Studyadz 3as-physique-c3-39-54

1. ‫ﺘﺒﺴــﺔ‬ ‫ﻟوﻻﻴﺔ‬ ‫اﻟﺘرﺒﻴــﺔ‬ ‫ﻤدﻴرﻴــﺔ‬
‫اﻟﺘﺎرﻴــﺦ‬
:
23
/
05
/
2019
‫اﻟﻤﺴﺘوى‬
:
‫ﺘﺠرﻴﺒﻴﺔ‬ ‫ﻋﻠوم‬ ‫اﻟﺜﺎﻟﺜـﺔ‬
‫اﻤﺘﺤـــــــﺎن‬
‫ﻤــﺎي‬ ‫اﻟﺘﺠرﻴﺒــــﻲ‬ ‫اﻟﺒﻛﺎﻟـــــورﻴﺎ‬
2019
‫اﻟﻤـــ‬
‫ـــــــ‬
‫ــدة‬
:
03
‫ﺴــﺎ‬
،
30
‫د‬
‫ﻤــــــﺎدة‬ ‫ﻓﻲ‬ ‫اﺨﺘﺒــــﺎر‬
:
‫اﻟﻔﻴزﻴﺎﺌﻴـــــــﺔ‬ ‫اﻟﻌﻠـــــــــوم‬
‫اﻟﺘﻔﺘﻴﺸﻴﺔ‬ ‫اﻟﻤﻘﺎطﻌـــﺔ‬
:
‫ﺘﺒﺴـــــﺔ‬
02
‫ﺻﻔﺤﺔ‬
1
‫ﻣﻦ‬
8
‫اﻟﺨﻴﺎر‬ ‫ﻋﻠﻰ‬ ‫ﻓﻘط‬ ‫اﺤدا‬‫و‬ ‫ﻤوﻀوﻋﺎ‬ ‫ﻋﺎﻟﺞ‬
‫اﻟﻤوﻀـــ‬
‫اﻷوﻝ‬ ‫ع‬
‫ــو‬
:
‫اﻟﺠ‬
‫ـــــ‬
‫اﻷوﻝ‬ ‫ء‬
‫ز‬
:
‫ﻴﺘﻛ‬
‫ـــ‬
‫ﻤ‬ ‫ون‬
‫ـــ‬
‫ن‬
‫ﺘﻤرﻴﻨﻴن‬
.
‫اﻟ‬
‫ﺘﻤرﻴ‬
‫ــــــ‬
‫ن‬
‫اﻷوﻝ‬
) :
00
.
06
‫ﻨﻘﺎط‬
(
Ι
–
‫اﻟﺒﻠوﺘوﻨﻴوم‬ ‫ﻤن‬ ‫ﻤﺸﻌﺔ‬ ‫ﻋﻴﻨﺔ‬
Pu
239
94
.
‫ﻛﺘﻠﺘﻬﺎ‬
g
m 1
0 =
‫ﻟﻨﺸﺎطﻬﺎ‬ ‫ﻤﺤﺎﻛﺎة‬ ‫اﺴطﺔ‬‫و‬‫وﺒ‬ ،
‫اﻟﺸﻛﻝ‬ ‫اﻟﺒﻴﺎن‬ ‫ﻋﻠﻰ‬ ‫اﻟﺤﺼوﻝ‬ ‫ﻤن‬ ‫ﺘﻤﻛﻨﺎ‬
-
1
‫أ‬
–
‫أن‬ ‫ﺒﻴن‬
t
e
m
t
m λ
−
= 0
)
(
‫ﻋﻼﻗﺔ‬ ‫ﻤن‬ ‫اﻨطﻼﻗﺎ‬
‫اﻻﺸﻌﺎﻋﻲ‬ ‫اﻟﺘﻨﺎﻗص‬
t
e
N
t
N λ
−
= 0
)
(
‫ﺤﻴث‬ ،
)
(t
m
‫اﻟﻠﺤظﺔ‬ ‫ﻋﻨد‬ ‫اﻟﻤﺘﺒﻘﻴﺔ‬ ‫اﻷﻨوﻴﺔ‬ ‫ﻛﺘﻠﺔ‬
t
‫ب‬
–
‫أن‬ ‫ﺒﻴن‬
t
m
m
λ
=
0
ln
‫اﻹﺸﻌﺎﻋﻲ‬ ‫اﻟﻨﺸﺎط‬ ‫ﺜﺎﺒت‬ ‫أﺤﺴب‬ ‫ﺜم‬
λ
‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ﺒ‬
1
−
s
‫ـ‬
‫ﺠ‬
‫ـ‬
-
‫اﻻﺒﺘداﺌﻴﺔ‬ ‫اﻷﻨوﻴﺔ‬ ‫ﻋدد‬ ‫أﺤﺴب‬
0
N
‫ﻓﻲ‬ ‫اﻟﻤوﺠودة‬
،‫اﻟﻌﻴﻨﺔ‬
‫اﺴﺘﻨﺘﺞ‬‫و‬
‫اﻟﻨﺸﺎط‬
‫اﻻﺒﺘداﺌﻲ‬
0
A
‫ﻟﻠﻌﻴﻨﺔ‬
.
‫د‬
–
‫أن‬ ‫ﺒﻴن‬ ‫ﺜم‬ ‫اﻟﻌﻤر‬ ‫ﻨﺼف‬ ‫زﻤن‬ ‫ﻋرف‬
λ
2
ln
2
/
1 =
t
،
‫ﻗﻴﻤﺘﻪ‬ ‫أﺤﺴب‬ ‫ﺜم‬
.
‫ه‬
–
‫أن‬ ‫ﺒﻴن‬
:
2
/
1
2
)
( 0
t
t
m
t
m =
،
‫اﻟﻠﺤظﺔ‬ ‫ﻋﻨد‬ ‫اﻟﻤﺘﺒﻘﻴﺔ‬ ‫اﻷﻨوﻴﺔ‬ ‫ﻛﺘﻠﺔ‬ ‫اﺴﺘﻨﺘﺞ‬ ‫ﺜم‬
2
/
1
2t
t =
.
‫و‬
–
‫اﻟﺘﻲ‬ ‫اﻟﻠﺤظﺔ‬ ‫أوﺠد‬
‫ﻷﻨوﻴﺔ‬ ‫اﻟﻤﺌوﻴﺔ‬ ‫اﻟﻨﺴﺒﺔ‬ ‫ﻓﻴﻬﺎ‬ ‫ﺘﻛون‬
‫اﻟﻤﺘﺒﻘﻴﺔ‬ ‫اﻟﺒﻠوﺘوﻨﻴوم‬
%
20
=
r
.
ΙΙ
–
‫اﻟﺒﻠوﺘوﻨﻴوم‬
239
‫اﻟﻨووﻴﺔ‬ ‫اﻟﻤﻔﺎﻋﻼت‬ ‫ﻓﻲ‬ ‫ﻨووي‬ ‫ﻛوﻗود‬ ‫ﺘﺴﺘﺨدم‬ ‫اﻟﺘﻲ‬ ‫اد‬‫و‬‫اﻟﻤ‬ ‫ﻤن‬ ‫ﻫو‬ ‫و‬ ‫اﻟﺒﻠوﺘوﻨﻴوم‬ ‫ﻨظﺎﺌر‬ ‫أﺤد‬ ‫ﻫو‬
‫ﺒﺎﺌﻴﺔ‬‫ر‬‫اﻟﻛﻬ‬ ‫اﻟطﺎﻗﺔ‬ ‫ﻹﻨﺘﺎج‬
،
‫ﻹﻨﺸطﺎر‬ ‫اﻟﻤﻤﻛﻨﺔ‬ ‫اﻟﺘﻔﺎﻋﻼت‬ ‫أﺤد‬ ‫ﻴﻨﻤذج‬
Pu
239
94
‫اﻟﺘﺎﻟﻴﺔ‬ ‫ﺒﺎﻟﻤﻌﺎدﻟﺔ‬
:
n
Te
Mo
n
Pu 1
0
135
52
102
42
1
0
239
94 3
+
+
→
+
1
-
‫اﻟﻨووي‬ ‫اﻻﻨﺸطﺎر‬ ‫ﺘﻔﺎﻋﻝ‬ ‫ﻋرف‬
.
2
-
‫اﻟﻨووي‬ ‫اﻟﺘﻔﺎﻋﻝ‬ ‫ﻫذا‬ ‫ﻤن‬ ‫اﻟﻨﺎﺘﺠﺔ‬ ‫اﻷﻨوﻴﺔ‬ ‫ﺒﻴن‬ ‫ﻤن‬ ‫ا‬
‫ر‬‫ا‬
‫ر‬‫اﺴﺘﻘ‬ ‫اﻷﻛﺜر‬ ‫اة‬‫و‬‫اﻟﻨ‬ ‫ﻤﺎﻫﻲ‬
)
‫اﻻﻨﺸطﺎر‬
(
.
3
-
‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ﺒ‬ ‫اﻟﺘﻔﺎﻋﻝ‬ ‫ﻫذا‬ ‫ﻤن‬ ‫ة‬
‫ر‬‫اﻟﻤﺤر‬ ‫اﻟطﺎﻗﺔ‬ ‫أﺤﺴب‬
)
Mev
(
‫ﺜم‬
‫ﺒﺎﻟﺠوﻝ‬
)
J
(
.
4
-
‫اﻨﺸطﺎر‬ ‫ﻋن‬ ‫ة‬
‫ر‬‫اﻟﻤﺤر‬ ‫اﻟطﺎﻗﺔ‬ ‫أﺤﺴب‬
1g
‫اﻟﺒﻠوﺘوﻨﻴوم‬ ‫ﻤن‬
Pu
239
94
‫ﺒﺎﻟﺠوﻝ‬
)
J
. (
e
n
c
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
‫اﻟﺠ‬
‫ـــــ‬
‫اﻷوﻝ‬ ‫ء‬
‫ز‬ ‫ﺠ‬
‫ـــــ‬
‫اﻷوﻝ‬ ‫ء‬
‫ز‬
a
m
:
‫ﻴﺘﻛ‬ :
‫اﻟ‬
‫ﺘﻤرﻴ‬
‫ــــــ‬
‫ن‬
‫اﻷوﻝ‬
) :
0 ‫رﻴ‬
‫ــــــ‬
‫ن‬
‫اﻷوﻝ‬
x
a
‫اﻟﺒﻠوﺘوﻨﻴوم‬ ‫ﻤن‬ ‫ﻤﺸﻌﺔ‬
‫اﻟﺒﻠوﺘوﻨﻴو‬ ‫ﻤن‬ ‫ﻌﺔ‬
g
m
m 1
0
0
‫اﺴطﺔ‬‫و‬‫وﺒ‬ ،
‫ا‬‫و‬‫وﺒ‬ ،
‫اﻟﺸﻛ‬ ‫اﻟﺒﻴﺎن‬ ‫ﻋﻠﻰ‬ ‫ﺤﺼوﻝ‬
‫اﻟﺒﻴﺎن‬ ‫ﻋﻠﻰ‬
t
m
m )
(
(t
‫ﻤن‬ ‫اﻨطﻼﻗﺎ‬
‫اﻨطﻼﻗﺎ‬
e
N
t
N N −
0
)
( )
(t
t
‫ﺤﻴ‬ ،
‫اﻟﻠﺤظﺔ‬ ‫ﻋﻨد‬ ‫ﻴﺔ‬
‫اﻟﻠﺤظﺔ‬
t
1
−
s
‫ـ‬
‫ﻲ‬
‫ﻗﻴﻤﺘﻪ‬ ‫أﺤﺴب‬ ‫ﺜم‬
‫ﻗﻴﻤﺘﻪ‬ ‫ﺤﺴب‬
.
‫ﻠﺤظﺔ‬
2
/
1 2
/
1
2t
t 2
2
.
=
r
.
‫اﻟﻨووﻴﺔ‬ ‫اﻟﻤﻔﺎﻋﻼت‬ ‫ﻲ‬
‫اﻟﻨووﻴﺔ‬ ‫ﻋﻼت‬
3as.ency-education.com

2. ‫ﺻﻔﺤﺔ‬
2
‫ﻣﻦ‬
8
5
-
‫ﺒﺎﺌﻴﺔ‬‫ر‬‫اﻟﻛﻬ‬ ‫اﺴﺘطﺎﻋﺘﻪ‬ ‫ﻨووي‬ ‫ﻤﻔﺎﻋﻝ‬ ‫ﻓﻲ‬ ‫ﺒﺎﺌﻴﺔ‬‫ر‬‫اﻟﻛﻬ‬ ‫اﻟطﺎﻗﺔ‬ ‫إﻨﺘﺎج‬ ‫ﻓﻲ‬ ‫اﻟﺴﺎﺒﻘﺔ‬ ‫اﻟطﺎﻗﺔ‬ ‫ﺘﺴﺘﻌﻤﻝ‬
MW
P 300
=
.
‫طﺎﻗوي‬ ‫ﺒﻤردود‬
%
30
=
r
،
‫اﻟﺴﺎﺒﻘﺔ‬ ‫اﻟﻛﺘﻠﺔ‬ ‫ﻻﺴﺘﻬﻼك‬ ‫اﻟﻼزﻤﺔ‬ ‫اﻟزﻤﻨﻴﺔ‬ ‫اﻟﻤدة‬ ‫أﺤﺴب‬
.
‫اﻟﻤﻌطﻴﺎت‬
:
‫اﻟﺜﺎ‬ ‫اﻟﺘﻤرﻴن‬
‫ﻨـــﻲ‬
) :
07.00
‫ﻨﻘﺎط‬
(
‫ﻤن‬ ‫ﻴﺘﻛون‬ ‫ﻴن‬‫ر‬‫اﻟﺘﻤ‬
‫أ‬
‫ز‬‫ﺠ‬
‫ﻴ‬
‫ن‬
‫ﻤﺴﺘﻘ‬
‫ﻠﻴ‬
‫ن‬
)
‫ﺘ‬
‫ﻌطﻲ‬
2
.
8
,
9 −
= s
m
g
(
‫اﻻوﻝ‬ ‫ء‬
‫اﻟﺠز‬
:
1
-
‫ﺒﺄﺤد‬
‫ﺤﻤوﻟﺔ‬ ‫ﺤرﻛﺔ‬ ‫ﺘﺼوﻴر‬ ‫ﺘم‬ ‫اﻟﺒﻨﺎء‬ ‫ورﺸﺎت‬
(C)
‫ﻋطﺎﻟﺘﻬﺎ‬ ‫ﻤرﻛز‬
G
‫وﻛﺘﻠﺘﻬﺎ‬
m= 400kg
‫أﺜﻨﺎء‬
‫رﻓﻌﻬﺎ‬
.
‫ﻋﻠﻰ‬ ‫ﻻذي‬
‫اﻟﻔو‬ ‫اﻟﺤﺒﻝ‬ ‫ﻴطﺒق‬ ‫اﻟﺤرﻛﺔ‬ ‫ﺨﻼﻝ‬
‫اﻟﺤﻤوﻟﺔ‬
)
C
(
‫اﻻﺤﺘﻛﺎﻛﺎت‬ ‫ﺠﻤﻴﻊ‬ ‫ﻨﻬﻤﻝ‬ ، ‫ﺜﺎﺒﺘﺔ‬ ‫ﻗوة‬
.
‫ﺤرﻛﺔ‬ ‫ﻴط‬‫ر‬‫ﺸ‬ ‫ﻤﻌﺎﻟﺠﺔ‬ ‫ﺒﻌد‬
)
C
(
‫ﻓﻲ‬ ‫اﻟﻤﻤﺜﻝ‬ ‫اﻟﻤﻨﺤﻨﻰ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺤﺼوﻝ‬ ‫ﺘم‬ ‫ﻤﻨﺎﺴب‬ ‫ﻨﺎﻤﺞ‬‫ر‬‫ﺒ‬ ‫اﺴطﺔ‬‫و‬‫ﺒ‬
‫اﻟﺸﻛﻝ‬
-
02
‫أ‬
-
‫ﻛﻝ‬ ‫ﻓﻲ‬ ‫اﻟﺠﺴم‬ ‫ﻋطﺎﻟﺔ‬ ‫ﻤرﻛز‬ ‫ﺤرﻛﺔ‬ ‫طﺒﻴﻌﺔ‬ ‫ﺤدد‬
‫طور‬
.
‫ب‬
-
‫اﻟﻘوة‬ ‫ﺸدة‬ ‫اوﺠد‬ ‫ﻟﻨﻴوﺘن‬ ‫اﻟﺜﺎﻨﻲ‬ ‫اﻟﻘﺎﻨون‬ ‫ﺒﺘطﺒﻴق‬
T
r
‫اﻟﺘﻲ‬
‫ﻛﻝ‬ ‫ﻓﻲ‬ ‫ﻻذي‬
‫اﻟﻔو‬ ‫اﻟﺤﺒﻝ‬ ‫ﻴطﺒﻘﻬﺎ‬
‫طور‬
.
2
-
‫ﻤﻌﻴن‬ ‫ﺘﻔﺎع‬‫ر‬‫ا‬ ‫ﻋﻨد‬ ‫اﻟﺤرﻛﺔ‬ ‫ﻋن‬ ‫اﻟﺤﻤوﻟﺔ‬ ‫ﺘﺘوﻗف‬
،
‫اﻟﻠﺤظﺔ‬ ‫ﻓﻲ‬
t=0
‫ء‬
‫ﺠز‬ ‫ﻤﻨﻬﺎ‬ ‫ﻴﺴﻘط‬
(S)
‫ﻛﺘﻠﺘﻪ‬
ms=30kg
‫دون‬
‫اﺒﺘداﺌﻴﺔ‬ ‫ﻋﺔ‬
‫ﺴر‬
،
‫اﻟﻌطﺎﻟﺔ‬ ‫ﻤرﻛز‬ ‫ﺤرﻛﺔ‬ ‫ﻨدرس‬
Gs
‫ء‬
‫ﻟﻠﺠز‬
)
S
(
‫اﻟﻠﺤظﺔ‬ ‫ﻋﻨد‬ ‫ﺤﻴث‬
t=0
‫ء‬
‫اﻟﺠز‬ ‫ﻴﻨطﻠق‬
)
S
(
‫ﻤن‬
‫اﻟﻨﻘطﺔ‬
O
‫ﻤﺘﺠﻬ‬
‫ﺎ‬
‫اﻟﺸﻛﻝ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬ ‫اﻷﺴﻔﻝ‬ ‫ﻨﺤو‬
3
.
‫ﺘ‬
‫ة‬
‫ر‬‫ﺒﺎﻟﻌﺒﺎ‬ ‫اء‬‫و‬‫اﻟﻬ‬ ‫ﻤﻊ‬ ‫اﻻﺤﺘﻛﺎك‬ ‫ﻗوة‬ ‫ﻌطﻲ‬
j
Kv
f
r
r 2
−
=
‫ﺤﻴث‬
K=2,7 SI
‫ارﺨﻤﻴدس‬ ‫داﻓﻌﺔ‬ ‫ﺘﺄﺜﻴر‬ ‫ﻨﻬﻤﻝ‬ ،
.
e
n
c
y
y
-
e
o
n
.
c
o
m
/
e
x
a
m
s
‫اﻟﺜﺎ‬ ‫ﻤرﻴن‬
‫اﻟﺜﺎﻨـــﻲ‬ ‫ن‬
‫ﻨـــﻲ‬
x
a
) :
7.00 ) :
‫ﻤن‬ ‫ﻴﺘﻛون‬
‫ﻤن‬ ‫ون‬
‫أ‬
‫ز‬‫ﺠ‬
‫ﻴ‬
‫ن‬
‫ﻤﺴﺘﻘ‬ ‫أ‬
‫ز‬‫ﺠ‬
‫ﻴ‬
‫ن‬
‫ﻤﺴﺘﻘ‬
‫ﺤرﻛﺔ‬ ‫ﺘﺼوﻴر‬ ‫ﺘم‬ ‫اﻟﺒﻨﺎء‬
‫ﺤ‬ ‫ﺘﺼوﻴر‬ ‫ﺘم‬
‫ﻋﻠﻰ‬ ‫ﻻذي‬
‫اﻟﻔو‬ ‫اﻟﺤﺒﻝ‬ ‫ق‬
‫ﻋ‬ ‫ﻻذي‬
‫اﻟﻔو‬ ‫ﺒﻝ‬
‫ﺔ‬
)
C
(
‫ﻤ‬ ‫ﻨﺎﻤﺞ‬‫ر‬‫ﺒ‬ ‫اﺴطﺔ‬‫و‬‫ﺒ‬
‫ﻨﺎﻤ‬‫ر‬‫ﺒ‬ ‫اﺴطﺔ‬‫و‬‫ﺒ‬
o
n
.
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3as.ency-education.com

