMARK [3][-]{section*.8}{Description de l'exp\351rience}{subsection.2.2.2}% 15
|
||||
\BOOKMARK [3][-]{section*.9}{Explications}{subsection.2.2.2}% 16
|
||||
\BOOKMARK [2][-]{subsection.2.2.3}{Les isolants}{section.2.2}% 17
|
||||
\BOOKMARK [2][-]{subsection.2.2.4}{L'\351lectroscope}{section.2.2}% 18
|
||||
\BOOKMARK [3][-]{section*.10}{Description de l'exp\351rience}{subsection.2.2.4}% 19
|
||||
\BOOKMARK [3][-]{section*.12}{Explications}{subsection.2.2.4}% 20
|
||||
\BOOKMARK [2][-]{subsection.2.2.5}{Charge par influence}{section.2.2}% 21
|
||||
\BOOKMARK [3][-]{section*.13}{Description de l'exp\351rience}{subsection.2.2.5}% 22
|
||||
\BOOKMARK [3][-]{section*.15}{Explications}{subsection.2.2.5}% 23
|
||||
\BOOKMARK [2][-]{subsection.2.2.6}{La machine de Van de Graaff}{section.2.2}% 24
|
||||
\BOOKMARK [3][-]{section*.16}{Description de l'exp\351rience}{subsection.2.2.6}% 25
|
||||
\BOOKMARK [3][-]{section*.18}{Explications}{subsection.2.2.6}% 26
|
||||
\BOOKMARK [2][-]{subsection.2.2.7}{La machine de Wimshurst}{section.2.2}% 27
|
||||
\BOOKMARK [2][-]{subsection.2.2.8}{La foudre}{section.2.2}% 28
|
||||
\BOOKMARK [2][-]{subsection.2.2.9}{Conservation et charge \351l\351mentaire}{section.2.2}% 29
|
||||
\BOOKMARK [2][-]{subsection.2.2.10}{La loi de Coulomb}{section.2.2}% 30
|
||||
\BOOKMARK [1][-]{section.2.3}{Le champ \351lectrique}{chapter.2}% 31
|
||||
\BOOKMARK [2][-]{subsection.2.3.1}{La loi de Coulomb}{section.2.3}% 32
|
||||
\BOOKMARK [2][-]{subsection.2.3.2}{La notion de champ}{section.2.3}% 33
|
||||
\BOOKMARK [2][-]{subsection.2.3.3}{D\351finition}{section.2.3}% 34
|
||||
\BOOKMARK [2][-]{subsection.2.3.4}{Corollaires}{section.2.3}% 35
|
||||
\BOOKMARK [2][-]{subsection.2.3.5}{La cage de Faraday}{section.2.3}% 36
|
||||
\BOOKMARK [2][-]{subsection.2.3.6}{Lignes de champ}{section.2.3}% 37
|
||||
\BOOKMARK [2][-]{subsection.2.3.7}{La notion de potentiel}{section.2.3}% 38
|
||||
\BOOKMARK [2][-]{subsection.2.3.8}{Potentiel}{section.2.3}% 39
|
||||
\BOOKMARK [2][-]{subsection.2.3.9}{Analogie gravitationnelle}{section.2.3}% 40
|
||||
\BOOKMARK [2][-]{subsection.2.3.10}{Tension et potentiel \351lectrique}{section.2.3}% 41
|
||||
\BOOKMARK [0][-]{chapter.3}{\311lectrocin\351tique}{}% 42
|
||||
\BOOKMARK [1][-]{section.3.1}{Circuit et courant}{chapter.3}% 43
|
||||
\BOOKMARK [2][-]{subsection.3.1.1}{Ordre de grandeur}{section.3.1}% 44
|
||||
\BOOKMARK [2][-]{subsection.3.1.2}{Mod\351lisation}{section.3.1}% 45
|
||||
\BOOKMARK [2][-]{subsection.3.1.3}{Estimation num\351rique}{section.3.1}% 46
|
||||
\BOOKMARK [3][-]{section*.41}{Vitesse thermique}{subsection.3.1.3}% 47
|
||||
\BOOKMARK [3][-]{section*.42}{Vitesse de d\351rive}{subsection.3.1.3}% 48
|
||||
\BOOKMARK [1][-]{section.3.2}{R\351sistance et loi d'Ohm}{chapter.3}% 49
|
||||
\BOOKMARK [1][-]{section.3.3}{S\351curit\351 \351lectrique}{chapter.3}% 50
|
||||
\BOOKMARK [2][-]{subsection.3.3.1}{Introduction}{section.3.3}% 51
|
||||
\BOOKMARK [2][-]{subsection.3.3.2}{L'\351lectricit\351 et l'homme}{section.3.3}% 52
|
||||
\BOOKMARK [3][-]{section*.47}{Facteurs importants}{subsection.3.3.2}% 53
|
||||
\BOOKMARK [3][-]{section*.48}{Sympt\364mes}{subsection.3.3.2}% 54
|
||||
\BOOKMARK [3][-]{section*.50}{Traitements}{subsection.3.3.2}% 55
|
||||
\BOOKMARK [2][-]{subsection.3.3.3}{Les mesures de s\351curit\351}{section.3.3}% 56
|
||||
\BOOKMARK [3][-]{section*.51}{Le r\351seau \351lectrique}{subsection.3.3.3}% 57
|
||||
\BOOKMARK [3][-]{section*.53}{Pour les b\342timents}{subsection.3.3.3}% 58
|
||||
\BOOKMARK [3][-]{section*.55}{Pour l'homme}{subsection.3.3.3}% 59
|
||||
\BOOKMARK [1][-]{section.3.4}{\311nergie et puissance \351lectrique}{chapter.3}% 60
|
||||
\BOOKMARK [1][-]{section.3.5}{Accumulateurs}{chapter.3}% 61
|
||||
\BOOKMARK [2][-]{subsection.3.5.1}{\311quations}{section.3.5}% 62
|
||||
\BOOKMARK [2][-]{subsection.3.5.2}{Types d'accumulateurs}{section.3.5}% 63
|
||||
\BOOKMARK [1][-]{section.3.6}{Circuits \351lectriques : les multiprises}{chapter.3}% 64
|
||||
\BOOKMARK [2][-]{subsection.3.6.1}{Circuit s\351rie}{section.3.6}% 65
|
||||
\BOOKMARK [2][-]{subsection.3.6.2}{Circuit parall\350le}{section.3.6}% 66
|
||||
\BOOKMARK [2][-]{subsection.3.6.3}{Lois de Kirchhoff}{section.3.6}% 67
|
||||
\BOOKMARK [3][-]{section*.65}{R\351sistance \351quivalente}{subsection.3.6.3}% 68
|
||||
\BOOKMARK [4][-]{section*.66}{Deux r\351sistances en s\351rie}{section*.65}% 69
|
||||
\BOOKMARK [4][-]{section*.68}{Deux r\351sistances en parall\350les}{section*.65}% 70
|
||||
\BOOKMARK [3][-]{section*.70}{Lois de Kirchhoff}{subsection.3.6.3}% 71
|
||||
\BOOKMARK [0][-]{chapter.4}{Magn\351tisme}{}% 72
|
||||
\BOOKMARK [1][-]{section.4.1}{Introduction}{chapter.4}% 73
|
||||
\BOOKMARK [1][-]{section.4.2}{Le champ magn\351tique}{chapter.4}% 74
|
||||
\BOOKMARK [2][-]{subsection.4.2.1}{D\351finitions}{section.4.2}% 75
|
||||
\BOOKMARK [2][-]{subsection.4.2.2}{Le champ magn\351tique terrestre}{section.4.2}% 76
|
||||
\BOOKMARK [2][-]{subsection.4.2.3}{L'exp\351rience d'\226rsted}{section.4.2}% 77
|
||||
\BOOKMARK [2][-]{subsection.4.2.4}{Le magn\351tisme dans la mati\350re}{section.4.2}% 78
|
||||
\BOOKMARK [2][-]{subsection.4.2.5}{Application}{section.4.2}% 79
|
||||
\BOOKMARK [1][-]{section.4.3}{La loi de Laplace}{chapter.4}% 80
|
||||
\BOOKMARK [2][-]{subsection.4.3.1}{D\351finition}{section.4.3}% 81
|
||||
\BOOKMARK [2][-]{subsection.4.3.2}{Exemple}{section.4.3}% 82
|
||||
\BOOKMARK [2][-]{subsection.4.3.3}{Applications}{section.4.3}% 83
|
||||
\BOOKMARK [3][-]{section*.89}{Galvanom\350tre}{subsection.4.3.3}% 84
|
||||
\BOOKMARK [3][-]{section*.91}{Moteur \351lectrique}{subsection.4.3.3}% 85
|
||||
\BOOKMARK [1][-]{section.4.4}{La loi de Lorentz}{chapter.4}% 86
|
||||
\BOOKMARK [2][-]{subsection.4.4.1}{D\351finition}{section.4.4}% 87
|
||||
\BOOKMARK [2][-]{subsection.4.4.2}{Applications}{section.4.4}% 88
|
||||
\BOOKMARK [3][-]{section*.94}{T\351l\351vision}{subsection.4.4.2}% 89
|
||||
\BOOKMARK [3][-]{section*.97}{Acc\351l\351rateur de particules : le cyclotron}{subsection.4.4.2}% 90
|
||||
\BOOKMARK [4][-]{section*.101}{Le cyclotron en \351quations}{section*.97}% 91
|
||||
\BOOKMARK [4][-]{section*.102}{Synchrocyclotron et synchrotron}{section*.97}% 92
|
||||
\BOOKMARK [3][-]{section*.103}{Spectrographe de masse}{subsection.4.4.2}% 93
|
||||
\BOOKMARK [0][-]{chapter.5}{\311lectromagn\351tisme}{}% 94
|
||||
\BOOKMARK [1][-]{section.5.1}{Nature de la lumi\350re}{chapter.5}% 95
|
||||
\BOOKMARK [1][-]{section.5.2}{Applications}{chapter.5}% 96
|
||||
\BOOKMARK [0][-]{appendix.A}{Unit\351s internationales}{}% 97
|
||||
\BOOKMARK [1][-]{section.A.1}{Introduction}{appendix.A}% 98
|
||||
\BOOKMARK [1][-]{section.A.2}{Les unit\351s choisies}{appendix.A}% 99
|
||||
\BOOKMARK [1][-]{section.A.3}{Exemple}{appendix.A}% 100
|
||||
\BOOKMARK [1][-]{section.A.4}{Conversions}{appendix.A}% 101
|
||||
\BOOKMARK [1][-]{section.A.5}{Multiples et sous-multiples}{appendix.A}% 102
|
||||
\BOOKMARK [1][-]{section.A.6}{Notation scientifique}{appendix.A}% 103
|
||||
\BOOKMARK [1][-]{section.A.7}{Valeurs importantes}{appendix.A}% 104
|
||||
\BOOKMARK [0][-]{appendix.B}{Petite histoire de l'\351lectricit\351}{}% 105
|
||||
\BOOKMARK [1][-]{section.B.1}{Introduction}{appendix.B}% 106
|
||||
\BOOKMARK [1][-]{section.B.2}{Le galvanisme}{appendix.B}% 107
|
||||
\BOOKMARK [1][-]{section.B.3}{Frankenstein}{appendix.B}% 108
|
||||
\BOOKMARK [1][-]{section.B.4}{La pile de Volta}{appendix.B}% 109
|
||||
\BOOKMARK [1][-]{section.B.5}{Courant et tension : Amp\350re}{appendix.B}% 110
|
||||
\BOOKMARK [1][-]{section.B.6}{Ohm et la notion de r\351sistance}{appendix.B}% 111
|
||||
\BOOKMARK [1][-]{section.B.7}{Conclusion}{appendix.B}% 112
|
||||
\BOOKMARK [0][-]{appendix.C}{La pile \351lectrique}{}% 113
|
||||
\BOOKMARK [0][-]{appendix.D}{Ligne \340 haute tension}{}% 114
|
||||
\BOOKMARK [0][-]{appendix.E}{Exercices}{}% 115
|
||||
\BOOKMARK [1][-]{section.E.1}{Probl\350mes}{appendix.E}% 116
|
||||
\BOOKMARK [2][-]{subsection.E.1.1}{Relatifs \340 la loi de Coulomb}{section.E.1}% 117
|
||||
\BOOKMARK [2][-]{subsection.E.1.2}{Relatifs au champ \351lectrique}{section.E.1}% 118
|
||||
\BOOKMARK [2][-]{subsection.E.1.3}{Relatifs \340 la s\351curit\351 \351lectrique}{section.E.1}% 119
|
||||
\BOOKMARK [2][-]{subsection.E.1.4}{Relatifs aux loi d'Ohm et de Pouillet, \340 la puissance et \340 l'\351nergie}{section.E.1}% 120
|
||||
\BOOKMARK [2][-]{subsection.E.1.5}{Relatifs au magn\351tisme}{section.E.1}% 121
|
||||
\BOOKMARK [1][-]{section.E.2}{Solutions}{appendix.E}% 122
|
||||
\BOOKMARK [0][-]{appendix*.141}{Bibliographie}{}% 123
|
||||
\BOOKMARK [0][-]{chapter.1}{\376\377\000I\000n\000t\000r\000o\000d\000u\000c\000t\000i\000o\000n}{}% 1
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\BOOKMARK [2][-]{subsection.E.1.4}{\376\377\000R\000e\000l\000a\000t\000i\000f\000s\000\040\000a\000u\000x\000\040\000l\000o\000i\000\040\000d\000'\000O\000h\000m\000\040\000e\000t\000\040\000d\000e\000\040\000P\000o\000u\000i\000l\000l\000e\000t\000,\000\040\000\340\000\040\000l\000a\000\040\000p\000u\000i\000s\000s\000a\000n\000c\000e\000\040\000e\000t\000\040\000\340\000\040\000l\000'\000\351\000n\000e\000r\000g\000i\000e}{section.E.1}% 120
|
||||
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|
||||
\BOOKMARK [1][-]{section.E.2}{\376\377\000S\000o\000l\000u\000t\000i\000o\000n\000s}{appendix.E}% 122
|
||||
\BOOKMARK [0][-]{appendix*.141}{\376\377\000B\000i\000b\000l\000i\000o\000g\000r\000a\000p\000h\000i\000e}{}% 123
|
||||
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||||
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@ -1,5 +1,4 @@
|
||||
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|
||||
\babel@toc {french}{}
|
||||
\babel@toc {french}{}\relax
|
||||
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|
||||
\contentsline {section}{\numberline {1.1}Introduction}{15}{section.1.1}%
|
||||
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|
||||
|
||||
@ -1,9 +1,6 @@
|
||||
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||||
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|
||||
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|
||||
\@writefile{toc}{\babel@toc {french}{}}
|
||||
\@writefile{lof}{\babel@toc {french}{}}
|
||||
\@writefile{lot}{\babel@toc {french}{}}
|
||||
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|
||||
\@writefile{lof}{\babel@toc {french}{}\relax }
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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||||
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||||
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||||
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||||
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||||
|
||||
0
Preambule.idx
Normal file
0
Preambule.idx
Normal file
942
Preambule.log
Normal file
942
Preambule.log
Normal file
@ -0,0 +1,942 @@
|
||||
This is pdfTeX, Version 3.141592653-2.6-1.40.22 (TeX Live 2022/dev/Debian) (preloaded format=latex 2024.8.20) 18 SEP 2024 11:30
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||||
entering extended mode
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||||
restricted \write18 enabled.
