Друштво физичара Србије
Српске физичке олимпијаде за средњишколце 2007-2011
takmicenja.dfs.rs
2007-2011
2007-2011,
:
, , .
, www.dfs.rs/takmicenja
:
: takmicenja.dfs.rs
, 2012.
. . , , , . , , . , , .
1. Српска физичка олимпијада, Београд 2007.
1. , , 2007.
-
1. , . . ϕ 1 = 180° , q1 = 7,2 mC . q2 = 1,8 mC ? (15) 2. M = 60 g/mol T = 0 o C . , , S = 10 −19 m 2 . p 0 = 100 Pa . , , . τ = 10 −3 s . . = 1012 . , , ε = 1 % p 0 . . (25) 3. m1 m2 k . , . , , v v 2 r
r
1
r
r
. . v1 = v 2 = v . . : ) ) . . (30)
4. . , ()
. . ν z θ -. u - ’. ’ θ ’ x’ ν’ – . ) , 0; (2) ) , ν’ ’ θ ’ ν u ( ); (13) ) ) π θ ’= π/2 (3) θ ’ =0, θ ’=
ν’
) ν’ (u << c). ν’ θ ’ =0, θ ’= π θ ’= π/2 (5) ) Ne U = 10 kV . λ = 500 nm. = 20,179 g/mol. (7) ( 30 )
, , . , . , . ,
1. , , 2007.
- -
1. ϕ = T /2
q=
T / 2
dt T / 2
ε
∫ Idt = ∫ Rdt = ∫ 0
ϕ 0
=
t , d
= ϕ 1 ,
0
0
q1
=
2ε 0 Rω
ε 0 sin ω t dt = R
=
ϕ 0
ε 0 sin ϕ ε d ϕ = − 0 cos ϕ 0 Rω Rω 0
ε 0
,
q2
cos ϕ 2
=1−
(
∫
Rω q1
− cos 0°) =
ε 0 Rω
(1 − cosϕ ) 0
(1 − cos ϕ ) = q (1 − cos ϕ )
2 q2
1
2
= 0,5
2
2
⇒
ϕ 2
= 60°
2. , , :
v=
< v2 > =
3 RT
≈ 337
m
, M s , , L = vτ = 3,4 × 10 −1 m. V ≈ 4 LS . p n = 0 A ≈ 2,7 × 10 22 m -3 , RT A R . nexc = τ ⋅ 10 6 ≈ 1015 m -3 . ν 1 , ν1
= Vnexc = 4vτSnexc .
1 2 ν = ν 1 nexc = 2vτ Sn exc ≈ 0,68 ⋅ 1011 m -3 2 . ½ . , ν 2 β = = 2vSnexc ≈ 6,8 × 1013 m -3s -1 . τ ε = 1% = 0,01 n.
t =
εn
≈ 4 × 10 6 s,
β 46 . , , .
< v 2 > , < v >
v n .
1,25, .
= m1 + m2 , p0 = M u 0 = m1 v1 + m2 v 2 u 0 . m2v 3. )
u r
r
r
uoy = 2
H max
r
m2v M
r
.
2
m2 = = . 2 g 2 g m1 + m2 uoy
v
2
) . : , , . x 2
( W c ), k
( x - )
2
1
Wk = m1 (ωl1 )2 + m2 (ω l2 ) 2 ω 2 , l 1 l 2 . l , , l 1 =
lm2 M
l 2 =
lm1 M
m m m m 1 2 2 m1m2 1 + Wk = ω 2 l 2 . = ω l M 2 2 M 2 2 m1 (ωl1 )l1 + m2 (ωl2 )l2 = m1ω0 l10 + m2ω0l20 l 10 l 20 , ω 0 . l 0 2 1 2 2
l 10
=
l0 m2 M
2 2 1 2
l 20
=
l0 m1 M
.
2 2 2 2 m2 2 l 0 m1 2 m2 m1 2 + = + = m m l m m l ω ω ω ω ω 0 1 1 2 2 0 2 . 00 M M M M l v1' v2' . u r
r
r
u 0 = u r
u r
u u r
v = v1 − u0 = ' 1
m2
r
m1 v1 + m2 v 2 M 1
u u r
(v − v ) M u r
u u r
u u r
u u r
uu r
v2' = v2 − u0 =
2
m1
(v M
u u r
2
− v1 ) . u r
ω 0 = 1
2
Wk = ω l 2
2 2 0 0
m1m2 l0 M
l
2
=
m1m2
2M
v2' y l 20
. ω 0 =
vm1 M l20
=
v l 0
2
v
2
l 0
2
l
.
x
l
l 02
<< 1 .
l2
=
(l − x) l2
2
=
l
2
− 2lx + x 2 l2
2
x
x
l
l2
= 1− 2 +
≈ 1
Wk = m1v12
Mv
2
= Wc +
2
kx
2
+
m2 v22
2
v
=
2
m1 m2
2
v
2
.
2
2
+ W k
2
W c -
. Wc
= W c 0 =
Wc
=
Mv
m12
+ m22
m12
=
u r
=
2
2
2
kM
M m1 v1 + m2 v2
v .
+ m22
m1 m2
x = v 5. )
2
2
2
2
Mu02
v
M
2
+
kx 2
u u r
2
.
2
+
m1m2 2 M
(m v + m v ) u r
1 1
u u r
2
2
2
= m12 v12 + m22 v22 = ( m12 + m22 ) v 2
2
v
.
0, r
φ = ω t − k ⋅ r x cos θ + z sin θ , φ = 2π ν t − λ ’ x ' cos θ '+ z ' sin θ ' φ ' = 2π ν ' t '− λ '
(1)
r
(1)
(2)
(1)
(3)
)
t = γ t '−
u c
2
x ' , x
= γ ( x ' − ut ' )
, y = y ' , z = ' (1)
(4)
γ = 1 / 1 − β 2 β = u / c , :
ν (t '−(u / c 2 ) x') ( x'−ut ' ) x' cos θ '+ z ' sin θ ' 2π ν ' t '− − = 2π 2 λ ' λ 1 − β 2 1 − β
cos θ −
z ' sin θ λ
, ’, y’ z’,
ν (1 + β cos θ ) cosθ + β sin θ x' cos θ '+ z ' sin θ ' 2π ν ' t '− t '− x '− z ' . = 2π 2 λ ' λ λ 1 − β 2 1 − β , ’ , y’ z’ , cos θ ' cos θ + β = (1) 2 λ ' λ 1 − β sin θ ' λ ' ν ' =
=
sin θ λ
ν (1 + β cos θ ) 1 − β 2
(3)
(5)
(6)
(1)
(7)
(1)
(8)
θ ’ 6 7, sin 2θ + cos2θ = 1, λ’ y S’ S: λ ' =
λ 1 − β 2
(2)
1 + β cos θ
(9)
6 λ’ν’=λν = θ : cos θ =
ν ' ν
cos θ ' 1 − β 2
− β .
(2)
(10)
10 8
ν ' = ν
1 − β 2 1 − β cosθ '
(2)
(11)
. ) θ ’ =0 ( 11) 1 + β c+u ν ' = ν = ν , (1) 1 − β c −u
(12)
θ ’ =π ( 11) 1 − β c−u = ν ν ' = ν . (1) (13) 1 + β c+u , θ ’= π/2, ν ' = ν 1 − β 2
(14)
(1)
≈ 1 − β 2 / 2 , , 1 1 − β 2 2 1 − β ν 2 ≈ ν ≈ ≈ ν (1 + β cos θ ') (2) ν ' = ν (15) 1 − β cos θ ' 1 − β cos θ ' 1 − β cos θ ' )
1 − β 2
. : ν ' = ν (1 + β ) θ ’= (16) 0 (1) ν ' = ν (1 − β ) θ ’= π (1)
(17)
ν ' = ν θ ’= π/2; . (1) ) Ne U u :
(18)
eU =
mu 2
(0,5) (19) 2 = 20,179 g/mol M m = 20 = 3.35 ⋅ 10 −26 kg (0,5) A
u
=
2eU
= 3,09 ⋅ 105
m
(21) (1). m s , θ ’= 0 c + u c c + u ∆ν 1 = ν '−ν = ν − 1 = − 1 = 618318,6 MHz (2) (22) λ − − c u c u .
, 2 2 c u u ∆ν 2 = ν '−ν = ν 1 − − 1 = 1 − − 1 = −318,3 MHz (2) (23) λ c c . ∆ν1 - ∆ν2 = 618636,9 MHz (1)
1. , , 2007.
- 1 . . . . . ( ). . . . (60 )
1. , . , 40 . - 20 - 7 - 7 . - 4 , 2 . 2. , . - 7 . - 10 , 3 . 3. 6 . 360°. 4. . 10 . 90°.
5. , . 6. 1% + . 1. : , , 2. 2
.
(15 )
. , . . ( ). 1 , 2. 1. 3
. . (25 ) 1000 kg/m3 1. 2.
1. , , 2007. - 1
Re φ , . φ
Re
360 450 540 630 720 810 900 990 1080 1170 1260 1350 1440 1530 1620 1710 1800 1890 1980 2070 2160 2250 2340 2430 2520 2610 2700 2790 2880 2970 3060 3150 3240
9,2 11,1 12,9 14,6 16,2 17,6 18,9 20,1 21,1 22,1 22,9 23,6 24,1 24,6 24,9 25,1 25,1 25,1 24,9 24,6 24,2 23,6 23,0 22,1 21,2 20,2 19,1 17,8 16,3 14,8 13,1 11,4 9,4
∆ Re 0,19 ≈ 0.2 0,21≈ 0.2 0,23 ≈ 0.3 0,25 ≈ 0.3 0,26 ≈ 0.3 0,28 ≈ 0.3 0,29 ≈ 0.3 0,30 ≈ 0.3 0,31 ≈ 0.3 0,32 ≈ 0.4 0,33 ≈ 0.4 0,34 ≈ 0.4 0,34 ≈ 0.4 0,35 ≈ 0.4 0,35 ≈ 0.4 0,35 ≈ 0.4 0,35 ≈ 0.4 0,35 ≈ 0.4 0,35 ≈ 0.4 0,35 ≈ 0.4 0,34 ≈ 0.4 0,34 ≈ 0.4 0,33 ≈ 0.4 0,32 ≈ 0.4 0,32 ≈ 0.4 0,30 ≈ 0.3 0,29 ≈ 0.3 0,28 ≈ 0.3 0,26 ≈ 0.3 0,25 ≈ 0.3 0,23 ≈ 0.3 0,22 ≈ 0.3 0,19 ≈ 0.2
Re / a
0,1 0,125 0,15 0,175 0,2 0,225 0,25 0,275 0,3 0,325 0,35 0,375 0,4 0,425 0,45 0,475 0,5 0,525 0,55 0,575 0,6 0,625 0,65 0,675 0,7 0,725 0,75 0,775 0,8 0,825 0,85 0,875 0,9
92 88,8 ≈ 89 86 83,4 ≈ 83 81 78,2 ≈ 78 75,6 ≈ 76 73,1 ≈ 73 70,3 ≈ 70 68 65,4 ≈ 65 62,9 ≈ 63 60,2 ≈ 60 57,88 ≈ 57,9 55,33 ≈ 55,3 52,84 ≈ 52,8 50,2 47,81 ≈ 47,8 45,27 ≈ 45,3 42,78 ≈ 42,8 40,33 ≈ 40,3 37,8 ≈ 37,8 35,38 ≈ 35,4 32,74 ≈ 32,7 30,29 ≈ 30,3 27,86 ≈ 27,9 25,47 ≈ 25,5 22,97 ≈ 23,0 20,38 ≈ 20,4 17,94 ≈ 17,9 15,41 ≈ 15,4 13,03 ≈ 13,0 10,44 ≈ 10,4
∆ ( Re / a ) 1,9 ≈ 2 1,7 ≈ 2 1,5 ≈ 2 1,4 ≈ 2 1,3 ≈ 2 1,3 ≈ 2 1,2 ≈ 2 1,1 ≈ 1 1,0 ≈ 1 0,98 ≈ 1 0,94 ≈ 1 0,9 ≈ 1 0,9 ≈ 1 0,81 ≈ 0.8 0,78 ≈ 0.8 0,74 ≈ 0.8 0,70 ≈ 0.7 0,67 ≈ 0.7 0,63 ≈ 0.7 0,60 ≈ 0.6 0,57 ≈ 0.6 0,54 ≈ 0.6 0,51 ≈ 0.5 0,48 ≈ 0.5 0,45 ≈ 0.5 0,42 ≈ 0.5 0,39 ≈ 0.4 0,36 ≈ 0.4 0,33 ≈ 0.4 0,30 ≈ 0.3 0,27 ≈ 0.3 0,24 ≈ 0.3 0,22 ≈ 0.3
3330 3420 3510 3600 Re
0,17 ≈ 0.2 0,15 ≈ 0.2 0,13 ≈ 0.2 0,11 ≈ 0.1
7,4 5,3 2,9 0,5
= f (φ )
0,925 0,95 0,975 1
8,0 5,58 ≈5,6 2,97 ≈ 3,0 0,5
0,19 ≈ 0.2 0,16 ≈ 0.2 0,13 ≈ 0.2 0,11 ≈ 0.1
. , . , . , (25.0 ± 0.4)Ω 25 , . . 20 15 . R a 10 , 5 50Ω (100.0 ± 1.6)Ω . 0 e
0
1000
2000
3000
4000
R = (100 ± 2 ) Ω
f
100
. Re = aR (1 − a )
Re
Re a
= f (a)
= R − Ra .
