Ly thuyet Vat ly 12 NC

8/4/2019 Ly thuyet Vat ly 12 NC

1/19

CHNG I. NG LC HC VT RN1. To gcL to xc nh v tr ca mt vt rn quay quanh mt trc c nh bi gc (rad) hp gia mt phng ng gn vi vt mt phng c nh chn lm mc (hai mt phng ny u cha trc quay)Lu : Ta ch xt vt quay theo mt chiu v chn chiu dng l chiu quay ca vt 02. Tc gcL i lng c trng cho mc nhanh hay chm ca chuyn ng quay ca mt vt rn quanh mt trc

* Tc gc trung bnh: ( / )tb rad st

* Tc gc tc thi: '( )d tdt

Lu: Lin h gia tc gc v tc di v = .r3. Gia tc gcL i lng c trng cho s bin thin ca tc gc

* Gia tc gc trung bnh: 2( / )tb rad st

* Gia tc gc tc thi:2

2'( ) ''( )

d dt t

dt dt

Lu: + Vt rn quay u th 0const + Vt rn quay nhanh dn u > 0

+ Vt rn quay chm dn u < 04. Phng trnh ng hc ca chuyn ng quay* Vt rn quay u ( = 0): = 0 + t

* Vt rn quay bin i u ( 0) = 0 + t; 201

2t t ; 2 20 02 ( )

5. Gia tc ca chuyn ng quay

* Gia tc php tuyn (gia tc hng tm) na

: c trng cho s thay i v hng ca vn tc di v

( na v

)2

2n

va r

r

* Gia tc tip tuynta

: c trng cho s thay i v ln ca v

(

ta

v v

cng phng) '( ) '( )tdv

a v t r t rdt

* Gia tc ton phn n ta a a

2 2n t

a a a

Gc hp gia a

vna

:

2tan t

n

a

a

Lu: Vt rn quay u th at = 0 a

=n

a

6. Phng trnh ng lc hc ca vt rn quay quanh mt trc c nh:M

M I hayI

Trong : + M = Fd (Nm)l mmen lc i vi trc quay (d l tay n ca lc)+ 2i i

i

I m r (kgm2)l mmen qun tnh ca vt rn i vi trc quay

Mmen qun tnh I ca mt s vt rn ng cht khi lng m c trc quay l trc i xng

- Vt rn l thanh c chiu di l, tit din nh: 21

12I ml ( trc quay i qua khi tm)

- Vt rn l vnh trn hoc tr rng bn knhR:I = mR2

- Vt rn l a trn mng hoc hnh tr c bn knhR: 21

2I mR

- Vt rn l khi cu c bn knh R: 22

5I mR

7. Mmen ng lng:L i lng ng hc c trng cho chuyn ng quay ca vt rn quanh mt trcL=I (kgm2/sLu: Vi cht im th mmen ng lngL = mr2= mvr (rl k/c t vtn trc quay)

Ti liu Vt ly n thi: TN - CD - DH

Tai liu Vt ly n thi: TN - CD - DH GV: Nguyn Xun Tri - 0168.509 4388

1


2/19

8. Dng khc ca phng trnh ng lc hc ca vt rn quay quanh mt trc c nhdL

Mdt

9. nh lut bo ton mmen ng lngTrng hp M = 0 thL = constNuI= const = 0 vt rn khng quay hoc quay u quanh trcNuIthay i thI11 = I22

10. ng nng ca vt rn quay quanh mt trc c nh 21

W ( )2I J

11. S tng t gia cc i lng gc v i lng di trong chuyn ng quay v chuyn ng thng

Chuyn ng quay(trc quay c nh, chiu quay khng i)

Chuyn ng thng(chiu chuyn ng khng i)

(rad) (m)(rad/s) (m/s)

(Rad/s2) (m/s2)(Nm) (N)

(Kgm2) (kg)(kgm2 /s)

To gc Tc gc Gia tc gc Mmen lc MMmen qun tnh IMmen ng lng L = I

ng nng quay 21

W2I (J)

To xTc vGia tc aLc FKhi lng mng lng P = mv

ng nng 21

W2

mv (J)

