Examen de Sciences Physiques - Baccalauréat Algérien

1 10
04  30 
2024
☞
     :
)05 () 1  10  5  10(
14 
04 
120
  .
⋅-2 = 9,80g ms
 :
m
0t=
O
() Oy
 1.
1. 
2. 
()ty
.
3.  
⋅
-1
72,11m s
.
h
.
ﺮﺋاﺰﺠﻟا ﻊﻣﺎﺟ
ّﺸﻟاﻞﻜ 1
•
O
y
h
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☞
  //   //  2024
2 10
:
0 t=
0
v
O
P
 2(.
( )
2
C
E =f t
 3( 
.
1. 
C
E (t)
22 2
0
11
22
c
E (t)= m g t + m v
 3 : 
= 100mg
2. 
O
P
 3 
h
( .
04 
201
.
201
.
( )
201
1 2 81
Tl 73t = heures
( )
202
1 2 81Tl = 294t' heures
:
23 -1
A= 6,02×10N mol
201
:
⋅
-1
= 201M g mol
1.  
201
+
β
Hg
γ
.
1.1 .
2.1. 
201
.
2. 
8
201
6
153,9×10 Bq
8
7
11×10 Bq
.
1.2 . 
()At
 .
2.2 .
 2
 3
0
10
40
22
()ts
J
c()E
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☞
  //   //  2024
3 10
3. 
8
202
02
A
202
01
A
201
02
01
= 0,005
A
A
.
1.3. 
202
81
201
81
( Tl)
( Tl)
A
A
  :
=
202
81
201
81
-6
1,982×10
( Tl) 0,005×
( Tl)
t
A
Ae
2.3. 
202
201
2%
.
.
:)06 
–
–
-1
Cu = 63,5M g mol
0t=
( ) ( )
( )
+-
3
Ag + NOaq aq
100V = mL
c
= 6, 35mg
( ) ( ) ( ) ( )
+ 2+
Cu + 2 Ag = Cu + 2 Ags aq aq s
1. 
2. 
( )
2+
Cu = f t
 4.(
1.2 .    .
2.2 .
3.2 .   .
3 .
0t=
.
-
2+ 2+
Pb + Sn = Pb + Snaq s s aq
= 2,18K
-1
Pb 207,2M = g mol
.
0
1
15
( )
t min
( )
 ⋅
2+ -1
Cu mmol L
 4
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☞
  //   //  2024
4 10
o
25 C
- 
Pb
2+ -
3
Pb 2NOaq + aq
50
1
V = mL
 ⋅
2+ -2 -1
Pb = 3 ×10 mol L
.
- 
Sn
2+ -
3
Sn 2NO+aq aq
2
= 50V mL
 ⋅
2+ -2 -1
Sn = 2 ×10 mol L
.
+-
3
K + NOaq aq
K
.
K
0t=
1. 
r ,i
Q
.
2. 
3.    .
4. 
5. 
Δt
 ⋅
2+ -2 -1
Pb = 1,572 ×10 mol L
 ⋅
2+ -2 -1
Sn = 3,428 × 10 mol L
5.1. 
r
Q
2.5. 
) :06 
)06 
01
02
:
()ut
b
E
1= 70ΩR
2= 80ΩR
1
L
1= 30Ωr
2
L
2= 20Ω
r
K
1
0
( )
Vub
( )
mst
2
1
2
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☞
  //   //  2024
5 10
1.   5.
K
0 t=
.
1.1 .
2.1. 
()it
3.1 . :
( )
⋅
0
1
1-t
τ
I
i t = -e
،
:
0
I
τ
.
( )
⋅
0b
1
-t
τ
IR u t = r+ e
2.  
RL
( )
b
ut
5 (01 
01 02 
01
( )
11 130LΩb ,r=
1
70=RΩ
(V)E
02
( )
22 220LΩ
b ,r =
2
= 80RΩ
(V)E
K
0 t =
( )
b
ut
( )
ExAO
   02(.
2.1.  02
( )
b
ut
 .
2.2 .
2.3.  
( )
1
b
ut
01(
4.2. 
-
E
- 
1
τ
 01( 
2
τ
  02.(
2.5. 
1
L
2
L
.
6.2.   02  01.
 5
( )
,Lr
R
E
K
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