3. ‫ﺻﻔﺤﺔ‬
3
‫ﻣﻦ‬
8
‫أ‬
-
‫اﻟﺒﻌدي‬ ‫اﻟﺘﺤﻠﻴﻝ‬ ‫ﻋﻠﻰ‬ ‫ﺒﺎﻻﻋﺘﻤﺎد‬
،
‫ﻟﻠﺜﺎﺒت‬ ‫اﻟدوﻟﻴﺔ‬ ‫اﻟوﺤدة‬ ‫اوﺠد‬
K
‫ب‬
-
‫اﺜﺒت‬
‫أن‬
‫ﻫﻲ‬ ‫ﻋﺔ‬
‫اﻟﺴر‬ ‫ﺒدﻻﻟﺔ‬ ‫ﻟﻠﺤرﻛﺔ‬ ‫اﻟﺘﻔﺎﻀﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬
:
8
,
9
10
9 2
2
=
×
+ −
v
dt
dv
‫ـ‬‫ﺠ‬
-
‫اﻟﺤدﻴﺔ‬ ‫ﻋﺔ‬
‫اﻟﺴر‬ ‫ﺤدد‬
lim
v
‫ﻟﻠﺤرﻛﺔ‬
.
‫د‬
-
‫أ‬
‫أن‬ ‫ﺜﺒت‬
‫اﻟﻠﺤظﺘﻴن‬ ‫ﺒﻴن‬ ‫اﻟﺠﺴم‬ ‫ﻋطﺎﻟﺔ‬ ‫ﻟﻤرﻛز‬ ‫اﻟوﺴطﻲ‬ ‫ع‬
‫اﻟﺘﺴﺎر‬ ‫ﻗﻴﻤﺔ‬
:
t2=τ , t1=0
‫ﻫﻲ‬
τ
β
=
m
a
‫ﺤﻴث‬
β
‫ﻗﻴﻤﺘﻪ‬ ‫ﺘﺤدﻴد‬ ‫ﻴطﻠب‬ ‫ﺜﺎﺒت‬
.
‫اﻟﺜﺎﻨﻲ‬ ‫ء‬
‫اﻟﺠز‬
:
‫ﻤرن‬ ‫اس‬‫و‬‫ﻨ‬ ‫ﻟﻴﻛن‬
‫أﻓﻘﻲ‬
‫ﺼﻠب‬ ‫ﺠﺴم‬ ‫ﻤن‬ ‫ﻴﺘﻛون‬
(S)
‫ﻛﺘﻠﺘﻪ‬
m
‫ﻋطﺎﻟﺘﻪ‬ ‫وﻤرﻛز‬
G
‫ﻏﻴر‬ ‫ﺤﻠﻘﺎﺘﻪ‬ ‫ﻨﺎﺒض‬ ‫ﺒطرف‬ ‫ﻤﺜﺒت‬ ،
‫ﻤﻬﻤﻠﺔ‬ ‫وﻛﺘﻠﺘﻪ‬ ‫ﻤﺘﻼﺼﻘﺔ‬
‫ﻤروﻨﺘﻪ‬ ‫ﺜﺎﺒت‬ ،
1
.
10 −
= m
N
K
‫اﻟطرف‬ ،
‫اﻵﺨر‬
‫ﻟق‬‫ز‬‫ﻴﻨ‬ ، ‫ﺜﺎﺒت‬ ‫ﺒﺤﺎﻤﻝ‬ ‫ﺘﺒط‬‫ر‬‫ﻤ‬ ‫ﻟﻠﻨﺎﺒض‬
‫اﻟﺠﺴم‬
(S)
‫اﻟﻤﺴﺘوي‬ ‫ﻓوق‬ ‫اﺤﺘﻛﺎك‬ ‫دون‬
‫اﻷﻓﻘﻲ‬
،
‫اﻟﺠﺴم‬ ‫ﻴﺢ‬‫ز‬‫ﻨ‬
)
S
(
‫أﻓﻘﻴﺎ‬
‫اﻟﻤوﺠب‬ ‫اﻻﺘﺠﺎﻩ‬ ‫ﻓﻲ‬ ‫ﻨﻪ‬‫ز‬‫ا‬‫و‬‫ﺘ‬ ‫وﻀﻊ‬ ‫ﻋن‬
‫ﺒﻤﺴﺎﻓﺔ‬
X0
‫ﻩ‬
‫ر‬‫وﻨﺤر‬
‫ﻤﺒدأ‬ ‫ﻫﺎ‬
‫ﻨﻌﺘﺒر‬ ‫ﻟﺤظﺔ‬ ‫ﻋﻨد‬ ‫اﺒﺘداﺌﻴﺔ‬ ‫ﻋﺔ‬
‫ﺴر‬ ‫دون‬
‫ﻟﻸزﻤﻨﺔ‬
.
‫ة‬
‫ز‬‫اﻟﻤﻬﺘ‬ ‫اﻟﺠﻤﻠﺔ‬ ‫طﺎﻗﺔ‬ ‫ات‬
‫ر‬‫ﺘﻐﻴ‬ ‫ﻤﺘﺎﺒﻌﺔ‬
)
‫ﺠﺴم‬
-
‫ﻨﺎﺒض‬
(
‫ﻤن‬ ‫ﻤﻛﻨﺘﻨﺎ‬
‫ﻤن‬ ‫ﻛﻝ‬ ‫ات‬
‫ر‬‫ﻟﺘﻐﻴ‬ ‫اﻟﻤﻤﺜﻠﻴن‬ ‫اﻟﻤﻨﺤﻨﻴﻴن‬ ‫ﻋﻠﻰ‬ ‫اﻟﺤﺼوﻝ‬
‫اﻟﺤرﻛﻴﺔ‬ ‫اﻟطﺎﻗﺔ‬
EC
‫اﻟﻤروﻨﻴﺔ‬ ‫اﻟﻛﺎﻤﻨﺔ‬ ‫اﻟطﺎﻗﺔ‬‫و‬
Epe
‫اﻟﺸﻛﻝ‬ ‫ﻓﻲ‬ ‫ﻛﻤﺎ‬
-
5
.
1
-
‫اﻟﻤﻨﺤﻨﻴﻴن‬ ‫ﻤن‬ ‫ﻋﻴن‬
)
‫أ‬
(
‫و‬
)
‫ب‬
(
‫اﻟﺤرﻛﻴﺔ‬ ‫اﻟطﺎﻗﺔ‬ ‫ات‬
‫ر‬‫ﺘﻐﻴ‬ ‫ﻴﻤﺜﻝ‬ ‫اﻟذي‬ ‫اﻟﻤﻨﺤﻨﻰ‬ ،
EC
.
‫ﻋﻠﻝ‬
‫إﺠﺎﺒﺘك‬
‫ﺤﻴ‬
‫اﻟﺜﺎﻨﻲ‬ ‫ء‬
‫اﻟﺠز‬
: ‫اﻟﺜﺎﻨﻲ‬ ‫ء‬
‫ﺠز‬
:
a
m
‫ﻤرن‬ ‫اس‬‫و‬‫ﻨ‬ ‫ﻟﻴﻛن‬
‫ﻤرن‬ ‫اس‬‫و‬‫ﻨ‬ ‫ن‬
‫أﻓﻘﻲ‬
‫ﻤﻬﻤﻠﺔ‬ ‫وﻛﺘﻠﺘﻪ‬ ‫ﺼﻘﺔ‬
‫ﻤﻬﻤﻠﺔ‬ ‫وﻛﺘﻠﺘﻪ‬
‫ﺜﺎ‬ ،
(
‫اﻟﻤ‬ ‫ﻓوق‬ ‫اﺤﺘﻛﺎك‬ ‫دون‬
‫ﻓوق‬ ‫اﺤﺘﻛﺎك‬ ‫ون‬
‫ﻩ‬
‫ر‬‫ﺤر‬
‫اﺒﺘداﺌﻴﺔ‬ ‫ﻋﺔ‬
‫ﺴر‬ ‫دون‬
‫اﺒ‬ ‫ﻋﺔ‬
‫ﺴر‬ ‫ن‬
‫ة‬
‫ز‬‫اﻟﻤﻬﺘ‬ ‫اﻟﺠﻤﻠﺔ‬
‫ة‬
‫ز‬‫اﻟﻤﻬﺘ‬
)
‫ﺠﺴم‬
- )
‫ﺠ‬
‫اﻟﻤروﻨﻴﺔ‬ ‫اﻟﻛﺎﻤﻨﺔ‬ ‫ﻗﺔ‬
‫اﻟﻤرو‬ ‫ﻟﻛﺎﻤﻨﺔ‬
pe
u
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a
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n
.
c
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c
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d
u
c
3as.ency-education.com

4. ‫ﺻﻔﺤﺔ‬
4
‫ﻣﻦ‬
8
2
-
‫اﺴﺘﻨﺘﺞ‬
‫ﻗﻴﻤﺔ‬
‫اﻟ‬
‫طﺎﻗﺔ‬
‫ﻟ‬ ‫اﻟﻛﻠﻴﺔ‬
‫ة‬
‫ز‬‫اﻟﻤﻬﺘ‬ ‫ﻠﺠﻤﻠﺔ‬
‫ﻟﺤظﺔ‬ ‫أي‬ ‫ﻋﻨد‬
t
.
3
-
‫ﻗﻴﻤﺔ‬ ‫ﺤدد‬
‫ﺴﻌﺔ‬
‫اﻟﺤرﻛﺔ‬
X0
.
4
-
‫أ‬
-
‫أﺤﺴب‬
‫اﻨﺘﻘﺎﻝ‬ ‫ﻋﻨد‬ ‫اﻟﺘوﺘر‬ ‫ﻗوة‬ ‫ﻋﻤﻝ‬
G
‫اﻟﻤوﻀﻊ‬ ‫ﻤن‬
xA=X0
‫إﻟﻰ‬
‫اﻟﻤوﻀﻊ‬
O
.
‫ب‬
-
‫اﻟﺘوﺘر‬ ‫ﻗوة‬ ‫ﺸدة‬ ‫ﺒدﻻﻟﺔ‬ ‫ﻟﻠﺤرﻛﺔ‬ ‫اﻟﺘﻔﺎﻀﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫أوﺠد‬
T
r
‫ﻟﻠﺤرﻛﺔ‬ ‫ذاﺘﻲ‬ ‫دور‬ ‫أﺠﻝ‬ ‫ﻤن‬ ‫ﺤﻠﻬﺎ‬ ‫وﻤﺜﻝ‬ ،
.
‫اﻟﺜﺎﻨــــــﻲ‬ ‫ء‬
‫اﻟﺠـــــز‬
:
‫ﺘﺠرﻴﺒﻲ‬ ‫اﺤد‬‫و‬ ‫ﺘﻤرﻴن‬ ‫ﻤن‬ ‫ﻴﺘﻛون‬
.
‫اﻟﺘﺠرﻴﺒـــﻲ‬ ‫اﻟﺘﻤرﻴــــــن‬
) :
07.00
‫ﻨﻘﺎط‬
(
‫ﻗﺴم‬ ‫ﻴﺔ‬‫ر‬‫اﻟﻤﺨﺒ‬ ‫اﻷﻋﻤﺎﻝ‬ ‫ﺤﺼﺔ‬ ‫ﻓﻲ‬
،‫ﻓوﺠﻴن‬ ‫إﻟﻰ‬ ‫اﻟﺘﻼﻤﻴذ‬ ‫اﻷﺴﺘﺎذ‬
‫ﻓوج‬ ‫ﻛﻝ‬ ‫ﻤن‬ ‫وطﻠب‬
‫اﻟ‬ ‫اﻨﺠﺎز‬
‫ﺒﺔ‬‫ر‬‫ﺘﺠ‬
‫اﻟﺘﺎﻟﻴﺔ‬
:
•
‫اﻷوﻝ‬ ‫اﻟﻔوج‬ ‫ﺒﺔ‬‫ر‬‫ﺘﺠ‬
:
‫اﻟﺒوﺘﺎﺴﻴوم‬ ‫ﺒﻴﻛروﻤﺎت‬ ‫ﻤﺤﻠوﻝ‬ ‫ﺒﻴن‬ ‫اﻟﺘﻔﺎﻋﻝ‬ ‫اﺴﺔ‬
‫ر‬‫د‬
( )
−
+
+
2
7
2
2 O
Cr
K
‫ﺤﺠﻤﻪ‬
ml
V 200
=
‫اﻟﻤوﻟﻲ‬ ‫ﻩ‬
‫ز‬‫وﺘرﻛﻴ‬
l
mol
C /
2
.
0
=
‫اﻟﻤﻴﺜﺎﻨوﻝ‬ ‫ﻤﻊ‬
OH
CH 3
‫ﻤﺎدﺘﻪ‬ ‫ﻛﻤﻴﺔ‬
mol
n 06
.
0
0 =
،
‫ﻫﻲ‬ ‫اﻟﺘﻔﺎﻋﻝ‬ ‫ﻓﻲ‬ ‫اﻟداﺨﻠﺔ‬ ‫اﻟﺜﻨﺎﺌﻴﺎت‬
:
)
/
(
),
/
( 3
2
7
2
3
+
−
Cr
O
Cr
OH
CH
HCOOH
•
‫اﻟﺜﺎﻨﻲ‬ ‫اﻟﻔوج‬ ‫ﺒﺔ‬‫ر‬‫ﺘﺠ‬
:
‫اﻹﻴﺜﺎﻨوﻴك‬ ‫ﺤﻤض‬ ‫ﺒﻴن‬ ‫اﻟﺘﻔﺎﻋﻝ‬ ‫اﺴﺔ‬
‫ر‬‫د‬
COOH
CH 3
‫ﻤﺎدﺘﻪ‬ ‫ﻛﻤﻴﺔ‬
mol
n 1
1 =
‫اﻟﻤﻴﺜﺎﻨوﻝ‬ ‫ﻤﻊ‬
OH
CH 3
‫ﻤﺎدﺘﻪ‬ ‫ﻛﻤﻴﺔ‬
2
n
‫اﻟذي‬
‫ﻋﻀوي‬ ‫وﻤرﻛب‬ ‫اﻟﻤﺎء‬ ‫ﻋﻨﻪ‬ ‫ﻴﻨﺘﺞ‬
E
.
‫اﻟﺒﻴﺎﻨﻴن‬ ‫رﺴم‬ ‫ﻤن‬ ‫اﻟﻔوﺠﻴن‬ ‫ﻟﻛﻼ‬ ‫ﻴﺒﻴﺔ‬‫ر‬‫اﻟﺘﺠ‬ ‫اﺴﺔ‬
‫ر‬‫اﻟد‬ ‫ﻤﻛﻨت‬
( )
t
f
=
τ
‫اﻟﺸﻛﻠﻴن‬ ‫ﻓﻲ‬ ‫اﻟﻤوﻀﺤﻴن‬
)
1
(
‫و‬
)
2
(
‫أﺴﻔﻠﻪ‬
.
1
-
‫اﻟﺘﻔﺎﻋﻝ‬ ‫ع‬
‫ﻨو‬ ‫ﻤﺤددا‬ ، ‫ﻓوج‬ ‫ﻛﻝ‬ ‫ﺒﺔ‬‫ر‬‫ﺘﺠ‬ ‫ﻓﻲ‬ ‫اﻟﺤﺎدث‬ ‫اﻟﺘﻔﺎﻋﻝ‬ ‫ﻤﻌﺎدﻟﺔ‬ ‫أﻛﺘب‬
.
2
-
‫ﻴر‬‫ر‬‫اﻟﺘﺒ‬ ‫ﻤﻊ‬ ‫اﻟﻤﻨﺎﺴﺒﺔ‬ ‫ﺒﺔ‬‫ر‬‫ﻟﻠﺘﺠ‬ ‫ﻤﻨﺤﻨﻰ‬ ‫ﻛﻝ‬ ‫أﻨﺴب‬
.
‫اﻟﻨﻬﺎﺌﻴﺔ‬ ‫اﻟﺘﻘدم‬ ‫ﻨﺴﺒﺔ‬ ‫ﺤدد‬ ‫ﺜم‬
f
τ
‫ﺘﻔﺎﻋﻝ‬ ‫ﻟﻛﻝ‬
.
3
-
‫ﻴﺞ‬‫ز‬‫اﻟﻤ‬ ‫أن‬ ‫وﺒﻴن‬ ، ‫اﻷوﻝ‬ ‫اﻟﻔوج‬ ‫ﺒﺔ‬‫ر‬‫ﻟﺘﺠ‬ ‫اﻟﺘﻔﺎﻋﻝ‬ ‫ﻟﺘﻘدم‬ ‫ﻻ‬
‫ﺠدو‬ ‫أﻨﺠز‬
‫اﻻﺒﺘداﺌﻲ‬
‫ﻗﻴﻤﺔ‬ ‫ﺤدد‬ ‫ﺜم‬ ‫ي‬
‫ﺴﺘوﻛﻴوﻤﺘر‬ ‫اﻟﻤﺴﺘﻌﻤﻝ‬
‫اﻷﻋظﻤﻲ‬ ‫اﻟﺘﻘدم‬
.
4
-
‫اﻟﺤﺠﻤﻴﺔ‬ ‫ﻋﺔ‬
‫اﻟﺴر‬ ‫ة‬
‫ر‬‫ﻋﺒﺎ‬ ‫أوﺠد‬
vol
v
‫ﺒدﻻﻟﺔ‬ ‫اﻷوﻝ‬ ‫اﻟﻔوج‬ ‫ﻟﺘﻔﺎﻋﻝ‬
τ
،
C
،
t
.
‫اﻟﻠﺤظﺔ‬ ‫ﻋﻨد‬ ‫أﺤﺴﺒﻬﺎ‬ ‫ﺜم‬
0
=
t
.
5
-
‫اﻟﻌﻀوي‬ ‫اﻟﻤرﻛب‬ ‫ﺴم‬
E
،‫اﻟﺜﺎﻨﻲ‬ ‫اﻟﻔوج‬ ‫ﺒﺔ‬‫ر‬‫ﺘﺠ‬ ‫ﻋن‬ ‫اﻟﻨﺎﺘﺞ‬
‫اﻟﻤﻴﺜﺎﻨوﻝ‬ ‫ﻤﺎدة‬ ‫ﻛﻤﻴﺔ‬ ‫اﺴﺘﻨﺘﺞ‬‫و‬
2
n
.
6
-
‫اﻟﻨﻬﺎﺌﻴﺔ‬ ‫اﻟﺘﻘدم‬ ‫ﻨﺴﺒﺔ‬ ‫ﻴﺎدة‬‫ز‬‫ﻟ‬
f
τ
‫اﻟﺸﻛﻝ‬ ‫ﻤﻨﺤﻨﻰ‬ ‫ﻓﻲ‬
)
1
(
،
‫ح‬
‫ﻨﻘﺘر‬
:
‫أ‬
-
‫اﻟﺘﻔﺎﻋﻠﻲ‬ ‫ﻴﺞ‬‫ز‬‫اﻟﻤ‬ ‫ة‬
‫ر‬‫ا‬
‫ر‬‫ﺤ‬ ‫ﻴﺎدة‬‫ز‬
.
‫ب‬
-
‫اﺘﺞ‬‫و‬‫اﻟﻨ‬ ‫أﺤد‬ ‫ﺤذف‬
.
‫ـ‬‫ـ‬‫ﺠ‬
-
‫اﻟﻤﺘﻔﺎﻋﻼت‬ ‫أﺤد‬ ‫ﺤذف‬
.
-
‫اﻟﺼﺤﻴﺢ‬ ‫اح‬
‫ر‬‫اﻻﻗﺘ‬ ‫اﺨﺘر‬
،
‫ﻴر‬‫ر‬‫اﻟﺘﺒ‬ ‫ﻤﻊ‬
.
‫اﻟﻨﺠﺎح‬‫و‬ ‫ﺒﺎﻟﺘوﻓﻴق‬
‫اﻟﻤﺎدة‬ ‫ﻤﻔﺘش‬ ‫اف‬
‫ر‬‫إﺸ‬ ‫ﺘﺤت‬ ‫اﻟﻤـــﺎدة‬ ‫أﺴﺎﺘذة‬
t (h)
τ (%)
20
1 t (h)
τ (%)
20
0.1
‫ﺍﻟﺸﻜﻞ‬
)
2
(
‫ﺍﻟﺸﻜﻞ‬
)
1
(
e
n
c
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
‫ﺠ‬
‫اﻟﺘﻤرﻴــــ‬
‫اﻟﺘ‬
s
‫ﺤﺼﺔ‬ ‫ﻓﻲ‬
‫ﺤﺼ‬ ‫ﻓﻲ‬
•
•
‫اﻟﻔوج‬ ‫ﺒﺔ‬‫ر‬‫ﺘﺠ‬
‫اﻟ‬ ‫ﺒﺔ‬‫ر‬‫ﺘﺠ‬
a
m
ml
V ml
mol
m
n 06
.
0 06
.
0
0
0 =
،
‫اﻟﺜﺎﻨﻲ‬ ‫ﻔوج‬
‫ﻟﺜﺎﻨﻲ‬
/
e
:
:
‫اﻟﺘﻔﺎ‬ ‫اﺴﺔ‬
‫ر‬‫د‬
‫اﺴ‬
‫ر‬‫د‬
OH
CH
CH 3
‫ﻤﺎدﺘ‬ ‫ﻛﻤﻴﺔ‬
‫ﻛﻤﻴﺔ‬
‫رﺴ‬ ‫ﻤن‬ ‫اﻟﻔوﺠﻴن‬ ‫ﻟﻛﻼ‬ ‫ﺒﻴﺔ‬
‫ﻤ‬ ‫اﻟﻔوﺠﻴن‬ ‫ﻛﻼ‬
‫ﺘﻔﺎﻋﻝ‬ ‫ﻟﻛﻝ‬
. ‫ﺘﻔﺎﻋﻝ‬
.
‫ﻗﻴﻤﺔ‬ ‫ﺤدد‬ ‫ﺜم‬ ‫ي‬
‫ﺴﺘوﻛﻴوﻤﺘر‬
‫ﺤدد‬ ‫ﺜم‬ ‫ي‬
‫ر‬
‫ﺤظﺔ‬
‫ظ‬
0
0
=
=
t
t
.
.
d
d
d
d
u
d
u
u
u
c
u
c
u
c
c
a
c
a
c
a
d
u
c
u
c
u
c
u
c
c
a
a
t
t
t
d
d
u
u
c
c
a
t
d
u
c
a
t
n
n
n
n
n
.
.
.
c
c
n
.
c
.
c
.
c
c
c
c
c
c
n
c
c
c
c
c
c
c
c
n
c
c
c
c
c
c
c
c
c
c
c
c
c
c
o
c
o
c
o
o
o
c
o
c
c
c
c
o
o
o
c
o
c
o
c
o
o
o
c
o
c
c
c
c
o
o
o
c
o
c
o
c
o
o
o
c
o
c
c
c
c
o
c
n
n
n
n
n
.
n
c
c
c
c
o
c
o
n
c
c
o
o
n
n
c
c
o
τ (%)
%)
o
c
a
t
c
a
t
‫ﺍﻟﺸ‬
3as.ency-education.com