80
60 a / e R
40
20
a . , . R = (101 ± 2) Ω .
0 0,0
0,2
0,4
0,6
0,8
1,0
a
2
50,4 Ω, , , .
1 , : ) , . ) 50,4Ω , . ) . , RRe 101 ⋅ 50.4 = Ω = 100.6 Ω : Rx = 50.6 R − Re
∆ R ∆ Re ∆ R + ∆ Re 2 0.61 2.61 Rx = ∆ Rx = + + + + ⋅ 100.6 Ω = 8.4 Ω ≈ 10 Ω R R R R 101 50 . 4 50 . 6 − e e R x = (100 ± 10 ) Ω 3
, l 0 . 180°. : l - h . . .
l 0 [cm]
l [cm]
h [cm]
1 2 3
64.9 64.9 64.9 64.9 0.2
63.5 63.4 63.6 63.5 0.2
24.4 24.4 24.4 24.4 0.2
:
p0 + ρ gh = p1 ,
p0 p1 - . , p0l 0 S = plS , . : l 63.5 kg m p 0 = ρ gh = 1000 3 9,81 2 0.244 m = 1.09 ⋅ 10 5 Pa l 0 − l 64.9 - 63.5 m s
= 1.1 ⋅ 10 5 Pa
∆l ∆l + ∆l ∆h 0.2 0.4 0.2 ∆ p0 = + 0 + p 0 = + + 1.09 ⋅ 10 5 Pa = 0.33 ⋅ 10 5 Pa ≈ 0.4 ⋅ 10 5 Pa l 0 − l h 63.5 1.4 24.4 l p 0 = (1.1 ± 0.4) ⋅ 10 5 Pa
1. -
.
.
.
1
. " "
100,0
100,0
I
2
"9. "
80,9
100,0
I
3
74,0
90,3
II
4
68,7
67,7
III
5
67,2
52,7
III
6
53,4
62,9
III
7
56,5
54,8
8
64,1
36,6
9
40,5
59,1
10
53,4
25,8
11
44,7
13
44,1
14
35,0
12
33,3
15
M
28,7
, , 2007.
1. (20 ) . , , , , , . , , . , h1 , . , , h2 ≈ h1 = h . . . , . . . “ ” h, 2h, 3h... , 20 m, . ) 20 m ∆T = 1 K H = 30 m , Q = 4 × 10 20 J. λ
, R = 6370 km. ) λ ) ρ = 5518 kg/m 3 c = 715 J/ (kg ⋅ K ), h , , .
2. (25 )
- ,, ” ( 1). -
. , . , , . ( 1). l . . , Nd:YAG 1064nm. ’. , Ar + 488nm ? ’ Nd:YAG .
3. (25 )
( I I R ) , k 1 = k 2 = k k 3 = 2k . . . t = 0 Φ 1
Φ 2 , . ) (6), ) (9), ) (3), ) (7),
4. (30 ) , , , , :
1 2
T
= A +
C
f ( x) , mgl
: T – , m – , l – –( ), g – g – , f ( x ) – x – x – . C Br , :
C ∝ µ Br . A A C . , . . : 1. . 2. 0.8 g 2. 4. 5. 6. – 7.
: , 1) . . 15mm, 15mm, 2) . , , – , , . . C x x,, 20 30 mm. f ( x ) . A A C .
1. 2. 3. 4.
. .
.
, , 2007.
1 t S ,
Q
= λ S
(T 2 − T 1 ) H
t = λ 4 π R 2
∆T H
t . (4)
, , λ =
HQ 4 π R
2
∆Tt
= 0,75
W
. (4) mK
) , ( H = h < 20 m ): λ =
hQ
⋅
m
S ∆Tt m
hmc
=
St
=
hρVc St
=
hρhSc St
=
h 2 ρc t
, (8)
, t = T / 2 = 365 ⋅ 24 ⋅ 3600 s/2 ≈ (3,1536 / 2) × 10 7 s ,
h=
λ ρc
t =
λ T ρc 2
≈ 1,73 m. (4)
2
0, . , . . Nd:YAG , posle p Nd = − p Nd . ∆ p = 2 p Nd (4). r
r
F Nd
=
∆ pukupno n ⋅ ∆ p n h n , n Φ = = = 2 p Nd = 2Φ λ Nd ∆t ∆t ∆t ∆t
(4). . ⊥ ⊥ M Nd = F Nd r Nd . r Nd
=
l 3 4
M Nd
=
l Φh 3 2λ Nd
(3). ( ). Ar + .
∆ p = 2
p Ar 3 2
= p Ar
3
(4).
F Ar =
∆ p ukupno n ⋅ ∆ p n = = p ∆t ∆t ∆t Ar
=
3
3Φ
h λ Ar
(4).
. . ⊥ ⊥ M Ar = F Ar r Ar . r Ar .
⊥ r Ar =
M Ar F Ar
=
M Nd
M Nd F Ar
d =
l 2
=
l Φ h 3 2λ Nd
⋅
= M Ar . λ Ar
l λ Ar
3Φh
= ⋅
2 λ Nd
(3).
l λ ⊥ + r Ar = 1 + Ar = 0,729 ⋅ l 2 2 λ Nd l
(3).
M Ar M Nd .
3 ) , (1). ϕ 1 ( – . )
1 , ϕ 2 2. l 1 1, l 2 2, l 3 3 : l 10 , l 20 l 30 . l 1 − l 10
= − Rϕ 1 ,
l 2
− l 20 = Rϕ 2 ,
l 3 − l 30
= R(ϕ 1 − ϕ 2 ) (1)
. F 11 F 11
= − kRϕ 1
M 11
= − kR 2ϕ 1 .
F 22
= −kRϕ 2 M 22 = − kR 2ϕ 2 . F 13 = −2kR(ϕ 1 − ϕ 2 ) M 23 = 2kR 2 (ϕ 1 − ϕ 2 )
F 22 F 13 M 13
= −2kR 2 (ϕ 1 − ϕ 2 ) ,
(2). () I ϕ 1 = −kR 2 (3ϕ 1 − 2ϕ 2 ) (1) t
t
I ϕ 2
= −kR 2 (3ϕ 2 − 2ϕ 1 ) (1)
) ϕ 1 (t ) = ϕ 10eiω t
ϕ 2 (t ) = ϕ 20eiω t (1). () , eiω t ,
2 kR 2 kR 2 ω − 3 ϕ 10 + 2 ϕ 20 = 0 I I 2
kR 2 I
2 kR 2 ϕ 20 = 0 ϕ 10 + ω − 3 I
ϕ 10 ϕ 20 (1). 2
ω
−3
2
kR 2
kR
I
2
I
2
kR 2
I kR 2 2 ω − 3 I
=0 2
(1)
2
2 kR 2 kR 2 ω − 3 − 2 = 0 I I
ω
=
kR 2
kR 2
ω = 5 (2). I I ω 12 (. ) ϕ 20 2 1
2 2
= ϕ 10
= −ϕ 10 (1). : ϕ 1(1) = A1 cos(ω 1t + δ 1 ) ϕ 2(1) = A1 cos(ω 1t + δ 1 ) ω 12 , (1)
(1), ω 22 ϕ 20 ϕ 1( 2)
= A2 cos(ω 2t + δ 2 ) ϕ 2( 2) = − A2 cos(ω 2t + δ 2 ) ω 22 . (1)
) ψ 1 = ϕ 1 + ϕ 2 ψ 2 = ϕ 1 − ϕ 2 (2) ψ 1 = −ω 12ψ 1 ψ 2 = −ω 22ψ 2 (1) . . t
t
) ϕ 1 (t ) = A1 cos(ω 1t + δ 1 ) + A2 cos(ω 2t + δ 2 )
ϕ 2 (t ) = A1 cos(ω 1t + δ 1 ) − A2 cos(ω 2t + δ 2 ) (1) d
1
dt d
(t ) = −ω 1 A1 sin (ω 1t + δ 1 ) − ω 2 A2 sin (ω 2t + δ 2 )
(t ) = −ω 1 A1 sin (ω 1t + δ 1 ) + ω 2 A2 sin (ω 2t + δ 2 ) (1) dt : Φ1 = A1 cos δ 1 + A2 cos δ 2 2
Φ 2 = A1 cos δ 1 − A2 cos δ 2
0 = −ω 1 A1 sin δ 1 − ω 2 A2 sin δ 2 0 = −ω 1 A1 sin δ 1 + ω 2 A2 sin δ 2 (2) : δ 1 A1
=
Φ1 + Φ 2 2
, A2
=
Φ1 − Φ 2 2
= δ 2 = 0 ,
(1). :
kR 2 Φ1 − Φ 2 kR 2 + ϕ 1 (t ) = cos t cos t 5 2 2 I I kR 2 kR 2 Φ1 − Φ 2 Φ1 + Φ 2 (2). ϕ 2 (t ) = cos t cos t 5 − I 2 2 I Φ1 + Φ 2
4
x. , f ( x ) , . 1. 2. 3. 4.
( T 0 ) x = 0.021 m ( T 1 ) x = 0.025 m ( T 2 ) x = 0.029 m ( T 3 )
f ( x) = 2.375 (1) f ( x) = 0.9 (1) f ( x) = 0.375 (1)
10 ( t 0 , t 1 , t 2 , t 3 ). 0.01 s. , , .
t 0 i [s ]
16.94 16.91 16.97 t 2 i [s ]
11.50 11.40 11.46
t 0 [s ]
16.94
t 2 [s]
11.453
∆t 0 [s
0.0 3
T 0 [s ]
1.694
∆t 2 [s] 0.053 0.06
T 2 [s]
1.1453
∆T 0 [s]
t 1i [s]
t 1 [s ]
∆t 1 [s]
T 1 [s]
∆T 1 [s]
0.003
8.00 8.10 8.06
8.053 8.05
0.053 0.06
0.8053 0.805
0.0053 0.006
∆T 2 [s]
t 3 i [s ]
t 3 [s]
∆t 3 [s]
T 3 [s]
∆T 3 [s]
0.0053 0.006
14.28 14.00 14.00
14.09
0.19 0.2
1.409
0.019 0.02
(.: 5mm 4 2, 5 do 10mm 4 1.8, 10mm 4 1.6) (2) :
A =
1 T 02
= 0.3485
1
≈ 0.348
s2
1 s2
,
:
∆ A = 2
∆T 0 T 0
= 0.0036
1 s
2
≈ 0.004
1 s2
,
: A = (0.348 ± 0.004)
1 s2
. (4+1 )
, : l =
T 2 g
= 0.713 m . (3, ) 4π 2 (2) C , x : mgl 1
2 0.0008 kg ⋅ 9.81 m/s 2 ⋅ 0.713 m 1 1 −3 kgm C 1 = − A = − 0.3485 2 ≈ 2.417 ⋅ 10 2 2 f 1 ( x) T 12 2 . 375 0 . 8053 s s s4
C 2
≈ 2.310 ⋅ 10
−3
kgm 2
s4 :
C 3
≈ 2.314 ⋅ 10
C ≈ ( 2.35 ± 0.07) ⋅ 10
−2
−3
kgm 2 s4
kgm 2 s4
(3 2 )
. (3+2 )
2007.