Chuyn ng quay u:

= const; = 0; = 0 + tChuyn ng quay bin i u: = const = 0 + t

20

1

2t t

2 20 02 ( )

Chuyn ng thng u:

v = cnt; a = 0; x = x0 + atChuyn ng thng bin i u:

a = constv = v0 + at

x = x0 + v0t +21

2at

2 20 02 ( )v v a x x

Phng trnh ng lc hcM

I

Dng khc dLMdt

nh lut bo ton mmen ng lng

1 1 2 2 iI I hay L const nh l v ng nng

2 2 1 2

1 1W

2 2I I A (cng ca ngoi lc)

Phng trnh ng lc hcF

am

Dng khcdtpdF

nh lut bo ton ng lngconstvmp

i

ii

i

i

.

nh l v ng nng

W =21

22 2

1

2

1mvmv = A (cng ca ngoi lc)

Cng thc lin h gia i lng gc v i lng di: s = r; v =r; at = r; an = 2rLu: Cng nh v, a, F, P cc i lng ; ; M; L cng l cc i lng vct

CHNG II. DAO NG C HCI. DAO NG IU HO1. Phng trnh dao ng: x = Acos(t + )2. Vn tc tc thi: v = -Asin(t + )

v

lun cng chiu vi chiu chuyn ng (vt chuyn ng theo chiu dng th v>0, theo chiu m th v


3/19

6. C nng: 2 21

W W W2t

m A

Vi 2 2 2 2 2

1 1W sin ( ) Wsin ( )

2 2mv m A t t v 2 2 2 2 2 2

1 1W ( ) W s ( )

2 2tm x m A cos t co t

7. Dao ng iu ho c tn s gc l , tn s f, chu k T. Th ng nng v th nng bin thin vi tn s gc 2, tn schu k T/28. ng nng v th nng trung bnh trong thi gian nT/2 ( nN*, T l chu k dao ng)

l: 2 2W 1

2 4m A

9. Khong thi gian ngn nht vt i t v tr c li x1 n x2

2 1t

vi

11

22

s

s

xco

A

xco

A

v ( 1 20 , )

10. Chiu di qu o: 2A11. Qung ng i trong 1 chu k lun l 4A; trong 1/2 chu k lun l 2A

Qung ng i trong l/4 chu k l A khi vt i t VTCB n v tr bin hoc ngc li12. Qung ng vt i c t thi im t1 n t2.

Xc nh: 1 1 2 21 1 2 2

Acos( ) Acos( )

sin( ) sin( )

x t x t v

v A t v A t

(v1 v v2 ch cn xc nh du)

Phn tch: t2 t1 = nT + t (n N; 0 t < T)Qung

ng i c trong thi gian nT l S

1= 4nA, trong th

i giant l S

2.

Qung ng tng cng l S = S1 + S2Lu:+ Nu t = T/2 th S2 = 2A

+ Tnh S2 bng cch nh v tr x1, x2 v chiu chuyn ng ca vt trn trc Ox+ Trong mt s trng hp c th gii bi ton bng cch s dng mi lin h gia dao ng iu ho v chuyn

trn u s n gin hn.+ Tc trung bnh ca vt i t thi im t1 n t2:

2 1tb

Sv

t t

vi S l qung ng tnh nh trn.

13. Bi ton tnh qung ng ln nht v nh nht vt i c trong khong thi gian 0 < t < T/2.Vt c vn tc ln nht khi qua VTCB, nh nht khi qua v tr bi n nn trong cng mt khong thi gian qung n

c cng ln khi vt cng gn VTCB v cng nh khi cng gn v tr bin.S dng mi lin h gia dao ng iu ho v chuyn ng trn u.Gc qut = t.

Qung ng ln nht khi vt i t M1 n M2i xng qua trc sin (hnh 1) ax 2Asin 2MS

Qung ng nh nht khi vt i t M1 n M2 i xng qua trc cos (hnh 2) 2 (1 os )2Min

S A c

Lu: + Trong trng hp t > T/2

Tch '2

Tt n t

trong *;0 '2

Tn N t

Trong thi gian 2

Tn qung ng lun l 2nA

Trong thi gian t th qung ng ln nht, nh nht tnh nh trn.

