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içindekiler
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5. Bölüm Değerlendirme Soruları (Çözümlü Test, Boşluk
1. Fiziğin Doğası .........................................................................7
Doldurma , Doğru - Yanlış Bulma , Uygulama Soruları,
2. Fiziğin Alt Dalları ...................................................................9
Cevaplı Tester ..................................................................... 164
3. Fiziğin Diğer Alanlarla İlişkisi ............................................17
II. Bölüm
4. Ölçme ......................................................................................19
1. Kuvvet .................................................................................. 187
5. Fiziksel Büyüklükler .............................................................20
2. Doğadaki Temel Kuvvetler ............................................... 189
6. Bilimsel Bilgi ..........................................................................28
3. Sürtünme Kuvveti ............................................................... 192
7. Fizikte Modelleme ve Matematik .......................................29
4. Newton’un Hareket Yasaları ............................................. 195
8. Ünite Değerlendirme Soruları (Çözümlü Test, Boşluk
5. Bölüm Değerlendirme Soruları (Çözümlü Test, Boşluk
Doldurma, Doğruyu - Yanlışı Bulma, Cevaplı Testler) ....32
Doldurma , Doğru - Yanlış Bulma , Uygulama Soruları,
Cevaplı Tester) .................................................................... 205
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I. Bölüm
1. İş ............................................................................................ 228
1. Maddenin Hâlleri .................................................................49
2. Enerji ................................................................................... 233
2. Kütlenin Ölçülmesi ...............................................................52
3. İş - Enerji İlişkisi ................................................................ 239
3. Hacmin Ölçülmesi ................................................................54
4. Sürtünme Kuvvetinin Yaptığı İş ....................................... 243
4. Çözümlü Test ........................................................................65
5. Enerji Dönüşümü ve Korunumu ..................................... 246
5. Özkütle ....................................................................................69
6. Güç........................................................................................ 251
6. Fiziksel ve Kimyasal Değişim ..............................................78
7. Verim .................................................................................... 253
7. Bölüm Değerlendirme Soruları (Çözümlü Test, Boşluk
8. Enerji Kaynakları ............................................................... 255
Doldurma, Doğru- Yanlış Bulma, Bulmaca Çözme ,
9. Enerji Tasarrufu ................................................................. 258
Uygulama Soruları, Cevaplı Test) ........................................80
10. Bölüm Değerlendirme Soruları (Çözümlü Test, Boşluk
II. Bölüm
Doldurma , Doğru - Yanlış Bulma , Uygulama Soruları,
1. Dayanıklılık ........................................................................ 100
Cevaplı Tester) .................................................................... 259
2. Canlılarda dayanıklılık ...................................................... 105
3. Bölüm Değerlendirme Soruları (Çözümlü Test
Uygulama Soruları) ............................................................ 107
h1ú7(,6,6,&$./,.*(1/(û0(
I. Bölüm
4. Sıvılarda Adezyon ve Kohezyon kuvveti ......................... 114
1. İç Enerji ............................................................................... 284
5. Yüzey Gerilimi .................................................................... 115
2. Sıcaklık ................................................................................ 284
6. Kılcallık ............................................................................... 116
3. Isı .......................................................................................... 287
7. Gazların Genel Özellikleri ................................................ 121
4. Bölüm Değerlendirme Soruları (Çözümlü Test, Boşluk
8. Plazma ................................................................................. 122
Doldurma , Doğru - Yanlış Bulma , Uygulama Soruları,
9. Bölüm Değerlendirme Soruları (Çözümlü Test,
Cevaplı Tester) .................................................................... 292
Boşluk Doldurma, Doğru - Yanlış Bulma, Bulmaca
II. Bölüm
Çözme, Uygulama Soruları, Cevaplı Testler) .................. 124
1. Hâl Değişimi ....................................................................... 306
2. Isı Aktarımı ......................................................................... 314
h1ú7(+$5(.(7
I. Bölüm
3. Isı Yalıtımı ve Enerji Tasarrufu ....................................... 317
4. Bölüm Değerlendirme Soruları (Çözümlü Test, Boşluk
1. Hareket ................................................................................ 141
Doldurma , Doğru - Yanlış Bulma , Uygulama Soruları,
2. Düzgün Doğrusal Hareket ................................................ 151
Cevaplı Tester) .................................................................... 318
3. Düzgün Hızlanan Hareket ................................................ 156
4. Düzgün Yavaşlayan Hareket ............................................. 159
1.
Bölüm
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VǦSR
:Û]Û
VǦSR
2H[Û
:Û]Û
2H[ÛJPZPT
Örnek
JT
Su
JT
Su
2
0
ǦsPUKL JT su buS\UHU RHIH 2 JPZTP IÛYHRÛSKÛȘÛUKH Z\`\U ZL]P`LZP cmK
aL`PULsÛRÛ`VY
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OHJTPRHsJT[
Y&
00
Çözüm
VCisim$=Son – VǦSR
VCisim $¶$JT
Örnek
)PSNP
)
[
U THKKLSLYPU [HULJPRSLYP
HYHZÛUKH T\[SHRH IVȴS\R I\S\U-
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aL`PJTK
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R\T\UOHJTPRHsJT[
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kum
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ULU OHJPTSL NLYsLROHJPTHYHZÛUKHMHYR]HYKÛY)\
ULKLUSL OHJPT N
]LUPSPY IPY Ss
KLȘPSKPY
Çözüm
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ULUOHJPTZVU
OHJPTNLYsLROHJPTsÛRHYÛSÛYZHIVȴS\Ș\UOHJTPI\S\U\Y
54
)ņ=ņ.
NRQXDQODW×PO×
V.Y
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V)VȴS\R$=.Y
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V)VȴS\R$¶$JT[
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Y
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aN
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ǦȴSLTSLYKLZÛRsHR\SSHUKÛȘÛTÛaIHaÛNLVTL[YPRJPZPTSLYPUOHJPTMVYT
S
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2
YL
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r
V=
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rr
3
Küp
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W
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V=a
a
a
3
a
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r
2
V = r r .h
h
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)V`\[SHYÛHIJVSHUKPRKY[NLUSLYWYPaTHZÛUÛUOHJTP!
b
c
V = a .b . c
a
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V=
h
r
r r2 . h
3
55
6×Q×I
2.
h1ņ7(
)ņ=ņ.
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NRQXDQODW×PO×
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T»L,ȴP[+LȘLYSLY
2PSVTL[YLR
WRT3)
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/LR[VTL[YLR
WOT3)
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HY[HY]L`HHaHSÛY
T
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KT
:HU[PTL[YLR
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6 cm
4PSPTL[YLR
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9 mm
Örnek
@HYÛsHWÛY$JTVSHUIPYKLTPYIPS`LLYP[PSPW`HYÛsHWÛY2$JTVSHUIPS`LSLY
`HWÛSTHRPZ[LUP`VY)\UHNYLRHs[HULIPS`L`HWÛSHIPSPY& (r = 3)
Çözüm
r`HYÛsHWSÛIPS`LUPUOHJTP!
Y2`HYÛsHWSÛIPYIPS`LUPUOHJTP!
V1 =
4 3
r .r
3 1
4
V1 = $ 3 $ 2 3
3
V2 =
V1 = 32 cm 3 )PS`LZH`ÛZÛUVSZ\U
4
.r.r 32
3
4
3
3
V2 = $ 3 $ 1 = 4 cm
3
V$U=2
56
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)ņ=ņ.
