Studyadz 3as-physique-c3-21-33

1. +
94 + 146
+ +
223383.02
223569.31
225376.42
E(Mev)
‫اﻟﺪراﺳﯿﺔ‬ ‫اﻟﺴﻨﺔ‬ ‫ﺧﺎﻟﺪ‬ ‫اﻷﻣﯿﺮ‬ ‫ﺛﺎﻧﻮﯾﺔ‬
2017
/
2018
‫اﻟﺸﻌﺒﺔ‬
3
‫رﯾﺎﺿﻰ‬
+
3
‫رﯾﺎﺿﻲ‬ ‫ﺗﻘﻨﻲ‬
‫اﻟﻤﺪة‬
4
‫ﺳﺎ‬
‫اﻟﻔﯿﺰﯾﺎﺋﯿﺔ‬ ‫اﻟﻌﻠﻮم‬ ‫ﻣﺎدة‬ ‫ﻓﻲ‬ ‫ﺗﺠﺮﯾﺒﻲ‬ ‫ﺑﻜﺎﻟﻮرﯾﺎ‬
‫اﻷول‬ ‫اﻟﺠﺰء‬
‫اﻟﺘﻤﺮﯾﻦ‬
‫اﻷول‬
04
‫ﻧﻘﺎط‬
‫اﻟﺒﻠﻮﺗﻮﻧﯿﻮم‬
239
‫اﻟﻄﺎﻗﺔ‬ ‫ﻹﻧﺘﺎج‬ ‫اﻟﻨﻮوﯾﺔ‬ ‫اﻟﻤﻔﺎﻋﻼت‬ ‫ﻓﻲ‬ ‫ﻧﻮوي‬ ‫ﻛﻮﻗﻮد‬ ‫ﺗﺴﺘﺨﺪم‬ ‫اﻟﺘﻲ‬ ‫اﻟﻤﻮاد‬ ‫ﻣﻦ‬ ‫ھﻮ‬ ‫و‬ ‫اﻟﺒﻠﻮﺗﻮﻧﯿﻮم‬ ‫ﻧﻈﺎر‬ ‫اﺣﺪ‬
‫اﻟﻜﮭﺮﺑﺎﺋﯿﺔ‬
.
‫ﻟﻠﯿﻮراﻧﯿﻮم‬ ‫ﻣﻦ‬ ‫اﻧﻄﻼﻗﺎ‬ ‫إﻧﺘﺎﺟﮫ‬ ‫ﯾﺘﻢ‬
238
‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫وﻓﻖ‬
‫أوﻻ‬
-
‫اﻟﺒﻠﻮﺗﻮﻧﯿﻮم‬
‫ﻟﺠﺴﯿﻤﺎت‬ ‫ﻣﺼﺪرا‬ ‫ﺗﻠﻘﺎﺋﯿﺎ‬ ‫ﯾﺘﻔﻜﻚ‬
α
-
‫و‬ ‫ﻧﻈﺎﺋﺮ‬ ‫ﻣﻦ‬ ‫ﻛﻼ‬ ‫ﻋﺮف‬
‫اﻟﺠﺴﯿﻤﺔ‬
α
-
‫ﻟﻠﺒﻠﻮﺗﻮﻧﯿﻮم‬ ‫اﻟﻨﻮوي‬ ‫اﻟﺘﻔﻜﻚ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫اﻛﺘﺐ‬
239
‫ﻋﻠﻤﺎ‬
‫أن‬
‫اﻟﻨﺎﺗﺠﺔ‬ ‫اﻟﻨﻮاة‬
‫اﻟﯿﻮراﻧﯿﻮم‬ ‫ﻧﻈﺎﺋﺮ‬ ‫اﺣﺪ‬ ‫ھﻲ‬
2
-
‫اﻟﺒﻠﻮﺗﻮﻧﯿﻮم‬ ‫ﻣﻦ‬ ‫ﻋﯿﻨﺔ‬
239
‫ﻛﺘﻠﺘﮭ‬
‫ﺎ‬
m0=1g
‫ﻣﺤﺎﻛﺎة‬ ‫ﺑﺮﻧﺎﻣﺞ‬ ‫ﺑﻮاﺳﻄﺔ‬
‫اﻟﻤﻘﺎﺑﻞ‬ ‫اﻟﺸﻜﻞ‬ ‫اﻟﺒﯿﺎن‬ ‫ﻋﻠﻰ‬ ‫اﻟﺤﺼﻮل‬ ‫ﻣﻦ‬ ‫ﺗﻤﻜﻨﺎ‬ ‫اﻹﺷﻌﺎﻋﻲ‬ ‫ﻟﻨﺸﺎطﮭﺎ‬
-
‫اﻻﻧﻮﯾﺔ‬ ‫ﻛﺘﻠﺔ‬ ‫ﻋﻦ‬ ‫ﺗﻌﺒﺮ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻌﻼﻗﺔ‬ ‫ھﻲ‬ ‫ﻣﺎ‬ ‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﻌﻼﻗﺎت‬ ‫ﺑﯿﻦ‬ ‫ﻣﻦ‬
‫اﻟﻌﯿﻨﺔ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺘﺒﻘﯿﺔ‬
-
‫اﻹﺷﻌﺎﻋﻲ‬ ‫اﻟﻨﺸﺎط‬ ‫ﺛﺎﺑﺖ‬ ‫اﺳﺘﻨﺘﺞ‬ ‫ﺛﻢ‬ ‫اﻟﺒﯿﺎن‬ ‫ﻋﺒﺎرة‬ ‫اﻛﺘﺐ‬
λ
-
‫اﻻﺑﺘﺪاﺋﻲ‬ ‫اﻟﻨﺸﺎط‬ ‫اﺣﺴﺐ‬
‫اﻟﺴﺎﺑﻘﺔ‬ ‫ﻟﻠﻌﯿﻨﺔ‬
‫ﺛﺎﻧﯿﺎ‬
-
‫ﻧﻮاة‬ ‫ﺑﻘﺬف‬ ‫ذﻟﻚ‬ ‫و‬ ‫ﻟﻠﺒﻠﻮﺗﻮﻧﯿﻮم‬ ‫ﻣﺘﺴﻠﺴﻞ‬ ‫اﻧﺸﻄﺎر‬ ‫ﻋﻤﻠﯿﺔ‬ ‫ﻋﻦ‬ ‫اﻟﻨﺎﺗﺠﺔ‬ ‫ﺑﺎﻟﻄﺎﻗﺔ‬ ‫اﻟﻨﻮوﯾﺔ‬ ‫اﻟﻐﻮاﺻﺎت‬ ‫إﺣﺪى‬ ‫ﻣﺤﺮﻛﺎت‬ ‫ﺗﺸﺘﻐﻞ‬
‫اﻟﺒﻠﻮﺗﺒﯿﻮم‬
239
‫ﺑﻨﯿﺘﺮون‬
1
-
‫ﺑﻤﺨﻄﻂ‬ ‫ذﻟﻚ‬ ‫ﻣﺜﻞ‬ ‫و‬ ‫اﻟﻤﺘﺴﻠﺴﻞ‬ ‫اﻻﻧﺸﻄﺎر‬ ‫ﻋﺮف‬
2
-
‫اﻟﺸﻄﻮر‬ ‫اﻟﺜﻘﯿﻠﺔ‬ ‫اﻻﻧﻮﯾﺔ‬ ‫ﻗﺬف‬ ‫ﻓﻲ‬ ‫اﻟﻨﯿﺘﺮون‬ ‫ﻧﺴﺘﻌﻤﻞ‬ ‫ﻟﻤﺎذا‬