|
||||
%&-line parsing enabled.
|
||||
**Preambule.tex
|
||||
(./Preambule.tex
|
||||
LaTeX2e <2021-11-15> patch level 1
|
||||
L3 programming layer <2022-01-21>
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
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||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.4 \usepackage[T1]{
|
||||
fontenc} % écrire avec les accents
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.5 \usepackage[utf8]{
|
||||
inputenc} % écrire avec les accents
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.7 \usepackage{
|
||||
natbib} % à utiliser impérativement avec apalike-fr pour...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.8 \usepackage[english,french]{
|
||||
babel} % pour le support des documents multi...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.10 \usepackage{
|
||||
anyfontsize}
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.11 \usepackage{
|
||||
lmodern}
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.16 \usepackage{
|
||||
makeidx} % pour permettre de construire un index
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
\@indexfile=\write3
|
||||
\openout3 = `Preambule.idx'.
|
||||
|
||||
Writing index file Preambule.idx
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.20 \usepackage{
|
||||
path} % pour mettre du code non interprété
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.21 \usepackage{
|
||||
fancybox} % pour faire des boites entourée de différent...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.22 \usepackage{
|
||||
framed} % pour mettre des sortes de minipages encadrées ...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.23 \usepackage[dvips]{
|
||||
color} % pour mettre du gris ou de la couleur dans ...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.24 \usepackage{
|
||||
pst-grad} % pour faire des dégradés en pstricks (! il f...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
! Undefined control sequence.
|
||||
l.25 \definecolor
|
||||
{shadecolor}{rgb}{.9,.9,.9} % pour définir la couleur de f...
|
||||
The control sequence at the end of the top line
|
||||
of your error message was never \def'ed. If you have
|
||||
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||
and I'll forget about whatever was undefined.
|
||||
|
||||
|
||||
! LaTeX Error: Missing \begin{document}.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.25 \definecolor{s
|
||||
hadecolor}{rgb}{.9,.9,.9} % pour définir la couleur de f...
|
||||
|
||||
You're in trouble here. Try typing <return> to proceed.
|
||||
If that doesn't work, type X <return> to quit.
|
||||
|
||||
Missing character: There is no s in font nullfont!
|
||||
Missing character: There is no h in font nullfont!
|
||||
Missing character: There is no a in font nullfont!
|
||||
Missing character: There is no d in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no c in font nullfont!
|
||||
Missing character: There is no o in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no o in font nullfont!
|
||||
Missing character: There is no r in font nullfont!
|
||||
Missing character: There is no r in font nullfont!
|
||||
Missing character: There is no g in font nullfont!
|
||||
Missing character: There is no b in font nullfont!
|
||||
Missing character: There is no . in font nullfont!
|
||||
Missing character: There is no 9 in font nullfont!
|
||||
Missing character: There is no , in font nullfont!
|
||||
Missing character: There is no . in font nullfont!
|
||||
Missing character: There is no 9 in font nullfont!
|
||||
Missing character: There is no , in font nullfont!
|
||||
Missing character: There is no . in font nullfont!
|
||||
Missing character: There is no 9 in font nullfont!
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.26 \usepackage{
|
||||
boxedminipage} % pour mettre des bordures aux minipages
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.27 \usepackage{
|
||||
relsize,fancyvrb} % pour mettre des textes non interprétÃ...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.28 \usepackage{
|
||||
array} % pour étendre les possiblités de tabular par dÃ...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.29 \usepackage{
|
||||
multirow} % pour faire une cellule de tableau sur plusieu...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.30 \usepackage{
|
||||
colortbl} % pour mettre de la couleur dans un tableau
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.31 \usepackage{
|
||||
hhline} % pour faire des filets spéciaux dans les tableaux
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.32 \usepackage{
|
||||
supertabular} % pour faire des tableaux sur plusieurs pages
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.33 \usepackage{
|
||||
float} % permet un placement obligatoire
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.34 \usepackage{
|
||||
url} % permet une bonne gestion de la césure des url
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.35 \usepackage{
|
||||
graphicx} % pour charger le module étendu graphicx du mo...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.37 \usepackage{
|
||||
wrapfig} % pour mettre des figures dans le flot du texte
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.38 \usepackage{
|
||||
psfrag} % pour mettre des équations latex dans des figur...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.39 \usepackage{
|
||||
pstricks} % pour étendre les possiblités de graphisme d...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.42 \usepackage{
|
||||
multido} % pour répéter des figures en pstricks
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.44 \usepackage{
|
||||
rotating} % permet de faire tourner des figures \sideways...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.45 \usepackage{
|
||||
subfigure} % pour faire des figures comprenant plusieures...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.46 \usepackage{
|
||||
fancyhdr} % pour des entêtes et pieds de page entièreme...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.47 \usepackage{
|
||||
fancyvrb} % pour mettre des \verb dans les minipages
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.48 \usepackage[Lenny]{
|
||||
fncychap} % pour des titres de chapitres différent...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.49 \usepackage{
|
||||
multicol} % pour mettre un \begin{multicols}...\end{multi...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.50 \usepackage{
|
||||
amsmath} % pour avoir beaucoup de symboles mathématiques...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.51 \usepackage{
|
||||
amssymb} % nécessaire pour avoir certains symboles mathÃ...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.52 \usepackage{
|
||||
amsfonts} % nécessaire pour avoir certains symboles math...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.54 \usepackage{
|
||||
epsfig} % pour inclure des fichiers .eps. Mais il semble ...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.56 \usepackage{
|
||||
caption} % pour gérer encore mieux les légendes
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: Command \captionfont undefined.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.57 \renewcommand{\captionfont}
|
||||
{\it \small} % pour avoir de l'italique pour...
|
||||
|
||||
Try typing <return> to proceed.
|
||||
If that doesn't work, type X <return> to quit.
|
||||
|
||||
|
||||
! LaTeX Error: Command \captionlabelfont undefined.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.58 \renewcommand{\captionlabelfont}
|
||||
{\it \bf \small}
|
||||
Try typing <return> to proceed.
|
||||
If that doesn't work, type X <return> to quit.
|
||||
|
||||
! Undefined control sequence.
|
||||
l.59 \DeclareCaptionLabelSeparator
|
||||
{endash}{ -- } % pour définir un tiret ap...
|
||||
The control sequence at the end of the top line
|
||||
of your error message was never \def'ed. If you have
|
||||
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||
and I'll forget about whatever was undefined.
|
||||
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no n in font nullfont!
|
||||
Missing character: There is no d in font nullfont!
|
||||
Missing character: There is no a in font nullfont!
|
||||
Missing character: There is no s in font nullfont!
|
||||
Missing character: There is no h in font nullfont!
|
||||
Missing character: There is no - in font nullfont!
|
||||
Missing character: There is no - in font nullfont!
|
||||
! Undefined control sequence.
|
||||
l.60 \captionsetup
|
||||
{labelsep=endash,justification=centering,belowskip=10pt} %...
|
||||
The control sequence at the end of the top line
|
||||
of your error message was never \def'ed. If you have
|
||||
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||
and I'll forget about whatever was undefined.
|
||||
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no a in font nullfont!
|
||||
Missing character: There is no b in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no s in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no p in font nullfont!
|
||||
Missing character: There is no = in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no n in font nullfont!
|
||||
Missing character: There is no d in font nullfont!
|
||||
Missing character: There is no a in font nullfont!
|
||||
Missing character: There is no s in font nullfont!
|
||||
Missing character: There is no h in font nullfont!
|
||||
Missing character: There is no , in font nullfont!
|
||||
Missing character: There is no j in font nullfont!
|
||||
Missing character: There is no u in font nullfont!
|
||||
Missing character: There is no s in font nullfont!
|
||||
Missing character: There is no t in font nullfont!
|
||||
Missing character: There is no i in font nullfont!
|
||||
Missing character: There is no f in font nullfont!
|
||||
Missing character: There is no i in font nullfont!
|
||||
Missing character: There is no c in font nullfont!
|
||||
Missing character: There is no a in font nullfont!
|
||||
Missing character: There is no t in font nullfont!
|
||||
Missing character: There is no i in font nullfont!
|
||||
Missing character: There is no o in font nullfont!
|
||||
Missing character: There is no n in font nullfont!
|
||||
Missing character: There is no = in font nullfont!
|
||||
Missing character: There is no c in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no n in font nullfont!
|
||||
Missing character: There is no t in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no r in font nullfont!
|
||||
Missing character: There is no i in font nullfont!
|
||||
Missing character: There is no n in font nullfont!
|
||||
Missing character: There is no g in font nullfont!
|
||||
Missing character: There is no , in font nullfont!
|
||||
Missing character: There is no b in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no o in font nullfont!
|
||||
Missing character: There is no w in font nullfont!
|
||||
Missing character: There is no s in font nullfont!
|
||||
Missing character: There is no k in font nullfont!
|
||||
Missing character: There is no i in font nullfont!
|
||||
Missing character: There is no p in font nullfont!
|
||||
Missing character: There is no = in font nullfont!
|
||||
Missing character: There is no 1 in font nullfont!
|
||||
Missing character: There is no 0 in font nullfont!
|
||||
Missing character: There is no p in font nullfont!
|
||||
Missing character: There is no t in font nullfont!