.
.
.
1
. " "
95.2
100,0
2
"9. "
69.6
76,5
3
78.5
67,7
4
76.8
50,6
5
78.2
47,5
6
" "
65.6
33,9
7
34.8
29,1
, .
2. Српска физичка олимпијада, Београд 2008.
2. , , 2008.
-
1. , C , . r , q0 , . .
2. . m ( m < M ) h ( 1). . :
(20 ) m h M
M
) , , ) (20 ) 3. V 1 = 20 l V 10
= 200
l .
: V [l] [l] 20 40 60 80 100 120 140 160 180 200 p[kPa] p[kPa] 100,00 35,40 19,20 12,58 9,00 7,80 6,95 6,30 5,75 5,30 ) V 2 = 40 l V 4 = 80 l . ? ) V 7 = 140 l V 9 = 180 l . e ? ) ) ). R = 8 ,3 J/(mol K) . (20 ) 4. (1-4)
, . (2 3) k . 3 4 L L . . 1 v 2. , . ) (2 3). ) 3 4 4 v. (20 )
5. R R = 30 cm, ω=10 ω=104 rad/s, 198Au. 198Au 198Hg γ . γ , , θ = 1200 – . - ∆ E γ k = , ∆ E γ a E γ ′ γ E γ ′
R
γ , k = 1,206. 0 198Hg u γ ; keV. e 0,511 MeV 931,4 MeV, , c = 2,998 ⋅ 10 8 m/s . (20 )
:
(2) (3) (5) (1,4)
θ
3. , , 2008.
-
1. (3), (3). , q0
−q
C
=
1
q
4πε 0 r
(6),
q
=
q0 1 + C / 4πε 0
(8).
2 ) x . ( h ) ( ). v , V L 0 = mv − MV L V L =
m
1
v (2).
1
mgh = mv 2 + MVL2 (2) 2
. v 2 =
2 2 gh
m + M
(2).
( ) ( H ) mv = (m + M )V (2) V .
1 2
mv
2
=
1 2
( m + M )V 2
+ mgH (2). 2
M h (2). H = m + M ) , , . v ( ) V L . v f > V L . ( ) mv = MVd − mv f (2) V d ,
1 2
mv 2
=
1 2
MV d 2
+
1 2
2 mv f (2).
( m + M ) v f2 + 2mvv f − (M − m )v 2 = 0 v f =
−mv ± Mv . +m
−m v (2) +m M − m m v . v f > V L . v> v M + m M 2 2m ± 8m M 2 − 2mM − m 2 > 0 . M > M v f
=
2
M > m(1 + 2) (2). 3. p = f (V ) . . p = f (V ) , , . 40 l 80 l 100
A40−80
= (546 + 317,8) J ≈ 864 J .
140 l 180 l
80
] a P k [ / p
60 40 20
A140−180
= (6,5 + 126 + 5,5 + 115) J ≈ 253 J .
. pV = RT , T 7
0 0
20
40
= pV / R = 35,4 kPa ⋅ 40 l / 8,3(J/molK ) ≈ 171 K , ≈ 117 K T 9 ≈ 125 K . T 2
60
80 100 120 140 160 180 200 220
V / [ l ]
T 4
≈ 120 K ,
, :
∆U 40−80 = ν
3 3 3 J R ⋅ ∆T 42 = ν R ⋅ (T 4 − T 2 ) = 1 mol ⋅ ⋅ 8,3 ⋅ (120 − 171) K ≈ -635 J , 2 2 2 mol ⋅ K
∆U 140−180 = ν
3 3 3 J R ⋅ ∆T 97 = ν R ⋅ (T 9 − T 7 ) = 1 mol ⋅ ⋅ 8,3 ⋅ (125 − 117 ) K ≈ 100 J . 2 2 2 mol ⋅ K
) , : Q40−80
= A40−80 + ∆U 40−80 ≈ 864 J − 635 J ≈ 230 J. (4 )
(2 ) , . 171 K 120 K.
)
= A140−180 + ∆U 140−180 ≈ 253 J + 100 J ≈ 353 J (4 )
Q140−180
(2 ), 117 K 125 K. ) C 40−80
=
Q40−80 ν∆T 42
=
230 J 1 mol ⋅ (- 51 K )
≈ −4,5
J
, (4 ) mol ⋅ K
C 140 −180
=
Q140−180
=
ν∆T 97
353 J 1 mol ⋅ 8 K
≈ 45
J
. (4 ) mol ⋅ K
4. , 2 , . (1). 2 v (2). ) v/2 (2). ) . , 2k , ω =
2 k
(3). m , 4 v, (2). 3, 3 , , 2 ( 1 2 (2). , L (1). 4 (1),
L v/2
= T (n +
1 2
) (3), n (2),
5. γ E γ 0
198
L = n +
1 m (3), n (2). π v 2 2k
Hg E 0 = E u + E γ 0 (1), E u
= hν 0 =
hc λ 0
=
p 2 2
(1)
(1) γ
. p = pγ 0 , p pγ 0
=
E γ 0
. E u
∆ν ν
=
v|| c
c
=
(1) γ ;
E γ 20 2 c2
(1) E 0
(1),
E = E γ 0 1 + γ 0 2 (1). 2 Mc
γ
E γ
v ω R cos ϕ (1), = E γ 0 1 + || = E γ 0 1 + c c
γ
φ
v||
=
R
cos ϕ (1) c γ – . R ≈ 10 −5 (1), c E γ = E γ 0 . γ θ ′ (1) λ ′ γ , λ =h/(mec) λ ′ − λ 0 = 2λ c sin 2 2 , θ ′ – . hc hc − ∆ E γ λ 0 λ ′ λ ′ − λ 0 E γ 0 θ ′ E θ ′ = = = ⋅ 2λ c sin 2 = γ 02 ⋅ 2 sin 2 = k (θ ′) hc E γ ′ λ 0 hc 2 2 me c λ ′ (3), k ′ = θ , . θ ′ . θ min k =
E γ 0 me c
2
⋅ 2 sin 2
θ 2
(1). E γ 0 = me c 2 ⋅
k 2 sin
2
θ
(1),
2
2 me c k k 2 1 + ≈ me c 2 ⋅ k (1), E 0 = me c ⋅ θ θ θ 2 sin 2 4 Mc 2 sin 2 2 sin 2 2 2 2 E u
=
(m c ) k 2 2
2
e
2
8 Mc sin
2
θ
(1). E 0 = 411 keV (1)
2 Eu = 4,6 ⋅ 10 keV=0,46 eV. (1) −4
2. ; -
. . X = X 0 ⋅ sin( k ⋅ x − 2πν ⋅ t + ϕ ). , x-. , X , . , , , , . , . p = p 0 ⋅ sin( k ⋅ x − 2πν ⋅ t + ϕ ) p 0 . . RT c = γ . M γ , R , T , . ( γ VAZ = 1, 4 , R = 8,315J ⋅ mol −1 ⋅ K −1 , M VAZ = 29,0g/mol ). . . . Q
E OSC
= 2π
E G
E OSC , E G . . : Q
=
ν REZ
∆ν REZ
.
ν REZ , ∆ν REZ , ( 1). A M
A
∆νREZ
a d u t i 2 l / p X A m M a A
νREZ
frekvencija
1. (2) , . . 250Hz. . , . . , . . . 70mV/Pa.
2. 1. . 2. () 250Hz. ( l = 60 cm, l = 65 cm, l = 70 cm). 1mm. 3. . 4. . 5. . 6. . 7. . 8. , , . ! , 0,5 cm, 1,5 cm 2 cm. 0,1 cm. 9. . 10. ( p atm = 1 ⋅ 10 5 Pa ) . .
. ( 3). . , 3. DCV. , . . 4. 7,5V 5V. 4. . : 5V. , . . . . , . . ( 5). ! ( ) , ( „ ”). . , , 5. . . (), , ,
. 6 ( FreqGenerator ) . . . Start/Stop. , Stop . : . . ,
: . ,
: . ,
6.
(60cm, 65cm 70cm). 70cm. . 1. . 200C, 293K. 300K . c=
∆c =
1
∆T
2 T
RT γ M
RT γ M
=
=
1,4
1 0.1 2 293
8,315mol −1 K −1 ⋅ 293K 0,029kg/mol 1,4
≈ 342.6
8,315mol −1 K −1 ⋅ 293K 0,029kg/mol
c = (342.6 ± 0.6)
m s
≈ 0.59
. m s
≈ 0.6
m s
m s
2. λ/ 4 = L, L . L = (70,0±0,1)cm. λ = 2,8m. c 343m/s , ν = = = 122,5 Hz . λ 2,8m . 3 λ/ 4 = L. λ = 0,933m, ν2 = 368Hz. 5 λ/ 4 = L, 7 λ/ 4 = L. ν3 = 612Hz, ν4 = 858Hz. 3. . ν2 = 368Hz, . 369Hz 367Hz ( 5V). . . 364Hz. 1Hz. „” . 1Hz, ν2 = (364±1)Hz. . ν3 = (608±1)Hz ν4 = (852±1)Hz. , , . 4. c2
c = λν . = λ 2ν 2 = 0,933m ⋅ 364Hz = 339,7m/s.
∆c
=
∆λ ∆ν , +
∆c = 1,4m/s ~2m/s. c λ ν c2 = (340±2)m/s. c3 = (340±1)m/s c4 = (341±1)m/s.
. . , . 5. 5V. . , . . 6
5
4 U
max
3
∆ν 2 U
/ 2
max
1
νREZ 0 340
350
360
370
380
390
610
620
630
ν [Hz] 6
5
4
3
2
1
0 580
590
600
ν [Hz]
6
5
4
3
2
1
0 800
820
840
860
880
900
ν [Hz]
6. . . . , 1Hz , , 1 Hz. ∆ν 2 = (7 ± 1) Hz , ∆ν 3 = (10 ± 1)Hz ∆ν 4 = ( 22 ± 1)Hz . 7.
ν Q = REZ . ∆ν REZ
∆Q Q
=
∆ (ν REZ ) ∆(∆ν REZ ) ∆(∆ν REZ ) . + ≈ ν REZ ∆ν REZ ∆ν REZ
: Q2 = 52±8, Q3 =
61±6, Q4 = 39±2.
8. , . . . a .
6
5
4
] V [
3
U
2
1
0
0
10
20
30
40
50
60
70
l [cm]
9. 10. . ( ) 48 cm . . . λ 2 = (96 ± 2) cm ν2 = (364±1)Hz c = λ 2ν 2 = 0,96m ⋅ 364Hz = 332,16m/s.
∆c
=
∆λ ∆ν ∆c + = 0,02358. =(332±8) m/s.
c λ ν c 11. . U=5,11V 22 cm . S=70 mV/Pa p0 = U/S=73 Pa. p 0 , = 0,00073 = 0,073% . p atm
2. , , 2008. - – 1. , l m , 1. g C , l C ““ ( ). m B . 1 , . θ 0
. θ , , . . (25 ) 2. , . ) , , 1 %. , , , ? , ?
(1 + x ) = 1 + nx x << 1 . ) , , ρ ≈ 3 × 10 −9 kg/m 3 v ≈ 8 km/s ? 1 × 103 kg r = 70 cm . . . . (25 ) n
3. . ν0 = 242,63 MHz. ( eV) . : R = 1,097 ⋅ 10 7 m -1 ,
c = 2,998 ⋅ 10 8 m/s , e = 1,602 ⋅ 10 −19 C , h = 6,626 ⋅ 10 −34 Js. (25 )
4. . . , . . (25 )
, , ,
, , ,
2. , , 2008. – - – 1. ,
ε
=
1 2
ω Bl 2 (2 ),
ω . , U = ε . , . . qU q . C =
q U
,
A = CU 2 (3 ).