A

-A

x1x2

M2M1

M'1

M'2

O

A-A

MM 12

O

P

x O

2

1

M

M

-A AP2 1P

P2

2

+ Tc trung bnh ln nht v nh nht ca trong khong thi gian t: axax

MtbM

Sv

t

v Min

tbMin

Sv

t

vi SMax; SMin tnh nh trn

13. Cc bc lp phng trnh dao ng dao ng iu ho:* Tnh * Tnh A* Tnh da vo iu kin u: lc t = t0 (thng t0 = 0) 0

0

Acos( )

sin( )

x t

v A t

Lu : + Vt chuyn ng theo chiu dng th v > 0, ngc li v < 0+ Trc khi tnh cn xc nh r thuc gc phn t th my ca ng trn lng gic

(thng ly - < )


3


4/19

14. Cc bc gii bi ton tnh thi im vt i qua v tr bit x (hoc v, a, Wt, W, F) ln th n* Gii phng trnh lng gic ly cc nghim ca t (Vi t > 0 phm vi gi tr ca k )* Lit k n nghim u tin (thng n nh)* Thi im th n chnh l gi tr ln th n

Lu :+ ra thng cho gi tr n nh, cn nu n ln th tm quy lut suy ra nghim th n+ C th gii bi ton bng cch s dng mi lin h gia dao ng iu ho v chuyn ng trn u

15. Cc bc gii bi ton tm s ln vt i qua v tr bit x (hoc v, a, Wt, W, F) t thi im t1 n t2.* Gii phng trnh lng gic c cc nghim* T t1 < t t2 Phm vi gi tr ca (Vi k Z)* Tng s gi tr ca k chnh l s ln vt i qua v tr .

Lu: + C th gii bi ton bng cch s dng mi lin h gia dao ng iu ho v chuyn ng trn u.+ Trong mi chu k (mi dao ng) vt qua mi v tr bin 1 ln cn cc v tr khc 2 ln.

16. Cc bc gii bi ton tm li , vn tc dao ng sau (trc) thi im t mt khong thi gian t.Bit ti thi im t vt c li x = x0.

* T phng trnh dao ng iu ho: x = Acos(t + ) cho x = x0Ly nghim t + = vi 0 ng vi x ang gim (vt chuyn ng theo chiu m v v < 0)

hoc t + = - ng vi x ang tng (vt chuyn ng theo chiu dng)* Li v vn tc dao ng sau (trc) thi im t giy l

x Acos( )

A sin( )

t

v t

hoc x Acos( )

A sin( )

t

v t

17. Dao ng c phng trnh c bit:* x = a Acos(t + ) vi a = const

Bin l A, tn s gc l , pha ban u ; x l to , x0 = Acos(t + ) l li .To v tr cn bng x = a, to v tr bin x = a AVn tc v = x = x0, gia tc a = v = x = x0

H thc c lp: a = -2x0 v 2 2 20 ( )v

A x

* x = a Acos2(t + ) (ta h bc) khi Bin l A/2; tn s gc 2, pha ban u 2.

II. CON LC L XO

1. Tn s gc: km

; chu k:2

2m

Tk

; tn s: 1 1

2 2

kf

T m

iu kin dao ng iu ho: B qua ma st, lc cn v vt dao ng trong gii

hn n hi2. C nng: 2 2 2

1 1W

2 2m A kA

3. * bin dng ca l xo thng ng khi vt VTCB:mg

lk

2l

Tg

* bin dng ca l xo khi vt VTCB vi con lc l xo nm trn mt phng nghing c gc nghing :sinmg

lk

2

sin

lT

g

+ Chiu di l xo ti VTCB: lCB = l0 + l (l0 l chiu di t nhin)+ Chiu di cc tiu (khi vt v tr cao nht): lMin = l0 + l A+ Chiu di cc i (khi vt v tr thp nht): lMax = l0 + l + A

lCB = (lMin + lMax)/2+ Khi A >l (Vi Ox hng xung):

l

gi

O

x

A

-An

lginO

x

A

-A

Hnh a (A l)

- Thi gian l xo nn 1 ln l thi gian ngn nht vt it v tr x1 = -l n x2 = -A.

