{"id":45759,"date":"2019-10-20T03:20:09","date_gmt":"2019-10-20T02:20:09","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/quelle-est-lenergie-cinetique-de-proton-definition\/"},"modified":"2020-02-26T11:05:20","modified_gmt":"2020-02-26T10:05:20","slug":"quelle-est-lenergie-cinetique-de-proton-definition","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/fr\/quelle-est-lenergie-cinetique-de-proton-definition\/","title":{"rendered":"Quelle est l\u2019\u00e9nergie cin\u00e9tique de Proton – D\u00e9finition"},"content":{"rendered":"
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Selon les relations relativistes, l\u2019acc\u00e9l\u00e9ration d\u2019un faisceau de protons \u00e0 5,7 GeV n\u00e9cessite des \u00e9nergies diff\u00e9rentes.\u00a0G\u00e9nie thermique<\/div>\n<\/div>\n
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Exemple: \u00e9nergie cin\u00e9tique de Proton<\/h2>\n

\"Qu'est<\/a>Un proton (\u00a0m = 1,67 x 10\u00a0-27<\/sup>\u00a0kg<\/strong>\u00a0) se d\u00e9place \u00e0 une vitesse\u00a0v = 0,9900c = 2,968 x 10\u00a08<\/sup>\u00a0m \/ s<\/strong>\u00a0.\u00a0Quelle est son\u00a0\u00e9nergie cin\u00e9tique<\/strong>\u00a0?<\/p>\n

Selon un calcul classique, qui n’est pas correct, on obtiendrait:<\/p>\n

K = 1 \/ 2mv\u00a02<\/sup>\u00a0= \u00bd x (1,67 x 10\u00a0-27<\/sup>\u00a0kg) x (2,968 x 10\u00a08<\/sup>\u00a0m \/ s)\u00a02<\/sup>\u00a0=\u00a07,355 x 10\u00a0-11<\/sup>\u00a0J<\/strong><\/p>\n

Avec correction\u00a0relativiste, l\u2019\u00e9nergie cin\u00e9tique relativiste<\/strong>\u00a0est \u00e9gale \u00e0:<\/p>\n

K = (\u0263 – 1) mc\u00a02<\/sup><\/strong><\/p>\n

o\u00f9 le facteur de Lorentz<\/p>\n

= 7,089<\/p>\n

par cons\u00e9quent<\/p>\n

K = 6,089 x (1,67 x 10\u00a0-27<\/sup>\u00a0kg) x (2,9979 x 10\u00a08<\/sup>\u00a0m \/ s)\u00a02<\/sup>\u00a0=\u00a09,139 x 10\u00a0-10<\/sup>\u00a0J = 5,701 GeV<\/strong><\/p>\n

C’est environ\u00a012 fois plus d’<\/strong>\u00a0\u00e9nergie que dans le calcul classique.\u00a0Selon cette relation, une acc\u00e9l\u00e9ration d’un faisceau de protons \u00e0 5,7 GeV n\u00e9cessite des \u00e9nergies de l’ordre diff\u00e9rent.<\/p>\n<\/div>\n<\/div>\n

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……………………………………………………………………………………………………………………………….<\/p>\n

Cet article est bas\u00e9 sur la traduction automatique de l’article original en anglais. Pour plus d’informations, voir l’article en anglais. Pouvez vous nous aider Si vous souhaitez corriger la traduction, envoyez-la \u00e0 l’adresse: translations@nuclear-power.com ou remplissez le formulaire de traduction en ligne. Nous appr\u00e9cions votre aide, nous mettrons \u00e0 jour la traduction le plus rapidement possible. Merci<\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Selon les relations relativistes, l\u2019acc\u00e9l\u00e9ration d\u2019un faisceau de protons \u00e0 5,7 GeV n\u00e9cessite des \u00e9nergies diff\u00e9rentes.\u00a0G\u00e9nie thermique Exemple: \u00e9nergie cin\u00e9tique de Proton Un proton (\u00a0m = 1,67 x 10\u00a0-27\u00a0kg\u00a0) se d\u00e9place \u00e0 une vitesse\u00a0v = 0,9900c = 2,968 x 10\u00a08\u00a0m \/ s\u00a0.\u00a0Quelle est son\u00a0\u00e9nergie cin\u00e9tique\u00a0? Selon un calcul classique, qui n’est pas correct, on obtiendrait: … Lire la suite<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[8],"tags":[],"yoast_head":"\nQuelle est l\u2019\u00e9nergie cin\u00e9tique de Proton - D\u00e9finition<\/title>\n<meta name=\"description\" content=\"Selon les relations relativistes, l\u2019acc\u00e9l\u00e9ration d\u2019un faisceau de protons \u00e0 5,7 GeV n\u00e9cessite des \u00e9nergies diff\u00e9rentes. 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