DS 09/10

Exercice 1 : Points
1. z
emggmP ------------------------------------------------------- 1
2. Fam ---------------------------------------------------------------- 1
Pam ---------------------------------------------------------------- 1
3. gtaz)( ---------------------------------------------------------------- 1
4. dttdztvz/)()( ------------------------------------------------------- 1
0 )( vgttvz ------------------------------------------------------- 1
5. 0)( 0 tvz ---------------------------------------------------------------- 1
gvt /
00 ---------------------------------------------------------------- 1
6. h
g
v
z2
0
max 2
1 ------------------------------------------------------- 1
7. 0
2
110
2
1htvgt ------------------------------------------------------- 1
eq. du 2nd ordre, discriminant : ghv 2
2
0---------------------------- 1
on ne garde que la solution 0
1
t ------------------------------------- 1
g
ghvv
t2
2
00
1 ------------------------------------------------------- 1
8. 2
0
ts ---------------------------------------------------------------- 1
5
1
ts ---------------------------------------------------------------- 1
45
max
zm ------------------------------------------------------- 1
Exercice 2 :
1. dttdqti /)()( ------------------------------------------------------- 1
)cos()( 0tqti ------------------------------------------------------- 1
2. )()( tCutq C ------------------------------------------------------- 1
)sin()/()( 0tCqtuC ---------------------------------------------- 1
3. dttLdituL/)()( ------------------------------------------------------- 1
)sin()( 2
0tLqtuL ---------------------------------------------- 1
4. )()( tRituR ---------------------------------------------------------------- 1
)cos()( 0tRqtuR ------------------------------------------------------- 1
5. )()()()( tutututu CLR ---------------------------------------------- 1
Car dipoles en série additivité des tensions ------------------- 1
)cos()sin()1()( 0
2
0tRqtLC
C
q
tu ---------------------------- 1
Note finale = total des points * 20/27 puis arrondi au 1/4 point
supérieur

ε0
FCq1q2r
FC=1
4πε0
q1q2
r2
[ε0] = M−1L−3T4I2
µ0
FLL
d I1I2
FL=µ0
4πI1I2
L
d
[µ0] = MLT −2I−2
ε0µ0c2= 1 c
M−1L−3T4I2MLT −2I−2LT −1= 1
O
h
g= 9,80 −2
α Oz
vlim
v=f(t)
v

a(t)a(t) = dv(t)
dt
v=f(t)t
v(t)
−αv
a=g−αv
mv
v
Oz dv
dt +α mv=g
v(t) = Ce−
α
mtv=mg
αv(t) = mg α+Ce−
α
mt
v(t)
v(t= 0) = 0 ⇒C=−mg
αv(t) = mg α³1−e−
α
mt´
vlim F
P
vlim
P
P(t)P0
dP (t)
dt =kP (t)
P(t) = P0exp(kt)9
k= 4,39.10−10 −1