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.61 \usepackage{
|
||||
verbatim} % pour faire des commentaires longs \begin{comm...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.62 \usepackage{
|
||||
endnotes} % pour faire des notes de fin
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.68 \usepackage{
|
||||
textcomp} % pour le symbole pourmille : \textperthousand ...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.69 \usepackage{
|
||||
boites,boites_exemples} % pour faire des paragraphes encadr...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.70 \usepackage{
|
||||
textpos} % pour positionner du texte ou des images à une...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.71 \usepackage{
|
||||
floatflt} % pour placer des tableaux entourés de texte
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.72 \usepackage{
|
||||
nextpage} % pour retirer les entêtes de la dernière...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
Overfull \hbox (20.0pt too wide) in paragraph at lines 25--75
|
||||
[]
|
||||
[]
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.82 \usepackage[derivedinbase,squaren,cdot,Gray]{
|
||||
SIunits} % pour utiliser ...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.90 \usepackage{
|
||||
lettrine} % pour mettre des lettrines
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.91 \usepackage{
|
||||
listings} % pour mettre du code informatique non interprÃ...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.92 \usepackage{
|
||||
xspace} % pour gérer les expaces nott. entre X et e pour...
|
||||
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
|
||||
! LaTeX Error: \usepackage before \documentclass.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.96 \usepackage[dvips,ps2pdf]{
|
||||
hyperref}
|
||||
\usepackage may only appear in the document preamble, i.e.,
|
||||
between \documentclass and \begin{document}.
|
||||
|
||||
! Undefined control sequence.
|
||||
l.98 \hypersetup
|
||||
{
|
||||
The control sequence at the end of the top line
|
||||
of your error message was never \def'ed. If you have
|
||||
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||
and I'll forget about whatever was undefined.
|
||||
|
||||
|
||||
! LaTeX Error: Missing \begin{document}.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.99 b
|
||||
reaklinks=true,
|
||||
You're in trouble here. Try typing <return> to proceed.
|
||||
If that doesn't work, type X <return> to quit.
|
||||
|
||||
Missing character: There is no b in font nullfont!
|
||||
Missing character: There is no r in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no a in font nullfont!
|
||||
Missing character: There is no k in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no i in font nullfont!
|
||||
Missing character: There is no n in font nullfont!
|
||||
Missing character: There is no k in font nullfont!
|
||||
Missing character: There is no s in font nullfont!
|
||||
Missing character: There is no = in font nullfont!
|
||||
Missing character: There is no t in font nullfont!
|
||||
Missing character: There is no r in font nullfont!
|
||||
Missing character: There is no u in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no , in font nullfont!
|
||||
Missing character: There is no p in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no a in font nullfont!
|
||||
Missing character: There is no i in font nullfont!
|
||||
Missing character: There is no n in font nullfont!
|
||||
Missing character: There is no p in font nullfont!
|
||||
Missing character: There is no a in font nullfont!
|
||||
Missing character: There is no g in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no s in font nullfont!
|
||||
Missing character: There is no = in font nullfont!
|
||||
Missing character: There is no f in font nullfont!
|
||||
Missing character: There is no a in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no s in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no , in font nullfont!
|
||||
Missing character: There is no c in font nullfont!
|
||||
Missing character: There is no o in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no o in font nullfont!
|
||||
Missing character: There is no r in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no i in font nullfont!
|
||||
Missing character: There is no n in font nullfont!
|
||||
Missing character: There is no k in font nullfont!
|
||||
Missing character: There is no s in font nullfont!
|
||||
Missing character: There is no = in font nullfont!
|
||||
Missing character: There is no t in font nullfont!
|
||||
Missing character: There is no r in font nullfont!
|
||||
Missing character: There is no u in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no , in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no i in font nullfont!
|
||||
Missing character: There is no n in font nullfont!
|
||||
Missing character: There is no k in font nullfont!
|
||||
Missing character: There is no c in font nullfont!
|
||||
Missing character: There is no o in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no o in font nullfont!
|
||||
Missing character: There is no r in font nullfont!
|
||||
Missing character: There is no = in font nullfont!
|
||||
Missing character: There is no b in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no u in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no , in font nullfont!
|
||||
Missing character: There is no c in font nullfont!
|
||||
Missing character: There is no i in font nullfont!
|
||||
Missing character: There is no t in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no c in font nullfont!
|
||||
Missing character: There is no o in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no o in font nullfont!
|
||||
Missing character: There is no r in font nullfont!
|
||||
Missing character: There is no = in font nullfont!
|
||||
Missing character: There is no b in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no u in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no , in font nullfont!
|
||||
Missing character: There is no u in font nullfont!
|
||||
Missing character: There is no r in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no c in font nullfont!
|
||||
Missing character: There is no o in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no o in font nullfont!
|
||||
Missing character: There is no r in font nullfont!
|
||||
Missing character: There is no = in font nullfont!
|
||||
Missing character: There is no b in font nullfont!
|
||||
Missing character: There is no l in font nullfont!
|
||||
Missing character: There is no u in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no , in font nullfont!
|
||||
Missing character: There is no h in font nullfont!
|
||||
Missing character: There is no y in font nullfont!
|
||||
Missing character: There is no p in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no r in font nullfont!
|
||||
Missing character: There is no t in font nullfont!
|
||||
Missing character: There is no e in font nullfont!
|
||||
Missing character: There is no x in font nullfont!
|
||||
Missing character: There is no n in font nullfont!
|
||||
Missing character: There is no a in font nullfont!
|
||||
Missing character: There is no m in font nullfont!
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\usepackage[derivedinbase,squaren,cdot,Gray]{SIunits} % pour utiliser des commandes relatives au unités du système internationnal ; il y a aussi les préfixes m M G ...
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296
Solutions.tex
296
Solutions.tex
@ -1,23 +1,23 @@
|
||||
\begin{Solution}{1}
|
||||
La force se calcule dans les unités du SI. On a donc :
|
||||
\begin{align*}
|
||||
q&=\SI{12}{\milli\coulomb}=\SI{12e-3}{\coulomb}\\
|
||||
Q&=\SI{4}{\milli\coulomb}=\SI{4e-3}{\coulomb}\\
|
||||
d&=\SI{3}{\centi\metre}=\SI{3e-2}{\metre}
|
||||
q&=\qty{12}{\milli\coulomb}=\qty{12e-3}{\coulomb}\\
|
||||
Q&=\qty{4}{\milli\coulomb}=\qty{4e-3}{\coulomb}\\
|
||||
d&=\qty{3}{\centi\metre}=\qty{3e-2}{\metre}
|
||||
\end{align*}
|
||||
Ainsi, on peut calculer la force comme suit :
|
||||
\begin{align*}
|
||||
F&=k\cdot \frac{q\cdot Q}{d^2}\\
|
||||
&=9\cdot 10^9\cdot \frac{12\cdot 10^{-3}\cdot 4\cdot 10^{-3}}{(3\cdot 10^{-2})^2}\\
|
||||
&=\SI{4,8e8}{\newton}
|
||||
&=\qty{4,8e8}{\newton}
|
||||
\end{align*}
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{2}
|
||||
Dans les unités du SI, on a :
|
||||
\begin{align*}
|
||||
F&=\SI{5}{\milli\newton}=\SI{5e-3}{\newton}\\
|
||||
d&=\SI{5}{\centi\metre}=\SI{5e-2}{\metre}
|
||||
F&=\qty{5}{\milli\newton}=\qty{5e-3}{\newton}\\
|
||||
d&=\qty{5}{\centi\metre}=\qty{5e-2}{\metre}
|
||||
\end{align*}
|
||||
Par ailleurs, les deux charges sont identiques et on peut poser : \(q=Q\).
|
||||
On peut écrire :
|
||||
@ -26,17 +26,17 @@
|
||||
9\cdot 10^9\cdot q^2&=5\cdot 10^{-3}\cdot (5\cdot 10^{-2})^2\;\Rightarrow \\
|
||||
q^2&=\frac{5\cdot 10^{-3}\cdot (5\cdot 10^{-2})^2}{9\cdot 10^9} \\
|
||||
&=1,38\cdot 10^{-15}\;\Rightarrow \\
|
||||
q&=3,73\cdot 10^{-8}=\SI{37,3}{\nano\coulomb}
|
||||
q&=3,73\cdot 10^{-8}=\qty{37,3}{\nano\coulomb}
|
||||
\end{align*}
|
||||
Ainsi, la charge totale de ces particules est :
|
||||
\[\sum q=q+Q=2\cdot q=\SI{74,6}{\nano\coulomb}\]
|
||||
\[\sum q=q+Q=2\cdot q=\qty{74,6}{\nano\coulomb}\]
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{3}
|
||||
Un problème identique serait : combien y a-t-il de pommes de \SI{2}{\kilo\gram} dans un panier de pommes d'une masse totale de \SI{24}{\kilo\gram} ? Évidemment, on divise la masse totale des pommes par la masse d'une pomme : \(24/2=12\,pommes\).
|
||||
Un problème identique serait : combien y a-t-il de pommes de \qty{2}{\kilo\gram} dans un panier de pommes d'une masse totale de \qty{24}{\kilo\gram} ? Évidemment, on divise la masse totale des pommes par la masse d'une pomme : \(24/2=12\,pommes\).
|
||||
|
||||
Dans notre problème, le panier d'électron a une charge de \SI{7}{\micro\coulomb} soit \SI{7e-6}{\coulomb} et chaque électron une charge, dite élémentaire, de \SI{1,6e-19}{\coulomb}. Ainsi, le nombre d'électrons vaut :
|
||||
\[n=\frac{7e-6}{1,6e-19}=\SI{4,375e13}{electrons}\]
|
||||
Dans notre problème, le panier d'électron a une charge de \qty{7}{\micro\coulomb} soit \qty{7e-6}{\coulomb} et chaque électron une charge, dite élémentaire, de \qty{1,6e-19}{\coulomb}. Ainsi, le nombre d'électrons vaut :
|
||||
\[n=\frac{7e-6}{1,6e-19}=\qty{4,375e13}{electrons}\]
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{4}
|
||||
@ -45,7 +45,7 @@
|
||||
Ainsi, la force qui s'exerce entre ces deux charges vaut :
|
||||
\begin{align*}
|
||||
F&=9\cdot 10^9\cdot \frac{1,6\cdot 10^{-19}\cdot 6\cdot 1,6\cdot 10^{-19}}{(10^{-10})^2}\\
|
||||
&=\SI{1,38e-7}{\newton}=\SI{138}{\nano\newton}
|
||||
&=\qty{1,38e-7}{\newton}=\qty{138}{\nano\newton}
|
||||
\end{align*}
|
||||
|
||||
\end{Solution}
|
||||
@ -54,13 +54,13 @@
|
||||
\begin{align*}
|
||||
F_e&=k\cdot \frac{q_{proton}\cdot q_{\acute electron}}{d^2}\\
|
||||
&=9\cdot 10^9\cdot \frac{(1,6\cdot 10^{-19})^2}{(10^{-10})^2}\\
|
||||
&=\SI{2,3e-8}{\newton}
|
||||
&=\qty{2,3e-8}{\newton}
|
||||
\end{align*}
|
||||
Pour la force de gravitation, on a :
|
||||
\begin{align*}
|
||||
F_g&=G\cdot \frac{m_{proton}\cdot m_{\acute electron}}{d^2}\\
|
||||
&=6,67\cdot 10^{-11}\cdot \frac{1,67\cdot 10^{-27}\cdot 9,1\cdot 10^{-31}}{(10^{-10})^2}\\
|
||||
&=\SI{10e-47}{\newton}
|
||||
&=\qty{10e-47}{\newton}
|
||||
\end{align*}
|
||||
Ainsi, le rapport vaut :
|
||||
\[r=\frac{2,3\cdot 10^{-8}}{10^{-47}}=2,3\cdot 10^{39}\]
|
||||
@ -74,7 +74,7 @@
|
||||
\[F=k\cdot \frac{e\cdot e}{d^2}=2\cdot m_e\cdot g\;\Rightarrow\]
|
||||
\[9\cdot 10^9\cdot \frac{(1,6\cdot 10^{-19})^2}{d^2}=2\cdot 9,1\cdot 10^{-31}\cdot 10\;\Rightarrow\]
|
||||
\[d^2=9\cdot 10^9\cdot \frac{(1,6\cdot 10^{-19})^2}{2\cdot 9,1\cdot 10^{-31}\cdot 10}=12,66\;\Rightarrow\]
|
||||
\[d=\SI{3,56}{\metre}\]
|
||||
\[d=\qty{3,56}{\metre}\]
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{7}
|
||||
@ -83,7 +83,7 @@
|
||||
Par ailleurs, sans que \(q\) et \(Q\) ne changent, lorsque la distance \(d\) entre les deux charges est diminuée d'un facteur neuf, soit pour la nouvelle distance \(d'=d/9\), on peut écrire pour la nouvelle force \(F'\) :
|
||||
\begin{align*}
|
||||
F'&=k\cdot \frac{q\cdot Q}{d'^2}=k\cdot \frac{q\cdot Q}{(\frac{d}{9})^2}=k\cdot \frac{q\cdot Q}{\frac{d^2}{81}}\\
|
||||
&=k\cdot \frac{q\cdot Q}{d^2}\cdot 81=400\cdot 10^{-3}\cdot 81=\SI{32,4}{\newton}
|
||||
&=k\cdot \frac{q\cdot Q}{d^2}\cdot 81=400\cdot 10^{-3}\cdot 81=\qty{32,4}{\newton}
|
||||
\end{align*}
|
||||
L'avant dernière égalité étant justifiée par la valeur de \(F\).
|
||||
|
||||
@ -105,7 +105,7 @@
|
||||
|
||||
Tout cela est valable quel que soit le signe de \(q'\) qui ne fait que changer le sens des forces qui s'exercent.