‫ة‬
3
-
‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫اﻟﻤﻮﺿﺢ‬ ‫اﻟﻄﺎﻗﺔ‬ ‫ﻣﺨﻄﻂ‬ ‫ﻋﻠﻰ‬ ‫ﺑﺎﻻﻋﺘﻤﺎد‬
‫ا‬
-
‫ﻣﺤﺪد‬ ‫اﻟﺤﺎدث‬ ‫اﻻﻧﺸﻄﺎر‬ ‫ﻟﺘﻔﺎﻋﻞ‬ ‫اﻟﻤﻨﻤﺬﺟﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫اﻛﺘﺐ‬
‫ﻗﯿﻤﺔ‬
a
‫ب‬
-
‫اﻟﻄﺎﻗﺔ‬ ‫ﺗﻤﺜﻞ‬ ‫ﻣﺎذا‬
E1
Δ
,
E2
Δ
E ,
Δ
‫ث‬
-
‫اﻟﺮﺑﻂ‬ ‫طﺎﻗﺔ‬ ‫اوﺟﺪ‬
El
‫اﻟﺒﻠﻮﺗﻮﻧﯿﻮم‬ ‫ﻟﻨﻮاة‬
‫ج‬
-
‫اﻟﻤﺤﺮرة‬ ‫اﻟﻄﺎﻗﺔ‬
Elib
‫ﻣﻦ‬ ‫واﺣﺪة‬ ‫ﻧﻮاة‬ ‫اﻧﺸﻄﺎر‬ ‫ﻣﻦ‬
‫اﻟﺒﻠﻮﺗﻮﻧ‬
‫ﯿﻮم‬
239
‫ب‬
Mev
‫اﻟﺠﻮل‬ ‫ﺛﻢ‬
‫د‬
-
‫اﻟﻨﯿﻮﺑﯿﻮم‬ ‫ﻟﻨﻮاة‬ ‫اﻟﻜﺘﻠﻰ‬ ‫اﻟﻨﻘﺺ‬ ‫ان‬ ‫ﻋﻠﻤﺖ‬ ‫اذا‬
‫ھﻮ‬
Δm=0.93119 u
-
‫اﻟﻄﺎﻗﺔ‬ ‫اﺣﺴﺐ‬
‫اﻟﯿﻮد‬ ‫ﻟﻨﻮاة‬ ‫اﻟﺮﺑﻂ‬
135
‫ﻧﻮاﺗﻲ‬ ‫اﺳﺘﻘﺮار‬ ‫ﻗﺎرن‬ ‫ﺛﻢ‬
‫اﻟﻨﯿﻮﺑﯿﻮم‬
102
‫واﻟﯿﻮد‬
135
4
-
‫ﺧﻼ‬ ‫اﻟﻤﺴﺘﮭﻠﻜﺔ‬ ‫اﻟﺒﻠﻮﺗﻮﻧﺒﻮم‬ ‫ﻛﺘﻠﺔ‬ ‫اﺣﺴﺐ‬
‫اﻟﻐﻮاﺻﺔ‬ ‫اﻧﺘﻘﺎل‬ ‫ل‬
‫اﺳﺘﻄﺎﻋﺔ‬ ‫ﻗﺪم‬ ‫ﻣﺤﺮﻛﮭﺎ‬ ‫أن‬ ‫ﻋﻠﻤﺎ‬ ‫ﻛﺎﻣﻞ‬ ‫ﺷﮭﺮ‬ ‫ﻟﻤﺪة‬
30Mw
‫ﺑﻤﺮدود‬
= 30%
1Mev=1.6*10-13
J 1u=931.5Mev/c2
Na=6.02*1023
‫ص‬ ‫اﻟﺜﺎﻧﻲ‬ ‫اﻟﻤﻮﺿﻮع‬
1

2. ‫اﻟﺜﺎﻧﻲ‬ ‫اﻟﺘﻤﺮﯾﻦ‬
04
‫ﻧﻘﺎط‬
‫ﻛ‬ ‫ﺟﺴﻢ‬ ‫ﻧﺘﺮك‬
‫ﺘﻠﺘﮫ‬
m=100g
‫اﻟﻨﻘﻄﺔ‬ ‫ﻣﻦ‬ ‫اﺑﺘﺪاﺋﯿﺔ‬ ‫ﺳﺮﻋﺔ‬ ‫ﺑﺪون‬ ‫ﺑﻨﺰل‬
A
‫اﻟﻤﺴﺘﻮي‬ ‫ﻋﻦ‬ ‫ﻣﺎﺋﻞ‬ ‫ﻟﻤﺴﺘﻮ‬ ‫اﻷﻋﻈﻢ‬ ‫اﻟﻤﯿﻞ‬ ‫ﺧﯿﻂ‬ ‫ﻋﻠﻰ‬
‫ﺑﺰاوﯾﺔ‬ ‫اﻷﻓﻘﻲ‬
α
)
‫اﻟﺸﻜﻞ‬
1
(
.
‫ﺑﻘﻮة‬ ‫اﻟﻤﺎﺋﻞ‬ ‫أﻟﻤﺴﺘﻮ‬ ‫ﻋﻠﻰ‬ ‫اﻻﺣﺘﻜﺎك‬ ‫ﻧﻨﻤﺬج‬
f
‫ﺛﺎﺑﺘﺔ‬ ‫ﺷﺪﺗﮭﺎ‬
.
‫اﻟﻨﻘﻄﺔ‬ ‫اﻟﻰ‬ ‫اﻟﺠﺴﻢ‬ ‫ﯾﺼﻞ‬ ‫ﻟﻤﺎ‬
B
‫ﻟﺜﻘﻠﮫ‬ ‫ﻟﻘﻮة‬ ‫ﻓﻘﻂ‬ ‫ﺧﺎﺿﻊ‬ ‫ﯾﺼﺒﺢ‬
.
‫ﻧﻌﺘﺒﺮ‬
t=0s
‫اﻟﺠ‬ ‫وﺟﻮد‬ ‫ﻟﺤﻈﺔ‬
‫ﻓﻲ‬ ‫ﺴﻢ‬
B
‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫ﻣﺜﻠﻨﺎ‬
2
‫اﻟﻤﺤﻮر‬ ‫وﻓﻖ‬ ‫اﻟﺠﺴﻢ‬ ‫ﺳﺮﻋﺔ‬
Bx
‫ﺳﺮﻋﺘﮫ‬ ‫و‬
‫اﻟﻤﺤﻮر‬ ‫وﻓﻖ‬
By
‫اﻟﺰﻣﻦ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫ذﻟﻚ‬ ‫و‬
.
1
-
‫اﻟﻨﻘﻄﺔ‬ ‫ﺑﻌﺪ‬ ‫اﻟﺠﺴﻢ‬ ‫ﺣﺮﻛﺔ‬ ‫ادرس‬
B
‫اﻟﻤﻌﻠﻢ‬ ‫ﻓﻲ‬
(Bx,By)
‫وﻓﻖ‬ ‫ﻟﻠﺴﺮﻋﺔ‬ ‫اﻟﺘﻔﺎﺿﻠﯿﺘﯿﻦ‬ ‫اﻟﻤﻌﺎدﻟﺘﯿﻦ‬ ‫اﻛﺘﺐ‬ ‫ﺛﻢ‬
‫اﻟﻤﺤﻮرﯾﻦ‬
.