- Thi gian l xo gin 1 ln l thi gian ngn nht vt it v tr x1 = -l n x2 = A,

Lu: Trong mt dao ng (mt chu k) l xo nn 2 lnv gin 2 ln

x

A

-A l

Nn 0Gin

Hnh v th hin thi gian l xo nn vgin trong 1 chu k (Ox hng xung)

4. Lc ko v hay lc hi phc F = -kx = -m2xc im: * L lc gy dao ng cho vt.

* Lun hng v VTCB* Bin thin iu ho cng tn s vi li


4


5/19

5. Lc n hi l lc a vt v v tr l xo khng bin dng.C ln Fh = kx* (x* l bin dng ca l xo)* Vi con lc l xo nm ngang th lc ko v v lc n hi l mt (v ti

VTCB l xo khng bin dng)* Vi con lc l xo thng ng hoc t trn mt phng nghing

+ ln lc n hi c biu thc:* Fh = kl + x vi chiu dng hng xung* Fh = kl - x vi chiu dng hng ln

+ Lc n hi cc i (lc ko): FMax = k(l + A) = FKmax (lc vt v tr thp nht)+ Lc n hi cc tiu:

* Nu A < l FMin = k(l - A) = FKMin* Nu A l FMin = 0 (lc vt i qua v tr l xo khng bin dng)

Lc y (lc nn) n hi cc i: FNmax = k(A - l) (lc vt v tr cao nht)6. Mt l xo c cng k, chiu di l c ct thnh cc l xo c cng k1, k2, v chiu di tng ng l l1, l2,

th c: kl = k1l1 = k2l2 = 7. Ghp l xo:

* Ni tip1 2

1 1 1...

k k k cng treo mt vt khi lng nh nhau th: T2 = T12 + T22 +

* Song song: k = k1 + k2 + cng treo mt vt khi lng nh nhau th: 2 2 21 2

1 1 1...

T T T

8. Gn l xo k vo vt khi lng m1 c chu k T1, vo vt khi lng m2c T2, vo vt khi lng m1+m2 c chuT3, vo v

t khi lng m1 m2 (m1 > m2

) c chu k T4.

Th ta c: 2 2 23 1 2T T T v2 2 2

4 1 2T T T 9. o chu k bng phng php trng phng: xc nh chu k T ca mt con lc l xo (con lc n) ngi ta so snh chu k T0 ( bit) ca mt con lc khc (TT0).

Hai con lc gi l trng phng khi chng ng thi i qua mt v tr xc nh theo cng mt chiu.Thi gian gia hai ln trng phng 0

0

TT

T T

Nu T > T0 = (n+1)T = nT0.Nu T < T0 = nT = (n+1)T0. vi n N*

III. CON LC N

1. Tn s gc: g

l

; chu k:2

2l

T

g

; tn s:1 1

2 2

gf

T l

iu kin dao ng iu ho: B qua ma st, lc cn v 0


6/19

8. Khi con lc n chu thm tc dng ca lc ph khng i:Lc ph khng i thng l:

* Lc qun tnh: F ma

, ln F = ma ( F a

)

Lu : + Chuyn ng nhanh dn u a v

( v

c hng chuyn ng)

+ Chuyn ng chm dn u a v

* Lc in trng: F qE

, ln F = qE (Nu q > 0 F E

; cn nu q < 0 F E

)

* Lc y csimt: F = DgV ( F

lung thng ng hng ln)Trong : D l khi lng ring ca cht lng hay cht kh; g l gia tc ri t do.

V l th tch ca phn vt chm trong cht lng hay cht kh .Khi : 'P P F

gi l trng lc hiu dng hay trong lc biu kin (c vai tr nh trng lc P

)

'F

g gm

gi l gia tc trng trng hiu dng hay gia tc trng trng biu kin.