( )
mgl (1 − cos θ )
= mgl (1 − cos θ 0 ) − CU 2 (5 ),
E p
= mgl (1 − cosθ 0 )
'
E p
= mgl (1 − cos θ )
.
mv 2 2
= mgl (1 − cos θ 0 ) (5 ),
v ,
ω =
v l
. ω
=
1 l
2 gl (1 − cos θ 0 ) (3 ).
ω mgl (1 − cos θ ) = mgl (1 − cosθ 0 ) −
1 2
Cl B g (1 − cosθ 0 ) 3
2
1 − cosθ
sin θ
= (1 − cosθ 0 )(1−
Cl 2 B 2 2m
) (3 ).
≈ θ 1 − cos θ = 2 sin
θ
2
2
,
θ
= θ
2
2 0
Cl 2 B 2 1 − 2m
(4 ).
2. ) r :
E r =
γ M = gR v 2
2
=
mv 2
Mm −γ 2 r
=−
mgR 2 2r
, (4 )
gR 2
, R , M . r ∆r . :
E r + ∆r = −
mgR2 2(r + ∆r )
= E r + ∆E . (3 )
−
mgR 2 2(r + ∆r )
mgR 2 ∆r =− 1 + ∆r 2r r r
mgR 2
=−
2r 1 +
−1
mgR 2 ∆r ∆r ≈− 1 − = E r 1 − , (1 ) 2r r r
(1 + x ) = 1 + nx x << 1 ( n = −1 ). : n
E r + ∆ E = E r 1 +
∆ E ∆r = E r 1 − , (1 ) E r r
∆r r
=−
∆ E E r
= −0,01.
(1 )
1 %. (1 ) , ,
(v + ∆v ) =
gR 2
2
(r + ∆r )
gR 2 ∆r = 1 + ∆r r r r
gR 2
=
r 1 +
−1
gR 2 ∆r ≈ 1 − , (1 ) r r
2
∆v ∆v (v + ∆v ) = v 1 + ≈ v 2 1 + 2 , (1 ) v v 2
2
∆v v
=−
∆r ∆ E = = 0,005. 2 r
2 E
(1 )
0,5 % . (1 ) ) , , v v .
∆ p = 2mm v = 2
M A
v . (2 )
∆t S = r 2 π V = r 2 π ⋅ v∆t . , , : F =
∆ p M v V Mv ρ V v ρ = 2 ⋅ ⋅ = 2nV = 2nr 2 π ⋅ v∆t 0 m ⋅ = 2r 2 πρv 2 , (4 ) A ∆t V A∆t A∆t ρ ∆t
∆t n . F = 2r πρv 2
2
= 2 ⋅ (0,7 m ) ⋅ 3,14 ⋅ 3 × 10 2
−9
2
8 × 103 m = 0,59 N. ⋅ m 3 s kg
(2 )
a=
F m
=
0,59 N 1 × 103 kg
= 5,9 × 10− 4
m s2
. (2 )
3. () n () k
1 − 1 . k 2 n 2
ν n ,k = cR
, n. n m (n > m) ( k ) (2). ,
1 − 1 (3) m 2 n 2
ν = ν n ,k −ν m ,k = cR
k . ν0 n = m + 1 ( 3). n+1 n ν 0
1 1 2n + 1 = cR 2 − = cR 2 2 2 n (n + 1) n ( n + 1 )
. (5)
n n + 1 ≈ n 2cR 2n + 1 ≈ 2 n , ν 0 = 3 (5) n n .
hν n
≈ hν n+1
1
2cR 3 ≈ 300 . (5) n = ν 0
3
E = hcR 1 −
1 300
2
≈ hcR ≈ 13,6 eV, (5)
.
2. -
.
.
.
.
1
100
99
40
97,8
97,4
2
85,8
64,1
44,5
82
71,2
3
96,3
78,8
4
-
94
79
5
73,1
77
26,52 87,2
62,4
6
82,8
89
30,25 61,2
61,3
7
85,1
57
28,25
41,5
8
56,7
89
23
9
65,7
65
26,9
10
73,1
86
1,5
11
53,7
71
26
12
48,5
50
13
58,2
37
18,25 81,5 26
66
47,3
66,7 62,5
39. International Physics Olympiad 5
3. Српска физичка олимпијада, Београд 2009.
3. 2008/2009.
-
07-08.05.2009.
1. ) m > m . H ( 1), . . . ? ) n ( 2 n = 4 ). m1 m2 , m2
1.
m3 . (. m1 >> m2 >> m3 >> ... >> mn ). H , n − H + l . . . , . ( n ). H = 1m l << H 1 km? (25 )
2.
2. S ( ). , . , . ν . L p0 .
∆Q . : ) ∆T ? ) C ? ) F ? ) ∆Qmax
L h
L
h = 2,5 L ? F ? R . . (25 ) 3. E L. , s = L/. ± s, () , . 1 . , , r = 0,2 cm m = 1,97 mg.
. λ/2 ( , ). , λ = 1064 nm ; P = 10W. . = 34,6s. θ0 = 50 . . (25 ) . 3 ⋅ 10 8 m/s .
4. f x , x> f . α . d , . (25 ) . .
: (1), (2), (3), (4) : (1), (2), (3), (4) : . , ,
3. 2008/2009.
07-08.05.2009.
1. ) H
v =
2 gH (1).
m . v − mv = Mv + mvm (2) ( ) v M , vm , . 1
v2 +
1
2 2
mv 2 =
1
v M = v −
Mv M2 +
1
m M
( v + vm )
(1),
mvm2 (2)
2 2 2 ( m + M ) vm − 2v ( M − m ) vm − ( 3M − m ) v 2 = 0 (2).
M − m ± 2M vm = 2 gH , ( + M m 3 M − m 2 gH (2). h M + m
) vm =
, mgh =
1 2
mvm2 (2)
9 2 − 6 Mm + m 2 h = H (1). 2 2 M 2 Mm m + + 8(1 − m M ) ∆h = h − H ⇒ ∆h = H (2). (1 + m M ) 2 ) v =
2 gH (1).
. Li vi Li −1 Li +1 Li . Li +1 Li v + vi ( , , , ...). Li vi ( mi >> mi +1 ) Li +1 (v + vi ) + vi vi +1 = 2vi + v (2). v1 = v v2 = 3v (
>> m ) , v3 = 7v , v4 = 15v .
n − vn = (2n − 1)v (4). h = l +
H = 1 m
1
km
n 2 2 (2 − 1) v
= l + (2n − 1)2 H (2) .
2 g
6
(1)
( 2 n − 1 > 1000 ⇒ n > ln 2 ( 1000 + 1) )
2. ) V 1 = SL , S
. ∆V 2 = S ∆ L . ∆Q ∆U A : 3 ∆U = 2 ν ⋅ R∆T , (2) 2 A = p0 ∆V 2 = ν R∆T . (2)
∆Q = ∆U + A = 4 ν R∆T , (2)
∆Q
∆T =
4 νR
. (2)
) C =
∆Q = 4 νR. (2) ∆T
) ( ) , ∆ p ν R∆T ∆Q ∆ p = = . . (3) ∆ pV 1 = ν R∆T , 4V 1 V 1 p0 , p0 , ∆ p . , F = S ∆ p =
V 1∆Q 4V 1 L
=
∆Q 4 L
. (3)
) p0 ∆V 2 max = p0 S ∆ Lmax = p0 S ⋅ 0,5 L = ν R∆T max (2)
∆T max =
p0 SL 2ν R
, (2)
∆Qmax = 4 ν R∆T max = 4 ν R
p0 SL 2ν R
= 2 p0 SL. (2)
,
F =
∆Qmax 4 L
=
2 p 0 SL 4 L
=
p 0 S . (3) 2
3. ω =
T = 2π
I k
k I
(2),
(1), I = mr 2 / 2 .
∆ E hc ∆ = ⋅ (3) ∆t λ ∆t T 2 ∆ P λ = (1). ∆t hc k =
2π 2 mr 2
(1). P =
, . 2 . . ∆ L ∆ = 2 sh M = (5). ∆t ∆t M = k θ 0 (3) ( M k θ 0 ); „ “ , . 2π 3 mr 2θ 0 c s = θ0 , T 2 λ P π 4 mr 2θ 0 c s = (5) θ0 . 90T 2 λ P s = 1,00314 ≈ 1 (4).
4. h = x tg α (1).
l =
xf x − f
1 f
=
1 x
1
+ (5). , l
(3). d ≤l ( 4a),
h1 =
l−d l
h=
l −d l
x tgα = ( x −
h1 =
d −l l
h=
d −l l
x tgα = (
d (x − f ) f
− x )tgα (3).
h1 h
=
h
=
l − d l
(5),
d (x − f ) )tgα (3). d >l ( 4), f
d − l l
h1
(5),
h
h1
α
d
x
4
l
h
α
d
x
h1
l 4
3. 2008/2009.
-
07-08.05.2009.
. p-n
-n . -n . -n : – (Light Emitting Diode), . , , ( ). . ( ) 1. -n , ( 1). -n , n , . , U 0, . , . U 0 -n . U 0 . (-n ) U . U ( ), . U U 0. I
I
U U 0
U
U
1. ) ; )
. 1 U 0 . 0,6V. , U 0 . – . ( ) eV , . -n . .
U 0, . -n : -n U 0 n . . . U 0 n , p . U 0
hν = eU 0
(1) U 0 . -n () . (1) . : *
hν = eU 0 + E
(2)
*
. . -n . , , -n -. -n . -n „” I-U ( 2). „” -n . . -n -n . , -n . -n : ( 2). ( ). (- ). -n . . , . U . , , I III I-U , , , II IV , . -n II , IV -n . -n .
I
II
I
( )
S=0 U
S 1 S 2 III
( )
IV
2. p-n , , , 1 < 2.
1. *
(2) - -. , - . , - . 1.1 -, 2 7, , U 0 . - : λ[nm]
2 594
3 652
4 630
5 610
6 570
7 940
. U 0 . 8 -19 c = 2,99810 m/s. = 1,602210 C. . ∆λ = 5nm. (30) : 3. 3. ( 9V ), prek1, R2 R3 , R1 . R1 -.
BAT +
U LED
prek 1
LED
R3 U R1
R1
R2
3. ) -; ) -.
: R1 Ω. prek1 0 . (). U R1 R1. DCV. , R2 R 3. R2 (0-1kΩ), R3 (0-250kΩ) . U 0 - (U LED ) . 50µ. 1.2 λ 8 λ 9 - 8 9, , U 0 . . (10) 1.3 3 I LED, ( 1.1) (U LED). -. 10 - . . . (10)
2. .
- , . . 2.1 , . I f - I LED. . (12)
: - 4. 4. , R5 prek2 . - R4 ().
BAT
U R5
+ -
LED
R5
prek 2
prek 2
U R4
R4
R4
a
LED
4. ) - ; ) - ; ) - .
: R4. prek 2. U R4 R4. 3. . 5. prek1 prek2. R2 R3 . 30 R4. - 10 .
a
5. ) - ; ) -, .
2.2 R5 , -. - 50mA . . . R5 . . -. -! , , ? (23) 2.3 2.2 η . -. . . (15) : 4. , R5 prek2 . - R5 . : R5. prek2 . . 3 . . 5. prek1. R2 R3 50mA. () R5 . - 2.1 : ( ) 1% .
: - . - 7 8 .
: , , : . , , : . , ,
3. 2008/2009.