2
-
‫اﻟﺰﻣﻨﯿﺘﯿﻦ‬ ‫اﻟﻤﻌﺎدﻟﺘﯿﻦ‬ ‫ﺟﺪ‬
y(t)
‫و‬
x(t)
‫ﻟﻠﺤﺮﻛﺔ‬
‫وﻓﻖ‬
‫اﻟ‬
‫اﻟﻤﺴﺎر‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫اﺳﺘﻨﺘﺞ‬ ‫ﺛﻢ‬ ‫ﻤﺤﻮرﯾﻦ‬
3
-
‫اﻟﺒﯿﺎن‬ ‫ﺑﺎﺳﺘﻐﻼل‬
-
‫اﻟﺠﺴﻢ‬ ‫ﺳﺮﻋﺔ‬ ‫ان‬ ‫ﺑﯿﻦ‬
3.7m/s2
VB=
‫اﻟﻘﺬف‬ ‫زاوﯾﺔ‬ ‫و‬
=20°
α
-
‫اﻷرﺿﻲ‬ ‫اﻟﺘﺴﺎرع‬ ‫ﻗﯿﻤﺔ‬ ‫اﺳﺘﻨﺘﺞ‬
g
4
--
‫اﻟﻤﺎﺋﻞ‬ ‫اﻟﻤﺴﺘﻮي‬ ‫ﻋﻠﻰ‬ ‫اﻟﺠﺴﻢ‬ ‫ﺗﺴﺎرع‬ ‫اﺣﺴﺐ‬
5
-
‫ﻋﺒﺎرة‬ ‫اوﺟﺪ‬ ‫ﻟﻨﯿﻮﺗﻦ‬ ‫اﻟﺜﺎﻧﻲ‬ ‫اﻟﻘﺎﻧﻮن‬ ‫ﺑﺘﻄﺒﯿﻖ‬
‫اﻻﺣﺘﻜﺎك‬ ‫ﻗﻮة‬ ‫ﺷﺪة‬
f
‫ﻋﻠﻰ‬
‫اﻟﻤﺎ‬ ‫اﻟﻤﺴﺘﻮي‬
‫ﺑﺪﻻﻟﺔ‬ ‫ﺋﻞ‬
α , g , m
‫و‬
a
‫ﻗﯿﻤﺘﮫ‬ ‫اﺣﺴﺐ‬ ‫ﺛﻢ‬
6
-
‫إ‬ ‫ﺟﺪ‬
‫ﻋﻨﺪﺋﺬ‬ ‫ﺳﺮﻋﺘﮭﺎ‬ ‫و‬ ‫اﻟﻜﺮة‬ ‫ﺳﻘﻮط‬ ‫ﺣﺪاﺛﯿﺎت‬
‫ص‬ ‫اﻟﺜﺎﻧﻲ‬ ‫اﻟﻤﻮﺿﻮع‬
2

3. ‫اﻟﺜﺎﻟﺚ‬ ‫اﻟﺘﻤﺮﯾﻦ‬
06
‫ﻧﻘﺎط‬
‫أوﻻ‬
‫اﻟﻠﯿﻜﻮل‬
lugol
‫اﻷ‬ ‫ﻣﻜﻮﻧﮭﺎ‬ ‫اﻟﺼﯿﺪﻟﯿﺎت‬ ‫ﻋﻨﺪ‬ ‫ﺗﺒﺎع‬ ‫ﻣﻄﮭﺮة‬ ‫ﻣﺎدة‬
‫اﻟﯿﻮد‬ ‫ﺛﺘﺎﺋﻲ‬ ‫ھﻮ‬ ‫ﺳﺎﺳﻲ‬
I2
‫اﻟﺰﻧﻚ‬ ‫ﻣﻦ‬ ‫ﺻﻔﯿﺤﺔ‬ ‫ﻧﻐﻤﺮ‬
Zn
‫ﻛﺘﻠﺘﮭﺎ‬
m0
‫ﺣﺠﻢ‬ ‫ﻋﻠﻰ‬ ‫ﯾﺤﺘﻮي‬ ‫ﻛﺎس‬ ‫ﻓﻲ‬
V
‫اﻟﯿﻮد‬ ‫ﻟﺜﻨﺎﺋﻲ‬ ‫اﻟﻤﻮﻟﻲ‬ ‫اﻟﺘﺮﻛﯿﺰ‬ ‫ﺣﯿﺚ‬ ‫اﻟﻠﯿﻜﻮل‬ ‫ﻣﻦ‬
C0
‫ﺗﺎم‬ ‫و‬ ‫ﺑﻄﻲء‬ ‫اﻟﺰﻧﻚ‬ ‫و‬ ‫اﻟﻠﯿﻜﻮل‬ ‫ﺑﯿﻦ‬ ‫اﻟﻜﯿﻤﯿﺎﺋﻲ‬ ‫اﻟﺘﺤﻮل‬
1
-
‫ﺑﻄﻲء‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫أن‬ ‫ﻣﻦ‬ ‫ﺗﺠﺮﯾﺒﯿﺎ‬ ‫ﻧﺘﺄﻛﺪ‬ ‫ﻛﯿﻒ‬
2
-
‫و‬ ‫اﻷﻛﺴﺪة‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫اﻛﺘﺐ‬
‫اﻟﺤﺎ‬ ‫اﻹرﺟﺎع‬
‫ﻟﺘﻘﺪم‬ ‫ﺟﺪوﻻ‬ ‫ﺿﻊ‬ ‫ﺛﻢ‬ ‫دث‬
‫اﻟﺘﻔﺎﻋﻞ‬
.
‫اﻟﺜﻨﺎﺋﯿﺘﯿﻦ‬ ‫ﺗﻌﻄﻰ‬
I2 / I-
‫و‬
Zn2+
/ Zn
3
-
‫أن‬ ‫ﺑﯿﻦ‬ ‫اﻟﺘﻘﺪم‬ ‫ﺟﺪول‬ ‫ﻋﻠﻰ‬ ‫اﻋﺘﻤﺎدا‬
= [ ] + −
4
-
‫اﻟﺘﺎﻟﯿﯿﻦ‬ ‫اﻟﺒﯿﺎﻧﯿﻦ‬ ‫اﻟﻤﻨﺤﻨﯿﯿﻦ‬ ‫رﺳﻢ‬ ‫ﻣﻦ‬ ‫ﺗﻤﻜﻨﺎ‬ ‫ﺧﺎﺻﺔ‬ ‫ﺗﻘﻨﯿﺔ‬ ‫ﺑﻮاﺳﻄﺔ‬
‫اﻟﺸﻜﻠﯿﻦ‬ ‫ﻋﻠﻰ‬ ‫اﻋﺘﻤﺎدا‬
1
‫و‬
2
-
‫اﻟ‬ ‫اﺳﺘﻨﺘﺞ‬
‫اﻟﻤﺤﺪ‬ ‫ﻤﺘﻔﺎﻋﻞ‬
-
‫اﻟﺒﯿﺎﻧﯿﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫اﻛﺘﺐ‬
nZn =f( I2 )
-
‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫ﻗﯿﻢ‬ ‫ﺣﺪد‬
C0 , V , x max
-
‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻧﺼﻒ‬ ‫زﻣﻦ‬
t1/2
-
‫ﺑﺎﻟﻌﺒﺎرة‬ ‫ﺗﻌﻄﻰ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫اﻟﺤﺠﻤﯿﺔ‬ ‫اﻟﺴﺮﻋﺔ‬ ‫أن‬ ‫ﺑﯿﻦ‬
= −
.