|
||||
|
||||
Pour déterminer la distance \(d\) qui sépare \(Q\) de \(q'\), on pose (avec \SI{10}{\centi\metre} soit \SI{0,1}{\metre}) :
|
||||
Pour déterminer la distance \(d\) qui sépare \(Q\) de \(q'\), on pose (avec \qty{10}{\centi\metre} soit \qty{0,1}{\metre}) :
|
||||
\begin{align*}
|
||||
F_{q\rightarrow q'}&=F_{Q\rightarrow q'}\;\Rightarrow\\
|
||||
k\cdot \frac{9\cdot 10^{-6}\cdot q'}{(0,1+d)^2}&=k\cdot \frac{2\cdot 10^{-6}\cdot q'}{d^2}\;\Rightarrow\\
|
||||
@ -120,7 +120,7 @@
|
||||
\begin{align*}
|
||||
d&=\frac{0,4\pm \sqrt{(-0,4)^2-4\cdot 7\cdot (-0,02)}}{2\cdot 7}\\
|
||||
&=\frac{0,4\pm 0,849}{14}\\
|
||||
&=\begin{cases}\SI{0,0892}{\metre}&q'\mathrm{\ \grave a\ l'ext\acute erieur}\\\SI{-0,0321}{\metre}&q'\mathrm{\ \grave a\ l'int\acute erieur}\end{cases}
|
||||
&=\begin{cases}\qty{0,0892}{\metre}&q'\mathrm{\ \grave a\ l'ext\acute erieur}\\\qty{-0,0321}{\metre}&q'\mathrm{\ \grave a\ l'int\acute erieur}\end{cases}
|
||||
\end{align*}
|
||||
Comme on a choisi d'écrire la distance entre \(q\) et \(q'\) comme \((0,1+d)\) avec \(q'\) à l'extérieur des deux charges \(q\) et \(Q\), la distance \(d\) est positive si la charge \(q'\) est à l'extérieur. C'est donc la solution positive qu'il faut retenir. L'autre solution correspond au cas où \(q'\) serait entre les deux charges. En effet, ce cas est possible, car la condition mathématique de départ impose seulement que les deux forces soient égales et non aussi qu'elles soient opposées. Ainsi, la solution négative correspond-elle au cas où la charge \(q'\) serait à l'intérieur des deux charge et subirait deux forces égales mais de même sens. Il faut la rejeter.
|
||||
|
||||
@ -145,10 +145,10 @@
|
||||
\begin{align*}
|
||||
&Q=\\
|
||||
&=\frac{7,02\cdot 10^6\pm \sqrt{(7,02\cdot 10^6)^2-4\cdot 9\cdot 10^9\cdot 30,86}}{2\cdot 9\cdot 10^9}\\
|
||||
&=\frac{7,02\cdot 10^6\pm6,94\cdot 10^6}{18\cdot 10^9}=\begin{cases}\SI{775,5}{\micro\coulomb}\\\SI{4,4}{\micro\coulomb}\end{cases}
|
||||
&=\frac{7,02\cdot 10^6\pm6,94\cdot 10^6}{18\cdot 10^9}=\begin{cases}\qty{775,5}{\micro\coulomb}\\\qty{4,4}{\micro\coulomb}\end{cases}
|
||||
\end{align*}
|
||||
Soit, pour l'autre charge :
|
||||
\[q=780-Q=780-\begin{cases}775,5\\4,4\end{cases}=\begin{cases}\SI{4,5}{\micro\coulomb}\\\SI{775,6}{\micro\coulomb}\end{cases}\]
|
||||
\[q=780-Q=780-\begin{cases}775,5\\4,4\end{cases}=\begin{cases}\qty{4,5}{\micro\coulomb}\\\qty{775,6}{\micro\coulomb}\end{cases}\]
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{10}
|
||||
@ -174,22 +174,22 @@
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{11}
|
||||
Le champ est produit par la charge \(Q=\SI{40e-5}{\coulomb}=\SI{400}{\micro\coulomb}\). Il faut le déterminer à une distance \(d=\SI{33}{\centi\metre}\) d'elle.
|
||||
Le champ est produit par la charge \(Q=\qty{40e-5}{\coulomb}=\qty{400}{\micro\coulomb}\). Il faut le déterminer à une distance \(d=\qty{33}{\centi\metre}\) d'elle.
|
||||
La direction et le sens sont donnés à la figure \ref{champ33cm}.
|
||||
La grandeur se calcule comme suit :
|
||||
\begin{align*}
|
||||
E&=9\cdot 10^9\cdot \frac{40\cdot 10^{-5}}{0,33^2}\\
|
||||
&=33'057'851\,N/C=\SI{33}{\mega\newton\per\coulomb}
|
||||
&=33'057'851\,N/C=\qty{33}{\mega\newton\per\coulomb}
|
||||
\end{align*}
|
||||
\begin{figure}[!ht]
|
||||
\centering
|
||||
\caption{Champ à \SI{33}{\centi\metre}}\label{champ33cm}
|
||||
\caption{Champ à \qty{33}{\centi\metre}}\label{champ33cm}
|
||||
\includegraphics{Champ33cm.eps}
|
||||
\end{figure}
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{12}
|
||||
Nous avons ici à déterminer un champ qui exerce une force donnée \(F=\SI{8e-4}{\newton}\) sur une charge \(q=\SI{2}{\micro\coulomb}=\SI{2e-6}{\coulomb}\). Comme \(q>0\)f, la relation :
|
||||
Nous avons ici à déterminer un champ qui exerce une force donnée \(F=\qty{8e-4}{\newton}\) sur une charge \(q=\qty{2}{\micro\coulomb}=\qty{2e-6}{\coulomb}\). Comme \(q>0\)f, la relation :
|
||||
\[\overrightarrow{F}=q\cdot \overrightarrow{E}\]
|
||||
implique que le champ est dans la même direction et le même sens que la force, comme indiqué sur la figure \ref{champforce}.
|
||||
\begin{figure}[!ht]
|
||||
@ -198,17 +198,17 @@
|
||||
\includegraphics{ChampForce.eps}
|
||||
\end{figure}
|
||||
Sa valeur est ainsi :
|
||||
\[E=\frac{F}{q}=\frac{8\cdot 10^{-4}}{2\cdot 10^{-6}}=\SI{400}{\newton\per\coulomb}\]
|
||||
\[E=\frac{F}{q}=\frac{8\cdot 10^{-4}}{2\cdot 10^{-6}}=\qty{400}{\newton\per\coulomb}\]
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{13}
|
||||
Le champ total est la somme vectorielle des champs \(\overrightarrow{E}_-\) et \(\overrightarrow{E}_+\) respectivement créés par les charges \(Q_-\) et \(Q_+\). On peut en calculer la valeur comme suit :
|
||||
\begin{align*}
|
||||
E_-&=9\cdot 10^9\cdot \frac{50\cdot 10^{-6}}{0,1^2}=\SI{45}{\mega\newton\per\coulomb}\\
|
||||
E_+&=9\cdot 10^9\cdot \frac{40\cdot 10^{-6}}{0,1^2}=\SI{36}{\mega\newton\per\coulomb}
|
||||
E_-&=9\cdot 10^9\cdot \frac{50\cdot 10^{-6}}{0,1^2}=\qty{45}{\mega\newton\per\coulomb}\\
|
||||
E_+&=9\cdot 10^9\cdot \frac{40\cdot 10^{-6}}{0,1^2}=\qty{36}{\mega\newton\per\coulomb}
|
||||
\end{align*}
|
||||
Comme ces champs sont dans la même direction et le même sens (voir figure \ref{champ2charges}), la valeur du champ total est la somme arithmétique de \(E_-\) et \(E_+\) :
|
||||
\[E_{tot}=E_-+E_+=45+36=\SI{81}{\mega\newton\per\coulomb}\]
|
||||
\[E_{tot}=E_-+E_+=45+36=\qty{81}{\mega\newton\per\coulomb}\]
|
||||
\begin{figure}[!ht]
|
||||
\centering
|
||||
\caption{Champ entre deux charges}\label{champ2charges}
|
||||
@ -220,7 +220,7 @@
|
||||
Pour que le proton soit en suspension, il faut que son poids \(P\) soit compensé par la force électrique \(F\) produite par le champ \(E\). Comme la charge du proton est positive, le champ \(\overrightarrow{E}\) est dans le même sens que la force \(\overrightarrow{F}\), c'est-à-dire opposé au poids, comme le montre la figure \ref{protonsuspension}. Numériquement, on a alors :
|
||||
\begin{align*}
|
||||
q\cdot E&=m\cdot g\;\Rightarrow\\
|
||||
E&=\frac{m\cdot g}{q}=\frac{1,67\cdot 10^{-27}\cdot 10}{1,6\cdot 10^{-19}}=\SI{1e-7}{\newton\per\coulomb}
|
||||
E&=\frac{m\cdot g}{q}=\frac{1,67\cdot 10^{-27}\cdot 10}{1,6\cdot 10^{-19}}=\qty{1e-7}{\newton\per\coulomb}
|
||||
\end{align*}
|
||||
\begin{figure}[!ht]
|
||||
\centering
|
||||
@ -236,7 +236,7 @@
|
||||
\begin{align*}
|
||||
F&=m\cdot a=q\cdot E\;\Rightarrow\\
|
||||
E&=\frac{m\cdot a}{q}=\frac{1,67\cdot 10^{-27}\cdot 74,3\cdot 10^3}{1,6\cdot 10^{-19}}\\
|
||||
&=\SI{7,76e-4}{\newton\per\coulomb}
|
||||
&=\qty{7,76e-4}{\newton\per\coulomb}
|
||||
\end{align*}
|
||||
\begin{figure}[!h]
|
||||
\centering
|
||||
@ -253,13 +253,13 @@
|
||||
\begin{align*}
|
||||
F&=e\cdot E=c^{ste}=m\cdot a\;\Rightarrow\\
|
||||
a&=c^{ste}=\frac{e\cdot E}{m}=\frac{1,6\cdot 10^{-19}\cdot 6,2\cdot 10^3}{9,1\cdot 10^{-31}}\\
|
||||
&=\SI{1,1e15}{\metre\per\second\squared}
|
||||
&=\qty{1,1e15}{\metre\per\second\squared}
|
||||
\end{align*}
|
||||
L'accélération est constante et de valeur déterminée. C'est en fait une décélération et sa valeur est donc \(a=\SI{-~1,1e15}{\metre\per\second\squared}\). C'est un MRUA et on peut écrire :
|
||||
L'accélération est constante et de valeur déterminée. C'est en fait une décélération et sa valeur est donc \(a=\qty{-1,1e15}{\metre\per\second\squared}\). C'est un MRUA et on peut écrire :
|
||||
\begin{align*}
|
||||
v_f^2&=v_i^2+2\cdot a\cdot d\;\Rightarrow\\
|
||||
0^2&=(2,6\cdot 10^6)^2-2\cdot 1,1\cdot 10^{15}\cdot d\;\Rightarrow\\
|
||||
d&=\frac{(2,6\cdot 10^6)^2}{2\cdot 1,1\cdot 10^{15}}=\SI{3e-3}{\metre}=\SI{3}{\milli\metre}
|
||||
d&=\frac{(2,6\cdot 10^6)^2}{2\cdot 1,1\cdot 10^{15}}=\qty{3e-3}{\metre}=\qty{3}{\milli\metre}
|
||||
\end{align*}
|
||||
\item On a encore :
|
||||
\[v_f=a\cdot t+v_i\]
|
||||
@ -270,27 +270,27 @@
|
||||
-v_i&=a\cdot t+v_i\;\Rightarrow\\
|
||||
-2\cdot v_i&=a\cdot t\;\Rightarrow\\
|
||||
t&=\frac{-2\cdot v_i}{a}=\frac{-2\cdot 2,6\cdot 10^6}{-1,1\cdot 10^{15}}\\
|
||||
&=\SI{4,73e-9}{\second}=\SI{4,73}{\nano\second}
|
||||
&=\qty{4,73e-9}{\second}=\qty{4,73}{\nano\second}
|
||||
\end{align*}
|
||||
\end{enumerate}
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{17}
|
||||
On a simplement :
|
||||
\[E=\frac{F_1}{q_1}=\frac{10^{-2}}{10^{-7}}=\SI{1e5}{\newton\per\coulomb}\]
|
||||
\[E=\frac{F_1}{q_1}=\frac{10^{-2}}{10^{-7}}=\qty{1e5}{\newton\per\coulomb}\]
|
||||
et pour la force :
|
||||
\[F_2=q_2\cdot E=10^{-8}\cdot 10^5=\SI{1e-3}{\newton}\]
|
||||
\[F_2=q_2\cdot E=10^{-8}\cdot 10^5=\qty{1e-3}{\newton}\]
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{18}
|
||||
Si on pose \(q_1=\SI{4}{\micro\coulomb}\) et \(q_2=\SI{6}{\micro\coulomb}\), \(\overrightarrow{E}_1\) le champ électrique créé par la charge \(q_1\), \(\overrightarrow{E}_2\) le champ électrique créé par la charge \(q_2\), \(\overrightarrow{F}_{1\rightarrow 2}\) la force créée par \(q_1\) qui s'exerce sur \(q_2\) et \(\overrightarrow{F}_{2\rightarrow 1}\) la force créée par \(q_2\) qui s'exerce sur \(q_1\), on a la situation présentée sur la figure \ref{lunlautre}. Par la troisième loi de Newton, on peut affirmer que :
|
||||
\[F_{1\rightarrow 2}=F_{2\rightarrow 1}=F=\SI{0,4}{\newton}\]
|
||||
Si on pose \(q_1=\qty{4}{\micro\coulomb}\) et \(q_2=\qty{6}{\micro\coulomb}\), \(\overrightarrow{E}_1\) le champ électrique créé par la charge \(q_1\), \(\overrightarrow{E}_2\) le champ électrique créé par la charge \(q_2\), \(\overrightarrow{F}_{1\rightarrow 2}\) la force créée par \(q_1\) qui s'exerce sur \(q_2\) et \(\overrightarrow{F}_{2\rightarrow 1}\) la force créée par \(q_2\) qui s'exerce sur \(q_1\), on a la situation présentée sur la figure \ref{lunlautre}. Par la troisième loi de Newton, on peut affirmer que :
|
||||
\[F_{1\rightarrow 2}=F_{2\rightarrow 1}=F=\qty{0,4}{\newton}\]
|
||||
Ainsi, chaque charge exerce sur l'autre la même force.