07-08.05.2009. 1.1 R1 100Ω (1). . - I LED = U R1 / R1 . 50µ, , R1 5mV(2). (2) U 0 = h ⋅
c 1
⋅
e λ
−
E * e
(1)
U 0 1/λ. 2 3 4 5 6 7
λ [nm] 594 652 630 61 570 940
U 0 [V] 1.67 1.52 1.53 1.66 1.73 0.90
∆U 0 [V] 0.02 0.02 0.02 0.02 0.02 0.02
1/ λ [1/nm] 1.6810 1.5310 1.5910 1.6410 1.7510 1.06410
∆(1/ λ) [1/nm] 0,0210 0,0110 0,0210 0,0210 0,0210 0,00610 (4)
1.8
- 1.6
1.4
] V [ 0
U 1.2
(6)
1.0
0.8 6 1.0x10
6
1.2x10
6
6
1.4x10
1.6x10
6
1.8x10
-1
1 / λ [m ] c
k = h ⋅ = 1,2264 ⋅ 10 −6 Jm/C (3) e
∆k = 0,057 ⋅ 10 −6 Jm/C (3). ∆h / h = ∆k / k . n = − n=−
E * e
E * e
h = (6,6 ± 0,3) ⋅ 10 −34 Js (4).
= −0,3912 V (3) ∆n = −0,089V (3) .
= ( −0,391 ± 0,08) V .
k n . : * - !
U 0 = k ⋅
1.2
1 λ
+ n λ =
k U 0 − n
(1) U 0
-. 8 U 0 = 0,927V ∆λ ∆k ∆U 0 + ∆n (2). λ 8 = 930nm . = + = 0,122 (1) k U 0 − n λ λ = (900 ± 100) nm (2). 9 U 0 = 2,44V (2) λ 9 = (430 ± 40)nm (2) . - , , - , ( -) 2.5
50
-
- 3
2.0
40
1.5 1.0
] 30 A m [ D E
I L
20
0.5
10 0.0 400
500
600
700
0
λ [nm]
1.5
1.6
1.7
1.8
1.9
2.0
U LED [V]
(5) 1.3 1.1 -. . ULED[V] UR1[V] ILED[mA]
1 1.498 0.003 0.031
2 1.617 0.05 0.5
3 1.643 0.1 1.0
4 1.658 0.15 1.501
5 1.669 0.2 2.002
6 1.686 0.3 3.003
7 1.707 0.5 5.005
8 1.748 1 10.01
9 1.781 1. 15.01
10 1.81 2 20.02
11 1.869 3 30.03
12 1.924 4.01 40.14
13 .982 5.05 50.55
(5) , .
2.1 R4 100kΩ (1). . . I f = U R 4 / R4 (1).
U R1[V] U R4[V] I LED[mA] I f[µ]
1 4.93 4.50 49.3 45.0
2 3.83 3.67 38.3 3 .7
3 3.08 3.09 30.8 30 9
4 2.55 2.63 25.5 26.3
5 2.08 2.19 20.8 21.9
6 1.73 1.85 17.3 18.5
7 1.52 1.65 15.2 16.5
8 1.26 1.37 12.6 13.7
9 1.01 1.09 10.1 10.9
10 0.75 0.8 7.5 8.0
11 0.50 0.52 5.0 5.2
12 0 0 0 0 (5)
50
( -) - 40
30
] A µ [ f
I 20
10
(5) 0 0
10
20
30
40
50
I LED [mA]
- . 2.2 ( R5) 10kΩ, 22kΩ, 30kΩ, 43kΩ, 56kΩ 100kΩ (2). .
P = U R2 5 / R5 (1). . (2.1). , - 2.1 -. - 49mA - 20,2mA (4). (5) 1 2 3 4 5 6 R5 [kΩ] 10 22 30 42.9 56.1 99.7 U R5, 44mA[V] 0.428 0.927 1.236 1.419 1.453 1.483 U R5, 20,3mA[V] 0.228 0.5 0.678 0.961 1.231 1.419 P 44mA[µW] 18.32 39.06 50.92 46.94 37.63 22.06 P 20,3mA[µW] 5.20 11.36 15.32 21.53 27.01 20.2
60
50
I LED= 49mA
40
] W30 µ [ P
I LED= 20,2mA
20
(5)
10
0 0
20
40
60
80
100
R5 [k Ω]
30kΩ - 49mA (2) 56kΩ - 20,2mA (2). (2). 2.3 -. - -. 1.3. - 1,953V (2) - 49mA 1,80V (2) - 20,2mA.
η 49mA =
P max P LED
η 20,2mA =
P max P LED
=
P max I LEDU LED
=
P max I LEDU LED
= =
50,92 µW 49mA ⋅ 1,953V
= 0,0532% (1)
27,06 µW 20,2mA ⋅ 1,80V
= 0,0744% (1)
∆η η
=
∆ P max P max
+
∆ P LED P LED
=2
∆U R5 U R5
+
∆ R5 R5
+
∆U R1 U R1
+
∆ R1 R1
+
∆U LED U LED
(3)
η 49mA = (0,053 ± 0,004)% (3) η 20,2mA = (0,074 ± 0,006)% (3).
2009.
18-29. 2009.
1. , , V 0 . m (m << M ) ,
, L . , , . . . ) ) (25 ) 2. ν . (1-2), (3-1) “” n (2-3). . k . )
. ) n = 10 k = 5 . k 1 / n .
kp0
p0
V 0
kV 0
(25 )
3. . m = 2 1030 kg d ≈ 10 km , eV. m0 = 1,67 10−27 kg, c = 3 10 8 m/s, h = 6,62 10 -34 Js. (25 )
4. () m = 50 t . t = 0 . , . t 1 = 2 s , . . . . (25 ) , , , .
] s [ t 2 1
0 1
8
6
4
2
] 0 2 1 . s / 0 m [
8 0 . 0
6 0 . 0
4 0 . 0
2 0 . 0
0
a
: (1), (2), (3), (4) : (1), (2), (3), (4)
: . , ,
2009.
18-29. 2009.
1. ) . V v , . l . l (v − V ) , . . : '
l . vt + Vt = 2l (3) ( l (v − V ) 2l ). l ' = l − Vt l ' = . , v + V ' ' v − V = v + V (3) ( ). , l (v − V ) = l ' (v ' − V ' ) . : >> m ) 2V 0 . V 0 ( L(2V0 − V0 ) = LV 0 (3). L
in
,
. v = V 0 Lmin ≈ L
m
M m
(3). , L
in LV0 ≈ Lmin (V 0
M m
− 0)
(3)
) : V (t ) , v (t ) ,
(t ) . 1 1 2 2 ( ) V0 = mv 2 2 v = V 0
. (t ) v(t ) . m , dn dt , vdt >> m ). 2V x dn = ( 2 x vVdt dx dv = 2Vdn . dv = , V = − x dt dv dx L = − . vx = const . v = 2V 0 x ≈ , v = 4V 0 v x 3 L L V 0 x ≈ , x ≈ 5 7 L x = , k , 2k − 1 L ≈ LV0 . v = 2kV 0 . vx = 2 kV0 2k − 1 xmin = L
m M
.
) V v . 2mv ( V v ). dn dt , dt dV = −2mvdn . 1 1 1 2 2 2 MV0 = MV + mv (3) 2 2 2
1 M 2
m
0
∫ 1
1 M
m ∫
2
V 0
dV V 0 1 −
V
2
= − ∫ dn = − (2),
y ≡
0
m
1−
V V 0
2
= − ∫ dn = − π
4
m
(2).
(
arcsin
−
π
0
≈
2
V 0
V 02
dy 1− y
0
v = V 0
V 2
2
)
(3).
2. )
A
Q+ . p0 V 0 , kp0 kV 0 . 1-2 “” 2-3. 1-2 : V ∆ p = 1,5V 0 (kp0 − p0 ) = 1,5V 0 p0 (k − 1) . (2) Q12 = νC v ∆T = 1,5 R 0 R 2-3 , p0 k p0 , p0 k ⋅ k (i −1) / n (2) ( i 1 n ). , V 0 V 0 k V 0 k i / n (2). i - . V 0 k (i
−1) / n
V 0 k i / n (1), p0 k 1
− ((i −1) / n )
(1).
( ) ∆ A = pi ∆V = p0 k 1− ((i −1) / n ) ⋅ V 0 k i / n − k (i −1) / n = kp0V 0 k 1 / n − 1 , (2)
(
)
(
)
( ) ∆U = νC v ∆T = 1,5 pi ∆V = 1,5∆ A = 1,5kp0V 0 k 1 / n − 1 , (2)
(
)
( ) ∆Q = νC p ∆T = 2,5 pi ∆V = 2,5∆ A = 2,5kp0V 0 k 1 / n − 1 . (2) . 3-1 A31 = − p0 ∆V = − p0 (V 0 k − V 0 ) = − p0V 0 (k − 1) . (2) 3 5 Q+ = Q12 + n∆Q = (k − 1) p0V 0 + nk k 1 / n − 1 p0V 0 , (2) 2 2 A = A31 + n∆ A = −(k − 1) p0V 0 + nk k 1 / n − 1 p0V 0 . (2)
(
(
)
)
η=
A Q+
=
− (k − 1) + nk (k 1 / n − 1) 3
(k − 1) +
2
5 2
(
1/ n
nk k
− 1)
. (3)
) n = 10 k = 5 η=
− (k − 1) + nk (k 1 / n − 1) 3 2
3.
(k − 1) +
5 2
(
1/ n
nk k
− 1)
3 2
≈ 2,29 ⋅ 10 45 m -3 (1), 4 3 r π 3 − − a = n 1 / 3 = 7,58 ⋅ 10 16 m (4). n=
=
− (5 − 1) + 10 ⋅ 5(51 / 10 − 1)
(5 − 1) +
2
(
10 ⋅ 5 5
=
5
m
1 / 10
− 4 + 50 ⋅ 0,1746 3 2
4+
5 2
= 1,2 ⋅ 1057 (1),
m0
− 1)
=
≈ 0,17. (2)
50 ⋅ 0,1746
a
v
v m0 c h − c (4), p ≈ ≈ 1,39 ⋅ 10 19 kg m/s (4). p = m0v = a v2 1− 2 c
c 1−
= v
2
p m0 c
= 0,277 (2).
c2 v/c v/c≈ 0,267 (2). v2 ( 2 = 0,077 << 1) c (2), T ≈
p
2
2m0
=
h
2
n2/3
2m0
= 5,77 ⋅ 10 −12 J ≈ 36MeV (3+2) .
4. F , ma = F − kv 2 . t 1 F max , a1 v1 , ma1 = F max − kv12 [3]. , , . v 2 , F max = kv 22 [3],
F max =
ma1
v1 v 2
2
[1].
1 − t 1
∫
v1 = a(t ) dt = S 1 - t 1 . S 1 = 98 x 0.4 x 0.004 0
m s
= 0,1568
∞
∫
v 2 = a(t ) dt = S 2 - . S 2 = 278 x.0.4 x 0.004 0
v1 = ( 0.157 ± 0.004) m/s [3+1]
m s
m s
= 0,4448
m s
v2 = ( 0.445 ± 0.004) m/s [3+1]
∆S 1 = ∆S 2 = 2 x(0.004 x0.4)
F max =
ma1
S 1 S 2
1 −
2
m s
= 0.0032
= 5.71 ⋅ 10 3 N , ∆ F max
m s
≈ 0.004
m s
a1 = (0.100 ± 0.001)
m s2
[2+1]
2 ∆S 1 ∆S 2 S 1 + 2 ∆a1 S S S 2 1 2 = F max + = 0.102 ⋅ 10 3 N 2 a1 S 1 − 1 S 2
F max = (5.7 ± 0.1) ⋅ 10 3 N [5+2]
3. 2008/2009.
.
.
.
.
.
99
67,8
86,9
97,5
92
60
59,4
67
74
49,6
74,7
58,9
86,5
58,5
70,7
58,5
1
2
3
4
5
95
51
66,8
56,2
6
72
49,6
73,7
55,1
7
51
33
87
48,7
8 9
. . ..
58,8 75
73 31,5
57,7
46
10
68,4
18
11
73
22,6
12
. . . . . .
69,4
13
9.
86
14
75,8
15
74,5
2
16
07-08.05.2009.
1-5
4. Српска физичка олимпијада, Београд 2010.
4. 2009/2010.
-
04-05.05.2010.
1. , , m (m << M ) . , v0 .