.
-
‫اﻟﻠﺤﻈﺔ‬ ‫ﻋﻨﺪ‬ ‫اﻟﺤﺠﻤﯿﺔ‬ ‫اﻟﺴﺮﻋﺔ‬ ‫ﻗﯿﻤﺔ‬ ‫اﺣﺴﺐ‬
t = 0s
‫ﺛﺎﻧﯿﺎ‬
‫ﯾﺤﺘﻮي‬ ‫ﻣﺤﻠﻮل‬ ‫ﻓﻲ‬ ‫اﻟﺘﺮﺗﯿﺐ‬ ‫ﻋﻠﻰ‬ ‫ﻣﻐﻤﻮرﺗﯿﻦ‬ ‫ﺑﻮﻓﺮة‬ ‫ﻧﻌﺘﺒﺮھﻤﺎ‬ ‫اﻟﺮﺻﺎص‬ ‫و‬ ‫اﻟﺰﻧﻚ‬ ‫ﻣﻦ‬ ‫ﺻﻔﯿﺤﺘﯿﻦ‬ ‫ﻣﻦ‬ ‫ﻛﮭﺮﺑﺎﺋﻲ‬ ‫ﻋﻤﻮد‬ ‫ﯾﺘﻜﻮن‬
‫ﺷﻮارد‬ ‫ﻋﻠﻰ‬ ‫ﯾﺤﺘﻮي‬ ‫ﻣﺤﻠﻮل‬ ‫و‬ ‫اﻟﺰﻧﻚ‬ ‫ﺷﻮارد‬ ‫ﻋﻠﻰ‬
.
‫ﻣﻨﮭﻤﺎ‬ ‫واﺣﺪ‬ ‫ﻛﻞ‬ ‫ﺣﺠﻢ‬
V=500ml
‫ﯾﺘﺤﻜﻢ‬ ‫اﻟﺬي‬ ‫اﻟﻜﯿﻤﯿﺎﺋﻲ‬ ‫اﻟﺘﺤﻮل‬ ‫ﯾﻨﻤﺬج‬
‫ﻓﻲ‬
‫ﺗﺸﻐﯿﻞ‬
‫اﻟﺘﺎﻟﻲ‬ ‫ﺑﺎﻟﺘﻔﺎﻋﻞ‬ ‫اﻟﻌﻤﻮد‬
Zn + Pb2+
= Zn2+
+ Pb
‫ﻟﻠﺘﻔﺎﻋﻞ‬ ‫اﻟﻤﻮاﻓﻖ‬ ‫ﺗﻮازن‬ ‫ﺛﺎﺑﺖ‬
K=4.6.1020
.
‫ﯾﻜﻮن‬ ‫اﻟﺒﺪاﯾﺔ‬ ‫ﻓﻲ‬
=0.05mol/l
[Zn2+
] = [Pb2+
]
1
-
‫اﻟﻌﻤﻮد‬ ‫ﻟﮭﺬا‬ ‫اﻻﺻﻄﻼﺣﻲ‬ ‫اﻟﺮﻣﺰ‬ ‫اﻛﺘﺐ‬ ‫ﺛﻢ‬ ‫اﻟﻜﯿﻤﯿﺎﺋﯿﺔ‬ ‫اﻟﺠﻤﻠﺔ‬ ‫ﺗﻄﻮر‬ ‫اﺗﺠﺎه‬ ‫ﻋﯿﻦ‬
2
-
‫ﺑﻨﺎﻗﻞ‬ ‫ﻣﻮﺻﻮﻻ‬ ‫ﯾﻜﻮن‬ ‫ﻋﻨﺪﻣﺎ‬ ‫اﻟﻌﻤﻮد‬ ‫ھﺬا‬ ‫ﺑﺎﻟﺮﺳﻢ‬ ‫ﻣﺜﻞ‬
‫أوﻣﻲ‬
R=200 Ω
‫ﻋﻠﯿ‬ ‫ﻣﺒﯿﻨﺎ‬
‫ﺣﺮﻛﺔ‬ ‫اﺗﺠﺎه‬ ‫و‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﺘﯿﺎر‬ ‫اﺗﺠﺎه‬ ‫ﮫ‬
‫اﻻﻟﻜﺘﺮوﻧﺎت‬
3
-
‫اﻟﻌﻤﻮد‬ ‫اﺷﺘﻐﺎل‬ ‫أﺛﻨﺎء‬ ‫اﻟﺤﺎﺻﻞ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫اﻟﺘﻘﺪم‬ ‫ﺟﺪول‬ ‫أﻧﺸﺊ‬
4
-
‫اﻟﻌﻤﻮد‬ ‫ﯾﻮﻟﺪھﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻟﻜﮭﺮﺑﺎء‬ ‫ﻛﻤﯿﺔ‬ ‫ھﻲ‬ ‫ﻣﺎ‬
5
-
‫اﻟﻤﺤﺮﻛﺔ‬ ‫ﻗﻮﺗﮫ‬ ‫ﻛﺎﻧﺖ‬ ‫إذا‬ ‫اﻟﻌﻤﻮد‬ ‫اﺷﺘﻐﺎل‬ ‫ﻣﺪة‬ ‫اﺣﺴﺐ‬
E = 2V
1F =96500 C M( Zn ) =56.4g /mol
‫اﻟﺜ‬ ‫اﻟﻤﻮﺿﻮع‬
‫ص‬ ‫ﺎﻧﻲ‬
3

4. ‫اﻟﺜﺎﻧﻲ‬ ‫اﻟﺠﺰء‬
‫ﺗﺠﺮﯾﺒﻲ‬ ‫ﺗﻤﺮﯾﻦ‬
06
‫ﻧﻘﺎط‬
‫اﻟﺴﻌﺔ‬ ‫ﻣﺠﮭﻮﻟﺔ‬ ‫ﻣﻜﺜﻔﺔ‬ ‫ﺛﺎﻧﻮﯾﺔ‬ ‫ﻣﺨﺒﺮ‬ ‫ﻓﻲ‬ ‫ﯾﻮﺟﺪ‬
.