|
||||
|
||||
Mais, comme la valeur de chaque charge est différente, le champ électrique produit l'est aussi. Le champ électrique produit par la charge \(q_1\) exerce son action, la force \(F\), sur la charge \(q_2\). De la même manière, le champ produit par la charge \(q_2\) exerce son action, l'autre force \(F\), sur la charge \(q_1\). Ainsi, par \(F=q\cdot E\), on a :
|
||||
\begin{align*}
|
||||
E_1&=\frac{F}{q_2}=\frac{0,4}{6\cdot 10^{-6}}=\SI{66666}{\newton\per\coulomb}\\
|
||||
E_2&=\frac{F}{q_1}=\frac{0,4}{4\cdot 10^{-6}}=\SI{100000}{\newton\per\coulomb}
|
||||
E_1&=\frac{F}{q_2}=\frac{0,4}{6\cdot 10^{-6}}=\qty{66666}{\newton\per\coulomb}\\
|
||||
E_2&=\frac{F}{q_1}=\frac{0,4}{4\cdot 10^{-6}}=\qty{100000}{\newton\per\coulomb}
|
||||
\end{align*}
|
||||
\begin{figure}[!ht]
|
||||
\centering
|
||||
@ -300,12 +300,12 @@
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{19}
|
||||
Les deux charges ont même valeur. On peut donc calculer le champ électrique à \SI{25}{\centi\metre} de l'une par :
|
||||
\[E_{25}=9\cdot 10^9\cdot \frac{10^{-6}}{0,25^2}=\SI{144}{\kilo\newton\per\coulomb}\]
|
||||
On peut aussi calculer le champ à \(75-25=\SI{50}{\centi\metre}\) de l'autre par :
|
||||
\[E_{50}=9\cdot 10^9\cdot \frac{10^{-6}}{0,5^2}=\SI{36}{\kilo\newton\per\coulomb}\]
|
||||
Les deux charges étant positives, chacune d'elle créée un champ de sens opposé à l'autre. Ce champ pointe à l'opposé de la charge qui l'a créé. Le plus grand est celui créé par la charge à \SI{25}{\centi\metre}. Le champ résultant pointe donc vers la charge à \SI{50}{\centi\metre} et sa valeur est :
|
||||
\[E_{total}=E_{25}-E_{50}=144-36=\SI{108}{\kilo\newton\per\coulomb}\]
|
||||
Les deux charges ont même valeur. On peut donc calculer le champ électrique à \qty{25}{\centi\metre} de l'une par :
|
||||
\[E_{25}=9\cdot 10^9\cdot \frac{10^{-6}}{0,25^2}=\qty{144}{\kilo\newton\per\coulomb}\]
|
||||
On peut aussi calculer le champ à \(75-25=\qty{50}{\centi\metre}\) de l'autre par :
|
||||
\[E_{50}=9\cdot 10^9\cdot \frac{10^{-6}}{0,5^2}=\qty{36}{\kilo\newton\per\coulomb}\]
|
||||
Les deux charges étant positives, chacune d'elle créée un champ de sens opposé à l'autre. Ce champ pointe à l'opposé de la charge qui l'a créé. Le plus grand est celui créé par la charge à \qty{25}{\centi\metre}. Le champ résultant pointe donc vers la charge à \qty{50}{\centi\metre} et sa valeur est :
|
||||
\[E_{total}=E_{25}-E_{50}=144-36=\qty{108}{\kilo\newton\per\coulomb}\]
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{20}
|
||||
@ -319,14 +319,14 @@
|
||||
\end{Solution}
|
||||
\begin{Solution}{21}
|
||||
Par \(F=m\cdot a\), on trouve la force nécessaire à accélérer la charge \(q\) :
|
||||
\[F=m\cdot a=10^{-3}\cdot 3=\SI{3e-3}{\newton}\]
|
||||
\[F=m\cdot a=10^{-3}\cdot 3=\qty{3e-3}{\newton}\]
|
||||
Le champ électrique nécessaire pour produire cette force est alors :
|
||||
\[E=\frac{F}{q}=\frac{3\cdot 10{-3}}{1\cdot 10^{-6}}=\SI{3000}{\newton\per\coulomb}\]
|
||||
\[E=\frac{F}{q}=\frac{3\cdot 10{-3}}{1\cdot 10^{-6}}=\qty{3000}{\newton\per\coulomb}\]
|
||||
La charge nécessaire pour créer ce champ électrique est finalement donnée par :
|
||||
\begin{align*}
|
||||
E&=k\cdot \frac{Q}{d^2}\;\Rightarrow\\
|
||||
Q&=E\cdot \frac{d^2}{k}=3000\cdot \frac{2^2}{9\cdot 10^9}\\
|
||||
&=\SI{1,33e-6}{\coulomb}=\SI{1,33}{\micro\coulomb}
|
||||
&=\qty{1,33e-6}{\coulomb}=\qty{1,33}{\micro\coulomb}
|
||||
\end{align*}
|
||||
|
||||
\end{Solution}
|
||||
@ -388,7 +388,7 @@
|
||||
La loi d'Ohm donne :
|
||||
\begin{align*}
|
||||
U&=R\cdot I\;\Rightarrow\\
|
||||
I&=\frac{U}{R}=\frac{230}{1500}=\SI{0,1534}{\ampere}=\SI{153,4}{\milli\ampere}
|
||||
I&=\frac{U}{R}=\frac{230}{1500}=\qty{0,1534}{\ampere}=\qty{153,4}{\milli\ampere}
|
||||
\end{align*}
|
||||
Ce courant est mortel. Il y a brûlures importantes et arrêt cardiaque.
|
||||
|
||||
@ -396,10 +396,10 @@
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{25}
|
||||
Supposons une personne de résistance corporelle faible \(R=\SI{100}{\kilo\ohm}\). La loi d'Ohm donne alors :
|
||||
Supposons une personne de résistance corporelle faible \(R=\qty{100}{\kilo\ohm}\). La loi d'Ohm donne alors :
|
||||
\begin{align*}
|
||||
U&=R\cdot I\;\Rightarrow\\
|
||||
I&=\frac{U}{R}=\frac{12}{100'000}=\SI{0,00012}{\ampere}=\SI{0,12}{\milli\ampere}
|
||||
I&=\frac{U}{R}=\frac{12}{100'000}=\qty{0,00012}{\ampere}=\qty{0,12}{\milli\ampere}
|
||||
\end{align*}
|
||||
Ce courant n'est absolument pas dangereux. Attention pourtant aux risques de brûlures par échauffement du fil lors du démarrage de la voiture. Il nécessite en effet un fort courant.
|
||||
|
||||
@ -421,11 +421,11 @@
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{30}
|
||||
À partir d'un courant de \SI{2}{\milli\ampere}. Il reste peu de risques d'électrocution. Mais il n'est pas impossible que le fusible ne déclenche pas lorsqu'il est derrière un transformateur, pour des raisons que nous ne développerons pas ici. Il faut donc garder à l'esprit que malgré son efficacité, il faut rester sur ses gardes.
|
||||
À partir d'un courant de \qty{2}{\milli\ampere}. Il reste peu de risques d'électrocution. Mais il n'est pas impossible que le fusible ne déclenche pas lorsqu'il est derrière un transformateur, pour des raisons que nous ne développerons pas ici. Il faut donc garder à l'esprit que malgré son efficacité, il faut rester sur ses gardes.
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{31}
|
||||
Le risque principal est que la phase vienne toucher le boîtier sans que celui-ci ne décharge vers la terre (l'appareil peut être sur une table isolée du sol, par exemple). L'appareil est branché, son boîtier est à \SI{230}{\volt} et on ne se rend compte de rien. Il suffit alors de toucher l'appareil pour être électrocuté.
|
||||
Le risque principal est que la phase vienne toucher le boîtier sans que celui-ci ne décharge vers la terre (l'appareil peut être sur une table isolée du sol, par exemple). L'appareil est branché, son boîtier est à \qty{230}{\volt} et on ne se rend compte de rien. Il suffit alors de toucher l'appareil pour être électrocuté.
|
||||
|
||||
Ce risque serait partiellement éliminé. En effet, avec le défaut décrit précédemment et une ligne de terre, au moment de brancher l'appareil, le boîtier aurait immédiatement déchargé vers le sol à travers la ligne de terre. Le courant de la phase aurait alors fortement augmenté et le fusible (ou le disjoncteur) aurait sauté. Pour autant que le fusible ait réagit immédiatement. Ce qui n'est pas toujours le cas et dépend des fusibles \dots\
|
||||
|
||||
@ -528,40 +528,40 @@
|
||||
\caption{Puissances diverses}
|
||||
\label{puissancesdiversessol}
|
||||
\end{table}
|
||||
\footnotetext{A une ampoule normale de \SI{40}{\watt} correspond une ampoule basse consommation de \SI{7}{\watt}.}
|
||||
\footnotetext{A une ampoule normale de \qty{40}{\watt} correspond une ampoule basse consommation de \qty{7}{\watt}.}
|
||||
|
||||
Il faut remarquer la forte consommation des sèche-cheveux comparé à une machine à laver la vaisselle (même si elle consomme aussi de l'eau). Il faut remarquer aussi qu'un four à micro-onde est un peu plus économique qu'une cuisinière. Enfin, il faut relever que la puissance d'un radiateur électrique tourne autour de \SI{1000}{\watt} pour une chambre de taille normale.
|
||||
Il faut remarquer la forte consommation des sèche-cheveux comparé à une machine à laver la vaisselle (même si elle consomme aussi de l'eau). Il faut remarquer aussi qu'un four à micro-onde est un peu plus économique qu'une cuisinière. Enfin, il faut relever que la puissance d'un radiateur électrique tourne autour de \qty{1000}{\watt} pour une chambre de taille normale.
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{35}
|
||||
On a simplement :
|
||||
\[P=U\cdot I\;\Rightarrow\;I=\frac{P}{U}=\frac{4}{12}=\SI{0,333}{\ampere}=\SI{333}{\milli\ampere}\]
|
||||
\[P=U\cdot I\;\Rightarrow\;I=\frac{P}{U}=\frac{4}{12}=\qty{0,333}{\ampere}=\qty{333}{\milli\ampere}\]
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{36}
|
||||
On a simplement :
|
||||
\[P=U\cdot I=9\cdot 0,4=\SI{3,6}{\watt}\]
|
||||
\[P=U\cdot I=9\cdot 0,4=\qty{3,6}{\watt}\]
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{37}
|
||||
On a simplement :
|
||||
\[P=\frac{U^2}{R}\;\Rightarrow\;U=\sqrt{P\cdot R}=\sqrt{0,25\cdot 800}=\SI{14,14}{\volt}\]
|
||||
\[P=\frac{U^2}{R}\;\Rightarrow\;U=\sqrt{P\cdot R}=\sqrt{0,25\cdot 800}=\qty{14,14}{\volt}\]
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{38}
|
||||
La principale difficulté de ce problème est de bien comprendre dans quelles unités on travaille.