NRQXDQODW×PO×
Örnek
@HYÛsHWÛ Y VSHU R
YL`SL`
RZLRSPȘPO$JT
6
r
O$JT
[HIHU`HYÛsHWÛYVSHURVr
ninin hacimleri birbirine
LȴP[[PY
)\UH NYL Y KLȘLYP
RHsJT»KPY&
Çözüm
VKüre = VKoni
2
r$r h
4
r $ r3 =
3
3
4r = h
4r = 8
r = 2 cm
Örnek
Y$JT
J$JT
I$JT
H$JT
@HYÛT R
YL ȴLRSPUKLRP Y $ JT `HYÛsHWSÛ PsP Z\ KVS\ RHWSH IV`\[SHYÛ
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RLaZ\RVU\S\YZHKPRKY[NLUSLYWYPaTHZÛȴLRSPUKLRPRHW[HTVSHYHRKVSK\Y\SHIPSPY& (r = 3)
Çözüm
@HYÛTR
YLUPUOHJTP!
+PRKY[NLUSLYWYPaTHUÛUOHJTP!
4 3
rr
2 3
VK = 3
= rr
2
3
VP = a . b . c
VK =
Vp = 6 $ 5 $ 4 = 120 cm
3
2
3
3
$ 3 $ 1 = 2 cm
3
+VSK\YTHZH`ÛZÛUVSZ\U
n=
VP 120
=
= 60 kez
VK
2
57
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NRQXDQODW×PO×
Örnek
)PYRLUHYÛJTVSHUR
W
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Y&
Çözüm
2
W
UOHJTP!
T$6 cm[
Y
=$H
+VȘY\VYHU[ÛR\YHSÛT
=$
6
3
10 cm
=$JT
1000 cm 3
x=
3
1m
x
1000
= 10 3 m 3
10 6
Örnek
JT
JTK
aL`PULRHKHYZ\PSL
KVS\SsLRSPRHIHKY[HKL[aKLȴIPS`LIÛYHRÛSÛUJHRHW[HU
cmZ\[HȴÛ`VY)\IPS`LSLYKLU
IPYP HSÛUHYHR [HIHU `HYÛsHWÛ
Y $ JT VSHU IPY ZPSPUKPY `HWÛSÛYZHZPSPUKPYPU`
RZLRSPȘPRHs
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Su
JT
Su
Çözüm
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VT$=b
$=b
Vb$JT
:PSPUKPYPUOHJTP!
V = r.r 2 h
16 = 3 . 2 2 . h
h=
58
4
cm
3
)ņ=ņ.
NRQXDQODW×PO×
Örnek
)V`\[SHYÛH$JTI$JTJ$JTVSHUKPRKY[NLUSLYWYPaTHZÛ
ȴLRSPUKLRPR\[\`HIPYRLUHYÛ , $JTVSHUR
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2
WȴLRLY
J$JT
JT
H$JT
$
I
, $JT
HRLUHYÛUH`LYSLȴ[PYPSLJLRȴLRLYZH`ÛZÛ
na =
a 20
=
= 10 a det
2
,
IRLUHYÛUH`LYSLȴ[PYPSLJLRȴLRLYZH`ÛZÛ
nb =
b 12
=
= 6 a det
,
,
JRLUHYÛUH`LYSLȴ[PYPSLJLRȴLRLYZH`ÛZÛ
nc =
c 10
=
= 5 a det
2
,
2\[\`HRVU\SHU[VWSHTȴLRLYZH`ÛZÛ
5$UaUbUcVS\Y
5$$HKL[
Örnek
:PSPUKPYȴLRSPUKLRPV`\UOHT\Y\[HIHU`HYÛsHWÛPSRK\Y\TKHRPUPU`HYÛZÛVSHJHRȴLRPSKLIPYZPSPUKPYLKU
ȴ[
Y
S
YZL`
RZLRSPȘPPSR`
RZLRSPȘPUPURHsRH[ÛVS\Y&
Çözüm
(`UÛV`\UOHT\Y\R\SSHUÛSKÛȘÛUHNYLPSR]LZVUOHJPTSLYLȴP[[PY
ri = 2r
rs = r
V i = VS
4 olsun.
2
2
rr i .h i = r r S . h S
2
2
4r .h i = r .h S (
hS
= 4 olur.
hi
59
6×Q×I
2.
h1ņ7(
)ņ=ņ.
0$''(9(g=(//ņ./(5ņ
NRQXDQODW×PO×
:Û]Û4HKKLSLYPU/HJPTkSs
T
:Û]ÛSHYÛUILSPYSPIPYȴLRSP`VR[\Y:Û]ÛSHYI\S\UK\RSHYÛRHIÛUȴLRSPUPHSKÛRSHYÛUKHU
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:Û]Û
JT
JT
JT
JT
kYULȘPU"ȴLRPSKLRPSsLRSPRHIHRVU\SHUZÛ]ÛUÛU
NLSKPȘPK
aL`LIHRHYHROHJTPUPUJTVSK\Ș\UH
RHYHY]LYLIPSPYPa
)PY RHIH RVU\SHU MHYRSÛ ZÛ]ÛSHYÛU H`YÛ H`YÛ OHJPTSLYP [VWSHTÛ ZÛ]ÛSHYÛU IPYIPYPUL RHYÛȴTHZÛ K\Y\T\UKHRP [VWSHT OHJTL LȴP[ KLȘPSKPY )\ K\Y\T
kSsLRSPRHW
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NZ[LYTLR[LKPY)\ULKLUSLOHJPTZÛ]ÛSHYPsPUKLN
]LUPSPYIPYSs
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Y:Û]ÛSHYIPYIPYPULRHYÛȴ[ÛYÛSKÛȘÛUKHǸLRPS00»KLOHJPTSLYP[VWSHTÛUÛU JTNLSKPȘPNY
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0
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RHYÛȴÛTÛ
00
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)PYPTP
3P[YL`L,ȴP[+LȘLYSLY
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10 ,
OLR[VSP[YLO , )
-2
10 ,
KLRHSP[YLKH , )
-1
10 ,
SP[YL
,
KLZPSP[YLK , )
10 d,
ZHU[PSP[YLJ , )
10 c,
TPSPSP[YLT , )
3
10 m ,
2
.HaSHYÛU/HJTP
Dikkat
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OHJTPULLȴP[[PY
60
2H[Û ]L ZÛ]ÛSHYKH VSK\Ș\ NPIP NHaSHYÛUKH
IPYOHJTP]HYKÛYkYULȘPU"IPYIHSVU\U]L`HIPZPRSL[[LRLYPUPUPsPOH]HNHaÛ`SHKVS\K\Y(UJHR
NHaSHYKH [HULJPRSLY HYHZÛ IVȴS\R sVR I
`
R
]L[HULJPRSLYIPYIPYPUKLUIHȘÛTZÛaKÛY.HaSHYÛU
OHJPTSLYP I\S\UK\RSHYÛ VY[HTÛU ZÛJHRSÛȘÛUKHU
]LIHZÛUJÛUKHUL[RPSLUPY:HIP[ZÛJHRSÛR]LIHZÛUs[HIPYRHIHRVU\SHUNHaÛUOHJTPI\S\UK\Ș\RHIÛUOHJTPULLȴP[[PY
)ņ=ņ.