‫اﻟﺘﺎﻟﯿﺔ‬ ‫اﻟﻮﺳﺎﺋﻞ‬ ‫ﺗﻼﻣﯿﺬه‬ ‫ﻣﺘﻨﺎول‬ ‫ﻓﻲ‬ ‫أﺳﺘﺎذ‬ ‫وﺿﻊ‬ ‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﻌﺮﻓﺔ‬ ‫ﻟﻐﺮض‬
-
‫اﻟﺜﺎﺑﺖ‬ ‫ﻟﻠﺘﻮﺗﺮ‬ ‫ﻣﻮﻟﺪ‬
E
-
‫أوﻣﻲ‬ ‫ﻧﺎﻗﻞ‬
R=1KΩ
‫ذاﺗ‬ ‫ﺻﺎﻓﯿﺔ‬ ‫وﺷﯿﻌﺔ‬
‫ﯿﺘﮭﺎ‬
L
-
‫ﺳﻌﺘﮭﺎ‬ ‫ﻣﻜﺜﻔﺔ‬
C
-
‫ﺑﺎدﻟﺔ‬
K
-
‫ﺗﻮﺻﯿﻞ‬ ‫أﺳﻼك‬
‫اﻟﺸﻜﻞ‬ ‫ﻓﻲ‬ ‫اﻟﻤﺒﯿﻨﺔ‬ ‫اﻟﺪارة‬ ‫ﺑﺘﺮﻛﯿﺐ‬ ‫اﻟﺘﻼﻣﯿﺬ‬ ‫ﻗﺎم‬
4
‫أ‬
‫وﻻ‬
‫ﻟﺤﻈﺔ‬ ‫ﻓﻲ‬
0s
t=
‫اﻟﺒﺎدﻟﺔ‬ ‫ﻧﻀﻊ‬
K
‫اﻟﻮﺿﻊ‬ ‫ﻓﻲ‬
1
‫ﻣﻦ‬ ‫ﻣﻜﻨﺖ‬ ‫اﻟﺘﺠﺮﯾﺒﯿﺔ‬ ‫اﻟﺪراﺳﺔ‬
‫اﻟﺒﯿﺎن‬ ‫ﻋﻠﻰ‬ ‫اﻟﺤﺼﻮل‬
‫ﻓﻲ‬ ‫اﻟﻤﻤﺜﻞ‬
)
‫اﻟﺸﻜﻞ‬
5
(
1
-
‫و‬ ‫اﻟﺘﻮﺗﺮات‬ ‫اﺗﺠﺎه‬ ‫اﻟﺪارة‬ ‫ﻋﻠﻰ‬ ‫ﺑﯿﻦ‬
‫ﺑﺎﻟﺪارة‬ ‫اﻟﻤﺎر‬ ‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﺘﯿﺎر‬
2
-
‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫ﻟﻠﺘﻮﺗﺮ‬ ‫اﻟﺘﻔﺎﺿﻠﯿﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺟﺪ‬ ‫اﻟﺘﻮﺗﺮات‬ ‫ﺟﻤﻊ‬ ‫ﻗﺎﻧﻮن‬ ‫ﺑﺘﻄﺒﯿﻖ‬
uc(t)
3
-
‫اﻟﺸﻜﻞ‬ ‫ﻣﻦ‬ ‫اﻟﺘﻔﺎﺿﻠﯿﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬
uc(t) =A+Ee - bt
‫ﻋﺒﺎرﺗﻲ‬ ‫اوﺟﺪ‬
A
‫و‬
B
‫ﺑﺪﻻﻟﺔ‬
R,C,E
4
-
‫اﻟﺰﻣﻦ‬ ‫ﺛﺎﺑﺖ‬ ‫وﺣﺪة‬ ‫ﺣﺪد‬ ‫اﻟﺒﻌﺪي‬ ‫اﻟﺘﺤﻠﯿﻞ‬ ‫ﺑﺎﺳﺘﻌﻤﺎل‬
τ
‫ﺛﻢ‬
‫ﻋﯿﻦ‬
‫ﻗﯿﻤﺘﮫ‬
5
-
‫ﻣﻦ‬ ‫ﻛﻞ‬ ‫اوﺟﺪ‬ ‫ﺑﺎﻟﺒﯿﺎن‬ ‫ﺑﺎﻻﺳﺘﻌﺎﻧﺔ‬
E
‫و‬
C
‫ﺛﺎﻧﯿﺎ‬
‫اﻟﻮﺿﻊ‬ ‫ﻓﻲ‬ ‫اﻟﺒﺎدﻟﺔ‬ ‫ﻧﻀﻊ‬ ‫ﻛﻠﯿﺎ‬ ‫اﻟﻤﻜﺜﻔﺔ‬ ‫ﺗﺸﺤﻦ‬ ‫ﻋﻨﺪﻣﺎ‬
2
1
-
‫ﺗﺤﺪث‬ ‫اﻟﺘﻲ‬ ‫اﻟﻈﺎھﺮة‬ ‫ھﻲ‬ ‫ﻣﺎ‬
2
-
‫اﻟﺘﻔﺎﺿﻠﯿﺔ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫ﺟﺪ‬ ‫اﻟﺘﻮﺗﺮات‬ ‫ﺟﻤﻊ‬ ‫ﻓﺎﻧﻮن‬ ‫ﺑﺎﺳﺘﻌﻤﺎل‬
‫اﻟﻜﮭﺮﺑﺎﺋﻲ‬ ‫اﻟﺘﻮﺗﺮ‬ ‫ﯾﺤﻘﻘﮭﺎ‬ ‫اﻟﺘﻲ‬
uc(t)
‫اﻟﻤﻜﺜﻔﺔ‬ ‫طﺮﻓﻲ‬ ‫ﺑﯿﻦ‬
3
-
‫اﻟﻌﺒﺎرة‬ ‫ان‬ ‫ﺗﺤﻘﻖ‬
( ) = ( + )
‫ﺣﯿﺚ‬ ‫اﻟﺘﻔﺎﺿﻠﯿﺔ‬ ‫ﻟﻠﻤﻌﺎدﻟﺔ‬ ‫ﺣﻞ‬ ‫ﺗﺸﻜﻞ‬
= 2 √ .
4
-
‫اﻟﻤﻨﺤﻨﻰ‬ ‫ﻋﻠﻰ‬ ‫اﻟﺤﺼﻮل‬ ‫ﻣﻦ‬ ‫ﻣﻜﻨﺘﻨﺎ‬ ‫اﻟﺘﺠﺮﯾﺒﯿﺔ‬ ‫اﻟﺪراﺳﺔ‬
‫اﻟﺸﻜﻞ‬ ‫اﻟﺰﻣﻦ‬ ‫ﺑﺪﻻﻟﺔ‬ ‫اﻟﻤﻜﺜﻔﺔ‬ ‫طﺮﻓﻲ‬ ‫ﺑﯿﻦ‬ ‫اﻟﺘﻮﺗﺮ‬
6
‫ا‬
-
‫ﻓﻲ‬ ‫اﻟﻤﺤﻨﻰ‬ ‫ﯾﺒﯿﻨﮭﺎ‬ ‫اﻟﺘﻲ‬ ‫اﻻھﺘﺰازات‬ ‫ﻧﻤﻂ‬ ‫ھﻮ‬ ‫ﻣﺎ‬
‫اﻟﺸﻜﻞ‬
6
‫ب‬
-
‫ﺳﺒﺐ‬ ‫ﻓﺴﺮ‬
‫اﻟﺪراﺳﺔ‬ ‫و‬ ‫اﻟﻨﻈﺮﯾﺔ‬ ‫اﻟﺪراﺳﺔ‬ ‫ﺑﯿﻦ‬ ‫اﻟﻔﺮق‬
‫اﻟﺘﺠﺮﯾﺒﯿﺔ‬
‫ج‬
-
‫اﻟﺪور‬ ‫ﺷﺒﮫ‬ ‫اﻋﺘﺒﺎر‬ ‫ﯾﻤﻜﻦ‬
T
‫ﻟﻠﺪور‬ ‫ﻣﺴﺎوﯾﺎ‬
‫اﻟﺬاﺗﻲ‬
T0
‫اﻟﺤﺎﻟﺔ‬ ‫ھﺬه‬ ‫ﻓﻲ‬
.