|
||||
\begin{itemize}
|
||||
\item Résolvons tout d'abord le problème dans les unités du système international. Dans celles-ci, le temps s'exprime en secondes. Ainsi, \(t=7\,h=7\cdot 3'600=\SI{25200}{\second}\). On peut ensuite calculer facilement l'énergie nécessaire pour laisser allumée la lampe :
|
||||
\[P=\frac{E}{t}\;\Rightarrow\;E=P\cdot t=200\cdot 25'200=\SI{5040000}{\joule}\]
|
||||
Mais le coût est donné en $cts/kWh$. Il faut donc convertir le résultat de joules en kilowattheures. Comme \SI{1}{\kilo\watt\hour}=\SI{3,6e6}{\joule}, on peut écrire :
|
||||
\[E=5'040'000\,J=\frac{5'040'000}{3,6\cdot 10^6}=\SI{1,4}{\kilo\watt\hour}\]
|
||||
\item Résolvons tout d'abord le problème dans les unités du système international. Dans celles-ci, le temps s'exprime en secondes. Ainsi, \(t=7\,h=7\cdot 3'600=\qty{25200}{\second}\). On peut ensuite calculer facilement l'énergie nécessaire pour laisser allumée la lampe :
|
||||
\[P=\frac{E}{t}\;\Rightarrow\;E=P\cdot t=200\cdot 25'200=\qty{5040000}{\joule}\]
|
||||
Mais le coût est donné en $cts/kWh$. Il faut donc convertir le résultat de joules en kilowattheures. Comme \qty{1}{\kilo\watt\hour}=\qty{3,6e6}{\joule}, on peut écrire :
|
||||
\[E=5'040'000\,J=\frac{5'040'000}{3,6\cdot 10^6}=\qty{1,4}{\kilo\watt\hour}\]
|
||||
Ainsi, le coût total est donné par :
|
||||
\[c=20\cdot 1,4=28\,cts\]
|
||||
|
||||
\item Mais, on peut résoudre bien plus facilement ce problème en n'utilisant pas les unités du système international. En effet, on peut convertir la puissance en \si{\kilo\watt} :
|
||||
\[P=\SI{200}{\watt}=\SI{0,2}{\kilo\watt}\]
|
||||
\item Mais, on peut résoudre bien plus facilement ce problème en n'utilisant pas les unités du système international. En effet, on peut convertir la puissance en \unit{\kilo\watt} :
|
||||
\[P=\qty{200}{\watt}=\qty{0,2}{\kilo\watt}\]
|
||||
et utiliser des heures pour le temps :
|
||||
\[P=\frac{E}{t}\;\Rightarrow\;E=P\cdot t=0,2\cdot 7=\SI{1,4}{\kilo\watt\hour}\]
|
||||
\[P=\frac{E}{t}\;\Rightarrow\;E=P\cdot t=0,2\cdot 7=\qty{1,4}{\kilo\watt\hour}\]
|
||||
Puis, comme précédemment, le coût total est donné par :
|
||||
\[c=20\cdot 1,4=28\,cts\]
|
||||
\end{itemize}
|
||||
@ -571,35 +571,35 @@
|
||||
\begin{Solution}{39}
|
||||
La principale difficulté de ce problème est de bien comprendre dans quelles unités on travaille.
|
||||
\begin{itemize}
|
||||
\item Résolvons tout d'abord le problème dans les unités du système international. Dans celles-ci la charge s'exprime en coulomb. Or, la charge nous est donnée ici par \(\Delta q=\SI{45}{\ampere\hour}\). Il s'agit bien d'une charge, puisque :
|
||||
\item Résolvons tout d'abord le problème dans les unités du système international. Dans celles-ci la charge s'exprime en coulomb. Or, la charge nous est donnée ici par \(\Delta q=\qty{45}{\ampere\hour}\). Il s'agit bien d'une charge, puisque :
|
||||
\begin{eqnarray*}
|
||||
I&=\frac{\Delta q}{t}\;\Rightarrow\;\Delta q=I\cdot t\;\Rightarrow\\
|
||||
\verb|[|\Delta q\verb|]|&=[I]\cdot [t]=\begin{cases}\si{\ampere\second} &SI\\\si{\ampere\hour}&pas\,SI\end{cases}
|
||||
\verb|[|\Delta q\verb|]|&=[I]\cdot [t]=\begin{cases}\unit{\ampere\second} &SI\\\unit{\ampere\hour}&pas\,SI\end{cases}
|
||||
\end{eqnarray*}
|
||||
Il s'agit donc de trouver la relation entre les \si{\ampere\hour} et les \si{\coulomb}. Pour cela, déterminons la charge \(\Delta q\) en \si{\coulomb} déplacée par un courant \(I\) de \SI{1}{\ampere} pendant un temps \(t\) d'une heure. On a :
|
||||
\[\Delta q=I\cdot t=1\cdot 3'600=\SI{3600}{\coulomb}\]
|
||||
Car \SI{1}{\hour}=\SI{3600}{\second}. Cette même charge peut être calculée en \si{\ampere\hour} de la manière suivante :
|
||||
\[\Delta q=I\cdot t=1\cdot 1=\SI{1}{\ampere\hour}\]
|
||||
Il s'agit donc de trouver la relation entre les \unit{\ampere\hour} et les \unit{\coulomb}. Pour cela, déterminons la charge \(\Delta q\) en \unit{\coulomb} déplacée par un courant \(I\) de \qty{1}{\ampere} pendant un temps \(t\) d'une heure. On a :
|
||||
\[\Delta q=I\cdot t=1\cdot 3'600=\qty{3600}{\coulomb}\]
|
||||
Car \qty{1}{\hour}=\qty{3600}{\second}. Cette même charge peut être calculée en \unit{\ampere\hour} de la manière suivante :
|
||||
\[\Delta q=I\cdot t=1\cdot 1=\qty{1}{\ampere\hour}\]
|
||||
Ainsi, on a l'équivalence :
|
||||
\[\SI{1}{\ampere\hour}=\SI{3600}{\coulomb}\]
|
||||
\[\qty{1}{\ampere\hour}=\qty{3600}{\coulomb}\]
|
||||
La charge considérée dans notre problème est donc :
|
||||
\[\Delta q=45\,Ah=45\cdot 3600=\SI{162000}{\coulomb}\]
|
||||
\[\Delta q=45\,Ah=45\cdot 3600=\qty{162000}{\coulomb}\]
|
||||
Cela permet de calculer l'énergie emmagasinée dans la batterie :
|
||||
\[E=\Delta q\cdot U=162'000\cdot 12=\SI{1944000}{\joule}\]
|
||||
\[E=\Delta q\cdot U=162'000\cdot 12=\qty{1944000}{\joule}\]
|
||||
|
||||
\item Mais, il y a beaucoup plus simple. On peut calculer l'énergie de la batterie en utilisant directement une charge en \si{\ampere\hour}. Ainsi, on a :
|
||||
\item Mais, il y a beaucoup plus simple. On peut calculer l'énergie de la batterie en utilisant directement une charge en \unit{\ampere\hour}. Ainsi, on a :
|
||||
\begin{align*}
|
||||
E&=\Delta q\cdot U=45\cdot 12=\SI{540}{\ampere\hour\volt}\\
|
||||
&=\SI{540}{\ampere\volt\hour}=\SI{540}{\watt\hour}
|
||||
E&=\Delta q\cdot U=45\cdot 12=\qty{540}{\ampere\hour\volt}\\
|
||||
&=\qty{540}{\ampere\volt\hour}=\qty{540}{\watt\hour}
|
||||
\end{align*}
|
||||
car :
|
||||
\[P=U\cdot I\,\Rightarrow\,[P]=\si{\watt}=[U]\cdot [I]=\si{\volt}\cdot \si{\ampere}=\si{\ampere}\cdot \si{\volt}\]
|
||||
\[P=U\cdot I\,\Rightarrow\,[P]=\unit{\watt}=[U]\cdot [I]=\unit{\volt}\cdot \unit{\ampere}=\unit{\ampere}\cdot \unit{\volt}\]
|
||||
Plus simplement, on peut alors exprimer l'énergie de la batterie par :
|
||||
\[E=\SI{540}{\watt\hour}=\SI{0,54}{\kilo\watt\hour}\]
|
||||
\[E=\qty{540}{\watt\hour}=\qty{0,54}{\kilo\watt\hour}\]
|
||||
\end{itemize}
|
||||
Évidemment, suivant les cas, le résultat n'est pas exprimé dans les mêmes unités. Mais, c'est bien le même. En effet, comme \SI{1}{\kilo\watt\hour}=\SI{3,6e6}{\joule}, on a :
|
||||
\[E=\SI{0,54}{\kilo\watt\hour}=0,54\cdot 3,6\cdot 10^6=\SI{1944000}{\joule}\]
|
||||
La solution n'utilisant pas les unités du système international est bien plus rapide. Si on peut se permettre d'exprimer le résultat dans une unité qui n'est pas celle du système international, elle est préférable. Notons que les \si{\kilo\watt\hour} sont une unité d'énergie assez répandue pour que, dans la plupart des cas, on puisse se permettre de l'utiliser.
|
||||
Évidemment, suivant les cas, le résultat n'est pas exprimé dans les mêmes unités. Mais, c'est bien le même. En effet, comme \qty{1}{\kilo\watt\hour}=\qty{3,6e6}{\joule}, on a :
|
||||
\[E=\qty{0,54}{\kilo\watt\hour}=0,54\cdot 3,6\cdot 10^6=\qty{1944000}{\joule}\]
|
||||
La solution n'utilisant pas les unités du système international est bien plus rapide. Si on peut se permettre d'exprimer le résultat dans une unité qui n'est pas celle du système international, elle est préférable. Notons que les \unit{\kilo\watt\hour} sont une unité d'énergie assez répandue pour que, dans la plupart des cas, on puisse se permettre de l'utiliser.
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{40}
|
||||
@ -607,24 +607,24 @@
|
||||
\begin{itemize}
|
||||
\item On trouve le temps grâce à la définition du courant :
|
||||
\[I=\frac{\Delta q}{t}\,\Rightarrow\;t=\frac{\Delta q}{I}\]
|
||||
Pour cela, il faut trouver le courant \(I\) nécessaire à faire briller une ampoule de \SI{60}{\watt} sous une tension de \SI{12}{\volt} :
|
||||
\[P=U\cdot I\;\Rightarrow\;I=\frac{P}{U}=\frac{60}{12}=\SI{5}{\ampere}\]
|
||||
Pour cela, il faut trouver le courant \(I\) nécessaire à faire briller une ampoule de \qty{60}{\watt} sous une tension de \qty{12}{\volt} :
|
||||
\[P=U\cdot I\;\Rightarrow\;I=\frac{P}{U}=\frac{60}{12}=\qty{5}{\ampere}\]
|
||||
On a alors :
|
||||
\[t=\frac{45}{5}=\SI{9}{\hour}\]
|
||||
\[t=\frac{45}{5}=\qty{9}{\hour}\]
|
||||
Remarquez que les unités du système international n'ont pas été utilisées.
|
||||
\item On peut calculer l'énergie contenue dans la batterie par :
|
||||
\[E=q\cdot U=45\cdot 12=\SI{540}{\ampere\hour\volt}=\SI{540}{\watt\hour}\]
|
||||
\[E=q\cdot U=45\cdot 12=\qty{540}{\ampere\hour\volt}=\qty{540}{\watt\hour}\]
|
||||
Puis, par définition de la puissance, trouver le temps :
|
||||
\[P=\frac{E}{t}\,\Rightarrow\,t=\frac{E}{P}=\frac{540}{60}=\SI{9}{\hour}\]
|
||||
\[P=\frac{E}{t}\,\Rightarrow\,t=\frac{E}{P}=\frac{540}{60}=\qty{9}{\hour}\]
|
||||
Dans ce cas aussi, les unités du système international n'ont pas été utilisées pour des raisons de simplicité.