NRQXDQODW×PO×
Örnek
;VWSHT`
aL`HSHUÛJT2VSHUIPYR
W
UPsPULRHsSP[YLZ\RVU\SHIPSPY&
)PSNP
Çözüm
a
@
aL`HSHUÛ
2
W
UOHJTP
6a2$
=$H
a2$
=$
H$JT
=$JT
a
a
)PYRLUHYÛHVSHUIPYR
W
U`
aL`HSHUÛH2KPY
JT$T , »KPY+VȘY\VYHU[ÛR\YHSÛT
1cm
3
512 cm
1 m,
3
x
4 ( x = 512 m, = 0, 512 L dir.
Örnek
@
RZLRSPȘPO[HIHU
r
2r
`HYÛsHWSHYÛY]LYVSHUZPSPUKPY
h
ȴLRSPUKLRPPRPRHWȴLRPSKLRPNPIP
IPSLȴPR[PY4\ZS\RHsÛSÛWKLUNL
Musluk
ZHȘSHUÛYZH
II
I
0RHW[HRPZÛ]ÛUÛU`
RZLRSPȘPPSR
K\Y\TKHRP`
RZLRSPȘPURHsRH[Û
VS\Y&
Çözüm
4S
x
h
(h – x
S
4x
Musluk
I
II
:PSPUKPYPRRHWSHYÛURLZP[SLYP:]L:VSZ\U:VU`
RZLRSPR
0RHW[H_`
RZLRSPȘPUKLZÛ]ÛIVȴHSÛYZH h-
h 4h
=
tir.
5
5
00RHW[HRP_RHKHY`
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6OoSKL
O¶_$_
x=
h
5
61
6×Q×I
2.
h1ņ7(
)ņ=ņ.
0$''(9(g=(//ņ./(5ņ
NRQXDQODW×PO×
Örnek
)PY RLUHYÛUÛU \a\US\Ș\ H $ JT VSHU R
W
Su
ǸLRLY
ȴLRSPUKLȴLRLYIPYRHIÛUPsPUKL`RLU
aLYPULJT
Z\ RVU\S\`VY ǸLRLY Z\KH sa
UK
Ș
UKL [VWSHT
hacim 92 cm VSK\Ș\UH NYL ȴLRLYKLRP IVȴS\Ș\U
OHJTPRHs[ÛY&
Çözüm
2
WȴLRLYPUNY
ULUOHJTP!
:\`\UOHJTP!
VSu$JT
=$H
V2$$JT
6STHZÛNLYLRLU[VWSHTOHJPTKLUNLYsLROHJPTsÛRHYÛSÛYZHIVȴS\Ș\UOHJTP
I\S\U\Y
V)VȴS\R$=.Y
ULU – V.LYsLR
V)VȴS\R$¶ V)VȴS\R$JT
+VȘY\VYHU[ÛR\YHSÛT
3
e ker de
64 cm 'ȴLRLYKL
3
'e ker de
100 cm ȴLRLYKL
x=
62
32.100
= % 50 dir.
64
32 cm
3
x
IVȴ\R]HYZH
IVȴS\R]HYKÛY
Çözümlü Test
(ȴHȘÛKHRPSLYKLUOHUNPZPTHKKLKLȘPSKPY?
(:\I\OHYÛ );VWYHR
*;Hȴ
+/H]H
,:LZ
JT
kum
kSsLRSPRHW[HJTK
aL`PULRHKHYR\Y\R\TI\S\UTHR[HKÛY2\T\U
aLYPULIPYTPR[HYZ\IVȴHS[ÛSÛUJH
Z\ZL]P`LZPJTK
aL`PULsÛRÛ`VY
2\T[HULJPRSLYPHYHZÛUKHRPIVȴS\RVSK\Ș\UH
NYLRVU\SHUZ\`\UOHJTPRHsJT3[
Y&
(
)
*
+
,
m2
m
m
ǸLRPSKLRPLȴP[RVSS\[LYHaPT1, m2, m3R
[SLSLYP`SL
KLUNLKLKPYT1R
[SLZPHSÛUÛW`LYPUL[HULT2 kom
U\SK\Ș\UKHKLUNLIVa\STHKÛȘÛUHNYL 1 VYHUÛ
m3
RHs[ÛY&
(
3
) 2
*
5
+ 2
,
:Û]Û
;HTHTLUZÛ]ÛKVS\RHIH=OHJPTSPIPYR
WIÛYHRÛSÛUJHR
WZÛ]Û`H[HTHTLUIH[Û`VY]LRHW[HUJT3
ZÛ]Û[HȴÛ`VY)\R
WSLYPR\SSHUHYHRIV`\[SHYÛJT
JTJTVSHUIPYWYPaTH`HWÛSÛYZHLUHaRHs
R
WR\SSHUÛSÛY&
(
/HJPT
-
Eylemsizlik
,ȴP[RVSS\[LYHaP2]L3JPZPTSLYP`SLKLUNLKLKPY
Binicinin kütlesi mI$NVSK\Ș\UHNYLIPUPJPRHsÛUJÛISTLKLKPY&
(
)
*
+
+
,
;HULJPRSP`HWÛ
3$ N
*
2
[SL
2$N
)
,
Renk
@\RHYÛKH ]LYPSLUSLYKLU OHUNPZP OLY THKKLKL I\S\UTHZÛNLYLRLUIPYaLSSPRKLȘPSKPY?
(2
[SL
)/HJPT
+;HULJPRSP`HWÛ
*,`SLTZPaSPR
,9LUR
63
6×Q×I
2.
Çözümlü Test
h1ņ7(
,ȴP[ Z
YLKL LȴP[ Z\ HRÛ[HU
T\ZS\RSHȴLRPSKLRPRHWKVSK\Y\S\`VY
r$JT
r2$JT
r
O$JT
Su
M
00
0
)\UHNYL"
2
[SL
/HJPT
Yükseklik
:PSPUKPY ȴLRSPUKLRP RHWSHY IPSLȴPR[PY 4\ZS\R HsÛSÛW
KLUNL ZHȘSHUÛYZH 0 ZPSPUKPYKLRP IVȴHSHU Z\`\U
OHJTPRHsJT3VS\Y& (r = 3)
(
0
Zaman
00
Zaman
Zaman
000
)
+
*
,
=LYPSLU NYHÄRSLYPUKLU OHUNPZP RV]H`Û KVSK\YHU
Z\`HHP[[PY&
(@HSUÛa0
)@HSUÛa00
+00]L000
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)PY R
W ȴLRSPUKLRP R\[\`H LU MHaSH HKL[ IPS`L
RVU\SHIPSP`VY)PS`LSLYKLUIPYPUPU`HYÛsHWÛY$JT
PZLR\[\U\UOHJTPRHsJT3[
Y&
(
)
*
+
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0 ]L 00 UVS\ ISTLSLYSL H`YÛSHU
K
aN
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IHYKHRSH KLMH Z\ RVU\S\UJH RHW KVSK\Y\SHIPSP`VY )HYKHȘÛU IV`\ O $ JT PZL [HIHU `HYÛsHWÛ
RHsJTKPY& (r = 3)
(
64
) 2 2 *
+ 4 2 ,
)\UHNYL2UVR[HZÛ
aLYPUKLRPZ\`
RZLRSPȘPOHUNPaHTHUHYHSÛRSHYÛUKHZHIP[[PY
(
)
*
+ ,
Çözümlü Test
h
J$JT
I$JT
2h
H$JT
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KPRKY[NLUSLYWYPaTHZÛȴLRSPUKLRPR\[\`H`HYÛsHWÛ
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(
)
*
+
2r
,
ǸLRPSKLRP RVUPUPU O `
RZLRSPȘPUKLRP [HYHSÛ RÛZTÛ
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(
1
2
)
r
2
3
*
3
4
+
5
6
2
,
7
8
L
L
M
0
00
h
2
M
000
@
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(
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*
+
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kaKLȴ[LYHaPSLYǸLRPS0]L00»KLKLUNLKLKPYǸLRPS
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(
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+
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6×Q×I
2.
ha
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;HȴHUZÛ]ÛOHJTPR
W
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2
W
V2
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m$T2TLȴP[SPȘP`HaÛSÛY
00KLUNLK\Y\T\PsPU
T2$T2TLȴP[SPȘP`HaÛSÛY
m1 = m2 + m3
2m 2 = m 3
4(
R
W
UIPYRLUHYÛ , ise
V = ,3
8 = , 3 ( , = 2 cm
m1 3
dir.