Chu k dao ng ca con lc n khi : ' 2'

lT

g

Cc trng hp c bit:* F

c phng ngang: + Ti VTCB dy treo lch vi phng thng ng mt gc c: tan F

P 2 2' ( )Fg g

m

* F

c phng thng ng th ' Fg gm

IV. CON LC VT L

1. Tn s gc: mgdI

; chu k: 2 ITmgd

; tn s1

2

mgdf

I

Trong : m (kg) l khi lng vt rnd (m) l khong cch t trng tm n trc quayI (kgm2) l mmen qun tnh ca vt rn i vi trc quay

2. Phng trnh dao ng = 0cos(t + )iu kin dao ng iu ho: B qua ma st, lc cn v 0


7/19

Nu dng my tnh FX 570 MS th bn thc hin nh sau:

Ch : - Bt k bi ton v tng hp 2, 3 hay nhiu dao ng iu ha no ta cng c th gii c bng ph

ng php ny. V

d trn y ch l ly ngu nhin v ti ly v d c bit mc ch cc bn gii bng cc phng php thng thng dnhm ra kt qu.

- Ngay t trc khi bt tay vo gii cc bn c th chuyn h ca my sang radian. Khi cc bn nhp pha ban di dng c s ca .

- Phng php ny cc bn c th m rng cho mt s bi ton v in xoay chiu.

V d: cho mch in gm R, L, C mc ni tip. Bit tcos1002220u AB ; )2t220cos100(uMB

tnh uAM=?

Nu dng my tnh FX 570 ES th bn thc hin thm nh sau:

- Thc hin nh i vi my FX 570 MS nhp cc gi tr bin v pha ban u ca cc dao ng. Kt thc bi du=

- Bm tun t cc phm sau: Shift 2 3 =Ta thu c cc kt qu ca A v

VI. DAO NG TT DN DAO NG CNG BC - CNGHNG1. Mt con lc l xo dao ng tt dn vi bin A, h s ma st .

* Qung ng vt i c n lc dng li l:2 2 2

2 2

kA AS

mg g

* gim bin sau mi chu k l:2

4 4mg gA

k

* S dao ng thc hin c:2

4 4

A Ak AN

A mg g

* Thi gian vt dao ng n lc dng li:.

4 2

AkT At N T

mg g

(Nu coi dao ng tt dn c tnh tun hon vi chu k

2T

)

3. Hin tng cng hng xy ra khi:f = f0 hay = 0 hay T = T0Vif, , Tvf0, 0, T0 l tn s, tn s gc, chu k ca lc cng bc v ca h dao ng.

T

x

O

CHNG III. SNG C HCI. SNG C HC1. Bc sng: = vT = v/f

Trong : : Bc sng; T (s): Chu k ca sng; f (Hz): Tn s ca sngv: Vn tc truyn sng (c n v tng ng vi n v ca )

2. Phng trnh sng

Ti im O: uO = acos(t + )Ti im M cch O mt on d trn phng truyn sng.

* Sng truyn theo chiu dng ca trc Ox th uM = aMcos(t + -d

v ) = aMcos(t + - 2

d

)

* Sng truyn theo chiu m ca trc Ox th uM = aMcos(t + +d

v ) = aMcos(t + + 2

d

)

3. lch pha gia hai im cch ngun mt khong d 1, d2 1 2 1 22d d d d

v

Nu 2 im nm trn mt phng truyn sng v cch nhau mt khong d th: 2d d

v

Lu :n v ca d, d1, d2, v v phi tng ng vi nhau4. Trong hin tng truyn sng trn si dy, dy c kch thch dao ng bi nam chm in vi tn s d ng in l f th

s dao ng ca dy l 2f.