‫اﻟﻮﺷﯿﻌﺔ‬ ‫ذاﺗﯿﺔ‬ ‫اﺳﺘﻨﺘﺞ‬
‫د‬
-
‫ﻧﮭﺎﯾﺔ‬ ‫ﻋﻨﺪ‬ ‫ﺟﻮل‬ ‫ﺑﻔﻌﻞ‬ ‫اﻟﻀﺎﺋﻌﺔ‬ ‫اﻟﻄﺎﻗﺔ‬ ‫اﺣﺴﺐ‬
‫زﻣﻦ‬ ‫ﺑﻌﺪ‬ ‫أي‬ ‫اﻟﺜﺎﻧﯿﺔ‬ ‫اﻻھﺘﺰازة‬
t=2T
‫ص‬ ‫اﻟﺜﺎﻧﻲ‬ ‫اﻟﻤﻮﺿﻮع‬
4
‫اﻟﺒﻜﺎﻟﻮرﯾ‬ ‫ﺷﮭﺎدة‬ ‫ﻓﻲ‬ ‫ﺑﺎﻟﺘﻮﻓﯿﻖ‬
‫ﺎ‬

5. ‫اﻷول‬ ‫اﻟﺗﻣرﯾن‬
‫اﻟﻧظﺎﺋر‬
‫اﻟﺷﺣﻧﻲ‬ ‫اﻟﻌدد‬ ‫ﻧﻔس‬ ‫ﻟﮭﺎ‬ ‫اﻧوﯾﺔ‬
Z
‫اﻟﻛﺗﻠﻲ‬ ‫اﻟﻌدد‬ ‫ﻓﻲ‬ ‫ﺗﺧﺗﻠف‬ ‫و‬
A
‫اﻟﺟﺳﯾﻣﺔ‬
α
‫اﻟﮭﯾﻠﯾوم‬ ‫ﻧواة‬ ‫ھﻲ‬
‫اﻟﺗﻔﻛك‬ ‫ﻣﻌﺎدﻟﺔ‬
‫اﻻﻧﺣﻔﺎظ‬ ‫ﻗواﻧﯾن‬ ‫ﺑﺗطﺑﯾق‬
+
‫اﻟﺻﺣﯾﺣﺔ‬ ‫اﻟﻌﻼﻗﺔ‬
m(t)=m0.e-λt
‫اﻟﺑﯾﺎن‬ ‫ﻋﺑﺎرة‬
= ∗
‫إﯾﺟﺎد‬
λ
‫اﻟﻧظرﯾﺔ‬ ‫اﻟﻌﻼﻗﺔ‬
m(t)=m0*e-λt
‫وﻣﻧﮫ‬
=
‫ﺑﺎﻟﺗﺎﻟﻲ‬ ‫و‬
= ∗
‫ﻧﺟد‬ ‫اﻟﺑﯾﺎﻧﯾﺔ‬ ‫و‬ ‫اﻟﻧظرﯾﺔ‬ ‫اﻟﻌﻼﻗﺗﯾن‬ ‫ﺑﻣطﺎﺑﻘﺔ‬
λ=a
=
4
14. 10
= 2.86 ∗ 10 = 9.07 ∗ 10
= ∗ = ∗ ∗ = 9.07 ∗ 10 ∗
6.02. 10
239
= 3.79 ∗ 10
‫ﻣﺗﺳﻠﺳ‬ ‫اﻧﺷطﺎر‬
‫ل‬
‫ﺑﻧﯾﺗرو‬ ‫ﻧواة‬ ‫ﻗذف‬ ‫ھو‬
‫ن‬
‫ﻧﯾﺗروﻧﯾ‬ ‫و‬ ‫ﻧواﺗﯾن‬ ‫ﻟﺗﻌطﻲ‬
‫ن‬
‫اﻟﺗﻲ‬ ‫ﺛﻼﺛﺔ‬ ‫أو‬
‫اﻟﺗﻔﺎﻋل‬ ‫ﯾﺗﺳﻠﺳل‬ ‫ﺑﺎﻟﺗﺎﻟﻲ‬ ‫و‬ ‫أﺧرى‬ ‫ﯾوراﻧﯾوم‬ ‫اﻧوﯾﺔ‬ ‫ﺗﻘذف‬ ‫ﺑدورھﺎ‬
‫ﻧﯾﺗرون‬ ‫ﻧﺳﺗﻌﻣل‬
‫اﻟﻘذف‬ ‫ﻓﻲ‬
‫ﻷﻧﮫ‬
‫ﻛﮭروﻣﻐﻧﺎطﯾﺳ‬ ‫ﺗﺄﺛﯾر‬ ‫ﯾﺣدث‬ ‫ﻻ‬ ‫ﺑﺎﻟﺗﺎﻟﻲ‬ ‫و‬ ‫اﻟﺷﺣﻧﺔ‬ ‫ﻋدﯾم‬
‫ﻲ‬
‫اﻻﻧﺣﻔﺎظ‬ ‫ﻗواﻧﯾن‬ ‫ﺑﺗطﺑﯾق‬
‫ﻧﺟد‬
a=3
E1
Δ
‫ﻟﻨﻮة‬ ‫اﻟﺮﺑﻂ‬ ‫طﺎﻗﺔ‬
E2
Δ
‫ﻟﻨﻮﺗﻲ‬ ‫اﻟﺮﺑﻂ‬ ‫طﺎﻗﺔ‬ ‫ﺳﺎﻟﺐ‬
‫و‬
E
Δ
‫اﻻﻧﺸﻄﺎر‬ ‫ﻣﻦ‬ ‫اﻟﻤﺤﺮرة‬ ‫اﻟﻄﺎﻗﺔ‬
( ) = =225376.42-
223569.31=1807.11Mev
= −[ + ( )] = −(223383.02 −225376.42)
+ ( ) = 1993.4
= 1993.4 − ( ) = 1993.4 − 0.93119 ∗ 931.5 = 1126
= == 223383.02 − 223569.31 = 186.29 = 2.98 ∗ 10
‫اﻟﻧواة‬
‫اﻷﻛﺛر‬
‫اﺳﺗﻘرار‬
( )
=
1126
135
= 8.34 /
( )
=
0.93119 ∗ 931.5
102
= 8.50 /
‫ھﻲ‬ ‫اﺳﺗﻘرار‬ ‫اﻷﻛﺛر‬ ‫اﻟﻧواة‬
‫ﻻن‬
( )
<
( )
‫اﻟﻣﺳﺗﮭﻠﻛﺔ‬ ‫اﻟﻛﺗﻠﺔ‬ ‫ﺣﺳﺎب‬
‫اﻟﻛﮭرﯾﺎﺋﯾﺔ‬ ‫اﻟطﺎﻗﺔ‬ ‫ﺣﺳﺎب‬
= ∗ = 30 ∗ 10 ∗ 30 ∗ 24 ∗ 3600 = 7.78 ∗ 10
‫اﻟﻛﻠﯾﺔ‬ ‫اﻟطﺎﻗﺔ‬ ‫ﺣﺳﺎب‬
=
30
∗ 100 = 2.59 ∗ 10
‫اﻻﻧوﯾﺔ‬ ‫ﻋدد‬ ‫ﺣﺳﺎب‬
ET=N*Elib
‫ﻣﻧﮫ‬ ‫و‬
= = 8.69 ∗ 10
‫اﻟﻛﺗﻠﺔ‬ ‫ﺣﺳﺎب‬
=
∗
= 3450 = 3.45

6. ‫اﻟﺛﺎﻧﻲ‬ ‫اﻟﺗﻣرﯾن‬
‫اﻟﺣرﻛﺔ‬ ‫دراﺳﺔ‬
‫ارﺿﻲ‬ ‫ﺳطﺣﻲ‬ ‫اﻟﻣرﺟﻊ‬
‫ﺟﺳم‬ ‫اﻟﻣدروﺳﺔ‬ ‫اﻟﺟﻣﻠﺔ‬
‫ﺑﺗطﺑﯾ‬
‫ق‬
‫ﻟﻧﯾوﺗن‬ ‫اﻟﺛﺎﻧﻲ‬ ‫اﻟﻘﺎﻧون‬
∑ =
P=ma
‫ﻣﺣور‬ ‫ﻋﻠﻰ‬
Bx
ax(t)=0
‫ﻣﻧﺗظﻣﺔ‬ ‫ﻣﺳﺗﻘﯾﻣﺔ‬ ‫ﺣرﻛﺔ‬
‫ﻣﺣور‬ ‫ﻋﻠﻰ‬
By
ay(t)= -g
‫ﺑﺎﻧﺗظﺎم‬ ‫ﻣﺗﻐﯾرة‬ ‫ﻣﺳﺗﻘﯾﻣﺔ‬ ‫ﺣرﻛﺔ‬
‫ﻟﻠﺳرﻋﺔ‬ ‫اﻟزﻣﻧﯾﺗﯾن‬ ‫اﻟﻣﻌﺎدﻟﺗﯾن‬
ax(t)=0