, , . . . µ . µ . . . : ) n − ) (20 )
2. ρ
ε r R2 R1 ( R1 ≈ R2 ) , . , H ( ). . C = 2 πε 0 ε r l / ln ( R2 / R1 ) , l . U hc , hd . )
ε 0
, hc ; − ) hc R1 = 10 mm , R2 = 11 mm , ε 0 = 9 × 10 12 F/m , ε r = 3 , U = 103 V ,
ρ = 103 kg/m3 , g = 10 m/s 2 , ln(1 + x ) ≈ x x ≤ 0,1 .
(20 )
3. , L , , . , m q . , B0 , . d < L , . . . . (20 ) 4. . (Charles Kao) . : , - 1. ( ) , . ) n j n ? ) ( ) Заштитни слој Омотач , Језгро „ “ . n. ) f d =1mm n j = 1,58 no = 1,53. n =1. ) ( ) L= 3km ), , . (20 ) 5. β
23 12
Mg t =0 .
∆t =2,0s β n1, , , β 1,26 . 23 12 Mg ? (20 ) : (1), (2), (3,5), (4) : (1), (2), (3,5), (4) : . , ,
4. 2009/2010.
04-05. 2010.
1. ) vi i - ,
mvi = MVi +1 − mvi +1 (3),
1 2
mvi2 =
1
MVi+21 +
1
mvi2+1 (3)
2 2 ( ) V i +1 vi +1
. vi = Vi +1 + vi +1 (2) (1 − α ) vi +1 = vi (3). (1 + α ) n
(1 − α ) , n vn = v0 + α ( 1 ) = m / M << 1 , (3). , , α = vi +1 ≈ (1 − 2α )vi Vi +1 ≈ 2α vi . n vn = (1 − 2α )n v0 . ) . , . . 1 2
mv02 = µ Mgd (4)
d =
mv02 2µ g
(2).
2. , ` , G , .
G = g ρπ R12 H
(1) (1)
, U , U , , () G + F = g ρπ R12 ( H + hc + hd ) , (2) (2) F F . , : π R12 hc = π R22 − R12 hd .
(3) (2)
(1), (2) (3) F = g ρπ
R12 R22 R22 − R12
hc .
(4) (1)
C C , . L ( ) U . U . C C : C 0 C d . , C = C 0 + C d . C 0 C 0 = 2πε 0
L − H − hd , R2 ln R1
(5) (2)
H + hc + hd C d = 2 πε 0 ε r . R2 ln R1
(6) (2)
(3) hc hd , (6) C = 2πε 0 ln
1 R2
R12 ( ) − ε 1 [(ε r − 1) H + L ] + ε r + 2 hc . r 2 − R R 2 1
(7) (2)
R1 F F :
∂W , h = hc , h ∂ U
F = W =
1 2
(8) (2)
CU 2 , U = const . (7) L H , H , (8) 1 2 2πε 0 ε r R22 − R12 ∂C . F = U = U R2 R22 − R12 2 ∂h U 2 ln 1
2
R1
(4) (9) 2 2 2 1 ε 0U ε r R2 − R1 hc = . (2) ρ g R2 R12 R22 ln R1
) , ln
R2 R1
= ln
1 1 = ln1 + ≈ = 10 −1 , 10 10 10 11
(9) (2)
hc = 10 3
1 kg m3
9 × 10
⋅ 10
m
−12
F
m − 10 1
⋅ 10 6 V 2
(3 ⋅ 11
− 10 2 ) ⋅ 10 − 6 m 2 = 1,96 × 10 -4 m ≈ 0,2 mm (2) −12 2 2 4 11 ⋅ 10 ⋅ 10 m 2
s2
3. , , , . , . ε = −
∆Φ ∆ B π d 2 ∆ B .(3) = − S =− ∆t ∆t 4 ∆t
. τ , B = B0 −
B0 τ
t ,
π d 2 B0 B0 ∆ B .(2), a ε = =− (3) τ ∆t 4τ , "" , ε =
As q
=
F s Lπ
= ELπ , (3)
q
E =
d 2 B 0 4τ L
.(3)
, a=
F m
=
qE m
=
qd 2 B 0 4mτ L
.(2)
τ v = aτ =
qd 2 B0 4 mL
,(2)
, ω =
2v L
=
qd 2 B0 2mL2
(2)
4. ) , n j > n. (1) ) - . n αg. sin α g = o . (2) n
омотач Угао прихватања
αg
β
језгро
n sin θ max = n sin(90o − α g ) (1).
sin(90o − α g ) = cos α g
n sin θ max = n j cos α g (1).
sin 2 α + cos2 α = 1 n sin θ max = n j 1 − sin α g (1). 2
2θ max = 2 arcsin
n 2j − no2 n
sin θ max =
n 2j − no2 n
(1).
(1)
) , 2ϕ 2θ ax ( ϕ ≤ θ ax )(1). ( ) . tgϕ = f ≥
d 2tgθ max
=
d
1 − sin 2 θ max
2
sin θ max
=
d
n2
2
n j2 − no2
d 2 f
(1),
− 1 (1).
f ≥ 1,17mm (1)
) φ . d /2 f . α . L L = (2), α = 90o − β L A = L (2) L B = cos β sin α . α , c (1). αg(1). v = n
Ln j
∆t =
L B − LA
no
− L
n j = 516ns (1). = v c 1s ≈ 1,93⋅ 106 (1) . 516ns α β
β β
β
α
L
5. = 0 e− λ ∆t (2) 2 = 0e −3λ ∆t (2), 1
( β ) ∆t n1 = 0 − 1 = 0 (1 − e − λ ∆ ) (2), 3∆t n2 = 1 − 2 = 0 (e − λ∆ − e −3 λ ∆ ) (2). t
t
t
n2 n1
=
e− λ∆t − e −3λ ∆t 1 − e− λ ∆t
n2 n1
=
x (1 − x 1− x
2
)=
x1,2 = x =
−1 + 1 + 4n2 / n1 2
(2)
= e − λ ∆t .
(1 + x ) (2)
−1 ± 1 + 4n2 / n1 2
x + x − 2
1
n1
= 0 (2).
(2),
(1).
λ = −
n2
1 + 4n2 / n1 − 1 (2), 2
ln ∆t
n2 / n1 = 1, 26 λ = 0,16s −1 (1)
4. 2009/2010.
, , 5.05.2010.
:
. , . . E ε . :
() L, a b (b < a < L) – 1.
L b
a
1. ( ) – 2, .1 ’ b. x=L
-
’ +b/2
y
x
F
2. F , b. . . ’ , . ’ . , , . 1
2 1 „“. 2 ; L b.
, y F , – 2. ( ) 6 F 2 x y ( x ) = x L − 3 Eab 3 y ( = L ) F
=
4 L3 Eab3
F .
. , . . ; σ = τ / E . σ , – . . : mala osa ε = velika osa . .
σ 1.0
0.5
τ/E
0.0 -1.0
-0.5
0.0
0.5
1.0
-0.5
-1.0
:
3.
8
3
2 1 6 5 4 7
( 3) : (1), (2) – , (3), (4), (5), (6),
(7) (8). : , . : ( 100 ; )
1) : a b. , . : M = ( 282,1 ± 0,1) g [10] 2) . [5] 3) L . [5] 4) L . [5] 5) L. 250mm < L < 400mm. . . . [30] 6) ε 1 . [15] 7) ε . [5] 8) ε L = 415mm . 1m . . 8-10cm. . A 2 n = 2 e. , A4, A6, A8, … . . ε . . [25] : ( )
B
l 0
B’
M
M’
A’
B’
B M
M’ A’
A R
)
) α O 4.
. AB A’B’ – 4; l 0 , MM’ . 4 – b, a, R α (α ); . R ,
MM’ . l 0= Rα MM’ . ρ M B; ρ=-b/2, ρ=0, ρ=+b/2. ρ, ; l 0. R+ρ l=(R+ρ)α ρ . ∆l=l-l 0= ρα , ∆l/l 0=ρα/R , τ=E∆l/l 0=Eρ/R, . F e AB τ : b /2 ρ Fe = τ dS = E ad ρ =0 . − b /2 R
∫
∫
∫
M e = ρτ dS =
Ea
∫
b /2
Eab3
, F e = 0 , ρ d ρ = R − b /2 12 R . x-y , R y " 1 , y ' " y = y ( x) . = R (1 + y '2 )3/ 2 2
R ≈ 1 / y " , M e ≈
Eab3
y " . 12 . F , . , F e = 0 M e , y F’ F ( L − x ) F "=
12 F Eab 3
e
.
( L − x ) . ,
y (0) = 0 y '(0) = 0 , y ( x ) =
6 F Eab 3
x2 L −
x
,
3
.
: , , : , ,
4. 2009/2010.
, , 5.05.2010.
1) ( 0,02mm), ( 0,01mm). () . . ( ). . : M = ( 282,1 ± 0,1) g : ( 0,02mm) a [mm]
42,06
42,04
42,14
42,18
42.08
3,14
3,12
a=(42,10±0,08) mm [5] : ( 0,01mm) b [mm]
3,13
3,05
3,07
b=(3,10±0,05)mm [5]
2) F =
Eab3
y 4 L3 . :
⋅ y&& = −
Eab3 4 L3
y
3) ω = T =
2π ω
3
T = 2π
4 ML Eab
3
[3]
4) :
T = 2
k =
16π 2 M Eab
3
L3 [1]
2
16π
Eab3
,
E =
16π 2 M kab3
[2]
Eab3 3
4 L
, [2]
∆ E
=
∆M
+
∆k
+
∆a
+3
∆b
[2] E M k a b 5) 6 L T . L 1mm; ∆ L = 1 mm. 3 10 t 10. 0,01s. , ∆t 10 t 10 , , . je 0,05s. t 10 3 10 , ∆t 10 , . 1. . 1 [15] i 3 3 2 2 ∆t 10 [s] t 10 [s] t 10 [s] L [m] ∆T [s] ∆T 2 [s2] T [s] L [m ] T [s ] 0,396
0,375
0,355
0,336
0,313
0,279
9,64 9,62 9,69 8,96 8,84 8,90 8,11 8,17 8,02 7,47 7,57 7,49 6,89 6,77 6,80 5,92 5,88 5,93
9,65
0,05
0,965
0,005
0,0621
0,93
0,01
8,90
0,06
0,890
0,006
0,05273
0,79
0,01
8,10
0,08
0,810
0,008
0,04474
0,656
0,013
7,51
0,06
0,751
0,006
0,03793
0,564
0,009
6,82
0,07
0,682
0,007
0,03066
0,46
0,01
5,91
0,05
0,591
0,005
0,02172
0,349
0,006
: ∆T = ∆ t 10 / 10 , ∆ ( L3 ) = 3L2 ∆L , ∆ ( T 2 ) = 2T ∆T . 2
3
5 (T ) ( L ). [10]
1.0
0.9
B(0,060; 0,89)
0.8
0.7 ]
2
s [
0.6
2
T 0.5
0.4
A(0,025; 0,39)
0.3 0.02
0.03
0.04 3
L
0.05
0.06
0.07
3
[m ]
5: (T 2) ( L3) A B k . : ∆k ∆ yA + ∆yB 0, 01 + 0, 01 ≈ = = 0,04 [1] − − k yA yB 0, 39 0, 89 :
k = (14,3 ± 0, 6)
s2 m3
.[2]
: E = ( 2,5 ± 0, 3) ⋅109 Pa . [2] 6) ξ , η . ξ = A cos ω t η = ε A sin ω t . [3]
= τ / E y = σ . x =
ξ − η 2
y =
≈ y + 2
ξ + η 2
ε
[3]. ε 1
y& [3] ω ε F ≈ M ω 02 y + 2 y& [3] ω 2 && + 2εω0 y & + ω 0 y = 0 [3] y .