|
||||
\end{itemize}
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{41}
|
||||
Il faut trouver le nombre de \si{\kilo\watt\hour} nécessaires pour cette action. On a :
|
||||
Il faut trouver le nombre de \unit{\kilo\watt\hour} nécessaires pour cette action. On a :
|
||||
\begin{align*}
|
||||
E&=P\cdot t=25\cdot 365\cdot 14\\
|
||||
&=\SI{127750}{\watt\hour}=\SI{127,75}{\kilo\watt\hour}
|
||||
&=\qty{127750}{\watt\hour}=\qty{127,75}{\kilo\watt\hour}
|
||||
\end{align*}
|
||||
Le coût est donc :
|
||||
\[c=127,75\cdot 0,20=25,55\,frs\]
|
||||
@ -634,66 +634,66 @@
|
||||
On peut résoudre le problème de deux manières :
|
||||
\begin{itemize}
|
||||
\item La puissance totale est de :
|
||||
\[P=2\cdot 40+2+2\cdot 6=\SI{94}{\watt}\]
|
||||
Le courant nécessaire pour obtenir une telle puissance sous une tension de \SI{12}{\volt} est de :
|
||||
\[P=U\cdot I\;\Rightarrow\;I=\frac{P}{U}=\frac{94}{12}=\SI{7,833}{\ampere}\]
|
||||
\[P=2\cdot 40+2+2\cdot 6=\qty{94}{\watt}\]
|
||||
Le courant nécessaire pour obtenir une telle puissance sous une tension de \qty{12}{\volt} est de :
|
||||
\[P=U\cdot I\;\Rightarrow\;I=\frac{P}{U}=\frac{94}{12}=\qty{7,833}{\ampere}\]
|
||||
Ainsi le temps de fonctionnement est donné par la définition du courant :
|
||||
\[I=\frac{\Delta q}{t}\;\Rightarrow\;t=\frac{\Delta q}{I}=\frac{40}{7,833}=\SI{5,1}{\hour}\]
|
||||
\[I=\frac{\Delta q}{t}\;\Rightarrow\;t=\frac{\Delta q}{I}=\frac{40}{7,833}=\qty{5,1}{\hour}\]
|
||||
\item L'énergie contenue dans la batterie est :
|
||||
\[E=\Delta q\cdot U=40\cdot 12=\SI{480}{\ampere\hour\volt}=\SI{480}{\watt\hour}\]
|
||||
\[E=\Delta q\cdot U=40\cdot 12=\qty{480}{\ampere\hour\volt}=\qty{480}{\watt\hour}\]
|
||||
Comme vu précédemment, la puissance totale est de :
|
||||
\[P=2\cdot 40+2+2\cdot 6=\SI{94}{\watt}\]
|
||||
\[P=2\cdot 40+2+2\cdot 6=\qty{94}{\watt}\]
|
||||
Et le temps de fonctionnement est donc :
|
||||
\[P=\frac{E}{t}\;\Rightarrow\;t=\frac{E}{P}=\frac{480}{94}=\SI{5,1}{\hour}\]
|
||||
\[P=\frac{E}{t}\;\Rightarrow\;t=\frac{E}{P}=\frac{480}{94}=\qty{5,1}{\hour}\]
|
||||
\end{itemize}
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{43}
|
||||
Pour trouver le diamètre du fil par la loi de Pouillet, il faut déterminer sa résistance :
|
||||
\[P=\frac{U^2}{R}\;\Rightarrow\;R=\frac{U^2}{P}=\frac{230^2}{2000}=\SI{26,45}{\ohm}\]
|
||||
La loi de Pouillet et la valeur de la résistivité du fer \(\rho=\SI{9,7e-8}{\ohm\metre}\), trouvée dans la table, donnent alors :
|
||||
\[P=\frac{U^2}{R}\;\Rightarrow\;R=\frac{U^2}{P}=\frac{230^2}{2000}=\qty{26,45}{\ohm}\]
|
||||
La loi de Pouillet et la valeur de la résistivité du fer \(\rho=\qty{9,7e-8}{\ohm\metre}\), trouvée dans la table, donnent alors :
|
||||
\begin{align*}
|
||||
R&=\rho\cdot \frac{L}{S}\;\Rightarrow\\
|
||||
S&=\rho\cdot \frac{L}{R}=9,7\cdot 10^{-8}\cdot \frac{1500}{26,45}=\SI{5,5e-6}{\metre\squared}
|
||||
S&=\rho\cdot \frac{L}{R}=9,7\cdot 10^{-8}\cdot \frac{1500}{26,45}=\qty{5,5e-6}{\metre\squared}
|
||||
\end{align*}
|
||||
Or, la surface est donnée par :
|
||||
\begin{align*}
|
||||
S&=\pi\cdot (\frac{d}{2})^2\;\Rightarrow\\
|
||||
d&=2\cdot \sqrt{\frac{S}{\pi}}=2\cdot \sqrt{\frac{5,5\cdot 10^{-6}}{\pi}}=\SI{2,65e-3}{\metre}
|
||||
d&=2\cdot \sqrt{\frac{S}{\pi}}=2\cdot \sqrt{\frac{5,5\cdot 10^{-6}}{\pi}}=\qty{2,65e-3}{\metre}
|
||||
\end{align*}
|
||||
Ainsi, le diamètre du fil est de :
|
||||
\[d=\SI{2,65e-3}{\metre}=\SI{2,65}{\milli\metre}\]
|
||||
\[d=\qty{2,65e-3}{\metre}=\qty{2,65}{\milli\metre}\]
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{44}
|
||||
L'énergie totale nécessaire chaque jour est de :
|
||||
\[E_{jour}=3,3\cdot 3+5\cdot 0,1\cdot 4+3\cdot 1,1+1,8=\SI{17}{\kilo\watt\hour}\]
|
||||
\[E_{jour}=3,3\cdot 3+5\cdot 0,1\cdot 4+3\cdot 1,1+1,8=\qty{17}{\kilo\watt\hour}\]
|
||||
Pendant un mois, l'énergie dépensée est de :
|
||||
\[E_{mois}=30\cdot 17=\SI{510}{\kilo\watt\hour}\]
|
||||
\[E_{mois}=30\cdot 17=\qty{510}{\kilo\watt\hour}\]
|
||||
Le coût est donc de :
|
||||
\[c=510\cdot 0,2=102.-\]
|
||||
|
||||
Pour une utilisation pendant une année, l'énergie nécessaire est de :
|
||||
\[E_{ann\acute ee}=365\cdot 17=\SI{6205}{\kilo\watt\hour}\]
|
||||
\[E_{ann\acute ee}=365\cdot 17=\qty{6205}{\kilo\watt\hour}\]
|
||||
Par ailleurs, chaque kilogramme de charbon produit une énergie de :
|
||||
\begin{align*}
|
||||
E&=\SI{7000}{\kilo cal}=7000\cdot 10^3\,cal=7\cdot 10^6\cdot 4,186\\
|
||||
&=\SI{29302}{\kilo\joule}=\frac{29,302\cdot 10^6}{3,6\cdot 10^6}=\SI{8,14}{\kilo\watt\hour}
|
||||
E&=\qty{7000}{\kilo cal}=7000\cdot 10^3\,cal=7\cdot 10^6\cdot 4,186\\
|
||||
&=\qty{29302}{\kilo\joule}=\frac{29,302\cdot 10^6}{3,6\cdot 10^6}=\qty{8,14}{\kilo\watt\hour}
|
||||
\end{align*}
|
||||
Comme seulement 35\% de cette énergie est exploitable, chaque kilogramme fournit en réalité :
|
||||
\[E_{r\acute eel}=0,35\cdot 8,14=\SI{2,85}{\kilo\watt\hour}\]
|
||||
Ainsi, le nombre de kilogrammes nécessaire à produire une énergie totale de \SI{6205}{\kilo\watt\hour} par année, est de :
|
||||
\[n=\frac{6205}{2,85}=\SI{2178,1}{\kilo\gram}\]
|
||||
\[E_{r\acute eel}=0,35\cdot 8,14=\qty{2,85}{\kilo\watt\hour}\]
|
||||
Ainsi, le nombre de kilogrammes nécessaire à produire une énergie totale de \qty{6205}{\kilo\watt\hour} par année, est de :
|
||||
\[n=\frac{6205}{2,85}=\qty{2178,1}{\kilo\gram}\]
|
||||
Soit plus de deux tonnes de charbon !
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{45}
|
||||
La puissance utilisée par le moteur est :
|
||||
\[P_{utilis\acute ee}=U\cdot I=230\cdot 4,5=\SI{1035}{\watt}\]
|
||||
\[P_{utilis\acute ee}=U\cdot I=230\cdot 4,5=\qty{1035}{\watt}\]
|
||||
C'est la puissance électrique qui rentre dans le moteur.
|
||||
|
||||
La puissance mécanique que le moteur fournit est de :
|
||||
\[P_{utile}=0,4\,CV=0,4\cdot 736=\SI{294,4}{\watt}\]
|
||||
\[P_{utile}=0,4\,CV=0,4\cdot 736=\qty{294,4}{\watt}\]
|
||||
C'est la puissance mécanique qui ressort du moteur.
|
||||
|
||||
Le rendement se calcule alors par :
|
||||
@ -701,41 +701,41 @@
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{46}
|
||||
La ville reçoit \SI{600}{\kilo\watt} directement exploitables. Pour les deux tensions données, on peut déterminer le courant nécessaire pour fournir cette puissance par :
|
||||
La ville reçoit \qty{600}{\kilo\watt} directement exploitables. Pour les deux tensions données, on peut déterminer le courant nécessaire pour fournir cette puissance par :
|
||||
\begin{align*}
|
||||
P&=U\cdot I\;\Rightarrow\\
|
||||
I&=\frac{P}{U}=\begin{cases}\frac{600\cdot 10^3}{50'000}\\
|
||||
\frac{600\cdot 10^3}{15'000}\end{cases}=\begin{cases}\SI{12}{\ampere}\\\SI{40}{\ampere}\end{cases}
|
||||
\frac{600\cdot 10^3}{15'000}\end{cases}=\begin{cases}\qty{12}{\ampere}\\\qty{40}{\ampere}\end{cases}
|
||||
\end{align*}
|
||||
On voit clairement que le courant nécessaire est plus important lorsque la puissance est fournie sous une basse tension.
|
||||
|
||||
Les pertes thermiques dans la ligne de \SI{4}{\ohm} sont alors données par :
|
||||
\[P=R\cdot I^2=\begin{cases}4\cdot 12^2\\4\cdot 40^2\end{cases}=\begin{cases}\SI{576}{\watt}\\\SI{6400}{\watt}\end{cases}\]
|
||||
On voit que les pertes sont plus importantes sous une basse tension. Cela est du au fait que le courant est plus important et que les fils chauffent plus. La différence est ici de \(6400-576=\SI{5824}{\watt}\).
|
||||
Les pertes thermiques dans la ligne de \qty{4}{\ohm} sont alors données par :
|
||||
\[P=R\cdot I^2=\begin{cases}4\cdot 12^2\\4\cdot 40^2\end{cases}=\begin{cases}\qty{576}{\watt}\\\qty{6400}{\watt}\end{cases}\]
|
||||
On voit que les pertes sont plus importantes sous une basse tension. Cela est du au fait que le courant est plus important et que les fils chauffent plus. La différence est ici de \(6400-576=\qty{5824}{\watt}\).
|
||||
|
||||
Cela explique pourquoi l'énergie électrique est transportée sur des lignes à haute tension : les pertes sont plus faibles. Avec des supraconducteurs (conducteurs de résistance nulle) cela ne serait pas nécessaire.
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{47}
|
||||
Pour trouver le rayon du fil par la loi de Pouillet, il faut déterminer sa résistance. On écrit ici :
|
||||
\[P=R\cdot I^2\;\Rightarrow\;R=\frac{P}{I^2}=\frac{1}{10^2}=\SI{0,01}{\ohm}\]
|
||||
\[P=R\cdot I^2\;\Rightarrow\;R=\frac{P}{I^2}=\frac{1}{10^2}=\qty{0,01}{\ohm}\]
|
||||
On applique ensuite la loi de Pouillet :
|
||||
\begin{align*}
|
||||
R&=\rho\cdot \frac{L}{S}\;\Rightarrow\\
|
||||
S&=\rho\cdot \frac{L}{R}=1,68\cdot 10^{-8}\cdot \frac{1}{0,01}\\
|
||||
&=\SI{1,68e-6}{\metre\squared}
|
||||
&=\qty{1,68e-6}{\metre\squared}
|
||||
\end{align*}
|
||||
Et le rayon est alors :
|
||||
\begin{align*}
|
||||
S&=\pi\cdot r^2\;\Rightarrow\\
|
||||
r&=\sqrt{\frac{S}{\pi}}=\sqrt{\frac{1,68\cdot 10^{-6}}{\pi}}\\
|
||||
&=\SI{7,31e-4}{\metre}\simeq\SI{1}{\milli\metre}
|
||||
&=\qty{7,31e-4}{\metre}\simeq\qty{1}{\milli\metre}
|
||||
\end{align*}
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{48}
|
||||
Le vecteur \(\overrightarrow L\) est orienté nord-sud, vers le sud. Le champ magnétique terrestre \(\overrightarrow B_{Terre}\) est lui orienté nord-sud, mais vers le nord. Ces deux vecteurs sont parallèles, mais de sens opposés. L'angle entre eux est donc de \SI{180}{\degree}. Ainsi, on a :
|
||||
\[F=I\cdot L\cdot B\cdot \sin(180)=\SI{0}{\newton}\]
|
||||
Le vecteur \(\overrightarrow L\) est orienté nord-sud, vers le sud. Le champ magnétique terrestre \(\overrightarrow B_{Terre}\) est lui orienté nord-sud, mais vers le nord. Ces deux vecteurs sont parallèles, mais de sens opposés. L'angle entre eux est donc de \qty{180}{\degree}. Ainsi, on a :
|
||||
\[F=I\cdot L\cdot B\cdot \sin(180)=\qty{0}{\newton}\]
|
||||
La force de Laplace est donc nulle.
