4.skraćenje duljine
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Skraćenje (kontrakcija) Skraćenje (kontrakcija) duljine duljine
Transcript of 4.skraćenje duljine
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Skraćenje (kontrakcija) duljineSkraćenje (kontrakcija) duljine
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2
2
11
2
2
22
11c
v
vtx
c
v
vtx
2
2
1212
1
)(
c
v
ttvxx
2
2
1122
1c
v
vtxvtx
Lo
y’ S’
x’
L
y S
x
Lo = x2’ – x1’
x1’ x2’
2
2
12
1c
v
xx
2
2
1c
v
L
2
2
0 1c
vLL
x2x1
Lo – vlastita duljina
L < Lo
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S’
Lo
y’
x’
y
x
S
L = L0
Budući da je: y’ = y i z’ = z , dužina okomita na smjer gibanja se ne skraćuje.
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Z: Tijelo ima duljinu 20 m u sustavu S´. Ako se sustav S´, u odnosu na sustav S giba brzinom 0.8c, koliku duljinu tijelo ima u sustavu S?
2
2
0 1c
vLL
2
2)8.0(120
c
cmL
2
264.0120
c
cmL
36.020mL
mL 12
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Z:Kolikom se brzinom mora gibati raketa da se skrati za 20 % vlastite duljine?
L = 0,8L0
v = ?
2
2
0 1c
vLL
2
2
00 18,0c
vLL
2
2
164,0c
v
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64.012
2
c
v
36.02
2
c
v
6.0c
v
v = 0,6 c
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