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06

I :m

0vO 10

)j,i,O(

10

OM

0

0)i,O(

)j,O(

3v0

0v.v0 x

0v

4v0 y

5OMh

7102ماي 3ASM)رايضيات( /اثنوي الثالثةاملس توى:

العلوم الفزيايئيةيف مادة امتحان الباكلوراي التجرييب 11سا10املدة:

V( m / S )

10,0

3,4

0,0 0,94 1,88

0v

i

j

x

y

O M

h

-10-الشكل -10-الشكل

t(s)

0/5الصفحة

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II m

h=830Km (O)

TMT=101min .

(C)

0

0

u

.

: 2

.

hR

MGa

T

T

.

3 , h, TTR

4 T ,G T, h, MTR

.

5

I

Lr

E= 1 2VR = 12ΩK

0

L, U RU E ,

0Kt = 0

RU

)t / B –e –(t) = A ( 1 RU

AB

RU

30

AB

rL

0/5الصفحة

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II

235

92U . 144

57 La 88

35 Br

0 Mev 235

92U

0

3 MeV

88 144 1 1

35 57 1 0

235

92

8,56 / , 8,28 / , 1,0073 , 1,0087

235,0134

l lE EBr Mev n La Mev n m p u m n u

A A

m U u

3

1 0

0 tuC

1

st 00 0

0

0

3 .

4 0T

- 0 –الشكل

3/5الصفحة

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2

1

L=1,0H , r =10

C=10µF

E=10V

i LU

CU

5

10²

6

I

0t mL500V1

1S

)OSK2()aq(

2

82)aq( L/mol105,1c 2

1

mL500V2

2S)IK( )aq()aq( 2C

]I[)aq(2

0

ox / red (aq) -2

4/ SO(aq) -2

8O2S (aq)-/ I2(aq) I

0

0

2 1

20 40 60 80 100

-15

-10

-5

0

15

10

5

t(ms)

(V)CU

0

1-mmoL .L ] 2[ I

4/5الصفحة

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3]I[)aq(22C2S

4 ½t

5t = 20 min

II

5

1

2

3

5/5الصفحة

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06) (

I :

0

0

GaF= m Σ

GaP= m

ox Gam 0== 0Ga

1cosα.t …0X(t)=v

oy Gam gm -g -= Ga

2sinα …0g.t + v-v(t)=

3sinα.t …0+ v 2½g.t-y(t)=

30v =10m/s0v=3.4m/s0xv

4α34.0cos0

0

v

xvα =70°

5

0t=1.88s 6.4m X(1.88)=

p

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amF .

3t=0.94s y(0.94)=4.41m

II :

m h=830Km

(O) TM

T=101min.

.1

.

2 / 2

T ST S

m xmF Gx x u

r

:

2

.

( )

T

T

G ma

R h

/ .T S SF m a

amF ST ./

rV

2 /a v r

3

r..2Tv2 ( )2

T T

TR hrv v

4 T

3

2

2

2 ( )T

T 2. . .

( )

T

T

T T T

T T T

R h

v R h

G m G mv G ma v

R h R h R h

3 2 22 2

3

T 4T 4

. .

T

T TT

R hCte

G m G mR h

5

I

0

0

= E R+ U LU

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d i / dt + ( R + r ) / L i = E / L

R

E / L = R R/ dt + [ ( R +r ) / L ] U Rd U

) t / B –e –(t) = A ( 1 RU

t / B -/ dt = ( A / B ) eR d U

R maxA = R E / R +r = U

B = L / R + r = τ RL63%

3A = 10 V B = 1.6ms

= 2 V R maxU –= E L U

0.83 A / R = 10 / 12 = R max= U 0I

= 2 / 0.83 = 2.4 Ω 0/ I Lr = U

= 0.023 H .14,4 3-( R + r ) = 1,6 .10 τL =

II

.( 1. Mev 235

92U

23592

. . 92.1,0073 235 92 .1,0087 235,0134

1,9023.931,5 1772 Mev

l p N U

l l

E Z m A Z m m u

E E

0 ( 235 1 144 88 1

92 0 57 35 04U n La Br n

(3 MeV

88 144 235

35 57 92 8,56.88 8,28.144 1772 173,6 Mevl l lE E Br E La E U E

06) (

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I

0 (aq) -2

4= 2 SO -+ 2e (aq) -2

8O2S

-+ 2 e 2 (aq)= I aq)( -2 I

(aq) 2+ I (aq) -24= 2 SO (aq)

-+ 2 I (aq) -2

8O2S

0

-24= 2 SO (aq)

-+ 2 I (aq) -2

8O2S

(aq) 2+ I (aq)

1 1= /0n

2V2C

1V1C 0n

= 3 -7,5 .10

lmo

1

t=0

X 2 X –2V2C

2 X

3-7,5.10

–X

X

t

maxX max2X-2V2C

max2X

-3-7,5.10

maxXmaxX

ft

3 1-mol .l 3-= 6 .10 f] 2[ I

) 2+V1(V .max ]2= [I max) = X2n(I

mol 3-= 6.10 maxX

mol 3-) = 1,5.10 3-( 6.10 – 3-= 7,5.10 f) -2

8O2n(S-2

8O2S-I

) = 0 3-.102( 6 – 2V2= C f) -n ( I

1-mol . l 2-= 2,4 .10 2C

4max = ½ X ½: X ½t = t

1-mol .l3-= 3 .10 max ] 2] = ½ [ I 2[ I

min =15 ½t

5( t ) = 1/V ( dx / dt ) vol V

) = x -n (I

] / dt 2= d [I vol Vt = 20 min1-.min1 -1.6 ) / 20 = 0.1 mol .l-= ( 3,6. α= tan voV

II

1

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2

3

0

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3

L , C

r

0

)01.(....................01

.

0.

0

2

2

CCC

C

LC

ULCdt

dU

L

r

dt

Ud

ridt

diLU

UU

3T = 20 mS

4)02.(....................01

2

2

C

C ULCdt

Ud

)03(..........).........cos()( 0max tUtUC

0 T13

)04.........(....................0

)cos(

2

02

2

2

0max

2

02

2

Ucdt

Ud

UctUdt

Ud

C

C

1314

LC10

CLT .22

0

0

50T = T

310.20.2 CL

FFC

10101.4

10.4 5

2

4