Opstinsko takmicenje iz fizike, 4. razred, 2011.

1. 2010/2011. .
IV 12.02.2009.
1. ( 96-2) 1 - 1, 2
∆t = 1 - 2, 2- 1=4,5 108m. '
V - , 2 ∆t' = 1
1. V. (15 )
2. Rs = 100 R
U = 100V. 0 1k .
R, ? (25 )
3. , r = 1 cm, .
λ = 200 nm?
Ai = 4,5 eV. (22 )
4. .
t 1.
n- ? . (15 )
5. m 1.
k . 1.
pV . ,
. (23 )
:
c = 3 108m/s
h = 6,63 10-34Js
e = 1,6 10-19C
ε0 = 8,85 10-12F/m
: , ,
: , ,
: . , ,
1

2. 2010/2011. .
IV 12.02.2009.
1.
t1′ = ( t1 − Vx1 c 2 ) 1 − V 2 c 2 (1 ) t2 = ( t2 − Vx2 c 2 )
′ 1 − V 2 c 2 (1 )
∆t ′ = t2 − t1′ =  ∆t − V ( x2 − x1 ) c 2  1 − V 2 c 2 (5 ).
′   (3 )
V V = 0 V = 12c 13 (3 ). , V = 2,8 10 m/s (2 ). 8
2. R I ( R) = U ( Rs + R ) (1 )
PS ( R) = I ( R ) Rs = U 2 Rs ( Rs + R ) 2 (3 )
2
PR ( R ) = I 2 ( R ) R = U 2 R ( Rs + R ) 2 (3 ), .
, R = 0 (4 ), PS max = U 2
Rs = 100W (3 ). R
dPR ( R ) dR = U 2
( Rs − R ) ( Rs + R ) = 0 (5 ),
3
R= R (3 ). PR max = U 2 4 Rs = 25W (3 ).
3.
mv 2 2 = hc l - Ai (3 ).
. q,
j = q 4peo r (3 ), E p = - ej = - eq 4pe0r (4 )
, E = m v 2 2 - eq 4pe0r (4 ).
.
. , ,
.
. (
q = 4pe0r hc l - Ai ) e = 1.9pC (8 ).
4. s = at 2 2 .
l . t1 = 2l a (2 ), (n - 1)-
tn- 1 = 2 (n - 1)l a = t 1 n - 1 (4 ), n- tn = 2nl a = t 1 n (4 ), n-
D t n = t n - t n - 1 = t1 ( n- )
n - 1 (5 ) .
5. pV (2 ). p
1 ∆U = nCV ∆T = 0 (4 )
3
p1 1
T3=T1. , 12=0 (2 ).
2-3
V = const
A = A23 = p2 (V3 − V2 ) = p1 (V3 − V1 ) k (2 ). 1
p1V1 = nRT1 (1 ) V1 = nRT1 p1 (1 ). 3 2 3
p2= p1/k
, p2V3 = nRT3 (1 ) p1V3 k = nRT1 (1 ), p = const
p2= p1/k T3=T1. V3 = k nRT1k p1 (1 ). V2 = V1 V3 V
A = nRT1 ( k − 1) k = mRT1 ( k − 1) Mk (4 ). I Q = ∆U + A ,
Q = mRT1 ( k − 1) Mk (4 ) ∆U = 0 .
2