‫وﻣﻧﮫ‬
= 0
‫اﻻﺑﺗداﺋﯾﺔ‬ ‫اﻟﺷروط‬ ‫اﺳﺗﻌﻣﺎل‬ ‫و‬ ‫ﺑﺎﻟﺗﻛﺎﻣل‬
vx(t)=v0cosα
ay(t)=-g
‫وﻣﻧﮫ‬
= 0
‫اﻻﺑﺗداﺋﯾﺔ‬ ‫اﻟﺷروط‬ ‫اﺳﺗﻌﻣﺎل‬ ‫و‬ ‫ﺑﺎﻟﺗﻛﺎﻣل‬
vY(t)=-gt+v0sinα
‫اﻟزﻣﻧﯾﺗﯾن‬ ‫اﻟﻣﻌﺎدﻟﺗﯾن‬
‫ﻟﻠﺣرﻛﺔ‬
vx(t)=v0cosα
‫وﻣﻧﮫ‬
=
‫اﻻﺑﺗداﺋﯾﺔ‬ ‫اﻟﺷروط‬ ‫اﺳﺗﻌﻣﺎل‬ ‫و‬ ‫ﺑﺎﻟﺗﻛﺎﻣل‬
x(t)=v0tcosα ( 1 )
vy(t)=-gt+v0sinα
‫وﻣﻧﮫ‬
= − −
‫و‬ ‫ﺑﺎﻟﺗﻛﺎﻣل‬
‫اﻻﺑﺗداﺋﯾﺔ‬ ‫اﻟﺷروط‬ ‫اﺳﺗﻌﻣﺎل‬
(2)
y(t)=-1/2 gt2
-v0t sinα
‫اﻟﻌﻼﻗﺔ‬ ‫ﻣن‬
1
=
( )
‫ﻓﻲ‬ ‫ﺑﺎﻟﺗﻌوﯾض‬
2
‫ﻧﺟد‬
( ) = − − ( )
‫اﻟﺳرﻋﺔ‬ ‫ﻗﯾﻣﺔ‬
+ = √3.5 + 1.3 = 3.7 /
‫اﻟزاوﯾﺔ‬ ‫ﻗﯾﻣﺔ‬
α
= =
.
.
= 0.37
‫ﻣﻧﮫ‬ ‫و‬
α=20°
‫ﻗﯾﻣﺔ‬
g
‫اﻟﺑﯾﺎن‬
2
‫ﯾواﻓق‬
vY(t)=-gt+v0sinα
‫ﯾﻣﺛل‬ ‫اﻟﻣﯾل‬ ‫ﻣﻧﮫ‬ ‫و‬
a=-g
=
.
. ∗ .
= 10
‫ﻣﻧﮫ‬ ‫و‬
g=10m/s2
‫اﻟﻣﺎﺋل‬ ‫اﻟﻣﺳﺗوى‬ ‫ﻋﻠﻰ‬ ‫اﻟﺗﺳﺎرع‬ ‫ﺣﺳﺎب‬
− = 2 .
‫ﻣﻧﮫ‬ ‫و‬
=
a =2.34m/s2
‫إﯾﺟﺎد‬
‫اﻻﺣﺗﻛﺎك‬ ‫ﻗوة‬ ‫ﺷدة‬
‫اﻟﻣﺳﺗو‬ ‫ﻋﻠﻰ‬
‫ى‬
‫اﻟﻣﺎﺋل‬
‫ارﺿﻲ‬ ‫ﺳطﺣﻲ‬ ‫اﻟﻣرﺟﻊ‬
‫ﺟﺳم‬ ‫اﻟﻣدروﺳﺔ‬ ‫اﻟﺟﻣﻠﺔ‬
‫ﺑﺗطﺑﯾ‬
‫ق‬
‫ﻟﻧﯾوﺗن‬ ‫اﻟﺛﺎﻧﻲ‬ ‫اﻟﻘﺎﻧون‬
∑ =
P+R+f=ma
‫ﻣﺣور‬ ‫ﻋﻠﻰ‬ ‫ﺑﺎﻹﺳﻘﺎط‬
xx’
Psinα-f=ma
‫ﺑﺎﻟﺗﺎﻟﻲ‬ ‫و‬
f=m(gsinα-a)
f =0.1N
‫ﻟﻸرض‬ ‫اﻟﻛرة‬ ‫وﺻول‬ ‫إﺣداﺛﯾﺎت‬ ‫إﯾﺟﺎد‬
‫ﻟﻸرض‬ ‫اﻟﻛرة‬ ‫وﺻول‬ ‫ﻟﺣظﺔ‬ ‫اﻟﺑﯾﺎن‬ ‫ﻣن‬
t=0.26s
‫ﻧﺟد‬ ‫ﻟﻠﺣرﻛﺔ‬ ‫اﻟزﻣﻧﯾﺔ‬ ‫اﻟﻣﻌﺎدﻟﺗﯾن‬ ‫ﻓﻲ‬ ‫ﺑﺎﻟﺗﻌوﯾض‬
(xc=0.91m , yc=-0.67m)
‫ﻋﻧد‬ ‫اﻟﺳرﻋﺔ‬
c
‫اﻟﺑﯾﺎن‬ ‫ﻣن‬
= 1.3 + (−3.9) = 4.11 /

7. ‫ﺗدرﯾﺟﯾﺎ‬ ‫اﻟﯾود‬ ‫ﺛﻧﺎﺋﻲ‬ ‫ﻟون‬ ‫ﻟﺗﻐﯾر‬ ‫ﺑطﻲء‬ ‫اﻟﺗﻔﺎﻋل‬ ‫أن‬ ‫ﻧﺗﺄﻛد‬
‫إرﺟﺎع‬ ‫اﻷﻛﺳدة‬ ‫ﻣﻌﺎدﻟﺔ‬
Zn2+
+2e-
=Zn
I2 +2e-
=2I-
I2+Zn2+
=2I-
+Zn
‫اﻟﻌﻼﻗﺔ‬ ‫إﯾﺟﺎد‬
( ) = −
[ ] = −
‫ﻣﻧﮫ‬ ‫و‬
= − [ ]
‫ﺑﺎﻟﺗﺎﻟﻲ‬ ‫و‬
( ) = − + [ ]
‫اﻟﺑﯾﺎن‬ ‫ﻣﻌﺎدﻟﺔ‬
( ) = [ ] +
‫ﺣﯾث‬
a =tanα =0.2
‫و‬
b=0.02
( ) = 0.2[ ] + 0.02
‫ﻧﺟد‬ ‫ﺑﺎﻟﻣطﺎﺑﻘﺔ‬
V=0.2l
‫اﻻﻋظﻣﻲ‬ ‫اﻟﺗﻘدم‬
( ) = −
‫وﻣﻧﮫ‬
( )
= −
‫وﻣﻧﮫ‬
= −
( )
=
4 ∗ 0.645
64.5
−
2 ∗ 0.645
64.5
= 0.02
‫اﻟﻣوﻟﻲ‬ ‫اﻟﺗرﻛﯾز‬
C0
− = 0.02
‫ﺑﺎﻟﺗﺎﻟﻲ‬ ‫و‬
= −0.02 +
‫وﻣﻧﮫ‬
= −0.02 + ∗
C0=0.1mol/l
‫اﻟﺗﻔﺎﻋل‬ ‫ﻧﺻف‬ ‫زﻣن‬
=
= 1.935
‫ﻧﺟد‬ ‫ﺑﺎﻹﺳﻘﺎط‬
20s
t1/2 =
‫ﻟﻠﺗﻔﺎﻋل‬ ‫اﻟﺣﺟﻣﯾﺔ‬ ‫اﻟﺳرﻋﺔ‬ ‫ﻋﺑﺎرة‬
=
1
( ) = −
‫وﻣﻧﮫ‬
= −
( )
‫ﺑﺎﻟﺗﻌوﯾض‬
= −
( )
‫وﻣﻧﮫ‬
= − ( )
‫ﻋﻧد‬ ‫اﻟﺣﺟﻣﯾﺔ‬ ‫اﻟﺳرﻋﺔ‬ ‫ﺣﺳﺎب‬
t=0s
= −
1
0.2 ∗ 64.5
−4 ∗ 0.645
3.2 ∗ 20
= 0.312 / .