7) A(t ) = A0 e− β t [1], β = εω 0 . n- ( tn = nT = 2π n / ω 0 ) An = A 0 e −2πε n [2] An n
ln
n
= ln A0 − ( 2πε ) n [2]
ε . 8) 6 . 3 . 2. . 2. n. A0 = 85 mm. [15] n 2 4 6 8 10 12 14 16 18 20
A [mm] 76 68.5 63.5 59 53 49 47 44 39 36
ln( A) 4.33073 4.22683 4.15104 4.07754 3.97029 3.89182 3.85015 3.78419 3.66356 3.58352
6. [8] 4.4 4.3 4.2 4.1
)
4.0
A
( 3.9 n l 3.8 3.7 3.6 3.5 0
2
4
6
8
10
12
14
16
18
20
22
n
k ' = −2πε -0,0402, ε = 0,0064 . [2]
2009/2010. 28.05.2010. 1. o , . , ( , ). ( ). . () : ) = 3 , ) . → ∞ . 1 3. σ = Z Φ1 L 1
IV
I 1 I 2 Φ1 = LI1 + MI 2 . P : L1ω = L2ω = 2 R 0 , M ω = 1 / Cω = R 0 , R = 3R 0
= 0cosωt
4. 1V : ) , ) 100keV . 1800 ; 0,511 MeV. 5.
I 0 m
, : I 0 - , m - . : , , , , : , . , . : , , : , , : . , ,
2009/2010.
IV
28.05.2010.
1. ) x . k - ( ) T k ( k k − 1 )
T k +1 ( k k + 1 ). k - ( k = 1,... ) mak = mg + Tk +1 − Tk
(1)
m , ak . () T k +1 = 0 ( ) (1) 2 ( ak T k ). . t k ak Sk = ak t 2 / 2 . vk = ak t , , ω k ω k R = vk − vk −1 . : S k (. ) T k +1 T k S k −1 . ( ) 1 2
mvk + 2
1 2
I ω k = mgS k + Tk +1S k − Tk S k −1 (2) 2
I = mR . Tk = m(ak − ak −1 ) k = 1,... (3) 2
(1) (3) 2 2 . (3) (1) . (4) − ak −1 + 3ak − ak +1 = g k = 1,... , a0 = 0 ( ), (3) a ( k = 1, 2,3 ) (4) 3a1 − a2 = g
+1
= a . = 3
−a1 + 3a2 − a3 = g − a2 + 2a3 = g
a1 =
8 13
g a2 =
11 13
g a3 =
12 13
g .
) , (4) . a1 = λ g λ ( ) . a1 , , g ' = g − a1 = (1 − λ ) g ( “ “ g ' = g − a1 ). ( ) . a1 ' = λ g ' . a2 = a1 '+ a1 = (2 − λ )λ g . (4) . k = 1 λ 2 + λ − 1 = 0
λ =
5 −1 2
(
). a1 =
5 −1 2
g .
: ( ),
= 3 0,5%. = 10 10−5 , . T k :
. Tk Rd ϕ S k . Tk S k . S k −1 Rd ϕ A = Tk Rd ϕ − Tk S k = −Tk S k −1 . : ( ), ( ) . 3. , 1. : ��� I
= I 3 + I 4
��� I 3
= I 2 + I C
��� I 4
= I1 + I R
= 0cosωt 2. : L1 I&1 + MI&2 = E0 cos ω t L2 I&2 + MI&1 = E0 cos ω t RI R = E0 cos ω t q / C = E0 cos ω t q& = I C . () ω . . E = E0e
iω t
,
iα . I 1 I1 = I10 eiω t . I10 = I10 e 1
, α 1 I 1 . I R I R 0 = E0 / R . I C C q / C = E0 cos ω t . I C = q& = CE& (t ) , I C = iCω E . : iω L1 I10 + iω MI 20 = E0 iω MI10 + iω L2 I 20 = E0 : I10 =
L2 − M iω ( L1 L2 − M
2
)
E0 , I 20 =
L1 − M iω ( L1L2 − M ) 2
E 0 .
I = I1 + I 2 + I C + I R , ,
L + L − 2M 1 1 2 I 0 = + iωC + E0 = σ E , 2 R iω ( L1L2 − M ) 1 L1 + L2 − 2M . σ = + i ω C − R ω ( L1 L2 − M 2 )
L1ω = L2ω = 2 R 0 , M ω = 1 / Cω = R 0 , R = 3R 0 σ = I =
2 3 R0
3 R0
, Z =
E0 cos (ωt + π / 4 ) , P (t ) = I (t ) E (t ) =
P =
P =
1− i
E 02 6 R0
1
2 3 R0
3 R0 2
e −iπ /4 .
E02 cos ωt cos (ωt + π / 4) .
T
∫ P(t )dt .
T 0
.
4.
hω 1 ,
hω 2 . E 1, p1 E 2, p2
( 1). 2
E 2 − p 2c 2 = ( m0c 2 ) ,
m0
(1)
,
100
keV
E1 = m0 c 2 + T = 0,611MeV . p1 c = 0,335 MeV , . . .
a)
)
ω1/c
ω1/c
1=0
1
ω2/c
2
ω2/c
2
: hω1 hω − p1 = p2 − 2 , c c hω1 + E1 = hω 2 + E 2 .
(2) (3)
) ) : E2 = h(ω1 − ω 2 ) + E 1 p2c = h(ω1 + ω 2 ) − p1c .
hω 2 .
(4)
: 4hω1hω2 = 2hω1 ( E1 + p1c ) − 2hω 2 (E1 − p1c ) . (5) 2h 2ω1ω 2 : E1 + p1c hω2
= 2+
E1 − p1c hω 1
.
(6)
) p1 = 0 ,
E1 = m0 c 2 = 0,511MeV (6) E1 hω2
= 2+
E 1 hω 1
(7)
1 hω2
=
2 E 1
+
1 hω 1
.
(8)
hω 2 = 0,204MeV . ) 100keV E1 = m0c 2 + T = 0,611MeV
p1c = 0,335MeV . (6) hω 2 = 0,416MeV . 5. , , . (3 , 2 70° , 1,5cm) (1) I 0 + ml 2
T = 2π I 0 m
=
mgl 2
T gl 4π
2
,
− l 2 (2)
(2) t [s] 1 8.16 2 8.18 3 8.24 t s = 8.193 s , t − t s
max
= 0.047 s , t = (8.19 ± 0.5) s , T = (0.819 ± 0.05) s (2)
l = (0.187 ± 0.001) m (2) I 0 m
=
T 2 gl 4π
2
− l 2 =
0.819 2 ⋅ 9.81 ⋅ 0.187 4π
2
m 2 − 0.187 2 m 2 ≈ 0.0308m 2 (4)
I T 2 gl 2∆T ∆l 2 ⋅ 0.0047 0.001 2 2 l l 0 . 031 ∆ 0 = + + ∆ = + m + 2 ⋅ 0.187 ⋅ 0.001m 2 ≈ 8.9 ⋅ 10 − 4 m 2 (2) 2 0.187 l 0.8193 m 4π T I 0 m I 0 m
= (0.0308 ± 0.0009)m 2 = (3.08 ± 0.09) ⋅ 10 −2 m 2 = (0.031 ± 0.001)m 2 = (3.1 ± 0.1) ⋅ 10 − 2 m 2 (1)
4. 2009/2010.
41. : , , , .
5. Српска физичка олимпијада, Београд 2011.
5. 2010/11
-
4. 5. 2011.
1. m α = 45o (). >> m . µ , . . (25 )
2. , , O 2 . O 2 O . 1 2 (ρ ,p ) , ρ - , p - . ρ/ρ 0 p / p 0 , ρ 0 p 0 . O 2 . , (O 2 ) (O ) . . . k max = T 2 max / T 1 max = 5,0 . ) ( ) . ) k min . p = p O + p O2 . (25 )
3. . P 1 . P 2. R P2. R, r , ε , L1 L2 , . (20 )
P2
P1 L1 L2
ε, r
R
4. . ( ) . - , , , (Maurice Paul Auguste Charles Fabry 1867- 1945), , (Jean-Baptiste Alfred Perot 1863- 1925). . : , (), , . . - . - , , . ) , λ, α n d . φ (1) (2) – 1. (1)
(2)
α n>1
α Μ
Ν d
Ο 1
) , I 0, , ( ). ,
( A
2
R
/ A0 ) = R .
I T () φ ( ) R. 2 . – . (: = kA0eiδ , δ ( ) , k 0 ;
I =
cε 0
2
A =
cε 0
AA∗ ,
2 2 ix . : e = sin x + i cosx ) ) I R φ R. ) ) () . ? ? ) ∆ν ( ). ? ) . . ) 30cm He-Ne 632,8nm ∆ν . n=1. ) δν ( ). . ( IT = I 0 / 2 - 2.)
. .
2
(α=0) R
2
= ( n − 1) / ( n + 1)
2
; R α, R≈const .
I 0
T
I 0/2
δν
ν ’1 /2 ν ’’ 1/2
; ν ν
2 ) ) δν 98%. . . ) - ∆ν/δν ∆ν δν. , . () . ) ) . -Ne )? (30 )
: , , : : . , ,
5. 2010/11
04. 05. 2011.
1. , . mA x = 1 sin α − Ftr (2) A x , 1 , Ftr = µ 2 (2) 2 = 1 cos α + mg (2). Ma x = − 1 sin α (2) Ma y = Mg − 1 cos α (2) a x a .
∆ (3) ∆ x ∆ y ∆ X − ∆x , ∆ X . tg 45 o =1 , a = A x − ax (3). tg α =
1 = M >> m 1 ≈ 2mg
(1 + µ ) (1 − µ )
2 Mmg (1 + µ ) 2m + M (1 − µ )
(3)
(1).
2. ν O 2 ,
.
O 2 ν O 2 = (1 − α ) ν (1 ), O ν O = 2αν (1 ). :
p O 2 V = ν O2 RT = (1 − α ) νRT , (1 )
p OV = ν O RT = 2αν RT . (1 )
p = p O + p O2 , pV = (1 + α ) ν RT =
m M O 2
RT (1 + α ) , (1 )
M O 2 - O 2 . p =
ρ M O 2
p RT 2 = (1 + α ) , (1 ) ρ M 2 O
RT (1 + α ) ,
2
2 . p RT 1 = . (1 ) ρ 1 M O2 p / ρ T . 1 2, M O2 1 M O2 ρ ρ 1 , (2 ) . (2 ) = = ( ) p R T p R 1 α T + 1,max 2,max 1, max 2, max ) (ρ / p )1, max (ρ / p )2,max 1 2, . (1 ) . r max
ρ/ρ 0 ρ 5,5 p / p 12 0 1, max p 1,max = = 2 = = 6 . (3 ) r max = 5,5 2 ρ ρ/ρ 0 p 2,max p / p0 2,max 12 M O ρ 1 R T 1,max T p 1,max r max = = = (1 + α ) 2,max = (1 + α )k max , (1 ) M O 1 T 1, max ρ p 2,max R(1 + α ) T 2,max 2
2
6 = (1 + α )k max , α=
6 k max
−1 =
6 5
− 1 = 0,2. (2 )
) r min (1 ) :
ρ/ρ 0 ρ 6 11 p 1,min p / p0 1,min = = 1 = = 11 . (3 ) r min = 6 ρ ρ/ρ 0 1 p 2,min p / p0 2,min 11 M O ρ 1 R T 1,min T p 1,min r min = = = (1 + α ) 2,min = (1 + α )k min , (1 ) M O 1 T 1,min ρ p 2,min R(1 + α ) T 2,min 2
2
11 = (1 + α )k min , k min =
11
(1 + α )
=
11 1,2
≈ 9,2. (2 )
3. L1 I 10 = P 2
L2 I 20 =
ε r
ε R + r
(2).
(1), L1 I 10 ,
L2 . R ∆q = I 1 ∆t (1), ∆ I ∆ I 2 . (3), I 1 R + L1 1 = L 2 R∆q + L1 ∆ I 1 = L 2 ∆I 2 .(1) ∆t ∆t ∑ ∆q = q (3),
∑ ∆ I 1 = − I 10 = −
q=
ε
R + r
(3)
∑ ∆ I 2 = I 20 =
ε
r
(3),
ε L1
L + 2 (3) R R + r r
4. ) 1 (1) (2)
∆ s = n(MO + ON) − MN sin α = 2nd / cos β − 2d ⋅ tgβ sinα , (1) sin α = n sin β ∆ s = 2nd cos β = 2nd 1− sin 2 β = 2d n 2 − sin2 α .(1) 2π ∆ s 2π nd 2π d 2 ⋅ = φ= 1 − sin 2 β = n − sin 2 α . (1) λ 2 λ λ ) , . I 0 , I R = RI 0 , IT
= I0 − I R = (1 − R) I0 = TI 0 ; R ,
R + T = 1 . 0 , t = rA0 , t = tA0 r =
R t = T = 1 − R .