5. ‫ﺻﻔﺤﺔ‬
5
‫ﻣﻦ‬
8
‫اﻟﺜﺎﻨــــــــﻲ‬ ‫ع‬
‫اﻟﻤوﻀــــــــــو‬
.
‫اﻷوﻝ‬ ‫ء‬
‫اﻟﺠـــــز‬
:
‫ﺘﻤرﻴﻨﻴن‬ ‫ﻤـــن‬ ‫ﻴﺘﻛـــون‬
.
‫اﻷوﻝ‬ ‫اﻟﺘﻤرﻴــــــن‬
) :
00
.
06
‫ﻨﻘﺎط‬
(
1
-
‫ﻤﺎﺌﻲ‬ ‫ﻤﺤﻠوﻝ‬
(S)
‫اﻟﻤوﻟﻲ‬ ‫ﻩ‬
‫ز‬‫ﺘرﻛﻴ‬ ‫ﻟﻠﻨﺸﺎدر‬
C =10-2
mol/L
‫وﺤﺠﻤﻪ‬
100mL
‫اﻟﻨﻬﺎﺌﻲ‬ ‫ﺘﻘدﻤﻪ‬ ‫وﻨﺴﺒﺔ‬
%
4
=
f
τ
.
‫أ‬
-
NH3
‫اﻟﻨﺸﺎدر؟‬ ْ
‫ي‬
‫ﺠز‬ ‫ﻓﻲ‬ ‫اﻷﺴﺎﺴﻴﺔ‬ ‫اﻟﺨﺎﺼﻴﺔ‬ ‫ﺘﻛﻤن‬ ‫أﻴن‬ ، ‫ﻀﻌﻴف‬ ‫أﺴﺎس‬
‫ب‬
-
‫ﺒدﻻﻟﺔ‬ ‫ﻟﻠﺘﻔﺎﻋﻝ‬ ‫ازن‬‫و‬‫اﻟﺘ‬ ‫ﺜﺎﺒت‬ ‫ﻋن‬ ‫ﻋﺒر‬ ‫ﺜم‬ ،‫اﻟﻤﺎء‬ ‫ﻤﻊ‬ ‫اﻟﻨﺸﺎدر‬ ‫ﺘﻔﺎﻋﻝ‬ ‫ﻤﻌﺎدﻟﺔ‬ ‫أﻛﺘب‬
C
،
f
τ
.
‫ـ‬‫ﺠ‬
-
‫ﻟﻠﺜﻨﺎﺌﻴﺔ‬ ‫اﻟﺤﻤوﻀﺔ‬ ‫ﺜﺎﺒت‬ ‫أن‬ ‫ﺒﻴن‬
NH3 )
/
( NH4
+
‫ﺒﺎﻟﻌﻼﻗﺔ‬ ‫ﻴﻌطﻰ‬
:
K
Ke
PKa log
−
=
‫ﻗﻴﻤﺘﻪ‬ ‫أﺤﺴب‬ ‫ﺜم‬
.
‫د‬
-
‫أن‬ ‫ﺒﻴن‬
PH
‫اﻟﻤﺤﻠوﻝ‬
(S)
‫ﺒﺎﻟﺸﻛﻝ‬ ‫ﻴﻛﺘب‬
:
)
1
(
log
f
f
PKa
PH
τ
τ
−
+
=
‫ـ‬‫ﻫ‬
-
‫اﻟﻤﺤﻠوﻝ‬ ‫ﻓﻲ‬ ‫اﻟﻐﺎﻟب‬ ‫اﻟﻛﻴﻤﻴﺎﺌﻲ‬ ‫ع‬
‫اﻟﻨو‬ ‫ﺤدد‬
(S)
،
‫ـ‬‫ﻟ‬‫ا‬ ‫ﻗﻴﻤﺔ‬ ‫أﺤﺴب‬‫و‬
PH
.
2
-
‫اﻟﻨوﻋﻴﺔ‬ ‫ﻨﺎﻗﻠﻴﺘﻪ‬ ‫وﻗﺴﻨﺎ‬ ‫ﻟﻠﻨﺸﺎدر‬ ‫ﻤﺎﺌﻴﺎ‬ ‫ﻻ‬
‫ﻤﺤﻠو‬ ‫ﻨﺎ‬‫ر‬‫ﺤﻀ‬
σ
‫ـ‬‫ﺒ‬
(S/m)
‫ـ‬‫ﻟ‬‫ا‬ ‫وﻗﻴﻤﺔ‬
PH
.
‫ﻟﻠﻤﺤﻠوﻝ‬ ‫اﻟﻤﻘطر‬ ‫اﻟﻤﺎء‬ ‫ﻤن‬ ‫ﻛﻤﻴﺔ‬ ‫ﻹﻀﺎﻓﺔ‬ ‫ﺒﻌد‬ ‫ات‬
‫ر‬‫ﻤ‬ ‫ﻋدة‬ ‫اﻟﻘﻴﺎﺴﻴن‬ ‫ﻫذﻴن‬ ‫ﻨﺎ‬‫ر‬‫ﻛر‬
‫ة‬
‫ر‬‫ﻤ‬ ‫ﻛﻝ‬ ‫ﻓﻲ‬
،
‫ﺒﻴﺎﻨﻴﺎ‬ ‫ﻤﺜﻠﻨﺎ‬ ‫ﺜم‬
)
(logσ
f
PH =
.
‫أ‬
-
‫ﻋن‬ ‫ﻋﺒر‬
PH
‫ﺒدﻻﻟﺔ‬
:
σ
،
+
−
4
, NH
OH
λ
λ
.
‫ب‬
-
‫اﻋﺘﻤﺎدا‬
‫ﻗﻴﻤﺔ‬ ‫أوﺠد‬ ‫اﻟﺒﻴﺎن‬ ‫ﻋﻠﻰ‬
+
4
NH
λ
.
3
-
، ‫ﺠدا‬ ‫ﻀﻌﻴف‬ ‫اﻷﺴﺎس‬ ‫ﻛﺎن‬ ‫إذا‬ ‫أﻨﻪ‬ ‫ﺒﻴن‬
‫ﺒﺎﻟﻌﻼﻗﺔ‬ ‫ﻴﻌطﻰ‬ ‫اﻟﻤوﻟﻲ‬ ‫ﻩ‬
‫ز‬‫ﺘرﻛﻴ‬ ‫ﻓﺈن‬
:
)
(
2
10 PKe
PKa
PH
b
C +
−
=
‫ﺘﺄﻛد‬ ‫ﺜم‬
‫ﺤﺴﺎﺒﻴﺎ‬ ‫ذﻟك‬ ‫ﻤن‬
.
‫ﻳﻌﻄﻰ‬
:
1
2
.
.
20 −
=
− mol
m
mS
OH
λ
،
PKe=14
4
-
‫ارد‬‫و‬‫ﺸ‬ ‫ﺒﻴن‬ ‫اﻟﻛﻴﻤﻴﺎﺌﻲ‬ ‫اﻟﺘﺤوﻝ‬ ‫ﻴﻨﻤذج‬
‫اﻟﻬﻴدروﻛﺴﻴد‬
‫اﻟﻤﺤﻠوﻝ‬ ‫ﻓﻲ‬ ‫اﻟﻨﺎﺘﺠﺔ‬
‫اﻟﺘﻔﺎﻋﻝ‬ ‫ﺒﻤﻌﺎدﻟﺔ‬ ‫ﻋﻀوي‬
‫أﺴﺘر‬‫و‬ ‫اﻟﺴﺎﺒق‬
‫اﻟﺘﺎﻟﻴﺔ‬
:
OH
CH
CH
COO
CH
CH
OH
R
COO
CH
CH 2
3
2
3
2
3 −
+
−
−
===
+
−
−
− −
−
‫أ‬
-
‫؟‬ ‫اﺼﻪ‬‫و‬‫ﺨ‬ ‫أﻫم‬ ‫وﻤﺎﻫﻲ‬ ‫؟‬ ‫اﻟﺤﺎدث‬ ‫اﻟﺘﻔﺎﻋﻝ‬ ‫ﻤﺎاﺴم‬
.
‫ب‬
-
‫اﺴﻤﻪ‬‫و‬ ‫اﻟﻤﺘﻔﺎﻋﻝ‬ ‫ﻟﻸﺴﺘر‬ ‫اﻟﻤﻔﺼﻠﺔ‬ ‫ﻨﺼف‬ ‫اﻟﺼﻴﻐﺔ‬ ‫اﺴﺘﻨﺘﺞ‬
.
‫ـ‬‫ﺠ‬
-
‫اﻟﻤﺎدة‬ ‫ﻛﻤﻴﺔ‬ ‫ﻓﻲ‬ ‫ﻤﺘﺴﺎوي‬ ‫اﻻﺒﺘداﺌﻲ‬ ‫ﻴﺞ‬‫ز‬‫اﻟﻤ‬ ‫ﻛﺎن‬ ‫اذا‬ ‫اﻟﻨﺎﺘﺞ‬ ‫اﻟﻛﺤوﻝ‬ ‫ﻛﺘﻠﺔ‬ ‫أﺤﺴب‬
.
‫د‬
-
‫؟‬ ‫اﻟﻴوﻤﻴﺔ‬ ‫اﻟﺤﻴﺎة‬ ‫ﻓﻲ‬ ‫ات‬
‫ر‬‫اﻷﺴﺘ‬ ‫أﻫﻤﻴﺔ‬ ‫ﺒﺎﺨﺘﺼﺎر‬ ‫أذﻛر‬
.
mol
g
H
M
mol
g
O
M
mol
g
C
M /
1
)
(
/
16
)
(
,
/
12
)
( =
=
=
‫اﻟﺜﺎﻨــــــــﻲ‬ ‫اﻟﺘﻤرﻴــــــن‬
) :
07.00
‫ﻨﻘﺎط‬
(
‫أوﻝ‬
‫ﻗﻤ‬
‫ا‬ ‫ر‬
‫ﺼ‬
‫ط‬
‫ﻨﺎﻋﻲ‬
‫رو‬
‫ﺴﻲ‬
Spoutnik
‫أط‬
‫ﻠ‬
‫ق‬
‫ﻓﻲ‬
‫أ‬
‫ﻛﺘ‬
‫و‬
‫ﺒ‬
‫ر‬
1957
‫م‬
‫ﺒﺤﻴ‬
‫ث‬
‫ﺘﺄﺨ‬
‫ا‬ ‫ذ‬
‫ﺒﻴ‬ ‫ﻟﻤﺴﺎﻓﺔ‬
‫ن‬
‫ﻤ‬
‫ر‬
‫ﻛ‬
‫ز‬
‫ﻋ‬
‫ط‬
‫ﺎﻟﺘ‬
‫ﻪ‬
‫و‬
‫ﺒﻴ‬
‫ن‬
‫ﻤ‬
‫ر‬
‫ﻛ‬
‫ا‬ ‫ز‬
‫ﻷ‬
‫ا‬ ‫رض‬
‫ﻟﻘﻴﻤﺘﻴ‬
‫ا‬ ‫ن‬
‫ﻟﻤ‬
‫ا‬‫و‬
‫ﻓﻘﺘﻴ‬
‫ن‬
‫ﻷ‬
‫د‬
‫ﺒﻌ‬ ‫ﻨﻰ‬
‫أ‬‫و‬ ‫د‬
‫ﻗﺼﺎ‬
‫ﻫ‬
‫ﺎ‬
‫ﻴﻠﻲ‬ ‫ﻛﻤﺎ‬
:
rP = 6610Km
‫و‬
rA = 7330Km
‫ﺒ‬ ‫ﻛﻤﺎ‬
‫ﺎ‬
‫ﻟﺸﻛ‬
‫ﻝ‬
01
PH
σ
log
15,563
e
n
c
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
‫أ‬
-
‫ب‬
-‫ب‬
-
‫أﻛﺘب‬
‫ـ‬‫ﺠ‬
-‫ـ‬‫ﺠ‬
-
‫ﺜﺎﺒت‬ ‫أن‬ ‫ﺒﻴن‬
‫أن‬ ‫ﺒﻴن‬
-
‫أن‬ ‫ﺒﻴن‬
‫أن‬ ‫ن‬
PH
PH
‫اﻟﻤﺤﻠوﻝ‬
‫اﻟﻤﺤ‬
‫ﻓﻲ‬ ‫اﻟﻐﺎﻟب‬ ‫اﻟﻛﻴﻤﻴﺎﺌﻲ‬ ‫ع‬
‫ﻨو‬
‫اﻟﻐ‬ ‫ﻟﻛﻴﻤﻴﺎﺌﻲ‬
‫ﻨﺎﻗﻠ‬ ‫وﻗﺴﻨﺎ‬ ‫ﻟﻠﻨﺸﺎدر‬ ‫ﻤﺎﺌﻴﺎ‬
‫وﻗﺴﻨﺎ‬ ‫ﻟﻠﻨﺸﺎدر‬
‫ﻛ‬ ‫ﻹﻀﺎﻓﺔ‬ ‫ﺒﻌد‬ ‫ات‬
‫ر‬‫ﻤ‬ ‫ﻋدة‬
‫ﻹﻀ‬ ‫ﺒﻌد‬ ‫ات‬
‫ر‬‫ﻤ‬
(log
f
PH f
PH
.
− ,
OH
λ
λ
λ − ,
OH
.
‫ﺒﺎﻟﻌﻼﻗ‬ ‫ﻴﻌطﻰ‬ ‫اﻟﻤوﻟﻲ‬ ‫ﻩ‬
‫ز‬
‫ﺒﺎﻟ‬ ‫ﻴﻌطﻰ‬ ‫ﻤوﻟﻲ‬
20
2
=
=
−
OH
OH
λ
λ
،
14
‫ﻤﺤﻠوﻝ‬
‫ﻋﻀ‬
‫أﺴﺘر‬‫و‬ ‫اﻟﺴﺎﺒق‬
‫أﺴ‬‫و‬ ‫اﻟﺴﺎﺒق‬
R
COO R
COO
2 R
COO R
COO
3as.ency-education.com

6. ‫ﺻﻔﺤﺔ‬
6
‫ﻣﻦ‬
8
-1
‫ﻤﺎ‬
‫ط‬
‫ﻤﺴﺎ‬ ‫ﺒﻴﻌﺔ‬
‫ا‬ ‫ر‬
‫ﻟﻘﻤ‬
‫ا‬ ‫ر‬
‫ﺼ‬
‫ط‬
‫ﻨﺎﻋﻲ‬
. Spoutnik
‫ﻤﺎﻫو‬
‫ﻤ‬
‫و‬
‫ﻗﻊ‬
‫ا‬
‫ﻷ‬
‫رض‬
‫ﻓﻲ‬
‫ﻫﺬ‬
‫ا‬ ‫ا‬
‫ﻟﻤﺴﺎ‬
‫ر‬
.
-2
‫ﻤﺎ‬
‫ذا‬
‫ﻴﻤﺜ‬
‫ا‬ ‫ﻝ‬
‫ﻟ‬
‫طوﻝ‬
2a
‫ا‬ ‫و‬
‫ﻟ‬
‫طوﻝ‬
2b
‫أ‬ ‫؟‬
‫ﺤﺴ‬
‫طوﻝ‬ ‫ب‬
‫ﻨﺼ‬
‫ا‬ ‫ف‬
‫ﻟﻤﺤ‬
‫ا‬ ‫ور‬
‫ﻟﻛﺒﻴ‬
‫ر‬
‫ﻟﻬذا‬
‫ا‬
‫ﻟﻤﺴﺎ‬
‫ر‬
.
-3
‫ﻓﻲ‬
‫أي‬
‫ﻨﻘ‬
‫ط‬
‫ﺘﻛ‬ ‫ﺔ‬
‫ون‬
‫ﺴ‬
‫ر‬
‫ﻋﺔ‬
‫ا‬
‫ﻟﻘﻤ‬
‫ا‬ ‫ر‬
‫ﻹﺼ‬
‫ط‬
‫ﻨﺎﻋﻲ‬
‫أ‬
‫ﺼﻐ‬
‫ر‬
‫ﻴﺔ‬
‫و‬
‫ﻓﻲ‬
‫أي‬
‫ﻨﻘ‬
‫ط‬
‫ﺘﻛ‬ ‫ﺔ‬
‫ون‬
‫ﺴ‬
‫ر‬
‫ﻋﺘ‬
‫ﻪ‬
‫أ‬
‫ﻋ‬
‫ظ‬
‫ﻤﻴﺔ‬
،
‫ﻤﻊ‬
‫ا‬
‫ﻟﺘﻌﻠﻴ‬
‫ﻝ‬
‫ﻤﺜ‬
‫ﻝ‬
‫ﻛﻼ‬
‫ﻫ‬
‫ﺒﺸﻛ‬ ‫ﻤﺎ‬
‫ﻝ‬
‫ﻋﻠﻰ‬ ‫ﻛﻴﻔﻲ‬
‫ا‬
‫ﻟ‬
‫ر‬
‫ﺴ‬
‫م‬
‫ﺒﻌ‬
‫د‬
‫ﻨﻘﻠﻪ‬
‫ﻋﻠﻰ‬
‫ور‬
‫ﻗﺔ‬
‫ا‬
‫ﻹﺠﺎﺒﺔ‬
.
-4
‫ﻨﻌﺘﺒ‬
‫ر‬
‫ﻗﻤ‬
‫إ‬ ‫ا‬
‫ر‬
‫ﺼ‬
‫ط‬
‫ﻨﺎﻋﻲ‬
S
‫ﻛﺘﻠﺘﻪ‬
m
‫ﻴ‬
‫دور‬
‫ﺤ‬
‫ا‬ ‫وﻝ‬
‫ﻷ‬
‫رض‬
‫ﺒﺤ‬
‫ر‬
‫ﻛﺔ‬
‫دا‬
‫ﺌ‬
‫ر‬
‫ﻤﻨﺘ‬ ‫ﻴﺔ‬
‫ظ‬
‫ﻤﺔ‬
‫و‬
‫ﻴ‬
‫ر‬
‫ﺴ‬
‫م‬
‫ﻤﺴﺎ‬
‫دا‬ ‫ا‬
‫ر‬
‫ﺌ‬
‫ر‬
‫ﻨﺼ‬ ‫ﻴﺎ‬
‫ف‬
‫ﻗ‬
‫ﻩ‬
‫ر‬‫ط‬
r = h + RT
‫و‬
‫ﻤ‬
‫ر‬
‫ﻛ‬
‫ﻩ‬
‫ز‬
O
‫ﻓﻲ‬
‫ا‬
‫ﻟﻤﻌﻠ‬
‫ا‬ ‫م‬
‫ﻟﺠﻴ‬
‫و‬
‫ﻤ‬
‫ر‬
‫ﻛ‬
‫ي‬
‫ز‬
)
‫ا‬
‫ﻟﺸﻜ‬
‫ﻞ‬
.(02
‫ا‬
-
‫أذ‬
‫ﻛ‬
‫ر‬
‫ﺸ‬
‫ا‬ ‫روط‬
‫ﻟﺤﺼ‬
‫وﻝ‬
‫ﺤ‬ ‫ﻋﻠﻰ‬
‫ر‬
‫ﻛﺔ‬
‫دا‬
‫ﺌ‬
‫ر‬
‫ﻤﻨﺘ‬ ‫ﻴﺔ‬
‫ظ‬
‫ﻤﺔ‬
.
‫ب‬
-
‫أ‬
‫ﻛﺘ‬
‫ﺐ‬
‫ا‬
‫ﻟﻌﺒﺎ‬
‫ا‬ ‫ة‬
‫ر‬
‫ﻟﺘﺴﺎ‬ ‫ﻟﺸﻌﺎﻋﻴﺔ‬
‫ع‬
‫ر‬
‫ﺤ‬
‫ﺮ‬
‫ﻤ‬ ‫ﻛﺔ‬
‫ﺮ‬
‫ﻛ‬
‫ﺰ‬
‫ﻋطﺎﻟﺔ‬
‫ا‬
‫ﻟﻘﻤ‬
‫ﺮ‬
‫ا‬
‫ﻹ‬
‫ﺼطﻨﺎﻋﻲ‬
‫ـ‬‫ﺠ‬
-
‫أ‬
‫ﻛﺘ‬
‫ب‬
‫ا‬
‫ﻟﻌﺒﺎ‬
‫ا‬ ‫ة‬
‫ر‬
‫ﻟﺸﻌﺎﻋﻴﺔ‬
FT/S
‫ﻟﻘ‬
‫وة‬
‫ﺠ‬
‫ا‬ ‫ذب‬
‫ﻷ‬
‫رض‬
‫ﻟﻠﻘﻤ‬
‫ا‬ ‫ر‬
‫ﻹﺼ‬
‫ط‬
‫ﻨﺎﻋﻲ‬
.
‫د‬
-
‫ﺒﺘ‬
‫ط‬
‫ﺒﻴ‬
‫ا‬ ‫ق‬
‫ﻟﻘﺎﻨ‬
‫ا‬ ‫ون‬
‫ﻟﻨﻴ‬ ‫ﻟﺜﺎﻨﻲ‬
‫و‬
‫ﺘ‬
‫أو‬ ‫ن‬
‫ﺠ‬
‫د‬
‫ﻋﺒﺎ‬
‫ة‬
‫ر‬
‫ﻛ‬
‫ﻝ‬
‫ﻤ‬
‫ن‬
:
‫ﺴ‬
‫ر‬
‫ﻋﺔ‬
‫ا‬
‫ﻟﻘﻤ‬
‫ر‬
v
‫ا‬ ‫و‬
‫ﻟ‬
‫دور‬
T
‫ﻟﺤ‬
‫ر‬
‫ﻛﺔ‬
‫ا‬
‫ﻟﻘﻤ‬
‫ر‬
‫ﺤ‬
‫ا‬ ‫وﻝ‬
‫ﻷ‬
‫رض‬
‫ﺒ‬
‫د‬
‫ﻻﻟﺔ‬
. G, MT , h, RT
‫ـ‬‫ﻫ‬
-
‫اﺴﺘﻨﺘﺞ‬
‫اﻟﻘﺎﻨون‬
‫اﻟﺜﺎﻟث‬
‫ﻟﻛﺒﻠر‬
.
-5
‫ﻴﺤﺘ‬
‫ا‬ ‫وي‬
‫ﻟﺠ‬
‫ا‬ ‫دوﻝ‬
‫ﻋﻠﻰ‬ ‫ﻟﺘﺎﻟﻲ‬
‫ا‬
‫ﻟﻘﻴ‬
‫ا‬ ‫م‬
‫ﻟﻌ‬
‫دد‬
‫ﻟﻠ‬ ‫ﻴﺔ‬
‫دور‬
T
‫ا‬‫و‬
‫ﻹ‬
‫ر‬
‫ﺘﻔﺎ‬
‫ع‬
h
‫ﻟﺒﻌ‬
‫ض‬
‫ا‬
‫ﻷﻗﻤﺎ‬
‫ر‬
‫ا‬
‫ﻹﺼ‬
‫ط‬
‫ﻨﺎﻋﻴﺔ‬
‫ﻟﻬﺎ‬
‫ﻤﺴﺎ‬
‫دا‬ ‫ات‬
‫ر‬
‫ﺌ‬
‫ر‬
‫ﻨﺼ‬ ‫ﻴﺔ‬
‫ف‬
‫ﻫﺎ‬
‫ﻗطر‬
r
‫ﻫﺎ‬
‫وﻤرﻛز‬
‫اﻻرض‬ ‫ﻤرﻛز‬
.
Astra
)
‫ﻗﻣ‬
‫ﺭ‬
‫ﺟﻳ‬
‫ﻭ‬
‫ﻣﺳﺗﻘ‬
‫ﺭ‬
(
Cosmos
Alsat1
‫ﺍ‬
‫ﻟﻘﻤﺮ‬
‫ﺍ‬
‫ﻹﺻﻄﻨﺎﻋﻲ‬
40,440
T(103
s)
0,708
r(107
m)
3,565
h(107
m)
)
.
( 3
2
3
2
−
= m
s
Cte
r
T
‫أ‬
-
‫أﻛﻤﻝ‬
‫اﻟﺠدوﻝ‬
.
‫ب‬
-
‫اﻷرض‬ ‫ﻟﻛﺘﻠﺔ‬ ‫اﻟﻌددﻴﺔ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫اﺴﺘﻨﺘﺞ‬
.
‫ﻤﻌطﻴﺎت‬
:
RT = 6380Km
،
G = 6.67x10-11
N.m2
/ Kg 2
،
1 jour = 23h56 min
e
n
c
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
‫ط‬
‫ﻤﺴﺎ‬ ‫ﺒﻴﻌﺔ‬
‫ا‬ ‫ر‬
‫ﻟﻘﻤ‬
‫ا‬ ‫ر‬
‫ﺼ‬
‫ط‬ ‫ﻤﺴﺎ‬ ‫ﺔ‬
‫ا‬ ‫ر‬
‫ﻟﻘﻤ‬
‫ا‬ ‫ر‬
‫ﺼ‬
‫ط‬
‫ا‬
‫ﻟ‬
‫طوﻝ‬
2a
‫ا‬ ‫و‬
‫ﻟ‬
‫طوﻝ‬ ‫وﻝ‬
2a
‫ا‬ ‫و‬
‫ﻟ‬
‫ط‬
b
‫ون‬
‫ﺴ‬
‫ر‬
‫ﻋﺔ‬
‫ا‬
‫ﻟﻘﻤ‬
‫ا‬ ‫ر‬
‫ﻹﺼ‬
‫ط‬ ‫ﺴ‬
‫ر‬
‫ﻋﺔ‬
‫ا‬
‫ﻟﻘﻤ‬
‫ا‬ ‫ر‬
‫ﻹ‬
‫ﻋﻠﻰ‬ ‫ﻲ‬
‫ا‬
‫ﻟ‬
‫ر‬
‫ﺴ‬
‫م‬
‫ﺒﻌ‬
‫د‬
‫ﻨﻘﻠﻪ‬ ‫ا‬
‫ﻟ‬
‫ر‬
‫ﺴ‬
‫م‬
‫ﺒﻌ‬
‫د‬
‫ﻛﺘﻠﺘﻪ‬
‫ﻪ‬
m
m
‫ﻴ‬
‫دور‬
‫ﺤ‬
‫ا‬ ‫وﻝ‬
‫ﻷ‬ ‫ﻴ‬
‫دور‬
‫ﺤ‬
‫ﻓﻲ‬
‫ا‬
‫ﻟﻤﻌﻠ‬
‫ا‬ ‫م‬
‫ﻟﺠﻴ‬
‫و‬
‫ﻤ‬
‫ر‬
‫ﻛ‬
‫ي‬
‫ز‬ ‫ا‬
‫ﻟﻤﻌﻠ‬
‫ا‬ ‫م‬
‫ﻟﺠﻴ‬
‫و‬
‫ﻤ‬
‫ر‬
‫ﻤﻨﺘ‬ ‫ﺔ‬
‫ظ‬
‫ﻤﺔ‬
. ‫ﺘ‬
‫ظ‬
‫ﻤﺔ‬
.
‫ﻋطﺎﻟﺔ‬
‫ا‬
‫ﻟﻘﻤ‬ ‫ﺔ‬
‫ا‬
‫ﻟﻘﻤ‬
‫ﺮ‬
‫ﺮ‬
‫ا‬
‫ﻹ‬
‫ﺼطﻨﺎﻋ‬ ‫ا‬
‫ﻹ‬
‫ﺼط‬
‫ﻤ‬
‫ا‬ ‫ر‬
‫ﻹﺼ‬
‫ط‬
‫ﻨﺎﻋﻲ‬
. ‫ﻹﺼ‬
‫ط‬
‫ﻨﺎﻋﻲ‬
.
‫ﺔ‬
‫ا‬
‫ﻟﻘﻤ‬
‫ﻤر‬
‫ر‬
v
v
. G
‫ﺍ‬
‫ﻟﻘﻤﺮ‬
‫ﺍ‬
‫ﻹﺻﻄﻨﺎﻋﻲ‬‫ﺍ‬
‫ﻹﺻﻄﻨﺎﻋﻲ‬
c
y
T(10
T(1 3
s)
s)
c
y
n
r(10
n
n
c
e
n
e
n
e
n
e
n
e
e
e
3as.ency-education.com