‫اﻟﻛﯾﻣﯾﺎﺋﯾﺔ‬ ‫اﻟﺟﻣﻠﺔ‬ ‫ﺗطور‬ ‫ﺟﮭﺔ‬ ‫ﺗﻌﯾﯾن‬
Q =
[ ]
[ ]
=
.
.
= 1 <
‫اﻟﻣﺑﺎﺷر‬ ‫اﻻﺗﺟﺎه‬ ‫ﻓﻲ‬ ‫ﺗﺗطور‬ ‫اﻟﺟﻣﻠﺔ‬ ‫وﻣﻧﮫ‬
Zn
+ Pb2+
=
Zn2+
+ Pb
‫اﻟﺗﻘدم‬
‫اﻟﺣﺎﻻت‬
n2
cv
cv
n 1
0
‫ح‬
.
‫ا‬
n2+x
Cv+x
cv-x
n1-X
x
‫و‬ ‫ح‬
xm
Cv+2xm
cv-x
n1-X
xm
‫ن‬ ‫ح‬
‫اﻟﻛﮭرﺑﺎء‬ ‫ﻛﻣﯾﺔ‬
Q=Z.Xm.F
Xm=cv=0.05*0.05=2.5*10-3
mol
‫وﻣﻧﮫ‬
*2.5*10-3
*96500=482.5 c
Q=2
‫اﻻﺷﺗﻐﺎل‬ ‫ﻣدة‬
Q=I*Δt
= = = 0.01
Δt = =
482.5
0.01
= 48250 = 13.7ℎ
I2 + Zn = 2I-
+
Zn 2+
‫اﻟﺗﻘدم‬
‫اﻟﺣﺎﻻت‬
0
0
n1
C 0V
0
‫ح‬
.
‫ا‬
x
2x
n1-x
C0V-X
x
‫و‬ ‫ح‬
xm
2xm
n1-x
C0V-X
xm
‫ن‬ ‫ح‬

8. ‫اﻟراﺑﻊ‬ ‫اﻟﺗﻣرﯾن‬
= ( ) + ( )
= ( ) + ( )
= ( ) +
( )
= ( ) +
( )
= ( ) +
( )
=
( )
+
( )
‫إﯾﺟﺎد‬
‫ﻋﺑﺎرة‬
A
‫و‬
b
‫ﻧﻌوض‬
uc(t) =A+E.e – bt
‫اﻟﺗﻔﺎﺿﻠﯾﺔ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫ﻓﻲ‬
=
+ –
+ ( + )
= + −
+ = +
‫ﻧﺟد‬ ‫ﺑﺎﻟﻣطﺎﺑﻘﺔ‬
A=E
‫و‬
=
‫اﻟﺑﻌدي‬ ‫اﻟﺗﺣﻠﯾل‬
τ = [ ][ ]
τ =
[ ] [ ]
=
[ ].[ ]
= [ ]
‫اﯾﺟﺎد‬
E
‫و‬
C
‫اﻟﺑﯾﺎن‬ ‫ﻣن‬
uc(max)=E=12V
UC(τ)=12*063=7.56V
‫ﺑﺎﻹﺳﻘﺎ‬
‫ط‬
‫اﻷزﻣﻧﺔ‬ ‫ﻣﺣور‬ ‫ﻋﻠﻰ‬
τ=0.04s
= =4*10-5
F
‫اﻟظﺎھرة‬
‫اھﺗزازات‬ ‫ھﻲ‬ ‫اﻟﺣﺎدﺛﺔ‬
‫ﻛﮭرﺑﺎﺋﯾﺔ‬
0 = ( ) + ( )
0 = ( ) +
( )
0 = ( ) +
( )
0 = ( ) +
( )
0 = ( ) +
( )
0 =
( )
+
( )
‫ان‬ ‫ﺗﺑﯾﺎن‬
( ) = ( + )
‫اﻟﺗﻔﺎﺿﻠﯾﺔ‬ ‫ﻟﻠﻣﻌﺎدﻟﺔ‬ ‫ﺣل‬
( ) =
2
+ =
1
√
+
0 =
1
√
+
+
1
√
+
0=0
‫اﻟﻣطﻠوب‬ ‫ھو‬ ‫و‬
‫دورﯾﺔ‬ ‫ﺷﺑﮫ‬ ‫ﻣﺗﺧﺎﻣدة‬ ‫ﺣرة‬ ‫اﻻھﺗزازات‬ ‫ﻧﻣط‬
‫اﻟطﺎﻗﺔ‬ ‫ﻣن‬ ‫ﺟزء‬ ‫ﻓﯾﮭﺎ‬ ‫ﯾﺻرف‬ ‫ﻣﻘوﻣﺔ‬ ‫ﻟوﺟود‬ ‫ذﻟك‬ ‫ﻧﻔﺳر‬
‫إﯾﺟﺎد‬
‫اﻟذاﺗﯾﺔ‬
‫اﻟﺑﯾﺎن‬ ‫ﻣن‬
6
T= 31.4*10-3
s
T=T0=2π*√
‫وﻣﻧﮫ‬
= = 0.6
‫اﻟﺿﺎﺋﻌﺔ‬ ‫اﻟطﺎﻗﺔ‬ ‫ﺣﺳﺎب‬
E =EC(max)-EC(2T0)= − (2 ) = 4 ∗ 10 12 − 4 ∗ 10 7.2 = 1.84 ∗ 10