=
m3 2
I$JT
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JT
J$JT
;LYHaPUPUKLUNLK\Y\T\PsPU
m
K = L + c m .a LȴP[SPȘP`HaÛSHIPSPY
n
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1
m .a
10 = 9, 6 + c
10
H$ISTL
+VȘY\JL]HW!+
$
H
HIJRLUHYSHYÛUHZÛȘHJHRR
WZH`ÛZÛ
na =
a 10
=
= 5 tane
,
2
nb =
b 12
=
= 6 tane
2
,
nc =
c 14
=
= 7 tane
2
,
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2
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Örnek
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TN
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C
=JT Çözüm
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V
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kaR
[SL d =
dA =
m A 40
=
= 4 g/cm 3
VA
10
dB =
m B 20
=
= 1 g/cm 3
VB
20
dC =
m C 10
=
= 0, 5 g/cm 3
VC
20
71
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)ņ=ņ.
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NRQXDQODW×PO×
Örnek
)VȴIPYRHWZ\PSLKVS\`RLUN(ZÛ]ÛZÛ`SHKVS\`RLUNNLSP`VY2HIÛU
IVȴRLUR
[SLZPT2$NPZL(ZÛ]ÛZÛUÛUaR
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Y&
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Çözüm
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:Û]Û
Su
mk$N
T2TSu$N
T2T:Û]Û$N
TSu$
T:Û]Û$
mSu$N
T:Û]Û$ N
mSu$=SuKSu
m:Û]Û$=:Û]ÛK:Û]Û
$=Su
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VSu$JT
K:Û]Û$NJT
Örnek
@HYÛsHWÛYVSHUR
YL`SL[HIHU
`HYÛsHWÛY`
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r
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d=
m
( m = V.d V
2
[SLSLYLȴP[PZL
m2$TS
V2K2$=SKS
4 3
2
r.r .d K = r.r .h.d S
3
4
2
2
r . r .d K = r . r .2r.d S
3
4 d K = 6d S (
72
dK 3
=
dS 2
)ņ=ņ.
NRQXDQODW×PO×
Örnek
(]L)ZÛ]ÛSHYÛUÛUR
[SLOHJPTNYHÄȘPȴLRPS-
TN
KLRPNPIPKPY(»KHUN)»KLUNHSÛUHYHR
(
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Y&
B
=JT Çözüm
(»UÛUaR
[SLZP!
mA
30 3
dA =
( dA =
=
g/ cm 3
VA
20 2
VA =
mA
60
3
( VA =
= 40 cm
dA
3
2
)»UPUaR
[SLZP!
m
10 1
=
dB = B ( dB =
g/cm 3
VB
30 3
VB =
mB
100
3
( VB =
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dB
1
3
Toplam hacim
VT$=(=B$$JT[
Y
Örnek
(
B
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V
V
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2
2
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olur.
73
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[SLSLYPUPUHYP[TL[PRVY[HSHTHZÛUHLȴP[[PY
74
d kar =
m 1 + m 2 V1 .d 1 + V2 .d 2 V.d 1 + V.d 2
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=
V1 + V2
V1 .V2
V+V
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2HYÛȴÛTHNPYLUPRPZÛ]ÛLȴP[R
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m+m
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(
V1 + V2
m
m
m (d 1 + d 2)
+
d1 d2
d 1 .d 2
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MVYT
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d1 + d2
Örnek
2
[SLZPT$NOHJTP=$JTVSHUZÛ]Û`SHR
[SLZPT2$N
hacmi V2$JTVSHUZÛ]ÛRHYÛȴ[ÛYÛSÛYZHRHYÛȴÛTÛUaR
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2HYÛȴÛTÛUaR
[SLZP!
m + m2
200 + 300 500 5
d Kar = 1
( d Kar =
=
=
g/cm 3 tür.
50 + 150
200 2
V1 + V2
Örnek
(ZÛ]ÛZÛUKHUN)ZÛ]ÛZÛUKHUNHSÛUHYHR
TN
`HWÛSHURHYÛȴÛTÛUaR
[SLZPRHsNJT[
Y&
(
B
=JT Çözüm
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[SLZP!
m
60
dA = A =
= 6 g/cm 3
VA
10
)»UPUaR
[SLZP!
m
30 3
=
dB = B =
g/cm 3
VB
20 2
N(»UÛUOHJTP!
m
30
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VA = A =
dA
6
N)»UPUOHJTP!
m
60
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VB = B =
dB
3
2
2HYÛȴÛTÛUaR
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m + mB
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( d Kar =
=
= 2 g/cm 3 tür.
d Kar = A
5 + 40
45
VA + VB
75
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0$''(9(g=(//ņ./(5ņ
NRQXDQODW×PO×
Örnek
kaKLȴ(]L)T\ZS\RSHYÛ`SHIVȴRHWKVSK\Y\S\YZH
RHYÛȴÛTÛUaR
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K
K
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[SLZP
HYP[TL[PRVY[HSHTH`SHI\S\UHIPSPY
dK =
dA + dB
d + 5d
( d Kar =
= 3d olur.
2
2
Örnek
2S
NJT
S
h
NJT
S
2S
0
00
+
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Örnek
2
[SLOHJPTNYHÄȘPȴLRPSKLRPNPIPVSHU(]L
TN
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d1
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B
=JT Çözüm
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[SLZP!
dA =
mA
30
( dA =
= 3 g/cm 3
VA
10
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dB =
mB
15
( dB =
= 1 g/cm 3
VB
15
,ȴP[OHJPTSPRHYÛȴÛTÛUaR
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d1 =
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2
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olur.
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77
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Fotosentez
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Y
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Fotosentez
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Fisyon
79
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2.
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h1ņ7(
/HJTP=aR
[SLZPKVSHUZÛ]Û`SHOHJTP=VSHUZÛ]Û
RHYÛȴ[ÛYÛSKÛȘÛUKHRHYÛȴÛTÛUaR
[SLZPKVS\`VY
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[SLZPRHsKKPY&
(
1
2
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+
5
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,
2
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(
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+ ,
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ZÛ]ÛSHYÛUÛU R
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R
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(
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JT
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aL`PUL RHKHY R\T I\S\UTHR[HKÛY 2\T\U
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2
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2
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h1ņ7(
TN
()*ZÛ]ÛSHYÛIPYRHW[HRHYÛȴTHKHU
ȴLRPSKLRPNPIPK\YTHR[HKÛY
*:Û]ÛZÛ
(
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2m
):Û]ÛZÛ
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m
V
2V
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())*]L(*LȴP[OHJPTSPRHYÛȴÛTSHYÛUÛU
aR
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1
1
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+
,
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4
(
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+K2%K%K
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B
TN
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NJT
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(
)
*
+
,
ha
TSLY
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m + m2
dK = 1
dir.