7. ‫ﺻﻔﺤﺔ‬
7
‫ﻣﻦ‬
8
‫اﻟﺜﺎﻨــــــﻲ‬ ‫ء‬
‫اﻟﺠـــــز‬
:
‫ﺘﺠرﻴﺒﻲ‬ ‫اﺤد‬‫و‬ ‫ﺘﻤرﻴن‬ ‫ﻤن‬ ‫ﻴﺘﻛون‬
.
‫اﻟﺘﺠرﻴﺒـــﻲ‬ ‫اﻟﺘﻤرﻴــــــن‬
) :
07.00
‫ﻨﻘﺎط‬
(
‫اﻟﺘﺎﻟﻴﺔ‬ ‫اﻟدﻻﻻت‬ ‫ﻤﻛﺜﻔﺔ‬ ‫ﺘﺤﻤﻝ‬
:
V
F 330
,
%)
10
160
( ±
μ
‫اﻟﺴﻌﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﻤن‬ ‫ﻟﻠﺘﺤﻘق‬
C
‫ﻤﻘﺎوﻤﺘﻪ‬ ‫أوﻤﻲ‬ ‫ﻨﺎﻗﻝ‬ ‫ﻋﺒر‬ ‫ﻨﺸﺤﻨﻬﺎ‬ ‫ﻟﻠﻤﻛﺜﻔﺔ‬
R=12,5KΩ
‫اﻟﻤﺤرﻛﺔ‬ ‫ﻗوﺘﻪ‬ ‫ﻤﺜﺎﻟﻲ‬ ‫ﻤوﻟد‬ ‫اﺴطﺔ‬‫و‬‫ﺒ‬
‫ﺒﺎﺌﻴﺔ‬‫ر‬‫اﻟﻛﻬ‬
E=300V
‫اﻟﺘوﺘر‬ ‫ﺘطور‬ ‫ﺒﺘﺴﺠﻴﻝ‬ ‫ﻨﻘوم‬ ‫ﻤﻌﻠوﻤﺎﺘﻴﺔ‬ ‫از‬
‫ر‬‫اﺤ‬ ‫ﺒﺒطﺎﻗﺔ‬ ‫ﻤزود‬ ‫آﻟﻲ‬ ‫اﻋﻼم‬ ‫ﺠﻬﺎز‬ ‫اﺴطﺔ‬‫و‬‫ﺒ‬ ،
uc
‫ﺒﻴن‬
‫اﻟﺘوﺘر‬‫و‬ ‫اﻟﻤﻛﺜﻔﺔ‬ ‫طرﻓﻲ‬
uR
‫اﻷوﻤﻲ‬ ‫اﻟﻨﺎﻗﻝ‬ ‫طرﻓﻲ‬ ‫ﺒﻴن‬
)
‫اﻟﺸﻛﻝ‬
02
. (
1
-
‫ات‬
‫ر‬‫اﻟﺘوﺘ‬ ‫ﺘطور‬
:
‫أ‬
-
‫ات‬
‫ر‬‫اﻟﺘوﺘ‬ ‫ﺒﻴن‬ ‫ﻤن‬
uc
‫و‬
uR
‫اﻟذي‬ ‫اﻟﺘوﺘر‬ ‫ﻤﺎﻫو‬
‫اﻟﺘﻴﺎر‬ ‫ﺸدة‬ ‫ﺘطور‬ ‫ﻴﺒرز‬
i(t)
‫ﻓﻲ‬ ‫اﻟﻤﺎر‬
‫؟‬ ‫ة‬
‫ر‬‫اﻟدا‬
‫ﻋﻠﻝ‬
.
‫ب‬
-
‫اﻟﺸﻛﻝ‬ ‫ﻋﻠﻰ‬ ‫اﻋﺘﻤﺎدا‬
02
‫اﻓق‬‫و‬‫اﻟﻤ‬ ‫اﻟﻤﻨﺤﻨﻰ‬ ‫اﺴﺘﻨﺘﺞ‬
‫اﻟﺘوﺘر‬ ‫ﻟﺘطور‬
uc
‫اﻟﺘﻌﻠﻴﻝ‬ ‫ﻤﻊ‬
‫ـ‬‫ﺠ‬
-
‫اﻟزﻤن‬ ‫ﺜﺎﺒت‬ ‫أن‬ ‫ﺒﻴن‬ ‫اﻟﺒﻌدي‬ ‫اﻟﺘﺤﻠﻴﻝ‬ ‫ﺒﺎﺴﺘﻌﻤﺎﻝ‬
τ
‫اﻟزﻤن‬ ‫ﻤﻊ‬ ‫ﻤﺘﺠﺎﻨس‬
.
2
-
‫اﻟﺘوﺘر‬ ‫ﻴﺤﻘﻘﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﺘﻔﺎﻀﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻋن‬ ‫اﻟﺒﺤث‬
R
u
:
-
‫اﻟﺘﺎﻟﻴﺔ‬ ‫ﺘﻔﺎﻀﻠﻴﺔ‬ ‫ﻤﻌﺎدﻻت‬ ‫ﺒﻊ‬‫ر‬‫اﻷ‬ ‫ح‬
‫ﻨﻘﺘر‬
:
)
02
..(
..........
0
)
(
.
)
(
)
01
..(
..........
0
)
(
.
)
(
=
+
=
+
t
u
C
dt
t
du
R
t
u
R
dt
t
du
R
R
R
R
)
04
..(
..........
0
)
(
)
(
)
03
..(
..........
0
)
(
.
)
(
=
+
=
+
t
u
dt
t
du
RC
t
u
RC
dt
t
du
R
R
R
R
‫أ‬
-
‫اﻟﺘﻔﺎﻀﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻫذﻩ‬ ‫ﺤدد‬ ‫اﻟﺒﻌدي‬ ‫اﻟﺘﺤﻠﻴﻝ‬ ‫ﻋﻠﻰ‬ ‫ﺒﺎﻻﻋﺘﻤﺎد‬ ، ‫ﺼﺤﻴﺤﺔ‬ ‫اﺤدة‬‫و‬ ‫ﺘوﺠد‬ ‫اﻟﺴﺎﺒﻘﺔ‬ ‫اﻟﻤﻌﺎدﻻت‬ ‫ﻤن‬
.
‫ب‬
-
‫ﻤن‬ ‫اﻟﺘﻔﺎﻀﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻫذﻩ‬ ‫ﺤﻝ‬ ‫ان‬
‫اﻟﺸﻛﻝ‬
:
τ
t
R Ee
t
u
−
=
)
(
-
‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻫذﻩ‬ ‫ﻛﺘﺎﺒﺔ‬ ‫ﻴﻤﻛن‬ ‫أﻨﻪ‬ ‫ﺒﻴن‬
‫ﺒﺎﻟﺸﻛﻝ‬
:
b
at
u
Ln R +
=
)
(
‫ﻤن‬ ‫ﻛﻝ‬ ‫ﺘﻲ‬‫ر‬‫ﻋﺒﺎ‬ ‫أوﺠد‬
b , a
‫ﺒدﻻﻟﺔ‬
E
‫و‬
τ
.
‫ـ‬‫ﺠ‬
-
‫ﺒرﺴم‬ ‫ﻤﻨﺎﺴﺒﺎ‬ ‫آﻟﻲ‬ ‫إﻋﻼم‬ ‫ﻨﺎﻤﺞ‬‫ر‬‫ﺒ‬ ‫ﺴﻤﺢ‬
‫اﻟﻤﻨﺤﻨﻰ‬
)
(
)
( t
f
u
Ln R =
‫ﺒﺎﻟﺸﻛﻝ‬ ‫اﻟﻤﺒﻴن‬
2
.
-
‫اﻟﺒﻴﺎن‬ ‫ﻤﻌﺎدﻟﺔ‬ ‫أﻋط‬
.
‫د‬
-
‫اﻟﻤﻛﺜﻔﺔ‬ ‫ﺴﻌﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫اﺴﺘﻨﺘﺞ‬
C
‫؟‬ ‫اﻟﺼﺎﻨﻊ‬ ‫طرف‬ ‫ﻤن‬ ‫اﻟﻤﻌطﺎة‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﻤﻊ‬ ‫اﻓق‬‫و‬‫ﺘﺘ‬ ‫وﻫﻝ‬
‫اﻟﺸﻛﻝ‬
01
‫اﻟﺸﻛﻝ‬
02
a
b
e
n
c
y
-
e
d
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
‫ﺒ‬‫ر‬‫اﻟﻛﻬ‬
‫اﻟﻛ‬
‫اﻟﻤﻛﺜﻔ‬ ‫طرﻓﻲ‬
‫طرﻓﻲ‬
1
-1
-
‫اﻟﺘوﺘ‬ ‫ﺘطور‬
‫ﺘطور‬
‫أ‬
-‫أ‬
-
‫اﻟﺘوﺘر‬ ‫ﺒﻴن‬ ‫ﻤن‬
‫ﺒﻴن‬ ‫ﻤن‬
‫اﻟﺘﻴ‬ ‫ﺸدة‬ ‫ﺘطور‬ ‫ﻴﺒرز‬
‫ﺸد‬ ‫ﺘطور‬ ‫ﺒرز‬
‫اﻟﺸﻛﻝ‬ ‫ﻋﻠﻰ‬ ‫ﻋﺘﻤﺎدا‬
‫اﻟﺸﻛﻝ‬ ‫ﻋﻠﻰ‬
02
02
‫اﻟﺘوﺘر‬ ‫ر‬
‫وﺘر‬
uc
‫اﻟﺘﻌﻠﻴﻝ‬ ‫ﻤﻊ‬
‫اﻟﺘﻌ‬ ‫ﻤﻊ‬
‫ﺜ‬ ‫أن‬ ‫ﺒﻴن‬ ‫اﻟﺒﻌدي‬ ‫ﻟﺘﺤﻠﻴﻝ‬
‫ﺒﻴن‬ ‫اﻟﺒﻌدي‬ ‫ﻴﻝ‬
‫ﻤن‬
.
‫اﻟﺘوﺘر‬ ‫ﻴﺤﻘﻘﻬﺎ‬ ‫ﺘﻲ‬
‫اﻟﺘ‬ ‫ﺤﻘﻘﻬﺎ‬
R
u
:
‫ﻟﻴﺔ‬
:
a
t
)
(
. (
.
)
( )
(
........
0
)
(
. ..
0
)
(
)
+
+
+ )
C
C (
.
.
C
C (
(
(
. (
.
dt
dt
t
du (
R
R
R (
. (
(
. (
R
R
R
R
c
a
)
( )
+
+
du
d (
RC
R
RC
R
dt
t
du
du (
(
R
R
R
‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻫذﻩ‬ ‫ﺤدد‬ ‫ﻟﺒﻌدي‬
‫اﻟﻤ‬ ‫ﻫذﻩ‬ ‫ﺤدد‬ ‫ي‬
3as.ency-education.com

8. ‫ﺻﻔﺤﺔ‬
8
‫ﻣﻦ‬
8
‫ـ‬‫ﻫ‬
-
‫اﻟﻘﻴﻤﺔ‬ ‫اﻷوﻤﻲ‬ ‫اﻟﻨﺎﻗﻝ‬ ‫ﻟﻤﻘﺎوﻤﺔ‬ ‫ﻨﻌطﻲ‬
2
R
R =
′
‫اﻟﺸﻛﻝ‬ ‫ﺒﻴﺎن‬ ‫ﻓﻲ‬ ‫ﻴﺘﻐﻴر‬ ‫ﻤﺎذا‬ ،
01
‫ﻋﻠﻝ‬ ‫؟‬
.
‫و‬
-
‫ﻤن‬ ‫اﻷوﻤﻲ‬ ‫اﻟﻨﺎﻗﻝ‬ ‫ﻤﻘﺎوﻤﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺘﻐﻴﻴر‬ ‫ﻋﻨد‬ ‫اﻟﻤﻛﺜﻔﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻌظﻤﻰ‬ ‫ﻨﺔ‬‫ز‬‫اﻟﻤﺨ‬ ‫اﻟطﺎﻗﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺘﺘﻐﻴر‬ ‫ﻫﻝ‬
R
‫إﻟﻰ‬
R′
‫؟‬
‫ﻋﻠﻝ‬
.
3
-
‫ذاﺘﻴﺘﻬﺎ‬ ‫ﻤﺜﺎﻟﻴﺔ‬ ‫وﺸﻴﻌﺔ‬ ‫ﻤﻊ‬ ‫اﻟﺘﺴﻠﺴﻝ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺴﺎﺒﻘﺔ‬ ‫اﻟﻤﺸﺤوﻨﺔ‬ ‫اﻟﻤﻛﺜﻔﺔ‬ ‫ﺒﺘوﺼﻴﻝ‬ ‫ﺒﺎﺌﻴﺔ‬‫ر‬‫ﻛﻬ‬ ‫ة‬
‫ر‬‫دا‬ ‫ﻨﺤﻘق‬
L
‫اﺴم‬
‫ر‬ ‫اﺴطﺔ‬‫و‬‫وﺒ‬ ،
‫اﻟرﻗﻤﻲ‬ ‫از‬
‫ز‬‫اﻻﻫﺘ‬
‫ﺒﺎﻟﺸﻛﻝ‬ ‫ﻛﻤﺎ‬ ‫اﻟﻤﻛﺜﻔﺔ‬ ‫طرﻓﻲ‬ ‫ﺒﻴن‬ ‫اﻟﺘوﺘر‬ ‫ﻤﺘﺎﺒﻌﺔ‬ ‫ﺘم‬
03
.
‫أ‬
-
‫اﻓﻘﺔ‬‫و‬‫اﻟﻤ‬ ‫ﺒﺎﺌﻴﺔ‬‫ر‬‫اﻟﻛﻬ‬ ‫ة‬
‫ر‬‫اﻟدا‬ ‫أرﺴم‬
.
‫ب‬
-
‫اﻟﺘوﺘر‬ ‫ﺒدﻻﻟﺔ‬ ‫ة‬
‫ر‬‫ﻟﻠدا‬ ‫اﻟﺘﻔﺎﻀﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫أوﺠد‬
‫طرﻓﻲ‬ ‫ﺒﻴن‬ ‫ﺒﺎﺌﻲ‬‫ر‬‫اﻟﻛﻬ‬
‫اﻟﻤﻛﺜﻔﺔ‬
‫؟‬ ‫ﺘﺴﺘﻨﺘﺞ‬ ‫وﻤﺎذا‬ ،
‫ـ‬‫ﺠ‬
-
‫ﻋﺒﺎ‬ ‫اﺴﺘﻨﺘﺞ‬
‫وﻗﻴﻤﺘﻪ‬ ‫اﻟذاﺘﻲ‬ ‫اﻟدور‬ ‫ة‬
‫ر‬
‫از‬
‫ز‬‫ﻟﻼﻫﺘ‬
‫اﻟﻤﺴﺠ‬
‫ﻝ‬
.
‫د‬
-
‫اﻟوﺸﻴﻌﺔ‬ ‫ذاﺘﻴﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫أﺴﺘﻨﺘﺞ‬
.
‫ـ‬‫ﻫ‬
-
‫ﺒﺎﺌﻴﺔ‬‫ر‬‫اﻟﻛﻬ‬ ‫ﻟﻠﺸﺤﻨﺔ‬ ‫اﻟزﻤﻨﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫أﻛﺘب‬
‫اﻟﻤﻛﺜﻔﺔ‬ ‫ﻓﻲ‬ ‫ﻨﺔ‬‫ز‬‫اﻟﻤﺨ‬
.
-
‫ﻨﻌﺘﺒر‬
10
² ≈
π
‫اﻟﻨﺠﺎح‬‫و‬ ‫ﺒﺎﻟﺘوﻓﻴق‬
‫اﻟﻤﺎدة‬ ‫ﻤﻔﺘش‬ ‫اف‬
‫ر‬‫إﺸ‬ ‫ﺘﺤت‬ ‫اﻟﻤـــﺎدة‬ ‫أﺴﺎﺘذة‬
( )
mS
t
( )
V
uC
‫ﺍﻟﺸﻜﻞ‬
)
3
(
5
0
120
e
n
c
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
‫أ‬
-
‫ب‬
-‫ب‬
-
‫أوﺠد‬
‫ﺒﺎﺌﻲ‬‫ر‬‫اﻟﻛﻬ‬
‫اﻟﻛﻬر‬
‫ـ‬‫ﺠ‬
-
‫ﻋﺒﺎ‬ ‫اﺴﺘﻨﺘﺞ‬ ‫ـ‬‫ﺠ‬
-
‫ﻋﺒﺎ‬ ‫اﺴﺘﻨﺘﺞ‬
‫اﻟد‬ ‫ة‬
‫ر‬
‫اﻟوﺸ‬ ‫ذاﺘﻴﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫أﺴﺘﻨﺘﺞ‬
‫ذاﺘﻴﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺴﺘﻨﺘﺞ‬
‫ﻟﻠﺸﺤ‬ ‫اﻟزﻤﻨﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺘب‬
‫ﻟﻠ‬ ‫اﻟزﻤﻨﻴﺔ‬ ‫ﻤﻌﺎدﻟﺔ‬
‫اﻟﻤﻛﺜﻔﺔ‬ ‫ﻓﻲ‬
‫ﺜﻔﺔ‬
.
.
-
‫ﻨﻌﺘﺒر‬ -
‫ﻨﻌﺘ‬
0
3as.ency-education.com

9. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
1
‫ﻣﻦ‬
18
‫ﻋﻨﺎﺻ‬
‫ـ‬‫ـ‬
‫اﻹﺟﺎﺑ‬ ‫ﺮ‬
‫ـ‬‫ـ‬
‫ﺔ‬
‫اﻟﻌﻼﻣﺔ‬
‫ﻣﺠﺰأة‬
‫ﻣﺠﻤﻮع‬
‫اﻷول‬ ‫ـﻮع‬‫ـ‬‫ـ‬‫ﺿ‬‫اﻟﻤﻮ‬
:
‫اﻷول‬ ‫ـﺰء‬‫ـ‬‫ﺠ‬‫اﻟ‬
)
‫ﺗﻤﺮﻳﻨﻴﻦ‬ ‫ﻣﻦ‬ ‫ﻳﺘﻜﻮن‬
(
‫اﻷول‬ ‫اﻟﺘﻤﺮﻳﻦ‬
) :
06.00
‫ـﺔ‬‫ـ‬‫ﻄ‬‫ﻧﻘ‬
(
Ι
–
‫أ‬
-
‫أن‬ ‫ﺗﺒﻴﺎن‬
:
t
e
m
t
m λ
−
= 0
)
(
‫ﻟﺪﻳﻨﺎ‬
:
)
01
........(
)
( 0
t
e
N
t
N λ
−
=
‫ﺣﻴﺚ‬
:
⎪
⎪
⎩
⎪
⎪
⎨
⎧
×
=
×
=
M
N
t
m
t
N
M
N
m
N
A
A
)
(
)
(
0
0
‫اﻟﻌﻼﻗﺔ‬ ‫ﻓﻲ‬ ‫ﺑﺎﻟﺘﻌﻮﻳﺾ‬
)
1
(
‫ﻧﺠﺪ‬
:
t
e
m
t
m λ
−
= 0
)
(
‫ب‬
–
‫أن‬ ‫ﺗﺒﻴﺎن‬
:
t
m
m
Ln ×
= λ
0
⎩
⎨
⎧
×
=
⇔
=
⇔
=
⇔
= −
−
t
t
m
m
Ln
e
t
m
m
e
m
t
m
e
m
t
m
t
t
t
λ
λ
λ
λ
)
(
)
(
)
(
)
(
0
0
0
0
-
‫اﻹﺷﻌﺎﻋﻲ‬ ‫اﻟﻨﺸﺎط‬ ‫ﺛﺎﺑﺖ‬ ‫ﺣﺴﺎب‬
λ
:
‫اﻟﺒﻴﺎﻧﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬
:
‫اﻟﺸﻜﻞ‬ ‫ﻣﻦ‬ ‫ﻣﻌﺎدﻟﺘﻪ‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﺧﻂ‬ ‫ﻋﻦ‬ ‫ﻋﺒﺎرة‬ ‫اﻟﺒﻴﺎن‬
:
)
01
.......(
)
(
0
t
a
t
m
m
Ln ×
=
‫اﻟﻨﻈﺮﻳﺔ‬ ‫اﻟﻌﻼﻗﺔ‬
:
)
02
.......(
)
(
0
t
t
m
m
Ln ×
= λ
‫اﻟﻌﻼﻗﺘﻴﻦ‬ ‫ﺑﻤﻄﺎﺑﻘﺔ‬
)
1
(
‫و‬
)
2
(
‫ﻧﺠﺪ‬
:
1
13
10
05
,
9 −
−
×
=
= s
a λ
00.50
00.50
00.25
00.25
00.25
06.00
e
n
c
y
-
e
00.25
00.25
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
‫اﻷ‬ ‫ـﺰء‬‫ـ‬‫ﺠ‬‫اﻟ‬
‫اﻟﺠ‬
s
m
s
‫اﻟﺘﻤﺮﻳﻦ‬
m
s
m
Ι
Ι
–
‫أ‬
- –
‫أ‬
-
‫أن‬ ‫ﺗﺒﻴﺎن‬
‫ﺗﺒﻴﺎن‬
:
‫ﻟﺪﻳﻨﺎ‬
‫ﻟﺪﻳﻨﺎ‬
:
)
01
..( )
01
‫ﺣﻴﺚ‬
: ‫ﺣﻴﺚ‬
:
/
e
m
=
=
M
M
N
t
m N
t
m ×
×
M
M
N
N
× A
)
( )
(t
t
)
(t
m(t
⎩
⎩
⎨
⎨
⎩
⎩
⎩
⎧
⎧
⎨⇔
⇔
−
m
e
m
m
=
=
m
m λ
)
( )
(t
t
t 0
00.2
0
e
n
c
y
-
e
m
s
m
s
3as.ency-education.com

10. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
2
‫ﻣﻦ‬
18
‫ﺟ‬
‫ـ‬
–
‫اﻻﺑﺘﺪاﺋﻴﺔ‬ ‫اﻷﻧﻮﻳﺔ‬ ‫ﻋﺪد‬ ‫ﺣﺴﺎب‬
0
N
‫اﻟﻌﻴﻨﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﻮﺟﻮدة‬
:
nouyaux
M
N
m
N A 21
23
0
0 10
51
,
2
239
10
02
,
6
1
×
=
×
×
=
×
=
-
‫اﻻﺑﺘﺪاﺋﻲ‬ ‫اﻟﻨﺸﺎط‬ ‫اﺳﺘﻨﺘﺎج‬
0
A
‫ﻟﻠﻌﻴﻨﺔ‬
:
Bq
N
A 9
21
13
0
0 10
27
,
2
10
51
,
2
10
05
,
9 ×
=
×
×
×
=
×
= −
λ
‫د‬
-
‫اﻟﻌﻤﺮ‬ ‫ﻧﺼﻒ‬ ‫زﻣﻦ‬ ‫ﺗﻌﺮﻳﻒ‬
:
‫ﻟﺘﻔﻜﻚ‬ ‫اﻟﻼزم‬ ‫اﻟﺰﻣﻦ‬ ‫ﻫﻮ‬
‫وﻧﻜﺘﺐ‬ ‫اﻟﻤﺸﻌﺔ‬ ‫اﻻﺑﺘﺪاﺋﻴﺔ‬ ‫اﻷﻧﻮﻳﺔ‬ ‫ﻋﺪد‬ ‫ﻧﺼﻒ‬
:
2
)
( 0
2
/
1
N
t
N =
-
‫أن‬ ‫ﺗﺒﻴﺎن‬
:
λ
2
2
/
1
Ln
t =
⎪
⎪
⎩
⎪
⎪
⎨
⎧
=
⇔
=
⇔
=
⇔
= −
−
λ
λ
λ
2
2
1
2
2
)
(
2
/
1
0
0
0
2
/
1
2
/
1
Ln
t
e
e
N
N
N
t
N t
t
‫ﻗﻴﻤﺘﻪ‬ ‫ﺣﺴﺎب‬
:
s
Ln
Ln
t 11
13
2
/
1 10
65
,
7
10
05
,
9
2
2
×
=
×
=
= −
λ
‫ـ‬‫ﻫ‬
-
‫أن‬ ‫ﺗﺒﻴﺎن‬
:
2
/
1
2
)
( 0
t
t
m
t
m =
⎪
⎪
⎩
⎪
⎪
⎨
⎧
=
⇔
=
⇔
=
=
=
×
−
2
/
1
2
/
1
2
/
1
2
)
(
)
(
)
(
0
2
0
2
0
0
0
t
t
Ln
t
Ln
t
t
t
m
t
m
e
m
t
m
e
m
e
m
e
m
t
m
t
t
λ
λ
-
‫اﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ ‫اﻟﻤﺘﺒﻘﻴﺔ‬ ‫اﻷﻧﻮﻳﺔ‬ ‫ﻛﺘﻠﺔ‬‫اﺳﺘﻨﺘﺎج‬
:
2
/
1
t
t =
g
m
e
m
t
m
t
t 25
,
0
4
)
( 0
0
2
/
1
2
/
1
2
/
1
2 =
⎪
⎩
⎪
⎨
⎧
=
=
00.25
00.25
00.50
00.25
00.25
00.50
00.25
e
n
c
y
-
e
0
e
n
c
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
-
‫اﺳﺘ‬
Bq
‫د‬
-‫د‬
-
‫ﻧﺼ‬ ‫زﻣﻦ‬ ‫ﺗﻌﺮﻳﻒ‬
‫ز‬ ‫ﺗﻌﺮﻳﻒ‬
‫أن‬
:
/
e
λ
λ
2
2
2
/
1
Ln
Ln
t =
=
=
⇔
⇔ −
− λ
λ
λ
2
2
1
0 e
N t
t
λ
λ
λ
n
.
Ln
Ln
13
10
05
,
9 10
05
,
9
2
Ln
Ln
=
= −
λ
⎪
⎪
⎪
⎨
⎧
⎧
⎪ )
(
m )
(
m
s
y
-
e
m
s
3as.ency-education.com

11. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
3
‫ﻣﻦ‬
18
‫و‬
–
‫اﻟﻤﺘﺒﻘﻴﺔ‬ ‫اﻟﺒﻠﻮﺗﻮﻧﻴﻮم‬ ‫ﻷﻧﻮﻳﺔ‬ ‫اﻟﻤﺌﻮﻳﺔ‬ ‫اﻟﻨﺴﺒﺔ‬ ‫ﻓﻴﻬﺎ‬ ‫ﺗﻜﻮن‬ ‫اﻟﺘﻲ‬ ‫اﻟﻠﺤﻈﺔ‬ ‫إﻳﺠﺎد‬
%
20
=
r
:
⎪
⎪
⎩
⎪
⎪
⎨
⎧
=
=
⇔
=
⇔
=
6
,
1
5
)
(
5
)
(
2
,
0
)
(
0
0
0
Ln
t
m
m
Ln
t
m
m
m
t
m
‫ﻣﺤﻮ‬ ‫ﻋﻠﻰ‬ ‫ﺑﺎﻹﺳﻘﺎط‬
‫ﻧﺠﺪ‬ ‫اﻟﻔﻮاﺻﻞ‬ ‫ر‬
:
ans
t 4
10
6
,
5 ×
=
ΙΙ
–
1
–
‫اﻟﻨﻮوي‬ ‫اﻻﻧﺸﻄﺎر‬ ‫ﺗﻌﺮﻳﻒ‬
:
‫ﻧﻮوي‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﻫﻮ‬
‫ﻣﻔﺘﻌﻞ‬
‫ﻓﺘﻨﺸﻄﺮ‬ ‫ﺑﻨﻴﺘﺮون‬ ‫ﺛﻘﻴﻠﺔ‬ ‫ﻧﻮاة‬ ‫ﺑﻘﺬف‬ ‫ﻳﺤﺪث‬
‫ﺧﻔﻴﻔﺘﻴﻦ‬ ‫ﻧﻮاﺗﻴﻦ‬ ‫إﻟﻰ‬
‫أﺧﺮى‬ ‫ﻧﻴﺘﺮوﻧﺎت‬ ‫إﺻﺪار‬ ‫ﻣﻊ‬ ‫اﺳﺘﻘﺮار‬ ‫أﻛﺜﺮ‬
‫ﻋﺎﻟﻴﺔ‬ ‫وﻃﺎﻗﺔ‬
.
2
–
‫ﺑ‬ ‫ﻣﻦ‬ ‫اﺳﺘﻘﺮار‬ ‫اﻷﻛﺜﺮ‬ ‫اﻟﻨﻮاة‬
‫اﻟﻨﺎﺗﺠﺔ‬ ‫اﻷﻧﻮﻳﺔ‬ ‫ﻴﻦ‬
:
‫ﻟﻜﻞ‬ ‫رﺑﻂ‬ ‫ﻃﺎﻗﺔ‬ ‫ﻟﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻨﻮاة‬ ‫ﻫﻲ‬
‫ﻧﻴﻜﻠﻴﻮن‬
‫أﻛﺒﺮ‬
.
nucleon
Mev
A
E
Mo
E l
/
568
,
8
102
981
,
873
)
(102
42 =
=
=
nucleon
Mev
A
E
Te
E l
/
345
,
8
135
674
,
1126
)
(135
52 =
=
=
‫ﻫﻲ‬ ‫اﺳﺘﻘﺮار‬ ‫اﻷﻛﺜﺮ‬ ‫اﻟﻨﻮاة‬ ‫وﻣﻨﻪ‬
:
Mo
102
42
3
–
‫اﻟﺒﻠﻮﺗﻮﻧﻴﻮم‬ ‫ﻣﻦ‬ ‫واﺣﺪة‬ ‫ﻧﻮاة‬ ‫اﻧﺸﻄﺎر‬ ‫ﻋﻦ‬ ‫اﻟﻤﺤﺮرة‬ ‫اﻟﻄﺎﻗﺔ‬ ‫ﺣﺴﺎب‬
:
J
E
Mev
E
Te
E
Mo
E
Pu
E
E
E
lib
lib
l
l
l
lib
11
10
1
,
3
739
,
193
674
,
1126
981
,
873
916
,
1806
)
(
)
(
)
(
−
×
=
⇔
=
−
−
=
⇔
−
−
=
Δ
=
4
–
‫اﻧﺸﻄﺎر‬ ‫ﻋﻦ‬ ‫اﻟﻤﺤﺮرة‬ ‫اﻟﻄﺎﻗﺔ‬ ‫ﺣﺴﺎب‬
1g
‫ﺑﺎﻟﺠﻮل‬ ‫اﻟﺒﻠﻮﺗﻮﻧﻴﻮم‬ ‫ﻣﻦ‬
:
‫ﻟﺪﻳﻨﺎ‬
‫ﻋﺪد‬
‫اﻷﻧﻮﻳﺔ‬
‫اﻟﻤﻮﺟﻮدة‬
‫ﻓﻲ‬
1g
:
J
E
N
E
nouyaux
M
N
m
N
lib
lib
A
T
10
11
21
0
21
23
0
0
10
8
,
7
10
1
,
3
10
51
,
2
10
51
,
2
239
10
02
,
6
1
×
=
×
×
×
=
×
=
×
=
×
×
=
×
=
−
5
–
‫اﻟﺴﺎﺑﻘﺔ‬ ‫اﻟﻜﺘﻠﺔ‬ ‫ﻻﺳﺘﻬﻼك‬ ‫اﻟﻼزﻣﺔ‬ ‫اﻟﺰﻣﻨﻴﺔ‬ ‫اﻟﻤﺪة‬ ‫ﺣﺴﺎب‬
:
‫ﻟﺪﻳﻨﺎ‬
:
min
13
780
10
30
10
34
,
2
10
34
,
2
10
8
,
7
30
,
0
6
10
10
10
=
=
×
×
=
=
⇔
=
×
=
×
×
=
×
=
⇔
=
s
P
E
t
t
E
P
j
E
r
E
E
E
r
ele
ele
lib
ele
lib
ele
T
T
00.25
00.25
00.25
00.25
00.25
00.25
00.25
00.25
e
n
c
y
-
e
d
u
c
a
c
c
a
a
c
a
E
E
E
E
E
E E
E
lib
lib
b
l
E
lib
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
‫ﻣﺤﻮ‬ ‫ﻋﻠﻰ‬ ‫ﺑﺎﻹﺳﻘﺎط‬
‫ﻋﻠﻰ‬ ‫ﺑﺎﻹﺳﻘﺎط‬
–
‫اﻟﻨﻮوي‬ ‫اﻻﻧﺸﻄﺎر‬ ‫ﺗﻌﺮﻳﻒ‬
‫اﻟﻨﻮوي‬ ‫اﻻﻧﺸﻄﺎر‬ ‫ﻒ‬
‫ﺧﻔﻴﻔﺘﻴﻦ‬ ‫ﻮاﺗﻴﻦ‬
‫ﺧﻔﻴﻔﺘﻴﻦ‬
‫اﺳﺘﻘﺮار‬ ‫أﻛﺜﺮ‬
‫اﺳ‬ ‫أﻛﺜﺮ‬
‫ﺑ‬ ‫ﻣﻦ‬ ‫اﺳﺘﻘﺮار‬ ‫ﺮ‬
‫ﻣﻦ‬ ‫ﺮار‬
‫اﻟ‬ ‫اﻷﻧﻮﻳﺔ‬ ‫ﻴﻦ‬
‫اﻷﻧ‬ ‫ﺑﻴﻦ‬
568
,
8 568
,
8 Mev
02
81
=
=
n
A
A
l
135
135
674
,
1126674
,
1126
=
=
l
‫ﻮﺗﻮﻧﻴﻮم‬
:
:
a
a
t
3
1806
1
(P
(
=
⇔
⇔ E
Elib
l
E
E
=
⇔ =
⇔ E
Elib
ib
E
Δ E
E E
E
E E (Pu
00.25
0
e
n
c
y
-
e
m
s
m
s
3as.ency-education.com

12. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
4
‫ﻣﻦ‬
18
‫اﻟﺘﻤﺮ‬
‫ﻳ‬
‫اﻟﺜﺎﻧﻲ‬ ‫ﻦ‬
) :
07.00
‫ـ‬‫ـ‬‫ﻃ‬‫ﻧﻘﺎ‬
(
‫اﻻول‬ ‫اﻟﺠﺰء‬
:
1
-
‫أ‬
-
‫ﻛﺔ‬
‫ﺣﺮ‬ ‫ﻃﺒﻴﻌﺔ‬ ‫ﺗﺤﺪﻳﺪ‬
G
:
-
‫اﻟﺰﻣﻨﻲ‬ ‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬
:
[0 , 3s]
‫ﻛﺔ‬
‫ﺣﺮ‬ ‫ﻓﺎن‬ ‫وﻣﻨﻪ‬ ‫ﻣﺘﺰاﻳﺪة‬ ‫ﺧﻄﻴﺔ‬ ‫داﻟﺔ‬ ‫ﻋﻦ‬ ‫ﻋﺒﺎرة‬ ‫اﻟﺴﺮﻋﺔ‬
G
‫ﺑﺎﻧﺘﻈﺎم‬ ‫ﻣﺘﺴﺎرﻋﺔ‬ ‫ﻣﺴﺘﻘﻴﻤﺔ‬
.
-
‫اﻟ‬ ‫ﻓﻲ‬
‫اﻟﺰﻣﻨﻲ‬ ‫ﻤﺠﺎل‬
:
[3s , 4s]
‫ﺛﺎﺑﺘﺔ‬ ‫اﻟﺴﺮﻋﺔ‬
vG = Cte
‫ﻛﺔ‬
‫ﺣﺮ‬ ‫ﻓﺎن‬ ‫وﻣﻨﻪ‬
G
‫ﻣﻨﺘﻈﻤﺔ‬ ‫ﻣﺴﺘﻘﻴﻤﺔ‬
.
‫ب‬
-
‫إﻳﺠﺎد‬
‫اﻟﺘﻮﺗﺮ‬ ‫ﻗﻮة‬ ‫ﺷﺪة‬
:
‫ﻟﻨﻴﻮﺗﻦ‬ ‫اﻟﺜﺎﻧﻲ‬ ‫اﻟﻘﺎﻧﻮن‬ ‫ﺑﺘﻄﺒﻴﻖ‬
:
a
m
T
P
r
r
r
=
+
‫ﺑﺎﻹﺳ‬
‫اﻟﻤﺤﻮر‬ ‫ﻋﻠﻰ‬ ‫ﻘﺎط‬
OZ
‫ﻧﺠﺪ‬
:
⎩
⎨
⎧
+
=
=
−
)
( a
g
m
T
ma
P
T
‫اﻟﻤﺮﺣﻠﺔ‬ ‫ﺧﻼل‬
‫اﻷوﻟﻰ‬
‫ﻟﺪﻳﻨﺎ‬
:
⎪
⎩
⎪
⎨
⎧
=
+
=
=
−
−
=
Δ
Δ
=
N
T
s
m
t
v
a
5520
)
4
8
,
9
(
400
/
4
0
1
0
4 2
‫ﻟﺪﻳﻨﺎ‬ ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﻤﺮﺣﻠﺔ‬ ‫ﺧﻼل‬
:
Cte
vG =
‫وﺑﺎﻟﺘﺎﻟﻲ‬
:
0
=
a
‫وﻣﻨﻪ‬
:
N
mg
P
T 3920
8
,
9
400 =
×
=
=
=
00.25
00.25
00.50
00.25
00.25
00.25
07.00
í
î
î⃗
ð
îî⃗
ñ
ò
e
n
c
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
a
t
i
o
n
.
c
o
m
/
e
a
t
i
o
n
.
c
o
m
/
e
s
s
‫اﻟﺠﺰء‬
s
1
-
‫أ‬
- 1
-
m
-
‫اﻟﻤﺠﺎل‬ ‫ﻓﻲ‬
‫ﻓﻲ‬
‫ﻣﺘﺴﺎرﻋ‬ ‫ﻣﺴﺘﻘﻴﻤﺔ‬
‫ﻣﺴﺘﻘﻴﻤﺔ‬
-
‫اﻟ‬ ‫ﻓﻲ‬
‫اﻟ‬ ‫ﻓﻲ‬
‫اﻟﺰﻣﻨﻲ‬ ‫ﻤﺠﺎل‬
: ‫اﻟﺰﻣﻨﻲ‬ ‫ﻟﻤﺠﺎل‬‫ﻟ‬‫ﻟ‬
‫ﻳﺠﺎد‬
‫اﻟﺘﻮﺗﺮ‬ ‫ﻗﻮة‬ ‫ﺷﺪة‬
: ‫اﻟﺘﻮﺗﺮ‬ ‫ﻗﻮة‬ ‫ﺪة‬
00.25
00
e
n
c
y
-
e
m
s
c
n
.
c
n
.
c
c
n
.
c
í
o
m
3as.ency-education.com

13. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
5
‫ﻣﻦ‬
18
2
–
‫أ‬
-
‫اﻟﺒﻌﺪي‬ ‫اﻟﺘﺤﻠﻴﻞ‬
:
[ ] [ ]
[ ]
[ ][ ]
[ ]
[ ]
[ ]
[ ]
[ ]
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎨
⎧
=
=
=
=
L
M
T
L
T
L
M
V
F
K
v
f
K
2
2
2
2
2
‫وﻣﻨﻪ‬
:
‫وﺣﺪة‬
K
‫ﻫﻲ‬
:
Kg/m
‫اﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ـ‬ ‫ب‬
:
‫ﻧﺠﺪ‬ ‫ﻟﻨﻴﻮﺗﻦ‬ ‫اﻟﺜﺎﻧﻲ‬ ‫اﻟﻘﺎﻧﻮن‬ ‫ﺑﺘﻄﺒﻴﻖ‬
:
G
S a
m
T
P
r
r
r
=
+
′
‫اﻟﻤﺤﻮر‬ ‫ﻋﻠﻰ‬ ‫ﺑﺎﻹﺳﻘﺎط‬
OY
‫ﻧﺠﺪ‬
:
G
S
S a
m
Kv
g
m =
− 2
‫وﻣﻨﻪ‬
:
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎨
⎧
=
×
×
+
⇔
×
×
−
=
⇔
−
=
−
=
−
−
8
,
9
10
9
10
9
8
,
9
30
7
,
2
8
,
9
2
2
2
2
2
v
dt
dv
v
dt
dv
v
dt
dv
v
m
K
g
dt
dv
S
00.50
00.25
00.50
00.50
ó
î⃗
ð`
îîî⃗
ñ
õ
‫وﻣﻨﻪ‬
:
‫وﺣﺪة‬ ‫وﻣﻨﻪ‬
:
‫وﺣﺪة‬
K
K
‫ﻫﻲ‬ K
K
‫ﺿﻠﻴﺔ‬
:
e
n
c
y
-
e
m
s
o
n
.
c
o
n
o
n
o
n
n
ó
ó
î
î⃗
⃗
î
î
î
ñ
o
3as.ency-education.com

14. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
6
‫ﻣﻦ‬
18
‫ـ‬‫ﺟ‬
-
‫اﻟﺤﺪﻳﺔ‬ ‫اﻟﺴﺮﻋﺔ‬ ‫اﻳﺠﺎد‬
L
v
:
‫ﻳﻜﻮن‬ ‫اﻟﺪاﺋﻢ‬ ‫اﻟﻨﻈﺎم‬ ‫ﻓﻲ‬
:
s
m
v
v
dt
dv
L
L
/
4
,
10
10
9
8
,
9
8
,
9
10
9
0
2
2
2
=
×
=
⇔
=
×
×
⇔
=
−
−
‫د‬
-
‫اﻳﺠﺎد‬
‫اﻟﻠﺤﻈﺘﻴ‬ ‫ﺑﻴﻦ‬ ‫اﻟﻮﺳﻄﻲ‬ ‫اﻟﺘﺴﺎرع‬ ‫ﻗﻴﻤﺔ‬
‫ﻦ‬
:
t2=τ , t1=0
:
‫ﻟﺪﻳﻨﺎ‬
:
)
01
...(
..........
1
2
1
2
t
t
v
v
am
−
−
=
‫ﺣﻴﺚ‬
:
⎩
⎨
⎧
≈
×
=
×
=
⇔
=
=
⇔
=
s
m
v
v
t
v
t
l /
6
,
6
4
,
10
63
,
0
63
,
0
0
0
2
2
1
1
τ
‫ﻓﻲ‬ ‫ﺑﺎﻟﺘﻌﻮﻳﺾ‬
)
01
(
‫ﻧﺠﺪ‬
:
)
/
(
6
,
6
0
0
6
,
6 2
s
m
am
τ
τ
=
−
−
=
‫ﺣﻴﺚ‬
:
s
m/
6
,
6
=
β
‫اﻟﺜﺎﻧﻲ‬ ‫اﻟﺠﺰء‬
:
1
-
‫اﻟﻤﻨﺤﻨﻰ‬ ‫ﻫﻮ‬ ‫ﻛﻴﺔ‬
‫اﻟﺤﺮ‬ ‫اﻟﻄﺎﻗﺔ‬ ‫ﺗﻐﻴﺮات‬ ‫ﻳﻤﺜﻞ‬ ‫اﻟﺬي‬ ‫اﻟﻤﻨﺤﻨﻰ‬
)
‫أ‬
. (
‫اﻟﺘﻌﻠﻴﻞ‬
:
‫ﺣﺴﺐ‬
‫ﻋﻨﺪ‬ ‫اﻻﺑﺘﺪاﺋﻴﺔ‬ ‫اﻟﺸﺮوط‬
t=0
‫اﻟﺠﺴﻢ‬ ‫ﺗﺤﺮﻳﺮ‬ ‫ﺗﻢ‬
‫اﺑﺘﺪاﺋﻴﺔ‬ ‫ﺳﺮﻋﺔ‬ ‫دون‬
‫وﻣﻨﻪ‬
:
0
=
O
Ec
2
-
‫اﻟﺠﻤﻠﺔ‬ ‫ﻃﺎﻗﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫اﻳﺠﺎد‬
:
‫ﻟﺪﻳﻨﺎ‬
:
Pe
c
T E
E
E +
=
‫ﻋﻨﺪ‬ ‫وﻟﺪﻳﻨﺎ‬
:
t=0
‫ﻧﺠﺪ‬
:
0
=
O
Ec
‫وﻣﻨﻪ‬
:
mJ
E
E Pe
T 2
max
=
=
00.50
00.25
00.25
00.25
00.25
e
n
c
e
c
y
-
e
00.25
00.25
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
s
/
‫اﻟﻠﺤﻈ‬ ‫ﺑﻴﻦ‬ ‫اﻟﻮﺳﻄﻲ‬ ‫اﻟﺘﺴﺎرع‬
‫ﺑﻴﻦ‬ ‫اﻟﻮﺳﻄﻲ‬ ‫رع‬
m
/
o
...(
..........
.........
1
2 1
1
1
t
t t
2
v
=
= 63
,
0 63
,
0 ×
×
0
2
n
c
y
-
e
m
s
m
s
3as.ency-education.com

15. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
7
‫ﻣﻦ‬
18
3
-
‫اﻟﻤﺴﺎﻓﺔ‬ ‫اﺳﺘﻨﺘﺎج‬
ö
:
m
K
E
X
KX
E
E
Pe
Pe
T
2
3
0
2
0
10
2
10
10
2
2
2
2
1
max
max
−
−
×
=
×
×
=
×
=
⇔
=
=
4
-
‫أ‬
-
‫اﻟﺘﻮﺗﺮ‬ ‫ﻗﻮة‬ ‫ﻋﻤﻞ‬ ‫اﻳﺠﺎد‬
:
{ j
E
E
E
T
W A
O Pe
Pe
Pe
AO
3
3
10
2
)
10
2
0
(
)
(
)
( −
−
×
=
×
−
−
=
−
−
=
Δ
−
=
r
‫ب‬
-
‫إﻳﺠﺎد‬
‫اﻟﺘﻮﺗﺮ‬ ‫ﻗﻮة‬ ‫ﺷﺪة‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﻛﺔ‬
‫ﻟﻠﺤﺮ‬ ‫اﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬
T
r
:
‫اﻟﺠﺴﻢ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺜﺎﻧﻲ‬ ‫ﻧﻴﻮﺗﻦ‬ ‫ﻗﺎﻧﻮن‬ ‫ﺑﺘﻄﺒﻴﻖ‬
:
a
m
T
R
P .
=
+
+
‫اﻟﻤﺤﻮر‬ ‫ﻋﻠﻰ‬ ‫ﺑﺎﻹﺳﻘﺎط‬ ‫و‬
OX
‫ﻧﺠﺪ‬
:
a
m
T .
=
−
‫وﻣﻨﻪ‬
2
2
.
dt
x
d
m
T =
−
‫ﺣﻴﺚ‬
:
k
T
x =
‫وﻣﻨﻪ‬
:
0
2
2
=
+ T
m
k
dt
T
d
‫اﻟﺸﻜﻞ‬ ‫ﻣﻦ‬ ‫ﺟﻴﺒﻲ‬ ‫ﺣﻠﻬﺎ‬ ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﺮﺗﺒﺔ‬ ‫ﻣﻦ‬ ‫ﺗﻔﺎﺿﻠﻴﺔ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫وﻫﻲ‬
:
)
cos(
)
( 0
max ϕ
ω +
= t
T
t
T
-
‫ﺗﻤﺜﻴﻞ‬
)
(t
f
T =
‫ﻛﺔ‬
‫ﻟﻠﺤﺮ‬ ‫ذاﺗﻲ‬ ‫دور‬ ‫أﺟﻞ‬ ‫ﻣﻦ‬
:
‫ﻧﺠﺪ‬ ‫اﻻﺑﺘﺪاﺋﻴﺔ‬ ‫اﻟﺸﺮوط‬ ‫ﻣﻦ‬
:
0
=
ϕ
00.50
00.50
00.25
00.50
00.25
→
i X
X’
O
+ Xm
- Xm
X
→
P
→
T
→
R
T0/2
t(ms)
1/15
T(N)
e
n
c
y
-
e
d
u
c
a
t
i
o
a
t
i
o
o
R
P
P R
)
(
T
T )
(
e
e
e
e
e
e
n
c
y
-
e
t
i
o
n
.
c
o
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
.
.
n
.
n
.
n
.
c
o
m
/
e
x
.
.
o
n
m
/
+ X
Xm
c
o
c
o
c
o
m
/
.
c
m
c
oX
X
→
T
/
e
x
a
m
s
4
-
‫أ‬
- 4
-
‫أ‬
-
‫ﻋﻤﻞ‬ ‫اﻳﺠﺎد‬
‫اﻳﺠﺎد‬
a
m
j
3
3
10
2
)
0 10
2 −
−
2
) 2
2
3
‫إﻳﺠﺎد‬
‫اﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬
‫اﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫ﻟﻤﻌﺎدﻟﺔ‬
m
s
t
i
o
n
.
c
o
m
/
e
x
3as.ency-education.com

16. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
8
‫ﻣﻦ‬
18
‫اﻟﺘﺠﺮﻳﺒﻲ‬ ‫اﻟﺘﻤﺮﻳﻦ‬
) :
07.00
‫ـ‬‫ـ‬‫ﻃ‬‫ﻧﻘﺎ‬
(
1
-
‫اﻟﻤﻌﺎدﻻت‬
:
‫اﻷول‬ ‫اﻟﻔﻮج‬
:
O
H
Cr
HCOOH
H
O
Cr
OH
CH 2
3
2
7
2
3 11
4
3
14
2
3 +
+
==
+
+ +
+
−
-
‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻧﻮع‬
) :
‫أﻛﺴﺪة‬
–
‫إرﺟﺎع‬
(
‫اﻟﺜﺎﻧﻲ‬ ‫اﻟﻔﻮج‬
:
O
H
COOCH
CH
OH
CH
COOH
CH 2
3
3
3
3 +
−
==
+
−
-
‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻧﻮع‬
) :
‫أﺳﺘﺮة‬
–
‫إﻣﺎﻫﺔ‬
(
2
-
‫اﻟﺸﻜﻞ‬
-
1
:
‫ﺗﺎم‬ ‫ﻏﻴﺮ‬ ‫ﻷﻧﻪ‬ ‫اﻷﺳﺘﺮة‬ ‫ﺗﺤﻮل‬ ‫ﻳﻮاﻓﻖ‬
،
%
66
% =
f
τ
.
-
‫اﻟﺸﻜﻞ‬
-
2
:
‫اﻷﻛﺴﺪة‬ ‫ﺗﺤﻮل‬ ‫ﻳﻮاﻓﻖ‬
-
‫ﺗﺎم‬ ‫ﻷﻧﻪ‬ ‫إرﺟﺎع‬
،
%
100
% =
f
τ
.
3
–
‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻘﺪم‬ ‫ﺟﺪول‬
:
O
H
Cr
HCOOH
H
O
Cr
OH
CH 2
3
2
7
2
3 11
4
3
14
2
3 +
+
==
+
+ +
+
−
‫اﻟﺘﻘﺪم‬
‫اﻟﺤﺎﻟﺔ‬
‫ﺑﻮﻓﺮة‬
0
0
‫ﺑﻮﻓﺮة‬
CV
n0
0
‫اﻻﺑﺘﺪاﺋﻴﺔ‬
‫ﺑﻮﻓﺮة‬
4x
3x
‫ﺑﻮﻓﺮة‬
CV-2x
- 3x
n0
x
‫اﻻﻧﺘﺘﻘﺎﻟﻴﺔ‬
‫ﺑﻮﻓﺮ‬
‫ة‬
4Xf
Xf
3
‫ﺑﻮﻓﺮة‬
CV-2Xf
- 3Xf
n0
f
x
‫اﻟﻨﻬﺎﺋﻴﺔ‬
-
‫ﻟﺪﻳﻨﺎ‬
:
2
3
02
,
0
06
,
0
0
=
=
×V
C
n
.
-
‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻣﻦ‬
:
2
3
0
=
×V
C
n
.
‫ﻛﻴﻮﻣﺘﺮي‬
‫ﺳﺘﻮ‬ ‫اﻟﻤﺘﻔﺎﻋﻞ‬ ‫اﻟﻤﺰﻳﺞ‬ ‫وﻣﻨﻪ‬
-
‫اﻷﻋﻈﻤﻲ‬ ‫اﻟﺘﻘﺪم‬ ‫ﺗﺤﺪﻳﺪ‬
:
mol
n
V
C
X 02
,
0
3
2
0
max =
=
×
=
.
4
-
‫اﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ ‫ﻟﻠﺘﻔﺎﻋﻞ‬ ‫اﻟﺤﺠﻤﻴﺔ‬ ‫اﻟﺴﺮﻋﺔ‬ ‫ﻋﺒﺎرة‬
(t=0)
:
)
01
.......(
1
dt
dx
V
vV ×
=
‫وﻟﺪﻳﻨﺎ‬
:
)
01
.......(
2
max V
C
x
X
x
×
=
=
τ
00.75
00.25
00.75
00.25
00.50
00.50
00.75
00.50
00.50
00.25
07.00
e
n
c
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
1
-
‫اﻷ‬ ‫اﻟﻔﻮج‬
‫اﻟﻔﻮج‬
m
s
O
H O
2
2
-
‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻧﻮع‬
) :
‫أﻛ‬ -
‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻧﻮع‬
‫اﻟﺜﺎﻧﻲ‬ ‫ﻟﻔﻮج‬
‫اﻟﺜﺎﻧﻲ‬
a
:
:
O
H O
H
OCH
CH 2
3
3 H
H
+
‫ﺘﻔﺎﻋﻞ‬
) :
‫أﺳﺘﺮة‬
–
‫إﻣﺎﻫﺔ‬
( ‫ﻞ‬
) :
‫أﺳﺘﺮة‬
–
‫إﻣﺎﻫ‬
‫ﻏﻴﺮ‬ ‫ﻷﻧﻪ‬ ‫اﻷﺳﺘﺮة‬ ‫ﺗﺤﻮل‬ ‫اﻓﻖ‬
‫ﻷﻧﻪ‬ ‫اﻷﺳﺘﺮة‬ ‫ﻮل‬
‫اﻷﻛﺴﺪة‬ ‫ﻮل‬
‫ﻛﺴﺪ‬
-
‫ﺗﺎم‬ ‫ﻷﻧﻪ‬ ‫إرﺟﺎع‬-
‫ﻷ‬ ‫إرﺟﺎع‬
H
H
O
Cr H
2
7
2O
r 3
H
H
O
Cr O
r H
H +
+
c
o
‫ﻮﻓﺮة‬
CV
CV
n
.
.
.
o
.
CV-2x
CV-
- 3x
o
o
o
n
i
o
CV-2
CV
- 3X
- 3Xf
f
X
X
X
n
n0
0
t
i
o
i
o
i
o
a
a
a
a
t
i
o
00.50
00.50
e
n
c
y
-
e
m
s
m
s
3as.ency-education.com

17. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
9
‫ﻣﻦ‬
18
)
02
.......(
2
τ
×
×
=
V
C
x
‫ﻧﻌﻮض‬
)
1
(
‫ﻓﻲ‬
)
2
(
‫ﻓﻨﺠﺪ‬
:
dt
d
C
vV
τ
×
=
2
-
‫ا‬ ‫ﻋﻨﺪ‬ ‫ﻟﻠﺘﻔﺎﻋﻞ‬ ‫اﻟﺤﺠﻤﻴﺔ‬ ‫اﻟﺴﺮﻋﺔ‬ ‫ﺣﺴﺎب‬
‫ﻟﻠﺤﻈﺔ‬
(t=0)
:
‫ﻣﻴﻞ‬
‫ﺍﻟﻤﻤﺎﺱ‬
C
dt
d
C
vV ×
=
×
=
2
2
τ
h
L
mol
vV .
/
10
5
,
1
80
12
,
0
2
2
,
0 4
−
×
×
=
5
-
‫ﻛﺐ‬
‫اﻟﻤﺮ‬ ‫إﺳﻢ‬
(E)
:
‫اﻟﻤﻴﺜﻴﻞ‬ ‫إﻳﺜﺎﻧﻮات‬
-
‫ﺣﺴﺎب‬
2
n
)
‫اﻟﻨﺎﺗﺞ‬ ‫اﻷﺳﺘﺮ‬ ‫ﻣﺎدة‬ ‫ﻛﻤﻴﺔ‬
E
(
‫ﻟﺪﻳﻨﺎ‬
:
mol
n
n
n
n
X
X
f
f
f
2
0
2
0
2
max
10
4 −
×
=
×
=
⇔
=
=
τ
τ
6
-
‫اﻟﻨﻮ‬ ‫أﺣﺪ‬ ‫ﻧﺤﺬف‬ ‫اﻟﻨﻬﺎﺋﻲ‬ ‫اﻟﺘﻘﺪم‬ ‫ﻧﺴﺒﺔ‬ ‫ﻟﺰﻳﺎدة‬
‫اﺗﺞ‬
.
-
‫اﻟﺘﻮازن‬ ‫ﻳﺨﺘﻞ‬ ‫اﻟﻨﻮاﺗﺞ‬ ‫أﺣﺪ‬ ‫ﻋﻨﺪﻧﺰع‬
)
‫ﻓﻲ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻳﻨﺰاح‬
‫اﻻﺗﺠﺎﻩ‬
‫اﻟﻤﺒﺎﺷﺮ‬
(
‫اﻟﺘﻘﺪم‬ ‫زﻳﺎدة‬ ‫ﻓﻲ‬ ‫ﻳﺆدي‬ ‫ﻣﻤﺎ‬
‫اﻟﻨﻬﺎﺋﻲ‬
f
X
‫اﻟﻨﻬﺎﺋﻲ‬ ‫اﻟﺘﻘﺪم‬ ‫ﻧﺴﺒﺔ‬ ‫ﺑﺬﻟﻚ‬ ‫ﻓﺘﺰداد‬
f
τ
.
00.50
00.50
00.25
00.25
00.25
00.25
‫ﻧﻌﻮ‬
-
-
‫اﻟﺴ‬ ‫ﺣﺴﺎب‬
‫ﺣﺴﺎب‬
‫ﻣﻴﻞ‬
‫ﻣﻴﻞ‬
h
L
l h
L.
/ L
ol L
(
:
‫اﻟﻤﻴﺜﻴﻞ‬ ‫إﻳﺜﺎﻧﻮات‬
‫اﻟﻤﻴﺜﻴﻞ‬ ‫ﺜﺎﻧﻮات‬
‫اﻟﻨﺎﺗﺞ‬ ‫اﻷﺳﺘﺮ‬ ‫ﻣﺎدة‬
‫اﻟﻨﺎﺗﺞ‬ ‫ﻷﺳﺘﺮ‬
E
E
(
n
n
n
n
X
X
X f
f
X
0
2
2
0
2
2
max
max
X
X
×
=
⇔ =
⇔ n
n n
n 0
2
2 n
n
=
= =
‫ﺒﺎﺷﺮ‬
(
‫ا‬ ‫زﻳﺎدة‬ ‫ﻓﻲ‬ ‫ﻳﺆدي‬ ‫ﻣﻤﺎ‬
‫ﻓﻲ‬ ‫ﻳﺆدي‬ ‫ﻣﻤﺎ‬
e
n
c
y
-
e
m
s
m
s
3as.ency-education.com

18. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
10
‫ﻣﻦ‬
18
‫ـﻲ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ﻧ‬‫اﻟﺜﺎ‬ ‫ـﻮع‬‫ـ‬‫ـ‬‫ـ‬‫ﺿ‬‫اﻟﻤﻮ‬
‫اﻷول‬ ‫اﻟﺠﺰء‬
:
‫اﻟ‬
‫اﻷول‬ ‫ﺘﻤﺮﻳﻦ‬
) :
06.00
‫ـﺔ‬‫ـ‬‫ﻄ‬‫ﻧﻘ‬
(
1
-
‫ﻣﺎﺋﻲ‬ ‫ﻣﺤﻠﻮل‬
(S)
‫اﻟﻤﻮﻟﻲ‬ ‫ﻛﻴﺰﻩ‬
‫ﺗﺮ‬ ‫ﻟﻠﻨﺸﺎدر‬
C =10-2
mol/L
‫اﻟﻨﻬﺎﺋﻲ‬ ‫ﺗﻘﺪﻣﻪ‬ ‫وﻧﺴﺒﺔ‬
%
4
=
f
τ
.
‫أ‬
-
‫اﻟﻨﺸﺎدر‬ ْ
‫ﺟﺰي‬ ‫ﻓﻲ‬ ‫اﻷﺳﺎﺳﻴﺔ‬ ‫اﻟﺨﺎﺻﻴﺔ‬ ‫ﺗﻜﻤﻦ‬
‫ﺗﻔﺎﻋﻠﻪ‬ ‫اﺛﻨﺎء‬ ‫ﻫﻴﺪروﺟﻴﻦ‬ ‫ﻟﺒﺮوﺗﻮن‬ ‫اﻛﺘﺴﺎﺑﻪ‬ ‫ﻓﻲ‬
‫اﻟﺘﺎﻟﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﺴﺐ‬
:
+
+
==
+ 4
3 NH
H
NH
‫ب‬
-
‫ﻛﺘ‬
‫ﺎ‬
‫ﺑ‬
‫ﺔ‬
‫اﻟﻤﺎء‬ ‫ﻣﻊ‬ ‫اﻟﻨﺸﺎدر‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﻣﻌﺎدﻟﺔ‬
:
−
+
+
==
+ OH
NH
O
H
NH 4
2
3
-
‫اﻟﺘ‬
‫ﻌﺒ‬
‫ﻴ‬
‫ﺑﺪﻻﻟﺔ‬ ‫ﻟﻠﺘﻔﺎﻋﻞ‬ ‫اﻟﺘﻮازن‬ ‫ﺛﺎﺑﺖ‬ ‫ﻋﻦ‬ ‫ﺮ‬
C
،
f
τ
:
[ ] [ ]
[ ] f
f
f
f
f C
NH
OH
NH
K
τ
τ
−
×
=
×
=
−
+
1
2
3
4
‫ـ‬‫ﺟ‬
-
‫إﺛﺒﺎت‬
‫أن‬
‫ﺛﺎﺑﺖ‬
‫ﻟﻠﺜﻨﺎﺋﻴﺔ‬ ‫اﻟﺤﻤﻮﺿﺔ‬
NH3 )
/
( NH4
+
‫ﺑﺎﻟ‬ ‫ﻳﻌﻄﻰ‬
‫ﻌﻼﻗﺔ‬
:
K
Ke
PKa log
−
=
‫ﻟﺪﻳﻨﺎ‬
:
[ ] [ ]
[ ]
K
K
Ka
PKa
K
Ke
OH
OH
NH
O
H
NH
Ka
e
f
f
f
log
log
4
3
3
−
=
−
=
⇔
=
×
×
= −
−
+
+
-
‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ﻟ‬‫ا‬ ‫ﻗﻴﻤﺔ‬ ‫ﺣﺴﺎب‬
PKa
‫ﻟﻠﺜﻨﺎﺋﻴﺔ‬
NH3 )
/
( NH4
+
:
‫ﻟﺪﻳﻨﺎ‬
:
5
2
2
2
10
67
,
1
04
,
0
1
)
04
,
0
(
10
1
−
−
×
=
−
×
=
−
×
=
f
f
C
K
τ
τ
‫وﻣﻨﻪ‬
:
2
,
9
10
67
,
1
10
log
log 5
14
=
×
−
=
−
= −
−
K
Ke
PKa
00.25
00.25
00.25
00.25
00.25
00.25
06.00
e
n
c
e
c
y
-
e
00.25
0.25
00.25
0
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
s
s
1
-
‫ﻣﺤﻠﻮل‬ 1
-
‫ﻣﺤ‬
‫أ‬
-‫أ‬
-
‫اﻟﺨ‬ ‫ﺗﻜﻤﻦ‬
‫ﺗﻜﻤﻦ‬
‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﺴﺐ‬
‫اﻟﻤ‬ ‫ﺣﺴﺐ‬
‫ﻛﺘ‬
‫ﺎ‬
‫ﺑ‬
‫ﺔ‬
‫اﻟﻨﺸﺎدر‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﻣﻌﺎدﻟﺔ‬
‫اﻟﻨﺸﺎدر‬ ‫ﺗﻔﺎﻋﻞ‬ ‫ﻣﻌﺎدﻟﺔ‬
−
−
OH
OH
+
‫ﺑﺪﻻﻟﺔ‬ ‫ﻟﻠﺘﻔﺎﻋﻞ‬ ‫اﻟﺘﻮازن‬ ‫ﺖ‬
‫ﺑﺪ‬ ‫ﻟﻠﺘﻔﺎﻋﻞ‬ ‫زن‬
]
]
o
o
m
f
f
f
f
f C
C
τ
τ
τ
−
−
×
×
=
=
1
1
2
2
3
‫ﻴﺔ‬
)
/
( NH )
/ NH
NH3
+
+
Ke
PKa
PKa log
−
= −
=
[
[ ]
] [
[
t
i
PK
NH
NH ]
]
Ka
Ka
f
3
⇔ P
=
=
n
c
y
-
e
m
s
m
s
3as.ency-education.com

19. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
11
‫ﻣﻦ‬
18
‫د‬
-
‫إﺛﺒﺎت‬
‫أن‬
PH
‫اﻟﻤﺤﻠﻮل‬
(S)
‫ﺑﺎﻟﺸﻜﻞ‬ ‫ﻳﻜﺘﺐ‬
:
)
1
(
log
f
f
PKa
PH
τ
τ
−
+
=
:
‫ﻟﺪﻳﻨﺎ‬
:
[ ]
[ ]
[ ]
[ ]
)
1
log(
)
log(
)
log(
)
(
log
4
3
f
f
f
f
f
f
f
f
PKa
C
C
C
PKa
PH
OH
OH
C
PKa
NH
NH
PKa
PH
τ
τ
τ
τ −
+
=
×
×
−
+
=
⇔
−
+
=
+
= −
−
+
‫ـ‬‫ﻫ‬
-
‫ﺗ‬
‫ﺤﺪ‬
‫ﻳ‬
‫اﻟﻤﺤﻠﻮل‬ ‫ﻓﻲ‬ ‫اﻟﻐﺎﻟﺐ‬ ‫اﻟﻜﻴﻤﻴﺎﺋﻲ‬ ‫اﻟﻨﻮع‬ ‫ﺪ‬
(S)
،
‫و‬
‫ﺣﺴ‬
‫ﺎ‬
‫اﻟ‬ ‫ﻗﻴﻤﺔ‬ ‫ب‬
‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬‫ـ‬
‫ـ‬
PH
:
‫ﻟﺪﻳﻨﺎ‬
:
[ ]
[ ]
⎪
⎩
⎪
⎨
⎧
×
=
×
−
=
×
−
=
×
=
×
=
×
=
−
−
−
−
−
+
L
mol
C
C
NH
L
mol
C
NH
f
f
f
f
/
10
6
,
9
)
04
,
0
10
(
10
/
10
4
04
,
0
10
3
2
2
3
4
2
4
τ
τ
‫أن‬ ‫ﺑﻤﺎ‬
:
[ ] [ ]f
f
NH
NH +
4
3 f
‫ﻫﻮ‬ ‫اﻟﻐﺎﻟﺐ‬ ‫اﻟﻔﺮد‬ ‫أي‬ ‫ﻏﺎﻟﺒﺔ‬ ‫اﻷﺳﺎﺳﻴﺔ‬ ‫اﻟﺼﻔﺔ‬ ‫ﻓﺎن‬
NH3
.
-
‫ـ‬‫ـ‬‫ـ‬‫ﻟ‬‫ا‬ ‫ﻗﻴﻤﺔ‬ ‫ﺣﺴﺎب‬
PH
‫اﻟﻤﺤﻠﻮل‬
:
6
,
10
)
04
,
0
04
,
0
1
log(
2
,
9
)
1
(
log ≈
−
+
=
−
+
=
f
f
PKa
PH
τ
τ
2
-
‫أ‬
-
‫ـ‬‫ﻟ‬‫ا‬ ‫ﻋﻦ‬ ‫اﻟﺘﻌﺒﻴﺮ‬
‫ـ‬‫ـ‬‫ـ‬
PH
‫ﺑﺪﻻﻟﺔ‬
σ
،
+
−
4
, NH
OH
λ
λ
:
‫ﻟﺤﻈﺔ‬ ‫أي‬ ‫ﻋﻨﺪ‬
t
‫ﻳﻜﻮن‬
:
[ ] [ ] [ ]
⎪
⎪
⎩
⎪
⎪
⎨
⎧
+
−
+
=
⇔
×
+
=
+
=
×
+
×
=
−
+
−
+
−
+
−
+
−
−
−
+
)
01
..(
).........
log(
14
log
10
)
(
)
(
)
(
)
(
4
4
4
4
14
4
OH
NH
PH
OH
NH
OH
NH
OH
NH
PH
t
OH
OH
NH
t
λ
λ
σ
λ
λ
σ
λ
λ
λ
λ
σ
‫ب‬
-
‫ﻗﻴﻤﺔ‬ ‫اﻳﺠﺎد‬
+
4
NH
λ
:
‫ﻧﺠﺪ‬ ‫اﻟﺒﻴﺎن‬ ‫ﻋﻠﻰ‬ ‫اﻋﺘﻤﺎدا‬
‫اﻟﺒﻴﺎن‬ ‫ﻣﻌﺎدﻟﺔ‬
:
)
02
(
..........
log b
a
PH +
×
= σ
‫ﺑﻤﻄﺎﺑﻘﺔ‬
)
01
(
‫و‬
)
02
(
‫ﻧﺠﺪ‬
:
⎪
⎩
⎪
⎨
⎧
+
−
=
=
−
+ )
03
)........(
log(
14
1
4 OH
NH
b
a
λ
λ
00.25
00.25
00.25
00.25
00.25
00.25
00.25
00.25
e
n
c
y
-
e
00
e
n
c
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
a
)
)
f
f
τ
‫ﺗ‬
‫ﺤﺪ‬
‫ﻳ‬
‫اﻟﻐﺎ‬ ‫اﻟﻜﻴﻤﻴﺎﺋﻲ‬ ‫اﻟﻨﻮع‬ ‫ﺪ‬
‫اﻟﻜﻴﻤﻴﺎﺋﻲ‬ ‫اﻟﻨﻮع‬ ‫ﻳﺪ‬
‫ﺪﻳﻨﺎ‬
:
L
L
9
)
04
,
0
10
( )
04
,
0 =
×
×
−
/
10
4 /
0
× −
mol
mol
2
4
[N
‫ﻫﻮ‬ ‫ﻐﺎﻟﺐ‬
‫ﻮ‬
NH
N 3
3
.
o
1
1
(
log (
log
−
−
+
+
f
PKa
Ka
τ
τ
[[
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎨
⎨
⎪
⎪
⎧
⎧
⎪
⎪
⎪
⇔
)
(
)
( )
( NH
[[
=
= [[
)
)
σ
σ )
(
( =
σ
σ
m
s
y
-
e
m
s
3as.ency-education.com

20. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
12
‫ﻣﻦ‬
18
‫ﻣﻦ‬
)
03
(
‫ﻧﺠﺪ‬
:
⎪
⎩
⎪
⎨
⎧
×
=
×
−
=
⇔
=
−
=
−
−
−
−
−
+
−
+
1
2
3
3
563
,
15
14
14
.
10
35
,
7
10
20
10
563
,
15
/
10
4
4
mol
m
S
b
NH
OH
b
NH
λ
λ
λ
3
-
‫إﺛﺒﺎت‬
‫ﺑﺎﻟﻌﻼﻗﺔ‬ ‫ﻳﻌﻄﻰ‬ ‫اﻟﻤﻮﻟﻲ‬ ‫ﻛﻴﺰﻩ‬
‫ﺗﺮ‬ ‫ﻓﺈن‬ ، ‫ﺟﺪا‬ ‫ﺿﻌﻴﻒ‬ ‫اﻷﺳﺎس‬ ‫ﻛﺎن‬‫إذا‬ ‫أﻧﻪ‬
:
)
(
2
10 PKe
PKa
PH
b
C +
−
=
:
‫اﻟﻤﻮﻟﻲ‬ ‫ﻛﻴﺰ‬
‫اﻟﺘﺮ‬ ‫أﻣﺎم‬ ‫ﻣﻬﻤﻼ‬ ‫اﻟﻨﺎﺗﺠﺔ‬ ‫ﻟﻠﺸﻮارد‬ ‫اﻟﻤﻮﻟﻲ‬ ‫ﻛﻴﺰ‬
‫اﻟﺘﺮ‬ ‫أن‬ ‫اﻟﺤﺎﻟﺔ‬ ‫ﻫﺬﻩ‬ ‫ﻓﻲ‬ ‫ﻧﻌﺘﺒﺮ‬
‫ﻟﻠﻤﺤﻠﻮل‬
Cb
.
‫ﻟﺪﻳﻨﺎ‬
:
[ ]
[ ]
[ ]
[ ]
)
(
2
4
3
10
10
10
)
10
log(
)
10
log(
)
log(
)
(
log
PKe
PKa
PH
b
PKa
PH
PKe
PH
b
PKe
PH
b
PKe
PH
b
f
f
b
f
f
C
C
PKa
PH
C
C
PKa
PH
OH
OH
C
PKa
NH
NH
PKa
PH
+
−
−
−
−
−
−
−
+
=
⇔
=
⇔
−
=
⇔
+
=
⇔
−
+
=
+
=
‫اﻟ‬
‫ﺣﺴﺎﺑﻴﺎ‬ ‫ﺘﺄﻛﺪ‬
:
‫ﻧﺠﺪ‬ ‫اﻟﻤﺒﺮﻫﻨﺔ‬ ‫اﻟﻌﻼﻗﺔ‬ ‫ﻓﻲ‬ ‫ﺑﺎﻟﺘﻌﻮﻳﺾ‬
:
L
mol
C PKe
PKa
PH
b /
10
10
10 2
)
14
2
,
9
(
6
,
10
2
)
(
2 −
+
−
×
+
−
=
=
=
4
-
‫أ‬
-
‫اﻟﺤﺎدث‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫اﺳﻢ‬
:
‫ﺗﻔﺎﻋ‬
‫اﻟﺘﺼﺒﻦ‬ ‫ﻞ‬
.
-
‫ﺧﻮاﺻﻪ‬ ‫أﻫﻢ‬
:
‫ﺑﻄﻲء‬ ، ‫ﺣﺮاري‬ ، ‫ﺗﺎم‬
.
‫ب‬
-
‫واﺳﻤﻪ‬ ‫اﻟﻤﺘﻔﺎﻋﻞ‬ ‫ﻟﻸﺳﺘﺮ‬ ‫اﻟﻤﻔﺼﻠﺔ‬ ‫ﻧﺼﻒ‬ ‫اﻟﺼﻴﻐﺔ‬ ‫اﺳﺘﻨﺘﺎج‬
:
3
2
2
3 CH
CH
COO
CH
CH −
−
−
−
‫اﻻﻳﺜﻴﻞ‬ ‫ﺑﺮوﺑﺎﻧﻮات‬
‫ـ‬‫ﺟ‬
-
‫اﻟﻤﺎدة‬ ‫ﻛﻤﻴﺔ‬‫ﻓﻲ‬ ‫ﻣﺘﺴﺎوي‬ ‫اﻻﺑﺘﺪاﺋﻲ‬ ‫اﻟﻤﺰﻳﺞ‬ ‫ﻛﺎن‬‫اذا‬ ‫اﻟﻨﺎﺗﺞ‬ ‫اﻟﻜﺤﻮل‬ ‫ﻛﺘﻠﺔ‬‫ﺣﺴﺎب‬
:
‫ﻣ‬
‫ﻣﺎدة‬ ‫ﻛﻤﻴﺔ‬‫ﺣﺴﺎب‬ ‫ﻳﻤﻜﻦ‬ ‫اﻷول‬ ‫اﻟﺠﺰء‬ ‫ﻦ‬
OH-
‫ﻓﻨﺠﺪ‬
:
[ ] mol
V
V
OH
OH
n PH 5
)
14
6
,
10
(
14
10
4
1
,
0
10
10
)
( −
−
−
−
−
×
=
×
=
×
=
×
=
‫ﻓﺎن‬ ‫ﺗﺎم‬ ‫واﻟﺘﻔﺎﻋﻞ‬ ‫ﺳﺘﻜﻴﻮﻣﺘﺮي‬ ‫اﻻﺑﺘﺪاﺋﻲ‬ ‫اﻟﻤﺰﻳﺞ‬ ‫أن‬ ‫ﺑﻤﺎ‬
:
mol
X
OH
H
C
n 5
max
5
2 10
4
)
( −
×
=
=
00.25
00.50
00.25
00.25
00.25
00.25
00.25
e
n
c
y
-
e
00.25
25
00.25
00.25
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
3
-
‫إﺛﺒﺎت‬ 3
-
‫إﺛﺒﺎت‬
‫أﻧﻪ‬
)
)
PKe
‫ا‬ ‫أن‬ ‫اﻟﺤﺎﻟﺔ‬ ‫ﻫﺬﻩ‬ ‫ﻓﻲ‬ ‫ﻧﻌﺘﺒﺮ‬
‫اﻟﺤﺎﻟﺔ‬ ‫ﻫﺬﻩ‬ ‫ﻓﻲ‬ ‫ﺮ‬
‫ﻮل‬
C
Cb
.
.
]
[ ]]
n
t
i
o
c
a
3
1
10
10
)
)
10
10
g(
10
0
log(
log
l
)
)
]
]]
3
PKe
PH PK
PH
b
b
PKe
PH PKe
PH
b
b
PK
PH
P
b
b
f
f
f
f
C
Cb
b
PH
C
C
C
C
PKa
PK
[
H
=
⇔
⇔
=
= PH
+
Ka
+
=
= PKa
PKa
C
Cb
C
00.25
00
e
n
c
y
-
e
m
s
m
s
3as.ency-education.com

21. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
13
‫ﻣﻦ‬
18
‫ﻫﻲ‬ ‫اﻟﻨﺎﺗﺠﺔ‬ ‫اﻟﻜﺤﻮل‬ ‫ﻛﺘﻠﺔ‬‫وﻣﻨﻪ‬
:
g
M
n
OH
H
C
m 3
5
5
2 10
84
,
1
46
10
4
)
( −
−
×
=
×
×
=
×
=
‫د‬
-
‫اﻟﻴﻮﻣﻴﺔ‬ ‫اﻟﺤﻴﺎة‬ ‫ﻓﻲ‬ ‫اﻷﺳﺘﺮات‬ ‫أﻫﻤﻴﺔ‬
:
، ‫اﻟﻐﺬاﺋﻴﺔ‬ ‫اﻟﻤﻮاد‬ ، ‫اﻟﻌﻄﻮر‬ ، ‫اﻷدوﻳﺔ‬ ‫ﺻﻨﺎﻋﺔ‬
.......................
‫ـﻲ‬‫ـ‬‫ﻧ‬‫اﻟﺜﺎ‬ ‫اﻟﺘﻤﺮﻳﻦ‬
) :
06.00
‫ـﺔ‬‫ـ‬‫ﻄ‬‫ﻧﻘ‬
(
-1
‫اﻻﺻﻄﻨﺎﻋﻲ‬ ‫اﻟﻘﻤﺮ‬ ‫ﻣﺴﺎر‬ ‫ﻃﺒﻴﻌﺔ‬
Spoutnik
:
‫اﻫﻠﻴﻠﻴﺠﻲ‬ ‫ﻣﺴﺎر‬
-
‫ﻣ‬
‫و‬
‫ﻗﻊ‬
‫ا‬
‫ﻷ‬
‫رض‬
‫ﻓﻲ‬
‫ا‬ ‫ﻫﺬا‬
‫ﻟﻣﺳﺎ‬
‫ر‬
:
‫ﻣﺤﺮﻗﻴﻪ‬ ‫اﺣﺪى‬ ‫ﻓﻲ‬ ‫اﻷرض‬ ‫ﺗﻘﻊ‬
)
‫اﻟﻤﺤﺮق‬ ‫ﻓﻲ‬
F1
(
.
-2
‫اﻟﻄﻮل‬
2a
:
‫اﻟﻜﺒﻴﺮ‬ ‫اﻟﻤﺤﻮر‬ ‫ﻃﻮل‬ ‫ﻳﻤﺜﻞ‬
.
-
‫اﻟﻄﻮل‬
2b
:
‫اﻟﺼﻐﻴﺮ‬ ‫اﻟﻤﺤﻮر‬ ‫ﻃﻮل‬ ‫ﻳﻤﺜﻞ‬
.
-
‫اﻟﻤﺤﻮر‬ ‫ﻧﺼﻒ‬ ‫ﻃﻮل‬ ‫ﺣﺴﺎب‬
‫اﻟﻜﺒﻴﺮ‬
‫ﻟﻬﺬا‬
‫اﻟﻤﺴﺎر‬
:
Km
r
r
a
a P
A
6970
2
6610
7330
2
2
2
=
+
=
+
=
=
-3
‫ﺗﻜﻮن‬
‫ﺳ‬
‫ر‬
‫ﻋﺔ‬
‫اﻟﻘﻤﺮ‬
‫اﻻﺻﻄﻨﺎﻋﻲ‬
‫اﺻﻐﺮﻳﺔ‬
‫اﻟﻨﻘﻄﺔ‬ ‫ﻓﻲ‬
A
‫ﺑﺴﺒﺐ‬
‫اﻷرض‬ ‫ﻋﻦ‬ ‫اﻻﺻﻄﻨﺎﻋﻲ‬ ‫اﻟﻘﻤﺮ‬ ‫ﺑﻌﺪ‬
‫ﻣﻤﺎ‬
‫ﻣﻦ‬ ‫ﻳﻨﻘﺺ‬
‫اﻻﺻﻄﻨﺎﻋﻲ‬ ‫ﻟﻠﻘﻤﺮ‬ ‫اﻷرض‬ ‫ﺟﺬب‬ ‫ﺗﺄﺛﻴﺮ‬
.
-
‫ﺗﻜﻮن‬
‫ﺳ‬
‫ر‬
‫ﻋﺔ‬
‫اﻟﻘﻤﺮ‬
‫اﻻﺻﻄﻨﺎﻋﻲ‬
‫اﻟﻨﻘﻄﺔ‬ ‫ﻓﻲ‬ ‫أﻋﻈﻤﻴﺔ‬
P
‫اﻷرض‬ ‫ﻣﻦ‬ ‫اﻻﺻﻄﻨﺎﻋﻲ‬ ‫اﻟﻘﻤﺮ‬ ‫ﻗﺮب‬ ‫ﺑﺴﺒﺐ‬
‫اﻻﺻﻄﻨﺎﻋﻲ‬ ‫ﻟﻠﻘﻤﺮ‬ ‫اﻷرض‬ ‫ﺟﺬب‬ ‫ﺗﺄﺛﻴﺮ‬ ‫ﻓﻲ‬ ‫ﻳﺰﻳﺪ‬ ‫ﻣﻤﺎ‬
.
-
‫اﻟﻤﻮﺿﻌﻴﻦ‬ ‫ﻓﻲ‬ ‫اﻟﺴﺮﻋﺔ‬ ‫ﺷﻌﺎﻋﻲ‬ ‫ﺗﻤﺜﻴﻞ‬
P , A
:
4
-
‫أ‬
-
‫ﺷ‬
‫ا‬ ‫روط‬
‫ﻟﺣﺻ‬
‫ول‬
‫ﺣ‬ ‫ﻋﻠﻰ‬
‫ر‬
‫ﻛﺔ‬
‫دا‬
‫ﺋ‬
‫ر‬
‫ﻴ‬
‫ﻣﻧﺗ‬ ‫ﺔ‬
‫ظ‬
‫ﻣﺔ‬
:
-
‫داﺋﺮي‬ ‫اﻟﻤﺴﺎر‬
.
–
‫واﻟﺠﻬﺔ‬ ‫اﻟﺤﺎﻣﻞ‬ ‫ﻓﻲ‬ ‫وﻣﺘﻐﻴﺮ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﻓﻲ‬ ‫ﺛﺎﺑﺖ‬ ‫اﻟﻠﺤﻈﻴﺔ‬ ‫اﻟﺴﺮﻋﺔ‬ ‫ﺷﻌﺎع‬
.
-
‫ﺛﺎﺑﺖ‬ ‫ﻧﺎﻇﻤﻲ‬ ‫ﺗﺴﺎرع‬ ‫ﻟﻬﺎ‬
.
-
‫ﻛﺰ‬
‫اﻟﻤﺮ‬ ‫ﻧﺤﻮ‬ ‫ﺟﺎذﺑﺔ‬ ‫ﺛﺎﺑﺘﺔ‬ ‫ﻟﻘﻮة‬ ‫ﺧﺎﺿﻊ‬ ‫ﻳﻜﻮن‬ ‫اﻟﻤﺘﺤﺮك‬
.
00.25
00.25
00.50
00.50
00.25
00.25
00.25
00.25
00.25
00.25
00.50
07.00
e
n
c
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
‫د‬
-
‫ﺻﻨﺎﻋﺔ‬
a
m
s
‫ـﻲ‬‫ـ‬‫ﻧ‬‫اﻟﺜﺎ‬ ‫اﻟﺘﻤﺮﻳﻦ‬
m
s
a
m
-1
-
‫اﻟﻘﻤﺮ‬ ‫ﻣﺴﺎر‬ ‫ﻃﺒﻴﻌﺔ‬
‫ﻣﺴﺎر‬ ‫ﻃﺒﻴﻌﺔ‬
-
‫ﻣ‬
‫و‬
‫ﻗﻊ‬
‫ا‬
‫ﻷ‬
‫رض‬
‫ﻓﻲ‬
‫ا‬ ‫ﻫﺬا‬
‫ﻟ‬ ‫و‬
‫ﻗﻊ‬
‫ا‬
‫ﻷ‬
‫رض‬
‫ﻓﻲ‬
‫ﻫ‬
‫ل‬
2a
:
‫اﻟﻤﺤﻮ‬ ‫ﻃﻮل‬ ‫ﻳﻤﺜﻞ‬
‫اﻟﻤﺤﻮ‬ ‫ﻃﻮل‬ ‫ﻳﻤﺜﻞ‬
2
:
‫اﻟﺼ‬ ‫اﻟﻤﺤﻮر‬ ‫ﻃﻮل‬ ‫ﻳﻤﺜﻞ‬
‫اﻟﻤﺤ‬ ‫ﻃﻮل‬ ‫ﻤﺜﻞ‬
‫اﻟﻤﺤﻮر‬ ‫ﺼﻒ‬
‫ﻤﺤﻮر‬
‫اﻟﻜﺒﻴﺮ‬
‫ﻟﻬﺬا‬
‫ا‬ ‫اﻟﻜﺒﻴﺮ‬
‫ﻟﻬ‬
o
o
m
rP
P
r
2
66
7330
7330
2
=
=
‫اﺻﻐﺮﻳﺔ‬
‫اﻟﻨﻘﻄﺔ‬ ‫ﻓﻲ‬
A
‫ﺑﺴ‬ ‫ﺮﻳﺔ‬
‫اﻟﻨﻘﻄﺔ‬ ‫ﻓﻲ‬
‫اﻻﺻﻄﻨﺎﻋﻲ‬ ‫ﻤﺮ‬
‫ﺻﻄﻨﺎﻋﻲ‬
.
.
‫ﻘﻄﺔ‬
P
P
‫اﻟﻘﻤﺮ‬ ‫ﻗﺮب‬ ‫ﺑﺴﺒﺐ‬
‫ﻗﺮب‬ ‫ﺑﺴﺒﺐ‬
e
n
c
y
-
e
m
s
m
s
3as.ency-education.com

22. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
14
‫ﻣﻦ‬
18
‫ب‬
-
‫ﻛﺘﺎﺑﺔ‬
‫اﻟﺸﻌﺎﻋﻴﺔ‬ ‫اﻟﻌﺒﺎرة‬
S
T
F /
r
‫ﻟﻘﻮة‬
‫ﺟﺬب‬
‫ا‬
‫ﻷ‬
‫رض‬
‫ﻟﻠﻘﻤﺮ‬
‫اﻻﺻﻄﻨﺎﻋﻲ‬
:
μ
r
r
2
/
)
( T
T
S
S
T
R
h
M
m
G
F
+
×
×
=
‫ـ‬‫ﺟ‬
-
‫ﻛﺘ‬
‫ا‬ ‫اﺑﺔ‬
‫ﻟﻌﺒﺎ‬
‫ا‬ ‫رة‬
‫ﻟﺸﻌﺎﻋ‬
‫ﻴ‬
‫ﻟﺘﺴﺎ‬ ‫ﺔ‬
‫رع‬
‫ﺣﺮﻛﺔ‬
‫ا‬ ‫اﻟﻘﻤﺮ‬ ‫ﻋﻄﺎﻟﺔ‬ ‫ﻛﺰ‬
‫ﻣﺮ‬
‫ﻹ‬
‫ﺻﻄﻨﺎﻋﻲ‬
:
‫ﻧﺠﺪ‬ ‫ﻟﻨﻴﻮﺗﻦ‬ ‫اﻟﺜﺎﻧﻲ‬ ‫اﻟﻘﺎﻧﻮن‬ ‫ﺑﺘﻄﺒﻴﻖ‬
:
μ
μ
r
r
r
r
r
r
2
2
)
(
)
(
T
T
S
T
T
S
S
ext
R
h
M
G
a
a
m
R
h
M
m
G
a
m
F
+
×
=
⇔
×
=
+
×
×
×
=
Σ
‫د‬
-
‫اﻳﺠﺎد‬
‫ﻋﺑﺎ‬
‫رة‬
‫ﺳ‬
‫ر‬
‫ﻋﺔ‬
‫ا‬
‫ﻟﻘﻣ‬
‫ر‬
‫اﻻﺻﻄﻨﺎﻋﻲ‬
v
:
‫ﺳﺒﻖ‬ ‫ﻣﻤﺎ‬ ‫ﻟﺪﻳﻨﺎ‬
:
)
01
.........(
)
( 2
μ
r
r
T
T
R
h
M
G
a
+
×
=
‫ﺑﺈﺳﻘﺎط‬
‫اﻟﻌﻼﻗﺔ‬
)
01
(
‫ﻧﺠﺪ‬ ‫اﻟﻨﺎﻇﻢ‬ ‫ﻋﻞ‬
:
)
02
....(
..........
)
(
)
(
2
2
T
T
T
T
T
n
R
h
M
G
v
R
h
v
R
h
M
G
a
a
+
×
=
⇔
+
=
+
×
=
=
-
‫اﻟﺪور‬ ‫ﻋﺒﺎرة‬ ‫إﻳﺠﺎد‬
T
‫ﻟﺣ‬
‫ر‬
‫ﻛﺔ‬
‫ا‬
‫ﻟﻘﻣ‬
‫ر‬
‫ﺣ‬
‫ا‬ ‫ول‬
‫ﻷ‬
‫رض‬
‫ﺑ‬
‫د‬
‫ﻻﻟﺔ‬
G, MT , h, RT
:
‫ﻟﺪﻳﻨﺎ‬
:
)
03
.........(
)
(
2
)
(
2 3
T
T
T
M
G
R
h
v
R
h
T
×
+
×
=
+
= π
π
00.50
00.50
00.50
00.25
FT/S
‫ا‬ ‫ﺔ‬
‫ﻟﻌﺒﺎ‬
‫ا‬ ‫رة‬
‫اﻟﺸﻌﺎﻋ‬ ‫رة‬
‫ﻟﺸﻌﺎﻋ‬
‫ﻴ‬
‫ﻴ‬‫ﻋ‬
‫ﻟﺘﺴﺎ‬ ‫ﺔ‬
‫ﻟﺘﺴﺎرع‬ ‫ﺔ‬
‫رع‬ ‫ﻴ‬
‫ﻧﺠﺪ‬ ‫ﻟﻨﻴﻮﺗﻦ‬ ‫اﻟﺜﺎﻧﻲ‬ ‫ﻧﻮن‬
‫ﻧﺠﺪ‬ ‫ﻟﻨﻴﻮﺗﻦ‬ ‫ﻟﺜﺎﻧﻲ‬
:
n
cμ
μ
r
r
r
r
2
2
(
(
)
)
T
S
S
T
T
M
M
G
a G
r
a
m
mS
S
T
×
= G
a G
×
= ×
m
mS
S
0.50
e
n
c
y
-
e
m
s
m
s
3as.ency-education.com

23. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
15
‫ﻣﻦ‬
18
‫ـ‬‫ﻫ‬
-
‫اﺳﺘﻨﺘ‬
‫ﺎ‬
‫ﻟﻜﺒﻠﺮ‬ ‫اﻟﺜﺎﻟﺚ‬ ‫اﻟﻘﺎﻧﻮن‬ ‫ج‬
:
‫اﻟﻌﺒﺎرة‬ ‫ﻣﻦ‬
)
03
(
‫ﻧﺠﺪ‬
:
)
04
.(
..........
4
)
(
2
3
2
T
T M
G
R
h
T
×
=
+
π
‫ﻣﺤﻘﻖ‬ ‫ﻟﻜﺒﻠﺮ‬ ‫اﻟﺜﺎﻟﺚ‬ ‫اﻟﻘﺎﻧﻮن‬ ‫وﻣﻨﻪ‬
.
- 5
‫أ‬
-
‫اﻟﺠﺪول‬ ‫أﻛﻤﺎل‬
.
Astra
)
‫ﺟﻴﻮ‬ ‫ﻗﻤﺮ‬
‫ﻣﺴﺘﻘﺮ‬
(
Cosmos
Alsat1
‫ا‬
‫ﻟﻘﻤﺮ‬
‫ا‬
‫ﻹﺻﻄﻨﺎﻋﻲ‬
86,2
40,440
5,96
T(103
s)
4,203
2,54
0,708
r(107
m)
3,565
1,9
0,07
h(107
m)
13
-
10
13
-
10
13
-
10
)
.
( 3
2
3
2
−
m
s
r
T
‫ب‬
-
‫اﺳﺘﻨﺘ‬
‫ﺎ‬
‫اﻷرض‬ ‫ﻟﻜﺘﻠﺔ‬ ‫اﻟﻌﺪدﻳﺔ‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ج‬
:
Kg
G
M
M
G
R
h
T
T
T
T
24
11
13
2
13
2
13
2
3
2
10
92
,
5
10
67
,
6
10
4
10
4
10
4
)
(
×
=
×
×
=
×
=
⇔
=
×
=
+
−
−
−
−
π
π
π
‫اﻟﺘﺠﺮﻳ‬ ‫اﻟﺘﻤﺮﻳﻦ‬
‫ﺒﻲ‬
) :
07.00
‫ـ‬‫ـ‬‫ﻃ‬‫ﻧﻘﺎ‬
(
1
-
‫اﻟﺘﻮﺗﺮات‬ ‫ﺗﻄﻮر‬
:
‫أ‬
-
‫اﻟﺘﻮﺗﺮ‬
uR
‫اﻟﺘﻴﺎر‬ ‫ﺷﺪة‬ ‫ﺗﻄﻮر‬ ‫ﻳﺒﺮز‬ ‫اﻟﺬي‬ ‫ﻫﻮ‬
i(t)
‫ﻷن‬ ‫اﻟﺪارة‬ ‫ﻓﻲ‬ ‫اﻟﻤﺎر‬
uR
‫ﻃﺮدﻳﺎ‬ ‫ﻳﺘﻨﺎﺳﺐ‬
‫اﻟﺘﻴﺎر‬ ‫ﺷﺪة‬ ‫ﻣﻊ‬
i(t)
‫ﺣﻴﺚ‬
:
i(t)
R
=
uR
.
‫ب‬
-
‫اﻟﺘﻮﺗﺮ‬ ‫ﻟﺘﻄﻮر‬ ‫اﻟﻤﻮاﻓﻖ‬ ‫اﻟﻤﻨﺤﻨﻰ‬
uc
‫اﻟﻤﻨﺤﻨﻰ‬ ‫ﻫﻮ‬
(a)
‫ﺣﻴﺚ‬ ‫ﻣﺘﺰاﻳﺪ‬ ‫أﺳﻲ‬ ‫ﻷﻧﻪ‬
:
E
u
t
u
t
C
C
=
∞
⇔
∞
=
=
⇔
=
)
(
0
)
0
(
0
‫ـ‬‫ﺟ‬
-
‫اﻟﺰﻣﻦ‬ ‫ﺛﺎﺑﺖ‬ ‫أن‬ ‫إﺛﺒﺎت‬
τ
‫اﻟﺰﻣﻦ‬ ‫ﻣﻊ‬ ‫ﻣﺘﺠﺎﻧﺲ‬
:
‫ﻟﺪﻳﻨﺎ‬
:
[ ] [ ] [ ] [ ]
[ ]
[ ]
[ ]
[ ] [ ]
[ ]
[ ]
T
I
T
I
U
Q
I
U
C
R
C
R
=
×
=
×
=
×
=
×
=
τ
τ
00.25
00.50
00.50
00.50
00.50
00.50
07.00
e
n
c
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
‫وﻣﻨﻪ‬
- 5
‫أ‬
-
‫اﻟﺠ‬ ‫أﻛﻤﺎل‬ - 5
‫أ‬
-
‫أﻛﻤﺎل‬
Astra
)
‫ﺟﻴﻮ‬ ‫ﻗﻤﺮ‬ Astra
)
‫ﻗﻤﺮ‬
‫ﻣﺴﺘﻘﺮ‬
( ‫ﻣﺴﺘﻘﺮ‬
(
a
a
e
a
e
x
6
40
0
e
e
e
x
/
e
e
m
/
54
/
e
/
e
/
e
/
e
m
/
1,9
m
/
/
/
o
/
o
m
10
o
o
o
m
c
o
c
o
c
o
c
o‫ض‬
:
o
n
a
t
M
M
M
G M
G
T
T
T
M
T
13
2
2
3
10
10
4
4
10
4
4
)
RT
=
⇔ =
M
MT
T
=
=
=
−
π
π
π
π
u
c
a
c
a
m
s
u
c
‫ﻃﺮدﻳﺎ‬ ‫ﺐ‬
‫ﻃﺮدﻳﺎ‬
00.50
50
0
e
n
c
y
-
e
m
s
3as.ency-education.com

24. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
16
‫ﻣﻦ‬
18
2
-
‫اﻟﺘﻮﺗﺮ‬ ‫ﻳﺤﻘﻘﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﻋﻦ‬ ‫اﻟﺒﺤﺚ‬
R
u
:
‫أ‬
-
‫ا‬ ‫اﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺗﺤﺪﻳﺪ‬
‫اﻟﺒﻌﺪي‬ ‫اﻟﺘﺤﻠﻴﻞ‬ ‫ﻋﻠﻰ‬ ‫ﺑﺎﻻﻋﺘﻤﺎد‬ ، ‫ﻟﺼﺤﻴﺤﺔ‬
:
‫ﻫﻲ‬ ‫اﻟﺼﺤﻴﺤﺔ‬ ‫اﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬
:
0
)
(
)
(
=
+ t
u
dt
t
du
RC R
R
‫ﻷن‬
:
[ ] [ ] [ ]
[ ]
[ ]
[ ]
[ ]
[ ]
[ ]
[ ]
[ ]
[ ]
[ ] [ ]
[ ]
[ ]
[ ]
[ ]
[ ] [ ] 0
0
0
0
2
0
0
0
=
⇔
=
⇔
=
⇔
=
+
×
×
⇔
=
+
×
×
⇔
=
+
×
×
U
U
U
T
U
I
T
I
U
T
U
U
Q
I
U
U
T
U
C
R
‫رﻗﻢ‬ ‫اﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫وﻣﻨﻪ‬
)
04
(
‫اﻟﺼﺤﻴﺤﺔ‬ ‫ﻫﻲ‬
.
‫ب‬
-
‫إﺛﺒﺎت‬
‫اﻟﺤ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﻛﺘﺎﺑﺔ‬‫ﻳﻤﻜﻦ‬ ‫أﻧﻪ‬
‫ﺑﺎﻟﺸﻜﻞ‬ ‫ﻞ‬
:
b
at
u
Ln R +
=
)
(
‫ﺣﻞ‬ ‫ﻟﺪﻳﻨﺎ‬
‫اﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬
:
τ
t
R Ee
t
u
−
=
)
(
)
(
1
)
( E
Ln
t
u
Ln R +
×
−
=
τ
-
‫ﻋﺒﺎرﺗﻲ‬ ‫اﻳﺠﺎد‬
b , a
‫ﺑﺪﻻﻟﺔ‬
E
‫و‬
τ
:
‫ﻧﺠﺪ‬ ‫ﺑﺎﻟﻤﻄﺎﺑﻘﺔ‬
:
⎪
⎩
⎪
⎨
⎧
=
−
=
)
(
1
E
Ln
b
a
τ
‫ـ‬‫ﺟ‬
-
‫ا‬
‫ﻋﻄ‬
‫ﺎء‬
‫اﻟﺒﻴﺎن‬ ‫ﻣﻌﺎدﻟﺔ‬
:
‫اﻟﺸﻜﻞ‬ ‫ﻣﻦ‬ ‫ﻣﻌﺎدﻟﺘﻪ‬ ‫ﺑﺎﻟﻤﺒﺪأ‬ ‫ﻳﻤﺮ‬ ‫ﻣﺴﺘﻘﻴﻢ‬ ‫ﻋﻦ‬ ‫ﻋﺒﺎر‬ ‫اﻟﺒﻴﺎن‬
:
b
at
u
Ln R +
=
)
(
‫ﺣﻴﺚ‬
:
⎩
⎨
⎧
=
−
=
7
,
5
504
,
0
b
a
00.50
00.25
00.50
00.25
e
n
c
e
c
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
‫ﻷن‬
: ‫ﻷن‬
:
]
[ ]
]
[ ]
[ ] 0
0
0
[ ]
]
0
0
0
⇔
] 0
0
0
0
]
]
[
[
[
[
=
=
]
]
+ [
=
[
[
[
[
[
[
[
04
(
‫اﻟﺼﺤﻴﺤﺔ‬ ‫ﻫﻲ‬
. ‫اﻟﺼﺤﻴ‬ ‫ﻲ‬
‫ﻟﺸﻜﻞ‬
:
:
b
at b
at
R at
at
at
)
(uR
R t
u
uR )
(t
n
c
y
-
e
m
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m
s
3as.ency-education.com

25. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
17
‫ﻣﻦ‬
18
‫د‬
-
‫اﺳﺘﻨﺘ‬
‫ﺎ‬
‫اﻟﻤﻜﺜﻔ‬ ‫ﺳﻌﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ج‬
‫ﺔ‬
C
:
‫ﻧﺠﺪ‬ ‫اﻟﺒﻴﺎن‬ ‫ﻣﻦ‬
:
⎪
⎪
⎩
⎪
⎪
⎨
⎧
=
×
=
×
=
=
≈
−
−
=
−
=
−
F
F
R
C
s
a
μ
τ
τ
160
10
6
,
1
10
5
,
12
2
2
)
504
,
0
(
1
1
4
3
-
‫اﻟﻤﻜﺜﻔﺔ‬ ‫ﺳﻌﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫وﻣﻨﻪ‬
‫اﻟﺼﺎﻧﻊ‬ ‫ﻃﺮف‬ ‫ﻣﻦ‬ ‫اﻟﻤﻌﻄﺎة‬ ‫اﻟﻘﻴﻤﺔ‬ ‫ﻣﻊ‬ ‫ﺗﺘﻮاﻓﻖ‬
.
‫ـ‬‫ﻫ‬
-
‫اﻷوﻣﻲ‬ ‫اﻟﻨﺎﻗﻞ‬ ‫ﻣﻘﺎوﻣﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺗﻜﻮن‬ ‫ﻋﻨﺪﻣﺎ‬
2
R
R =
′
‫اﻟﺰﻣﻦ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻓﺎن‬ ،
τ
‫ﻳﻘﻞ‬
)
‫ﺗﻨﺎﺳﺐ‬
‫واﻟﻤﻘﺎوﻣﺔ‬ ‫اﻟﺰﻣﻦ‬ ‫ﺛﺎﺑﺖ‬ ‫ﺑﻴﻦ‬ ‫ﻃﺮدي‬
(
‫اﻟﺸﻜﻞ‬ ‫ﺑﻴﺎﻧﻲ‬ ‫ﻓﺎن‬ ‫وﻋﻠﻴﻪ‬ ،
01
‫اﻟﻨﻈﺎم‬ ‫إﻟﻰ‬ ‫ﻳﺼﻼن‬
‫أﺳﺮع‬ ‫ﺗﻜﻮن‬ ‫اﻟﻤﻜﺜﻔﺔ‬ ‫ﺷﺤﻦ‬ ‫ﻋﻤﻠﻴﺔ‬ ‫أن‬ ‫أي‬ ‫أﻗﻞ‬ ‫زﻣﻦ‬ ‫ﻓﻲ‬ ‫اﻟﺪاﺋﻢ‬
.
‫و‬
-
‫اﻟﻨﺎﻗﻞ‬ ‫ﻣﻘﺎوﻣﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺗﻐﻴﻴﺮ‬ ‫ﻋﻨﺪ‬ ‫اﻟﻤﻜﺜﻔﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻌﻈﻤﻰ‬ ‫اﻟﻤﺨﺰﻧﺔ‬ ‫اﻟﻄﺎﻗﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺗﺘﻐﻴﺮ‬ ‫ﻻ‬
‫ﻣﻦ‬ ‫اﻷوﻣﻲ‬
R
‫إﻟﻰ‬
R′
‫اﻷوﻣﻲ‬ ‫اﻟﻨﺎﻗﻞ‬ ‫ﻣﻘﺎوﻣﺔ‬ ‫وﻟﻴﺲ‬ ‫اﻟﻤﻜﺜﻔﺔ‬ ‫ﺑﺴﻌﺔ‬ ‫ﺗﺘﻌﻠﻖ‬ ‫اﻟﻤﻜﺜﻔﺔ‬ ‫ﻃﺎﻗﺔ‬ ‫ﻷن‬ ،
‫ﺣﻴﺚ‬
:
2
2
1
)
( C
C Cu
t =
ξ
3
-
‫أ‬
-
‫اﻟﻤﻮاﻓﻘﺔ‬ ‫اﻟﻜﻬﺮﺑﺎﺋﻴﺔ‬ ‫اﻟﺪارة‬ ‫رﺳﻢ‬
:
‫ب‬
-
‫اﻳﺠﺎد‬
‫اﻟ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ﻟﻠﺪارة‬ ‫اﻟﺘﻔﺎﺿﻠﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬
‫ﺘﻮﺗﺮ‬
‫ﻃﺮﻓﻲ‬ ‫ﺑﻴﻦ‬ ‫اﻟﻜﻬﺮﺑﺎﺋﻲ‬
‫اﻟﻤﻜﺜﻔﺔ‬
:
‫ﻧﺠﺪ‬ ‫اﻟﺘﻮﺗﺮات‬ ‫ﺟﻤﻊ‬ ‫ﻗﺎﻧﻮن‬ ‫ﺑﺘﻄﺒﻴﻖ‬
:
0
)
(
)
(
0
)
(
)
(
=
+
=
+
dt
t
di
L
t
u
t
u
t
u
C
b
C
‫ﺣﻴﺚ‬
:
⎪
⎪
⎩
⎪
⎪
⎨
⎧
=
=
2
2
)
(
)
(
)
(
)
(
dt
t
u
d
C
dt
t
di
dt
t
du
C
t
i
C
C
00.25
00.25
00.25
00.25
00.50
00.25
00.25
L ub
uc C
e
n
c
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
F
μF
-
-
‫ﺳﻌﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫وﻣﻨﻪ‬
‫ﻗ‬ ‫وﻣﻨﻪ‬
‫اﻟﻨﺎ‬ ‫ﻣﻘﺎوﻣﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺗﻜﻮن‬ ‫ﻨﺪﻣﺎ‬
‫اﻟﻨﺎ‬ ‫ﻣﻘﺎوﻣﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ﺗﻜﻮن‬
‫واﻟﻤﻘﺎوﻣﺔ‬ ‫اﻟﺰﻣﻦ‬ ‫ﺛﺎﺑﺖ‬ ‫ﻦ‬
‫واﻟﻤﻘﺎو‬ ‫اﻟﺰﻣﻦ‬ ‫ﺖ‬
(
‫اﻟ‬ ‫ﺷﺤﻦ‬ ‫ﻋﻤﻠﻴﺔ‬ ‫أن‬ ‫أي‬ ‫أﻗﻞ‬
‫ﺷﺤ‬ ‫ﻋﻤﻠﻴﺔ‬ ‫أن‬ ‫ي‬
‫اﻟﻤﻜﺜﻔ‬ ‫ﻓﻲ‬ ‫اﻟﻌﻈﻤﻰ‬ ‫اﻟﻤﺨﺰﻧﺔ‬
‫اﻟ‬ ‫ﻓﻲ‬ ‫اﻟﻌﻈﻤﻰ‬ ‫ﺰﻧﺔ‬
‫ا‬ ‫ﺑﺴﻌﺔ‬ ‫ﺗﺘﻌﻠﻖ‬ ‫اﻟﻤﻜﺜﻔﺔ‬ ‫ﻃﺎﻗﺔ‬
‫ﺑﺴ‬ ‫ﺗﺘﻌﻠﻖ‬ ‫ﻤﻜﺜﻔﺔ‬
)
(
C
C )
ξ
ξ
00.25
0.
e
n
c
y
-
e
m
s
uc
o
t

26. ‫اﻹﺠﺎﺒﺔ‬
‫اﻟﻨﻤوذﺠﻴﺔ‬
‫وﺴﻠم‬
‫اﻟﺘﻨﻘﻴط‬
‫ع‬
‫ﻟﻤوﻀو‬
‫ﻴﺒﻲ‬‫ر‬‫اﻟﺘﺠ‬ ‫ﻴﺎ‬‫ر‬‫اﻟﺒﻛﺎﻟو‬
‫ﻟ‬
‫ﻤﻘﺎطﻌﺔ‬
‫ﺘﺒﺴﺔ‬
02
-
‫ﻤﺎي‬
-
2019
-
)
‫ﻴﺎﺌﻴﺔ‬‫ز‬‫ﻓﻴ‬ ‫ﻋﻠوم‬
(
‫اﻟﻤﺴﺘوى‬
:
‫ـﺔ‬‫ـ‬‫ﻴ‬‫ﻴﺒ‬‫ر‬‫ﺘﺠ‬ ‫ـوم‬‫ـ‬‫ـ‬‫ﻠ‬‫ﻋ‬ ‫ـﺔ‬‫ـ‬‫ﺜ‬‫اﻟﺜﺎﻟ‬
‫ﺻﻔﺤﺔ‬
18
‫ﻣﻦ‬
18
‫ﻧﺠﺪ‬ ‫ﺑﺎﻟﺘﻌﻮﻳﺾ‬
:
0
)
(
1
)
(
0
)
(
)
(
2
2
2
2
=
+
⇔
=
+
t
u
LC
dt
t
u
d
dt
t
u
d
LC
t
u
C
C
C
C
‫ﻣﻌﺎدﻟﺔ‬ ‫وﻫﻲ‬
‫اﻟﺸﻜﻞ‬ ‫وﻣﻦ‬ ‫ﺟﻴﺒﻲ‬ ‫ﺣﻠﻬﺎ‬ ‫اﻟﺜﺎﻧﻴﺔ‬ ‫اﻟﺮﺗﺒﺔ‬ ‫ﻣﻦ‬ ‫ﺗﻔﺎﺿﻠﻴﺔ‬
:
)
cos(
)
( 0 ϕ
ω +
= t
E
t
uC
-
‫وﻣﻨﻪ‬
‫ﺗﺴﺘﻨﺘﺞ‬
‫ﻣﺘﺨﺎﻣﺪ‬ ‫ﻏﻴﺮ‬ ‫دوري‬ ‫ﺑﻨﻈﺎم‬ ‫ﻣﻬﺘﺰة‬ ‫اﻟﺪارة‬ ‫أن‬
.
‫ـ‬‫ﺟ‬
-
‫اﺳﺘﻨﺘ‬
‫ﺎ‬
‫اﻟﺬاﺗﻲ‬ ‫اﻟﺪور‬ ‫ﻋﺒﺎرة‬ ‫ج‬
:
LC
T π
ω
π
2
2
0
0 =
=
-
‫ﻗﻴﻤ‬
‫اﻟﺬاﺗﻲ‬ ‫اﻟﺪور‬ ‫ﺔ‬
‫ا‬ ‫ﻟﻼﻫﺘﺰاز‬
‫ﻟﻤﺴﺠﻞ‬
:
‫ﻧﺠﺪ‬ ‫اﻟﺒﻴﺎن‬ ‫ﻣﻦ‬
:
s
ms
T 01
,
0
10
0 =
=
‫د‬
-
‫ا‬
‫ﺳﺘﻨﺘ‬
‫ﺎ‬
‫اﻟﻮﺷﻴﻌﺔ‬ ‫ذاﺗﻴﺔ‬ ‫ﻗﻴﻤﺔ‬ ‫ج‬
:
‫ﻟﺪﻳﻨﺎ‬
:
H
C
T
L
LC
T
0156
,
0
10
160
40
)
01
,
0
(
4
2
6
2
2
2
0
0
=
×
×
=
⎩
⎨
⎧
=
⇔
=
−
π
π
‫ـ‬‫ﻫ‬
-
‫ﻛﺘ‬
‫ﺎ‬
‫ﺑ‬
‫ﺔ‬
‫اﻟﻤﻜﺜﻔﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺨﺰﻧﺔ‬ ‫اﻟﻜﻬﺮﺑﺎﺋﻴﺔ‬ ‫ﻟﻠﺸﺤﻨﺔ‬ ‫اﻟﺰﻣﻨﻴﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬
.
)
cos(
)
( 0
0 ϕ
ω +
= t
Q
t
Q
‫ﺣﻴﺚ‬
:
C
C
E
C
Q 2
0 10
8
,
4
48000
300
160 −
×
=
=
×
=
×
= μ
s
rad
T
/
200
01
,
0
2
2
0
0 π
π
π
ω =
=
=
0
1
cos
)
0
(
0 0 =
⇔
=
⇔
=
⇔
= ϕ
ϕ
Q
q
t
‫وﻣﻨﻪ‬
:
)
200
cos(
10
8
,
4
)
( 2
t
t
Q π
−
×
=
00.25
00.25
00.25
00.25
00.25
00.25
00.25
00.25
e
n
c
e
c
y
-
e
00.25
25
00.25
0.
y
-
e
d
u
c
a
t
i
o
n
.
c
o
m
/
e
x
a
m
s
‫ﻣﻌﺎدﻟﺔ‬ ‫وﻫﻲ‬
‫ﻣ‬ ‫وﻫﻲ‬
)
)
0 ϕ
ϕ)
)
0
-
‫وﻣﻨﻪ‬
‫ﺗﺴﺘﻨﺘﺞ‬ ‫وﻣﻨﻪ‬
‫ﺗﺴﺘﻨﺘﺞ‬
‫اﻟﺪار‬ ‫أن‬
‫اﻟﺬاﺗﻲ‬ ‫اﻟﺪور‬ ‫ﻋﺒﺎرة‬ ‫ج‬
‫اﻟﺬاﺗﻲ‬ ‫اﻟﺪور‬ ‫ﺒﺎرة‬ ‫ﺎ‬
:
m
LC
C
‫ا‬ ‫ﻫﺘﺰاز‬
‫ﻟﻤﺴﺠﻞ‬
:‫ﻟﻤﺴﺠﻞ‬
:
T 01
,
0
10 0
10ms
ms
ms
0
T ms
ms
t
i
t
i
o
i
o
C
T
T
L
LC
LC
4
4 2
2
2
0
0
T
T
=
=
⇔ =
L π
‫ﺔ‬
.
E
C
Q C
Q0
Q E
C
C
e
e
n
c
y
-
e
m
s
m
s