التصحيح النموذجي لاختبار الثلاثي الأول

B
 E B E A
B B A A
B
BB
m v m g z m v m g z
m v m g AB sin
v g AB sin v g AB sin
22
2
2
½ . . . ½ . . .
½ . 0 . . .
2 . . 2 . .
0
C C 
1
2 10 0,50 0,5 2,2 .
C
v m s
B
v
. . .      
G G G G
ext
F ma P ma mg ma a g
2
0
2. . .sin
02. . .s n
½
i . 1
2
x 0x C C
dx
v = = v = v v
dt
dy
v = = -gt + v -gt
dt
x
x
z
zzy
dv g AB
ax t g AB t
dt
dv z gt
ag
dt
2. . .sin
x tg AB
2
1
22. . .sin
x
zg
g AB
: هنمو
2
4. .sin
x
zAB
C
zh
 .
f xx
f
C
-x ²
-h = 4.AB.sin
0, 50 0, n² 4 40. . . si 30 2 ,
fC
x AB h sin x      
f0 63 m
0,60
f
xm
AB 
22
DD
C
C
xx
h = AB
4.AB.sin 4.h .sin
m
( , )²
, sin = ,
0 57
4 0 4
0
0
41
03
max
x
33
max max
2 10 0 2 10x x mol
aq aq g l
CaCO H O Ca CO H O
2
3(s) 3 ( ) ( ) 2( ) 2 ( )
2 3
mmol
M
3
02 10
CV
0
0
3
2 10 x
CV 2x
x
x
3
2 10 f x
C V 2 f x
f
x
f
x

3
21
max
max
2 2 2 10
C V 2 0 C 2 10 .
0, 2
x
x mol L
V
1
vol
dx
vV dt
0
00
mm
n n x x n n M
0
1 1 1
vol
dx d m m dm
vV dt V dt M M V M dt
vol
vol
dm
v mol L s
V M dt
dm
v mol L s
V M dt
3 4 1 1
0
3 4 1 1
30
11
(0) ( 16 10 ) 8 10 . .
. 0, 2 100
11
(30) ( 6 10 ) 3 10 . .
. 0, 2 100
ext G
F = m.a
AG
P+π + f = m.a
AG
P-π - f = m.a
t=0v=0
0f
A0
mg - π = m.a
( , ) , N
310 6 3 0 2
A0
π = m.(g - a ) = 54 10
K 
lim
v = v
adt dv =0
limA
mg - π - k v = 0 2
lim
A
mg - π
k =
v
lim
v
K
lim
lim ,.
, ( , )
LKg m
t
23
11
22
26 10 02
0 2 13
A
P-π 54 10 10 - 0,2
v = = 1,3m.s k = =
v
AG
P-π - f = m.a
A
v
mg - π - K v = m. d
dt
2
A
vπ
-v
dK
g
dt m m 2
A= ,
A
-3
π 0,2
- 10 - 54×10 gm63
,, K
Bm3
02 37
54 10
P
f
·
G
A
π
k
O

, ,
vv d
dt 2
6 3 3 7
0,2ts
2
a 2,8m.s
, , , v 2 2 8 6 3 3 7
1
v = 0,97m.s
3
54 10 2,8 0,15      F m a N