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Equa dimension

Go to solution Solved by GregDKO,

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  • Solution


ça donne

lambda^A . M^B . rho^C . c^D = t

(m.kg.s^-3.K^-1)^A . kg^B . (kg.m^-3)^C . (m^2.s^-2.K^-1)^D = s


A+B+C=0    pour kg

A-3C+2D = 0  pour m

-3A-2D = 1  pour s

-A-D = 0  pour K 



Donc -3A+2A = 1     soit A=-1  et D=1

Donc A-3C-2A = 0  soit C = -A/3 = 1/3

Donc A+B-A/3 = 0  soit B = -2A/3 = 2/3


Et donc ça fait 

c/lambda . rho^1/3 . M^2/3


Donc pour moi je dirais e, aucune réponse exacte!

(D'ailleurs il a mis des m dans les propositions alors que ce serait M normalement)

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