Φυσική Κατεύθυνσης Απαντήσεις Θέματων Πανελληνίων 2004
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Transcript of Φυσική Κατεύθυνσης Απαντήσεις Θέματων Πανελληνίων 2004
-
3 2004 :
1 1 2 3 4 5. , , , , . 2 1. .
:
m1: 121
211 mm
mm +
=
m2: 121
12 mm
m2 +=
.
==+=+= 211211
21
11
21
2121 mm3m2mmmm
m2mmmm
31
mm
2
1 =
2. . . : 900 crit .
3. . -
fz
-
(1) ff SA1 +=
(2) ff SA1 =
=+=10fff
10fff SSS
)1(
)2(S
AA 21
=++101
201
==
1012
4. .
( = 0), . /4. :
====1
2
2
1
2
1
2
1
2
1
2
1
2
1
2
1kk
tt
km2
km2
tt
TT
tt
4T4T
tt
.tt1tt2
tt
kk2
tt
212
1
2
1
1
1
2
1 >>== 3 .
2 . t = T 2 t = 1 s 10
s 0,2T ==102
1T
. ,
/2 .
AO
L
24
(x = 0)
-
/4.
m 4,01,04
== L :
m 0,9L ===+=4
4,09L4
9L42
4L
. 4.
. m 05.0A21,0A4 == :
== t2,0
2x4,0
205,0yt2x2A2y
(S.I.)10t 5x0,05y = .
: 2mD
222 )A2(D21Dy
21m
21EUK
==+=+
=+= 22222mD )A2(mym m2
= 22222 y )A2(
= 22 y )A2( = 22-22 )103( )105(01
.m/s 0,4 =
4 . , :
(1) Rcm == . (2) Rcm ==
cm = = 8 m/s (1) : .1,08R rad/s 80 ===
. :
+=== cmcmxcm mmgTmwTmF (3) 1056T1056,01010T cmcm +=+=
-
wN
T
O
+
+x
ww y
- :
== 2)( mR52R
)2(mR
52 =
= cm)2(
1052
(4) 4T cm= (3) (4) :
.m/s 4561410564 2cmcmcmcm2
cm m/s 4 ===+=
. (4) : N. 16T)4(4T ==
:
== RdtdL
dtdL
)(
= 1,016dtdL
.22/sKgm 1,6dtdL =
.
30N = ,
m. 6x1,0230xR2Nx ===
. .
+=+= 22cm I21m
21KK
+= 222cm mR52
21m
21
)1(22
cm )R(m51m
21 +=
+= 2cm2cm)1(
m51m
21
-
(5) m107 2
cm = :
=xwWK
xmg m107m
107 2
o2cm =
656,010 8107
107 22
cm = = 162cm
.m/s 4cm = ( ).