V1 + V2
dK =
)Vȴ
V1. d 1 + V2 .d 2
V1 + V2
T$N
V.d + 2V.d 2
2d =
V + 2V
6 V d = V ( d + 2d 2 )
d2 =
5d
2
TTSu$N
TTSu$
TSu$
dir.
K
Su
TT:Û]Û$N
mSu$N
+VȘY\JL]HW!+
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dA =
)»UPUaR
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m A 60
=
= 20 cm 3
dA
3
K
d=
m
V
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m B = VB .d B = 40 $
2HYÛȴÛTKH(»UÛUOHJTP!
VA =
TT:Û]Û$
V:Û]ÛK:Û]Û$
m A 30
=
= 3 g/cm 3
VA
10
m
10 1
dB = B =
g/cm 3
=
VB
20 2
0
1
= 20 g
2
00
2HYÛȴÛTÛUaR
[SLZP!
m
m + mB
60 + 20 80
4
dK = A
g/cm 3
(
=
=
VA + VB
20 + 40 60 3
+VȘY\JL]HW!*
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[SL HY[TÛȴ HUJHR aR
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2
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2
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4
3
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3
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mK = mS
T$=K
4
3
2
r . r .3d = r . r .h. d
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h1ņ7(
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m
30 3
3
=
dK = K =
g/cm
VK
20 2
(»UÛUaR
[SLZP!
V)VȴS\R$=.Y
ULU – V.LYsLR
V)VȴS\R$¶$JT
2\T\UNLYsLROHJTP
V2\T$=.Y
ULU – V)VȴS\R
V2\T$¶$JT
:\`\UR
[SLZP!
T$=K
T$$N
2\T\UR
[SLZP!
dA =
m A 40
3
=
= 2 g/cm
VA
20
dK =
m A + m B VA .d A + VB .d B
=
VA + VB
VB + VB
3 2V.2 + 3V.d B
=
2
2V + 3V
3 V (4 + 3d B)
=
2
5V
7
3
g/cm
6
dB =
m2\T$TT – mSu
+VȘY\JL]HW!+
m2\T$¶$N
2\T\UaR
[SLZP
d Kum =
m Kum
VKum
d Kum =
16
8
d Kum = 2 g/cm 3
+VȘY\JL]HW!+
(
(
B
N
NZÛ]Û
=
K
=
K
:Û]Û
(»KHRPZÛ]ÛR
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2HW[HRPHȘÛYSHȴTHNPZL[HȴHUZÛ]ÛN»KÛY:Û]ÛUÛU
aR
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Y
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m
32
3
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2
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m($=K$=K
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V.d = V'.4d ( V' =
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84
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m
40 5
3
=
dc = c =
g/cm »[
Y
Vc
16 2
+VȘY\*L]HW!+
ha
TSLY
2
YLUPUR
[SLZP
m K = VK .d K
mK =
4
3
$r$r $d
3
mK =
4
3
$ 3 $ 2 $ 2 = 64 g
3
2
[SLHY[ÛȴÛ!
T$TT – m2
T$¶$N
2
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dK =
m A + m B VA .d A + VB .d B
=
VB + V A
VA + VB
dK =
3V.1, 2 + V.2, 8
= 1, 6 g/cm 3
4V
+VȘY\JL]HW!*
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dA =
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m
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V
dC =
m
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2V
(»UÛUaR
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m
40
dA = A =
= 4 g/cm 3
VA
10
2HYÛȴÛTÛUaR
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m
20
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= 1g/cm 3
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s
R[
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mA + mC
VA + VC
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VA .4d + VC .d
V
1
( A =
olur.
VA + VC
VC 2
KB#K2HYÛȴÛT#K(
#K2HYÛȴÛT#
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+VȘY\JL]HW!,
85
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0$''(9(g=(//ņ./(5ņ
NRQXDQODW×PO×
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2
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:Û]ÛSHYÛUILSPYSPIPY`VR[\Y
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@LY+LȘPȴ[PYTL!)PYJPZPTRVU\TKLȘPȴ[PYP`VYZHOHYLRL[LKP`VYKLTLR[PY*PZTPUZVURVU\T\`SHPSRRVU\T\HYHZÛUKHRP]LR[YLSMHYR`LYKLȘPȴ[PYTLZPULLȴP[[PY
(»UVR[HZÛUKHU)UVR[HZÛUHNPKLUOHYLRL[SPUPU`LYKLȘPȴ[PYTLZP
Tx = x 2 - x 1 KPY
ǸLRPS00»KLNY
SK
Ș
NPIP`LYKLȘPȴ[PYTLRH]YHTÛOHYLRL[PU`U
]LOHYLRL[SPUPUZVUI\S\UK\Ș\UVR[HOHRRÛUKHIPSNPPsLYKPȘPUKLUULTSPKPY
@LY KLȘPȴ[PYTL ]LR[YLS IPY I
`
RS
R VSK\Ș\UKHUPȴHYL[P]L`H¶VSHIPSPY
(SÛUHU `VS PSL `LY KLȘPȴ[PYTL HYHZÛUKHRP
MHYRÛKHOHP`PHUSH`HIPSTLRPsPU2UVR[HZÛUKHU
3UVR[HZÛUHNZ[LYPSLU`VSKHNPKLUOHYLRL[SP`L
KPRRH[LKPUPa
y
A
Tx
x1
)
2 UVR[HZÛUKHU 3 UVR[HZÛUH RLZPRSP sPaNPx2
SLYP [HRPW LKLYLR NPKLU OHYLRL[SPUPU HSKÛȘÛ `VS 6
ǸLRPS00
`Y
UNL \a\US\Ș\UH LȴP[[PY @LY KLȘPȴ[PYTLZP
PZLȴLRPSKLNZ[LYPSKPȘPNPIP23UVR[HSHYÛHYHZÛUKHRPLURÛZH\aHRSÛR[ÛY
x
+PRRH[
xT(SÛUHU`VS
(SÛUHU`VSZRHSLY"`LYKLȘPȴ[PYTL]LR[YLSI
`
RS
R[
Y
2
Tx
@LYKLȘPȴ[PYTL
L
145
6×Q×I
3.
h1ņ7(
)ņ=ņ.
HAREKET
NRQXDQODW×PO×
Örnek
A
)
C
¶
¶
¶
+
,
-
)HȴSHUNÛsRVU\T\UKHI\S\UHUIPYOHYLRL[SPKVȘY\ZHS`VSKHUJL-»`LKHOH
ZVUYH)»`LNPKP`VY
)\UHNYLOHYLRL[SPUPU"
H(SKÛȘÛ`VSRHsT»KPY&
I@LYKLȘPȴ[PYTLRHsT»KPY&
Çözüm
H_;VWSHT$_6-_-)
_;VWSHT$
_;VWSHT$T
I
Tx = x B - x O
Tx = - 20 - 0
Tx = - 20 m
Örnek
)
T
T
C
( UVR[HZÛUKH I\S\UHU IPY OHYLRL[SP
KVȘY\ZHS `VSKH UJL )»`L ZVUYH *»`L
T
KHOHZVUYH+»`LNLSP`VY
+
A
)\UHNYLOHYLRL[SPUPU"
H(SKÛȘÛ`VSRHsT»KPY&
I@LYKLȘPȴ[PYTLZPRHsT»KPY&
146
)ņ=ņ.
