Exercice 1
M : masse du noyau de l'atome de cuivre
M = 29 x 1.673 x 10^-27 + 34 x 1.675 x 10^-27 (^ se lit puissance)
M = 48,517 x 10^-27 + 56,95 x 10^-27
M = (48,517 + 56,95) x 10^-27
M = 105,467 x 10^-27
M = 1,05467 x 10^2 x 10^-27 ( rappel : 10^a x 10^b =10^(a+b))
M = 1,05467 x 10^(2-27)
M = 1,05467 x 10^(-25)
La masse d'un noyau de cuivre est de 1,05467 x 10^(-25) kg
Exercice 2
1. V(13+V(7+V(3+V(1)))) (V se lit racine de)
or V(1) = 1
V(13+V(7+V(3+V(1)))) = V(13+V(7+V(3+1)))
V(13+V(7+V(3+V(1)))) = V(13+V(7+V(4)))
or V(4) = 2
V(13+V(7+V(3+V(1)))) = V(13+V(7+2))
V(13+V(7+V(3+V(1)))) = V(13+V(9))
or V9 = 3
V(13+V(7+V(3+V(1)))) = V(13+3)
V(13+V(7+V(3+V(1)))) = V(16)
V(13+V(7+V(3+V(1)))) = 4
2.
V(x+V(13+V(7+V(3+V(1))))) = 5
or V(13+V(7+V(3+V(1)))) = 4 donc
V(x+4) = 5
donc x+4 = 25
x= 25-4
x = 21
Exercice 3
Hauteur de l'immeuble = BH
BH = BI + IH
BI = 1,70 m
Calcul de IH
Dans le triangle OIH rectagle en I, l'angle O = 76,8°
Tan(ô) = IH/OI or OI = PB = 5
tan(76,8°) = IH / 5
IH = tan(76,8°) x 5
IH = 21,32 m
BH = 1.70 + 21,32
BH = 23,02
L'immeuble mesure 23 m de haut