|
||||
|
||||
\begin{figure}[!ht]
|
||||
@ -743,10 +743,10 @@
|
||||
\caption{Fil droit orienté est-ouest}\label{fildroitestouest}
|
||||
\includegraphics{FilDroitEstOuest.eps}
|
||||
\end{figure}
|
||||
Par contre, si le vecteur \(\overrightarrow L\) est orienté est-ouest, vers l'ouest, l'angle qu'il fait avec le vecteur \(\overrightarrow B_{Terre}\), orienté lui nord-sud vers le nord, est de \SI{90}{\degree}. Ainsi, on a :
|
||||
Par contre, si le vecteur \(\overrightarrow L\) est orienté est-ouest, vers l'ouest, l'angle qu'il fait avec le vecteur \(\overrightarrow B_{Terre}\), orienté lui nord-sud vers le nord, est de \qty{90}{\degree}. Ainsi, on a :
|
||||
\begin{align*}
|
||||
F&=I\cdot L\cdot B\cdot \sin(90)\\
|
||||
&=10\cdot 20\cdot 21,594\cdot 10^{-6}=\SI{0,00432}{\newton}
|
||||
&=10\cdot 20\cdot 21,594\cdot 10^{-6}=\qty{0,00432}{\newton}
|
||||
\end{align*}
|
||||
La figure \ref{fildroitestouest} présente la situation. La règle du tire-bouchon détermine le sens du vecteur \(\overrightarrow F\), à partir de l'équation de Laplace :
|
||||
\[\overrightarrow F=I\cdot \overrightarrow L \times \overrightarrow B\]
|
||||
@ -760,14 +760,14 @@
|
||||
\caption{Fils sur des pylônes}\label{pylones}
|
||||
\includegraphics{Pylones.eps}
|
||||
\end{figure}
|
||||
Comme l'angle \(\beta=\SI{30}{\degree}\), on peut écrire :
|
||||
Comme l'angle \(\beta=\qty{30}{\degree}\), on peut écrire :
|
||||
\begin{align*}
|
||||
F&=I\cdot L\cdot B\cdot \sin(90-\beta)\\
|
||||
&=20\cdot 200\cdot 21,594\cdot 10^{-6}\cdot \sin(90-30)\\
|
||||
&=\SI{0,0748}{\newton}
|
||||
&=\qty{0,0748}{\newton}
|
||||
\end{align*}
|
||||
Soit pour cinq fils, la force totale entre les deux pylônes est de :
|
||||
\[F_{tot}=5\cdot 0,0748=\SI{0,374}{\newton}\]
|
||||
\[F_{tot}=5\cdot 0,0748=\qty{0,374}{\newton}\]
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{50}
|
||||
@ -784,9 +784,9 @@
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{51}
|
||||
Le poids est une force qui s'exerce vers le bas. La force de Laplace agit donc vers le haut. L'angle entre le fil et le champ étant de \SI{90}{\degree}, sa valeur est donnée par :
|
||||
\[F=I\cdot L\cdot B=10\cdot 2\cdot 0,1=\SI{2}{\newton}\]
|
||||
Le poids qu'il faut suspendre est donc de \SI{2}{\newton}. La masse correspondante est donc d'environ \(2/10=\SI{0,2}{\kilo\gram}=\SI{200}{\gram}\) (\(g\simeq\SI{10}{\metre\per\second\squared}\)).
|
||||
Le poids est une force qui s'exerce vers le bas. La force de Laplace agit donc vers le haut. L'angle entre le fil et le champ étant de \qty{90}{\degree}, sa valeur est donnée par :
|
||||
\[F=I\cdot L\cdot B=10\cdot 2\cdot 0,1=\qty{2}{\newton}\]
|
||||
Le poids qu'il faut suspendre est donc de \qty{2}{\newton}. La masse correspondante est donc d'environ \(2/10=\qty{0,2}{\kilo\gram}=\qty{200}{\gram}\) (\(g\simeq\qty{10}{\metre\per\second\squared}\)).
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{52}
|
||||
@ -794,7 +794,7 @@
|
||||
\begin{align*}
|
||||
F&=I\cdot L\cdot B\cdot \sin(\alpha)\\
|
||||
&=0,1\cdot 0,1\cdot 100\cdot 10^{-6}\cdot \sin(30)\\
|
||||
&=\SI{5e-7}{\newton}
|
||||
&=\qty{5e-7}{\newton}
|
||||
\end{align*}
|
||||
|
||||
\end{Solution}
|
||||
@ -824,29 +824,29 @@
|
||||
\begin{align*}
|
||||
F&=q\cdot v\cdot B\cdot \sin(90-40)\\
|
||||
&=10\cdot (343+\frac{50}{3,6})\cdot 21,594\cdot 10^{-6}\cdot \sin(50)\\
|
||||
&=\SI{0,059}{\newton}
|
||||
&=\qty{0,059}{\newton}
|
||||
\end{align*}
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{55}
|
||||
Comme les vecteurs vitesse et champ magnétique font un angle de \SI{90}{\degree} entre eux, la force est simplement donnée par :
|
||||
Comme les vecteurs vitesse et champ magnétique font un angle de \qty{90}{\degree} entre eux, la force est simplement donnée par :
|
||||
\begin{align*}
|
||||
F&=q\cdot v\cdot B\\
|
||||
&=1,6\cdot 10^{-19}\cdot 0,01\cdot 3\cdot 10^8\cdot 5\cdot 10^{-9}\\
|
||||
&=\SI{2,4e-21}{\newton}
|
||||
&=\qty{2,4e-21}{\newton}
|
||||
\end{align*}
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{56}
|
||||
La vitesse du proton est :
|
||||
\[v=108\cdot 10^6\,km/h=\SI{3e7}{\metre\per\second}\]
|
||||
\[v=108\cdot 10^6\,km/h=\qty{3e7}{\metre\per\second}\]
|
||||
De l'expression de la force de Lorentz, on tire :
|
||||
\begin{align*}
|
||||
F&=q\cdot v\cdot B\cdot \sin(\alpha)\;\Rightarrow\\
|
||||
sin(\alpha)&=\frac{F}{q\cdot v\cdot B}\\
|
||||
&=\frac{0,6\cdot 10^{-12}}{1,6\cdot 10^{-19}\cdot 3\cdot 10^7\cdot 0,9}\\
|
||||
&=0,139\;\Rightarrow\\
|
||||
\alpha&=\sin^{-1}(0,139)=\SI{8}{\degree}
|
||||
\alpha&=\sin^{-1}(0,139)=\qty{8}{\degree}
|
||||
\end{align*}
|
||||
|
||||
\end{Solution}
|
||||
@ -858,13 +858,13 @@
|
||||
\includegraphics{EclairVoiture.eps}
|
||||
\end{figure}
|
||||
Pour déterminer sa valeur, il faut trouver l'angle \(\beta\) entre \(\overrightarrow L\) et \(\overrightarrow B\). On a que :
|
||||
\[\tan(\beta)=\frac{100}{200}\;\Rightarrow\;\beta=\SI{26,57}{\degree}\]
|
||||
\[\tan(\beta)=\frac{100}{200}\;\Rightarrow\;\beta=\qty{26,57}{\degree}\]
|
||||
La force se calcule alors par :
|
||||
\begin{align*}
|
||||
F&=I\cdot L\cdot B\cdot \sin(\beta)\\
|
||||
&=10'000\cdot \sqrt{200^2+100^2}\cdot 21,594\\
|
||||
&\cdot 10^{-6}\cdot \sin(26,57)\\
|
||||
&=\SI{21,6}{\newton}
|
||||
&=\qty{21,6}{\newton}
|
||||
\end{align*}
|
||||
|
||||
\end{Solution}
|
||||
@ -880,18 +880,18 @@
|
||||
\begin{align*}
|
||||
F&=q\cdot v\cdot B\cdot \sin(90)\;\Rightarrow\\
|
||||
B&=\frac{F}{q\cdot v}=\frac{10^{-15}}{1,6\cdot 10^{-19}\cdot 10^3}\\
|
||||
&=\SI{6,25}{\tesla}
|
||||
&=\qty{6,25}{\tesla}
|
||||
\end{align*}
|
||||
|
||||
\end{Solution}
|
||||
\begin{Solution}{59}
|
||||
L'énergie électrique fournie par le champ électrique à l'électron de charge \(q=\SI{1,6e-19}{\coulomb}\) sous la tension \(U=\SI{1000}{\volt}\) vaut :
|
||||
\[E=q\cdot U=1,6\cdot 10^{-19}\cdot 1000=\SI{1,6e-16}{\joule}\]
|
||||
L'énergie électrique fournie par le champ électrique à l'électron de charge \(q=\qty{1,6e-19}{\coulomb}\) sous la tension \(U=\qty{1000}{\volt}\) vaut :
|
||||
\[E=q\cdot U=1,6\cdot 10^{-19}\cdot 1000=\qty{1,6e-16}{\joule}\]
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Cette énergie est transformée en énergie cinétique et on a :
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||||
\begin{align*}
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E&=\frac{1}{2}\cdot m\cdot v^2\;\Rightarrow\\
|
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v&=\sqrt{\frac{2\cdot E}{m}}=\sqrt{\frac{2\cdot 1,6\cdot 10^{-16}}{9,1\cdot 10^{-31}}}\\
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&=\SI{18752}{\kilo\metre\per\second}=6,25\%\,c
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||||
&=\qty{18752}{\kilo\metre\per\second}=6,25\%\,c
|
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\end{align*}
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Soit 6,25\% de la vitesse de la lumière.
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@ -900,7 +900,7 @@
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Comme le champ est perpendiculaire à la vitesse, le sinus vaut un et on a simplement :
|
||||
\begin{align*}
|
||||
F&=q\cdot v\cdot B\\
|
||||
&=1,6\cdot 10^{-19}\cdot 18'752\cdot 10^3\cdot 10^{-3}=\SI{3e-15}{\newton}\\
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&=1,6\cdot 10^{-19}\cdot 18'752\cdot 10^3\cdot 10^{-3}=\qty{3e-15}{\newton}\\
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\end{align*}
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Comme on le voit sur la figure \ref{deviationmagnetiquesurecran}, le mouvement de l'électron est un mouvement circulaire uniforme (MCU). En effet, la force étant perpendiculaire à la vitesse, elle ne fait que dévier l'électron. Pour un tel mouvement, l'accélération prend la forme suivante :
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\[a=\frac{v^2}{R}\]
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@ -909,7 +909,7 @@
|
||||
F&=q\cdot v\cdot B=m\cdot \frac{v^2}{R}\;\Rightarrow \nonumber \\
|
||||
R&=\frac{m\cdot v}{q\cdot B}\\
|
||||
&=\frac{9,1\cdot 10^{-31}\cdot 18'752\cdot 10^3}{1,6\cdot 10^{-19}\cdot 10^{-3}} \nonumber \\
|
||||
&=\SI{0,1067}{\metre} \nonumber
|
||||
&=\qty{0,1067}{\metre} \nonumber
|
||||
\end{align}
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||||
\begin{figure}[!ht]
|
||||
\centering
|
||||
@ -920,13 +920,13 @@
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Par ailleurs, à partir de la figure \ref{deviationmagnetiquesurecran}, on peut écrire :
|
||||
\begin{align*}
|
||||
L&=R\cdot \sin(\beta)\;\Rightarrow\\
|
||||
\beta&=\arcsin(\frac{L}{R})=\arcsin(\frac{0,03}{0,1067})=\SI{16,33}{\degree}
|
||||
\beta&=\arcsin(\frac{L}{R})=\arcsin(\frac{0,03}{0,1067})=\qty{16,33}{\degree}
|
||||
\end{align*}
|
||||
Et finalement, toujours d'après la figure \ref{deviationmagnetiquesurecran}, on a :
|
||||
\begin{align*}
|
||||
d&=R-R\cdot \cos(\beta)\\
|
||||
&=0,1067-0,1067\cdot \cos(16,33)\\
|
||||
&=\SI{0,0043}{\metre}=\SI{4,3}{\milli\metre}
|
||||
&=\qty{0,0043}{\metre}=\qty{4,3}{\milli\metre}
|
||||
\end{align*}
|
||||
|
||||
\end{Solution}
|
||||
@ -952,8 +952,8 @@
|
||||
\begin{align*}
|
||||
R&=\frac{m\cdot v}{q\cdot B}\;\Rightarrow\\
|
||||
q&=\frac{m\cdot v}{R\cdot B}=\frac{9,1\cdot 10^{-31}\cdot 500'000'000/3,6}{0,01\cdot 10\cdot 10^{-3}}\\
|
||||
&=\SI{1,26e-18}{\coulomb}
|
||||
&=\qty{1,26e-18}{\coulomb}
|
||||
\end{align*}
|
||||
Or, la charge de l'électron \(e=\SI{1,6e-19}{\coulomb}\). Il ne s'agit donc pas d'électrons.
|
||||
Or, la charge de l'électron \(e=\qty{1,6e-19}{\coulomb}\). Il ne s'agit donc pas d'électrons.
|
||||
|
||||
\end{Solution}
|
||||
|
||||
21
texput.log
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21
texput.log
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@ -0,0 +1,21 @@
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This is pdfTeX, Version 3.141592653-2.6-1.40.22 (TeX Live 2022/dev/Debian) (preloaded format=latex 2024.8.20) 18 SEP 2024 09:25
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entering extended mode
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restricted \write18 enabled.
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%&-line parsing enabled.
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**CoursElectriciteMagnetisme.tex.tex
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! Emergency stop.
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||||
<*> CoursElectriciteMagnetisme.tex.tex
|
||||
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||||
*** (job aborted, file error in nonstop mode)
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Here is how much of TeX's memory you used:
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3 strings out of 478288
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141 string characters out of 5849315
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289000 words of memory out of 5000000
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18302 multiletter control sequences out of 15000+600000
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469259 words of font info for 28 fonts, out of 8000000 for 9000
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0i,0n,0p,1b,6s stack positions out of 5000i,500n,10000p,200000b,80000s
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No pages of output.
|
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