NRQXDQODW×PO×
Çözüm
H_;VWSHT$_()_)*_*+
_;VWSHT$$T
I
)
T
+
C
Tx
T
T
+
A
Tx
A
,
=
50
m
T
T
,
Tx = x AE + x ED
2
Tx 2 = x 2AE + x ED
2
2
2
Tx = 30 + 40
2
Tx = 2500
Tx = 50 m
SÜRAT
)PYUVR[HKHUKPȘLYPULNPKLUIPYOHYLRL[SP UL RHKHY RÛZH Z
YLKL PRPUJP UVR[H`H ]HYÛYZH Z
YH[P V RHKHY I
`
R KLTLR[PY )HȴRH IPY
KL`PȴSL `VS\U UL RHKHY Z
YLKL HSÛUKÛȘÛUÛU IPY
Ss
Z
K
Y(UJHRZ
YH[OHYLRL[PU`U
]LRVU\T\OHRRÛUKHIPSNPPsLYTLakYULȘPU¸ZHH[[L
RTO OÛaSH NPKLU HYHIH¹ PMHKLZPUKL HYHIHUÛUOHUNP`UKLNP[[PȘP`HKHOHUNPUVR[HSHY
HYHZÛUKH `LY KLȘPȴ[PYKPȘP HUSHȴÛSHTHa `HSUÛaJH
IPYZHH[SPRaHTHUKPSPTPUKLRTSPR`VS\HSHJHȘÛHUSHȴÛSÛY:
YH[POLZHWSHTHRPsPU
HSÛUHU`VS\]LNLsLUOHYLRL[Z
YLZPUPIPSTLR`L[LYSPKPY:
YH[ZRHSLYIPYI
`
RS
R[
Y
Sürat =
Alınan yol
Hareket süresi
:
YH[¸]¹ZLTIVS
`SLNZ[LYPSPY)PYPTPTZKPY
x $ $ Yol
x
1 ( v = dir.
t
t $ Süre
/Ûa!/HYLRL[PU`U
]LOHYLRL[SPUPUI\S\UK\Ș\RVU\TOHRRÛUKHIPSNPZHOPIP
VSTHRPZ[LKPȘPTPaKLOHYLRL[SPUPU`LYKLȘPȴ[PYTLZPULIHRHYÛa@LYKLȘPȴ[PYTLUPUOHYLRL[Z
YLZPULVYHUÛOÛaÛ]LYPY)HȴRHIPYKL`PȴSLOÛaIPYPTaHTHUKHRP`LYKLȘPȴ[PYTLKPY " j" PSLNZ[LYPSPY=LR[YLSIPYI
`
RS
R[
Y
/Ûa$
j=
@LYKLȘPȴ[PYTL
AHTHU
Yol
Süre
:
YH[
x
[
v
T
Z
TZ
:
YH[IPYPT[HISVZ\
+PRRH[
Yol
Zaman
@LYKLȘPȴ[PYTL
Sürat =
/Ûa$
AHTHU
Tx
x Son - x ilk
=
Tt
t Son - t ilk
147
6×Q×I
3.
h1ņ7(
)ņ=ņ.
HAREKET
NRQXDQODW×PO×
Örnek
)PYOHYLRL[SPKVȘY\ZHS`VSKHZHIP[OÛaSHUJLR\aL`LTKHOHZVUYHKVȘ\`HKVȘY\T`VS\[VWSHTZHUP`LKLNPKP`VY
)\UHNYLOHYLRL[SPUPU"
H:
YH[PRHsTZKPY
I/ÛaÛRHsTZKPY
Çözüm
H(SKÛȘÛ`VS! x = x AB + x BC
T
)
T
I@LYKLȘPȴ[PYTL! Tx = x AB + x BC
2
2
2
Tx = x AB + x BC
2
2
Tx = 15 + 20
2
Tx = 525
Tx = 25 m
Hız: j =
(USÛR OÛa VY[HSHTH OÛa RH]YHTSHYÛUÛ
HsÛRSHY]LYULR]LYPY
A
Tx
Tt
j=
2HaHUÛT
2
C
Tx
x = 15 + 20
x = 35m
x 35
= 7 m /s
Sürat: v = =
t
5
25
= 5m/s
5
6Y[HSHTH:
YH[!)PYOHYLRL[Z
YLZPUJLZ
YH[
KLȘPȴRLUVSHIPSPY)HȴRHIPYPMHKL`SLOHYLRL[PUMHYRSÛHUSHYÛUKHZ
YH[MHYRSÛKLȘLYSLYHSHIPSPYkYULȘPU
IPY H[ `HYÛȴÛUKH Z[HY[ ]LYPSPYRLU Z
YH[P ZÛMÛY VSHU H[
`HYÛȴIV`\UJHOÛaSHUÛY`H]HȴSHY]L`HZHIP[Z
YH[SL
OHYLRL[LKLIPSPY)\K\Y\TKHIHȴ[HUZVUHZHOPW
VS\UHUKLȘLYSLYPUVY[HSHTHZÛI\S\UHIPSPY6Y[HSHTH
Z
YH[HSÛUHU[VWSHT`VS\U[VWSHTOHYLRL[Z
YLZPULVYHUÛKÛY
6Y[HSHTHZ
YH[$
;VWSHT`VS
;VWSHTOHYLRL[Z
YLZP
v ort =
x Toplam
t Toplam
6Y[HSHTH/Ûa!)PYOHYLRL[SPPRPUVR[HHYHZÛUKHMHYRSÛOÛaSHYSHOHYLRL[L[[PȘPUKL
OÛaÛUIPYVY[HSHTHKLȘLYSLYPUKLUZaLKPSLIPSPY
;VWSHT`LYKLȘPȴ[PYTLUPU[VWSHTOHYLRL[Z
YLZPULVYHUÛVY[HSHTHOÛaÛ]LYPY
6Y[HSHTHOÛa$
j Ort =
148
Tx
dir.
Tt
;VWSHT`LYKLȘPȴ[PYTL
;VWSHTOHYLRL[Z
YLZP
)ņ=ņ.
NRQXDQODW×PO×
Örnek
+VȘY\ZHS `VSKH j 1 = 20 m/s OÛaSH ZHUP`L KHOH ZVUYH KH H`UÛ `UKL
j 2 = 5 m/s OÛaSHZHUP`LNPKLUHYHJÛU
H6Y[HSHTHZ
YH[PRHsTZ»KPY
I6Y[HSHTHOÛaÛRHsTZ»KPY
Çözüm
H :
YH[$
;VWSHT`VS
;VWSHTZ
YL
x 1 = j 1 .t 1 = 20.5 = 100 m
x 2 = j 2 .t 2 = 5.10 = 50 m
v=
x1 + x2
t1 + t2
v=
100 + 50 150
=
= 10 m/s'dir.
5 + 10
15
I+VȘY\ZHS`VSKHH`UÛ`UKLNPKLUHYHsSHYÛUHSKÛRSHYÛ`VS`LYKLȘPȴ[PYTLSLYPULLȴP[VSHJHȘÛUKHUVY[HSHTHZ
YH[SLYPVY[HSHTHOÛaSHYÛUHLȴP[VS\Y
j=
Tx
= 10 m/s'dir.