Φ = 0
α ��� ���
t Φ = 0
r 4 t e i 4
r 2 t e i 2
r 6 t e i 6
Φ
Φ
Φ
� t e i
Φ
r 2 t e i 3
Φ
Φ
Φ
r 2 t 2 e i 3
t 2 e i Φ
r 6 t e i 7
r 4 t e i 5
r 6 t 2 e i
r 4 t 2 e i 5
Φ
7 Φ
Φ
∞
AT = A0 (t 2 e iφ + t 2 r 2ei 3φ + t 2 r 4 ei 5φ + .... + t 2 eiφ ( rei 2φ )q + .... = t 2 ei φ ∑ ( r 2 ei 2φ )q . (3) q =0
i 2φ re < 1
∞
∑ x = q
q =0 2
AT = A0
t e
1 1 − x
,
= re i 2φ . :
iφ
. (2) 2 i 2φ 1− r e I T () :
I T = I 0 I 0 =
cε 0 2
t 4 1 + r 4 − 2r 2 cos 2φ
, (1)
2 2 2 A02 . cos 2φ = cos φ − sin φ , r 2 = R t = T = 1 − R , :
I T = I 0
(1 − R )2
. (2) (1 − R )2 + 4R sin 2 φ ) I 0 . , 4 R sin 2 φ I R = I 0 . (1) (1 − R ) 2 + 4 R sin 2 φ )
(1 − R )2 (1 − R ) 2 + 4 R sin 2 φ
,
sin φ = 0 , φ = mπ (1). IT = I 0 I R ) φ = ν
(
φm = mπ )
2π d
= 0 (0,5 +0,5).
n 2 − sin 2 α (1),
c
ν m
νm
2π d
n 2 − sin 2 α = mπ (1). ∆ν
c ( ) c ∆ν = ν m +1 −ν m = , (1) 2d n2 − sin 2 α (α=0) c ∆ν α = 0 = . (1) 2nd ) λ ′ λ m′ d , d = m m (1). λ m′ 2 v c = ν m = (1), v = c / n λm′ nλ m′
n , ∆ν = ν m +1 −ν m =
c
(1). 2nd () . m 3 ⋅108 s ) ∆ν = 2 ⋅ 1 ⋅ 0, 3m = 500MHz (0,5)
He-Ne . He-Ne ν He− e
∆ν ν He − e
= 474,0834THz ,
= 1,055 ⋅ 10−6 (0,5).
) 2 I 0 (1 − R) = I 0 , (1) 2 (1 − R) 2 + 4 R sin 2 φ 1/2 (1 − R)2 + 4 R sin 2 φ 1/2 = 2(1 − R) 2 , (0,5)
sin φ 1/2 =
1 − R
. (0,5) 2 R 1 − R << R , φ 1/2 , φ 1/2 ≈
1 − R
. (1) 2 R ′ −ν 1/′ 2 . δν =ν1/2 ) 2π nd c φ = ν = . λ λ c ν = φ (0,5) , 2π nd c ��� δν = (φ1/′′2 − φ1/′ 2 ) ����п� 2π nd
φ1/′ 2
= mπ − φ 1/ 2 φ1/′′2 = mπ + φ 1/ 2 (0,5), (1) δν = δν =
c
⋅
1− R
c 2π nd
2φ 1/ 2 (0,5) ,
, (1) 2nd π R . . , , . ) m 3 ⋅ 108 s ⋅ 1 − 0, 98 = 3,217MHz . (0,5) δν = 2 ⋅1⋅ 0,3m π 0,98 He-Ne δν = 6, 786 ⋅ 10−9 . (0,5) ν He − e
) F =
∆ν δν
=
π R 1 − R
≈ 155 . (0,5 +0,5)
: , -e , 1MHz!
[ :)] :
-Ne :
T I
∆ν = 500Hz
∆ν = 500Hz
@ λHe-Ne=632,8nm νHe-Ne =474,0834THz
@ λHe-Ne=632,8nm νHe-Ne =474,0834THz
∆ν/ νHe-Ne=0,000001055
∆ν/ νHe-Ne=0,000001055
; ν ν δν = 3,2Hz
@ λHe-Ne=632,8nm νHe-Ne =474,0834THz δν/ νHe-Ne=0,00000000678
!!! δν ∆ν!!!
~ 1GHz≈0,001nm
T I
@ λHe-Ne=632,8nm νHe-Ne =474,0834THz He-Ne 1GHz, 2-3 . (400-700nm) ! 400
500
600
; λ [nm]
632,8
700
5. 2010/11
4. 5. 2011.
. , S S 0 : k =
S S 0
.
1. . 2. . D0 = 1mm
. : D1 = ( 89.0 ± 0.5) mm
.
: - - . . - - -
: . : . : .
5. 2010/11
4. 5. 2011.
S 1 v1 = Sv
ρ gh = ρ v 2 / 2 . (2)
v1 = − dh / dt , (2)
S 1 dh / dt = S 2 gh
h=−
, (2)
S g S 1
2
−
.
t + const = k
dh
=
h
S 0 g S 1
2
S S 1
2 g dt (3)
t + const . (8)
, . , . a = k
S 0
g
S 1
2
⇒
k = a
S 1
2
S 0 g
(3)
h2 − h1 .. − .. cm a= = = −1.91⋅10−3 (20 - ) t2 − t 1 .. − .. s ∆a a
=
∆
(
) ( h ) + ∆t
h2 + ∆
h2 − h1
k=a
S 1 S0
1
+ ∆t 1 .. + .. .. + .. = + = 0.01 (5) t2 − t 1 .. − .. .. − .. 2
8.9 = −1.91⋅ 10 g 0.1 2
2
−3
2 981
≈ 0.68 (15)
∆a ∆S1 ∆a ∆D 0.5 ∆k = k + + 2 1 = 0.682 0.012 + 2 = k ≈ 0.016 (5) S1 D1 89 a a k = 0.682 ± 0.016
: 20 - 6 , 1 20% : 15 - 6 , 1 20% 1, .
h [cm]
h
[ cm ]
9.5
3.08
9.0
3.00
8.5
2.92
8.0
2.83
7.5
2.74
7.0
2.65
6.5
2.55
6.0
2.45
5.5
2.35
5.0
2.24
4.5
2.12
4.0
2.00
t i [s] 37.24 40.04 38.52 79.86 84.24 77.74 124.74 124.72 125.08 169.74 170.00 169.72 214.74 213.88 212.86 263.68 263.44 258.72 318.08 310.04 304.24 366.76 366.32 356.74 422.96 410.66 413.74 458.00 481.32 473.82 548.92 535.86 532.22 612.94 615.04 604.32
t [s] 38.6
80.6
124.87
169.82
213.8
261.9
310.8
363.2
415.8
480.0
539.0
610.8
Dt [s]
2010/11
-
20.05.2011.
1. . , . F mg . m. µ . . (20) 2. R0
m . − q , + q . , ( ). , , R1 O , R1 < R0 . O . . ) q Q . Q . q Q ? ; ) τ R1 O . (20)
3. d =0,21nm. v =1,33 103 m/s. ) ? ) ? ) ? h = 6,62510-34Js, a 1,67510-27kg. (20) 4. 24 Na 1/2=14,9h. 23 m=5mg ) 1 µ ) 10h. A = 6,022 ⋅10 . (20)
5. . . (20)
: .
: , . ,
: . , ,
2010/11
20.05.2011.
1. , : ) ) . . , . F mg , F sin α 0 , µ F sin α 0 .
F cos α 0 = µ F sin α 0 , tgα 0 = 1/ µ . α < α 0 : , a1 =
µ m
=
µ F sin α m
α ≥ α 0 : : a2 =
F cos α m
, α , α π / 2 . α = α 0 , . . , F µ amax = m 1 + µ 2 2. ) . F1 , F2 F3 ( ).
F 1 = F 2 =
1
q
F3
2
4 πε 0 2 r 2
, (2 )
F1 F 3 =
1
q
F2
2
4 πε 0 4r 2
. (2 )
F (r ) O .
q 2 q 2 1 q 2 2 2 − 1 . 2 2 − 2 = F (r ) = 2 4 πε 0 2 r 4r 4 πε 0 r 4 1
(2 )
Q , (2 ), .
2 2 − 1 4 . (2 )
Q = q
, , a(t ) =
2 ⋅ r (t ) , r (t ) .
F (r ) ~ 1 / r 2 , R0 + R1 . O . (2 ) ) τ R1 O τ = T/ 2 , T . ( ) 2
3
T R0 + R1 R + R1 = T = T 0 0 2 2 T R R 0 0 0
3/ 2
, (2 )
T 0 m R0 . , ,
T 0 =
2π R0 v0
, (2 )
v0 R0 . F (r ) mv02 R0
= F (r ) =
q 2 2 2 − 1
1
4 πε 0 r 2
= 1 qQ 4πε r 2 . (2 ) 0
4
v0 =
1
⋅
qQ
4 πε 0 mR0
=q
1 4 πε 0
⋅
(2
) . (2 )
2 −1
4mR0
R1
T 0 =
2π q
4 πε 0 ⋅
4mR03
(2
)
2 −1
, (2 )
πε 0 m( R0 + R1 )
3
T = π
2
⋅
qQ
=
π q
m( R0 + R1 )
3
2ππ0 ⋅
(2
2 − 1)
. (3 )
3. ) v =1,33 10 m/s (Louis-Victor-Pierre-Raymond de Broglie 1892 – 1987) h = 2,97 ⋅10−10 m , λ = mv . , .
2d sin θ , θ . ( ) , 2d sin θ = nλ , (Sir William Lawrence Bragg 1890 – 1971) , . ) , , λ = n ⋅ 0, 708 sin θ = n 2d n 1. θ = arcsin 0, 707 ≈ 450 . ) , 3λ = n ⋅ 0, 236 , sin θ = n 2d 1 nmax = [ ]= 4 . 0,236 , θ = arcsin( n ⋅ 0, 236) ≈ 70, 70 .
4. ) ☺ . d = −λ dt d dt , λ , t . , ∆ = − 0
0 . ≈ 0, : − 0 = −λ 0 ∆t ,
0 − =
ln 2 T 1/2
0 ∆t .
0 =
= 24 g/mol Na, 24
0 − =
ln 2 m
m M
A = 1,255 ⋅1020 ,
A ∆t = 1, 62 ⋅109 .
T1/2 M ) = 0e − λ t , 0 − = 0 (1 − e
− λ t
) = 0 (1 − e
−
ln 2 T 1/ 2
t
) = 4, 65⋅ 1019 .
5. , : mg = ρ gV / = 2γ Rπ , γ =
ρ gV 1
. (3) 2π R . V = 2 ml : 130, 124, 128, 125, 134,
(5 3 1ml)
s= 128.2 . γ =
ρ gV 1 2π R
=
103 ⋅ 9.81⋅ 2 ⋅ 10−6
1
−4 2π ⋅ 4 ⋅10
128.2
≈ 0.0609
N m
(8)
∆V = 0.1 ml , ∆ = 5 (1 ∆V = 0.1 ml 0.2 ml) 5 N N N ∆V ∆ 0.1 ∆γ = γ + = + = ⋅ ≈ (3) 0.0609 0.0609 0.089 0.006 2 128.2 m m m V N N γ = ( 0.061 ± 0.006) = ( 6.1 ± 0.6) 10−2 . m m
5. 2010/2011.
.
.
.
1
.
. , .
73,5
100
91,5
376,96
2
.
. , .
83
83
73
328,61
3
77,8
76
65
316,46
4
.
. , .
95,5
28
77,5
299,45
5
.
73
69
58
263,05
6
.
. , .
80
17
75
259,85
7
.
. , .
67,9
17
76
257,91
8
. ". "
58,5
14
33
142,05
9
1..
28
36
24
133,12
1-4 8