Tt
Örnek
)PY `VS\U `HYÛZÛUÛ j 1 = 100 km/h KPȘLY `HYÛZÛUÛ j 2 = 60 km/h OÛaSH NPKLU
HYHJÛUVY[HSHTHOÛaÛULKPY&
Çözüm
@VS\U`HYÛZÛ_VSZ\U
x
x
j 1 = 100 km/h
j 2 = 60 km/h
/HYLRL[Z
YLZP
x _bb
b
j1 b
2x
x+x
x+x
`( j =
(j=
=
( j = 75 km/h
x bbb
t1 + t2
x
x
x
x
t2 =
+
+
bb
j2
j1 j2
100 60
a
t1 =
149
6×Q×I
3.
)ņ=ņ.
HAREKET
h1ņ7(
NRQXDQODW×PO×
(UP/Ûa(UP:
YH[!)PYOHYLRL[SPUPUOLYOHUNPIPYHUKHRPOÛaÛUHHUSÛRHUP
OÛaKLUPY
+\]HY
)L`aIVS
;VW\
$ j = 10 m/s
ı
j = - 6 m/s
2
KPY
(UPOÛaI
`
RS
RVSHYHRHUPZ
YH[LLȴP[[PY(UJHROÛaÛU]LR[YLSIPYI
`
RS
RVSK\Ș\
`U
U
U ULTSP VSK\Ș\ KPRRH[L HSÛUTHSÛKÛY
kYULȘPU ȴLRPSKLRP [LUPZ [VW\ j = 10 m/s OÛaSH 2 OPaHZÛUKHU NLsPW K\]HYH sHYWHYHR
j ı = 6 m/s OÛaSHKUT
ȴVSZ\U
hHYWTHKHU UJL 2 OPaHZÛUKHRP HUSÛR
OÛaÛ j K = + 10 m/s Z
YH[P]2$TZ»KPY
(UJHRKULYRLU2OPaHZÛUKHRPHUSÛROÛaÛ j K = - 6m/s, HUSÛRZ
YH[PPZL]2$TZ
.Y
SK
Ș
NPIPOÛa`US
IPYI
`
RS
RVSK\Ș\UKHU¶PȴHYL[PHSTÛȴ[ÛY(UJHR
I
`
RS
RVSHYHR`PULVHUKHRPZ
YH[LLȴP[[PY
/Ûa+LȘPȴPTP
AHTHU
Ǧ]TL
6v
6[
a
TZ
Z
m
s2
Ǧ]TLIPYPT[HISVZ\
Ǧ]TL!/HYLRL[LKLUIPYJPZPTOÛaÛUÛKLȘPȴ[PYP`VYZH P]TLZP ]HY KLTLR[PY )\ KLȘPȴPT
ULRHKHYRÛZHZ
YLKLNLYsLRSLȴPYZLP]TLVRHKHYI
`
RVS\YǦ]TLIPYPTaHTHUKHRPOÛaKLȘPȴPTPVSHYHR[HUÛTSHUÛY¸H¹OHYÄ`SLNZ[LYPSPY
]L]LR[YLSIPYI
`
RS
R[
Y
a=
Tj
j Son - j ilk
[PY
=
t Son - t ilk
Tt
Örnek
+VȘY\ZHS`VSKHZHIP[TZOÛaSHNPKLUHYHsOÛaÛUÛK
aN
UHY[ÛYHYHRZHUP`LKLTZOÛaH\SHȴÛ`VY(YHJÛUP]TLZPRHsTZ2KPY
Çözüm
P]TL
+PRRH[
:HIP[OÛaSHOHYLRL[LKLUJP-
a=
Tj
j - j1
= 2
Tt
t2 - t1
a=
12 - 2 10
=
= 2m /s 2
5-0
5
ZPTSLYPUP]TLZPZÛMÛYKÛY
Örnek
+VȘY\ZHS`VSKHRTOOÛaSHNPKLUHYHsMYLULIHZÛUJH]LK
aN
U`H]HȴSH`HYHRZHUP`LKLK\Y\`VY(YHJÛUP]TLZPRHsTZ2KPY
Çözüm
1 km/h =
10
m/s
36
RTO$TZ
150
a=
Tj j 2 - j 1 0 - 10
=
=
= –2m/s 2
t2 - t1
5-0
Tt
)ņ=ņ.
NRQXDQODW×PO×
+
aN
U+VȘY\ZHS:HIP[/ÛaSÛ/HYLRL[
j = j
+PRRH[
j = j
j2 = j
j = j
Tx 1 = Tx 2 = Tx 3
[
Tx 1 = x
Tx 2 = x
[
Tx 2 = x
[2
[
+VȘY\ZHS`VS\UI
[
UUVR[HSHYÛUKHH`UÛOÛaSHNLsLUJPZTPUOHYLRL[PKPY*PZPT
LȴP[Z
YLKLLȴP[TPR[HY`VSHSÛY/ÛaKLȘPȴPTPVSTHKÛȘÛUKHUP]TLZPZÛMÛYKÛY
Bilgi
/Ûa AHTHU .YHÄȘP ! /Ûa aHTHUSH KLȘPȴTLKPȘPUKLU NYHÄR ȴLRPSKLRP NPIP
VS\Y.YHÄȘPUHS[ÛUKHRHSHUISNLUPUHSHUÛ`LYKLȘPȴ[PYTL`P]LYPY
Tx
/Ûa
[
[
Tx = T j.Tt
[
AHTHU
.YHÄȘPU HS[ÛUKH RHSHU ISNL
HSHUÛ`LYKLȘPȴ[PYTL`P]LYPY
2VU\TAHTHU.YHÄȘP
/HYLRL[SPLȴP[aHTHUHYHSÛRSHYÛUKHLȴP[TPR[HY`VSHSÛY/HYLRL[PURVU\TaHTHUNYHÄȘPȴLRPSKLRPNPIPKPY
2VU\T
x
Bilgi
2VU\T aHTHU NYHÄȘPUKL
LȘPTOÛaHLȴP[[PY
x
[
AHTHU
x
x
6x
x
_
6[
[Z
[
Ǧ]TLAHTHU.YHÄȘP
/Ûa KLȘPȴPTP VSTHKÛȘÛUKHU P]TL KHPTH ZÛMÛYKÛYǦ]TLaHTHUNYHÄȘPȴLRPSKLRPNPIPKPY
Ǧ]TL
[
Tx
Tt
x - x0
j=
dır.
t - t0
tan a = j =
AHTHU
151
6×Q×I
3.
h1ņ7(
)ņ=ņ.
HAREKET
NRQXDQODW×PO×
Örnek
2HaHUÛT
AHTHUZ
2
+
aN
U KVȘY\ZHS OHYLRL[ PsPU RV-
2VU\TT
2
6
U\T OÛa ]L aHTHU RH]YHTSHYÛUÛ
+VȘY\ZHS `VSKH RVU\T\U
aHTHUH IHȘSÛ KLȘPȴPTP
[HISVKHRPNPIPVSHUHYHJÛU"
H2VU\TaHTHUNYHÄȘPUPsPaPUPa
HsÛRSHY
I/ÛaaHTHUNYHÄȘPUPsPaPUPa
JǦ]TLaHTHUNYHÄȘPUPsPaPUPa
Çözüm
H;HISVKHRPKLȘLYSLY`LYPUL`HaÛSKÛȘÛUKH2VU\TaHTHUNYHÄȘPȴLRPSKLRPNPIPVS\Y
_T
6
2
2
[Z
I_[NYHÄȘPUKLLȘPTOÛaÛ]LYPY
TZ
aHTHUHYHSÛȘÛUKHRPOÛa!
4
x1 - x0
t1 - t0
6-2
= 4m/s
j=
1-0
j=
aHTHUHYHSÛȘÛUKHRPOÛa! j =
/ÛaKLȘPȴPTPZÛMÛYVSK\Ș\UKHUP]TLZÛMÛYKÛY
2
HTZ 152
[
[Z
x 3 - x 2 14 - 10
=
= 4m / s
t3 - t2
3-2
JǦ]TLaHTHUNYHÄȘP!
2
x 2 - x 1 10 - 6
=
= 4m/s
t2 - t1
2-1
AaHTHUHYHSÛȘÛUKHRPOÛa! j =
.YHÄRȴLRPSKLRPNPIPKPY
[Z
)ņ=ņ.
NRQXDQODW×PO×
Örnek
+PRRH[
j L = 4m/s
j K = 2m/s
+VȘY\ZHS `VSKH aÛ[ `UKL OHYLRL[ LKLU OHYLRL[SPSLYPU
HYHZÛUKHRP\aHRSÛR
x = 300 m
x = (j K + j L ) [ PMHKLZP`SL
I\S\U\Y
)
A
+VȘY\ZHS`VSKHZHIP[OÛaSHYSHOHYLRL[LKLU2]L3HYHsSHYÛH`UÛHUKH(]L
)UVR[HSHYÛUKHUNLsP`VY)\HYHsSHYRHsZZVUYHRHYȴÛSHȴÛY&
Çözüm
[Z
YLZVUYH*UVR[HZÛUKHRHYȴÛSHȴTÛȴVSZ\USHY
x = x AC + x BC
x = j K .t + j L .t
x = (j K + j L) .t
300 = (2 + 4) .t
t = 50 s
C
A
)
Örnek
, K = 100 m
, L = 200 m
j k = 6 m/s
A
j L = 4m / s
x = 300 m
)
2[YLUPT3[YLUPT\a\US\Ș\UKHKÛY+VȘY\ZHS`VSKH j K = 6 m/s j L = 4 m/s OÛaSHOHYLRL[LKLU2]L3[YLUSLYP(]L)KLUH`UHUKHNLs[PR[LURHs
ZHUP`LZVUYHHYRH\sSHYÛ`HU`HUHNLSPY&
Çözüm
x = (j K + j L) .t
600 = (6 + 4) . t
t = 60 s
153
6×Q×I
3.
h1ņ7(
)ņ=ņ.
HAREKET
NRQXDQODW×PO×
Örnek
j L = 2m/s
j k = 4m/s
A
x = 200 m
+PRRH[
+VȘY\ZHS `VSKH H`UÛ `UKLNPKLUOHYLRL[SPSLYHYHZÛUKHRP\aHRSÛR
x = (j K - j L) .t LȴP[SPȘP`SL
I\S\U\Y
A
)
A
+VȘY\ZHS`VSKHH`UÛ`UKLZHIP[OÛaSHYSHOHYLRL[LKLU2]L3HYHsSHYÛRHs
ZHUP`LZVUYH`HU`HUHNLSPY&
Çözüm
[Z
YLZVUYH*UVR[HZÛUKH`HU`HUHVSZ\USHY
x = x AC - x BC
x = j K .t - j L .t
x = (j K - j L) .t
200 = (4 - 2) .t
t = 100 s
A
)
C
Örnek
2VU\TaHTHUNYHÄȘP]LYPSLUOHYLRL[SPSLYKLU
4»UPU OÛaÛ j VSK\Ș\UH NYL 2 ]L 3 UPU OÛaÛ RHs j KPY&
_T
2
L
2x
M
x
[
[Z
[
Çözüm
/ÛaIHȘÛU[ÛZÛUKHKLȘLYSLYP`LYPUL`HaHSÛT
j=
154
Tx
(
Tt
jM =
x-0
x
=
=j
2t - 0 2t
jL =
x-0 x
= = 2j
t-0
t
jK =
2x - 0 2x
=
= 4j
t-0
t
)ņ=ņ.
NRQXDQODW×PO×
Örnek
2VU\TaHTHUNYHÄȘP]LYPSLU
OHYLRL[SPUPUOÛaaHTHUNYHÄȘPUPsPaPUPa
_T
2
4
6
[Z
¶
Çözüm
2VU\TAHTHUNYHÄȘPUKLLȘPTOÛaÛ]LYPY
Tx x 1 - x 0 20 - 0
HYHSÛȘÛUKHRPOÛa $ j =
=
=
= 10m/s
Tt
t1 - t0
2-0
HYHSÛȘÛUKHRPOÛa $ j =
Tx x 2 - x 1 20 - 20
=
=
=0
Tt
t2 - t1
4-2
HYHSÛȘÛUKHRPOÛa $ j =
Tx x 3 - x 2 0 - 20
=
=
= - 10 m/s
Tt
t3 - t2
8-6
HYHSÛȘÛUKHRPOÛa $ j =
Tx x 4 - x 3 - 20 - 0
=
=
= - 10 m/s
Tt
t4 - t3
8-6
TZ
2
4
6
[Z
¶
/ÛaaHTHUNYHÄȘP
155
6×Q×I
3.
h1ņ7(
)ņ=ņ.
HAREKET
NRQXDQODW×PO×
Örnek
TZ
/ÛaaHTHUNYHÄȘP]LYPSLUOHYLRL[SPUPURVU\T
aHTHUNYHÄȘPUPsPaPUPa
[Z
2
¶
Çözüm
/ÛaaHTHUNYHÄȘPUPUHS[ÛUKHRPISNLUPUHSHUÛ
`LYKLȘPȴ[PYTL`LLȴP[[PY
TZ
Tx 1 = 8.1 = + 8 m
Tx 2 = 8.1 = - 8 m
Tx 1
Tx 2
[Z
2
¶
_T
,SKL LKPSLU KLȘLYSLYL NYL RVU\T aHTHU
NYHÄȘPȴLRPSKLRPNPIPVS\Y!
+PRRH[
[Z
2
+
aN
U/ÛaSHUHU:HIP[Ǧ]TLSP/HYLRL[
Tx 1 < Tx 2 < Tx 3
j
Tx 1
[
j2
j
Tx 2
[
j
Tx 3
[2
[
,ȴP[aHTHUHYHSÛRSHYÛUHRHYȴÛSÛROÛaÛUÛLȴP[TPR[HYKHHY[ÛYHUJPZTPU`HW[ÛȘÛOHYLRL[LZHIP[P]TLSPKLUPY/ÛaKLȘPȴPTPOLYHYHSÛR[HZHIP[VSK\Ș\UKHUP]TLZPZHIP[[PY
/HYLRL[SPUPU`LYKLȘPȴ[PYTLZPIPYUJLRPaHTHUHYHSÛȘÛUHNYLKHOHI
`
RVSHJHR[ÛY
]LI\I
`
TLK
aLUSPVSHYHRHY[HJHR[ÛY
156
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:
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Örnek
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Örnek
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Q = m.c.Tt
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Q = m.c.Tt
Q = 100.0, 1 (10 - 50)
Q = - 400 cal
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Q = m.c.Tt
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