O M

d


7


8/19

II. GIAO THOA SNGGiao thoa ca hai sng pht ra t hai ngun sng kt hp cch nhau mt khong l:Xt im M cch hai ngun ln lt d1, d2Gi x l s nguyn ln nht nh hn x (v d: 535.5 )1. Hai ngun dao ng cng pha:

Bin dao ng ca im M: AM = 2aMcos( 1 2d d

)

* im dao ng cc i: d1 d2 = k (kZ)

S im hoc s ng (khng tnh hai ngun): l lk

hoc NC = 12

l

* im dao ng cc tiu (khng dao ng): d1 d2 = (2k+1)2

(kZ)

S im hoc s ng (khng tnh hai ngun): 1 12 2

l lk

hoc NCT =

2

12

l

2. Hai ngun dao ng ngc pha:

Bin dao ng ca im M: AM = 2aMcos( 1 22

d d

)

* im dao ng cc i: d1 d2 = (2k+1)2

(kZ)

S im hoc s ng (khng tnh hai ngun):

1 1

2 2

l l

k hoc NC =

2

1

2

l

* im dao ng cc tiu (khng dao ng): d1 d2 = k (kZ)

S im hoc s ng (khng tnh hai ngun): l lk

hoc NC = 12

l

3. Hai ngun dao ng vung pha:

Bin dao ng ca im M: AM = 2aMcos( 1 24

d d

)

S im (ng) dao ng cc i bng s im (ng) dao ng cc tiu (khng tnh hai ngun):1 1

4 4

l lk

Ch : Vi bi ton tm s ng dao ng cc i v khng dao ng gia hai im M, N cch hai ngun ln lt l d1M, d1N, d2N. t dM = d1M - d2M ; dN = d1N - d2N v gi s dM < dN.

+ Hai ngun dao ng cng pha: Cc i: dM < k < dN & Cc tiu: dM < (k+0,5) < dN+ Hai ngun dao ng ngc pha: Cc i:dM < (k+0,5) < dN & Cc tiu: dM < k < dNS gi tr nguyn ca k tho mn cc biu thc trn l s ng cn tm.III. SNG DNG1. * Gii hn c nh Nt sng

* Gii hn t do Bng sng* Ngun pht sng c coi gn ng l nt sng* B rng bng sng 4a (vi a l bin dao ng ca ngun)

2. iu kin c sng dng gia hai im cch nhau mt khong l:

* Hai im u l nt sng: *( )2

l k k N

S bng sng = s b sng = k

S nt sng = k + 1* Hai im u l bng sng: *( )

2l k k N

S b sng nguyn = k 1S bng sng = k + 1S nt sng = k

* Mt im l nt sng cn mt im l bng sng: (2 1) ( )4

l k k N

S b sng nguyn = kS bng sng = s nt sng = k + 1

3. Trong hin tng sng dng xy ra trn si dy AB vi u A l nt sng

Bin dao ng ca im M cch A mt on d l: 2 sin(2 )M

dA a

vi a l bin dao ng ca ngun.


8


9/19

IV. SNG M

1. Sng m, phn loi, cc c trng: cao, to v m sc.

2. Cng m:S

P

tS

WI

.

Vi W (J), P (W) l nng lng, cng sut pht m ca ngunS (m2) l din tch mt vung gc vi phng truyn m (vi sng cu th S l din tch mt cu S=4R2)

3. Mc cng m

0

( ) lg IL BI

Hoc0

( ) 10.lg IL dBI

(cng thc thng dng)

Vi I0 = 10-12 W/m2 f = 1000Hz: cng m chun.

4. Hiu ng DoppleHiu ng Doppler l s thay i tn s sng(m my thu ghi nhn c) khi c s chuyn ng tng i gia m

thu v ngun pht.Cc h thc gia tn s my thu thu c v tn s do ngun pht ra:

fv

vvf M

' : nu my thu chuyn ng vi tc vM li gn ngun pht ng yn.

fvvvf M'' : nu my thu chuyn ng vi tc vM ra xa ngun pht ng yn.

fvv

vf

S

' : nu ngun pht chuyn ng vi tc vS li gn my thu ng yn.

fvv

vf

S

'' : nu ngun pht chuyn ng vi tc vS ra xa my thu ng yn.

Lu: fvv

vvf

S

M

' : nu my thu chuyn ng vi tc vM, ngun pht chuyn ng vi tc vS

CHNG IV. DAO NG V SNG IN T1. Dao ng in t

* in tch tc thi q = Q0cos(t + )* Dng in tc thi i = q = -Q0sin(t + ) = I0sin(t + )

* in p tc thi )cos()cos( 00 tUt

C

Q

C

qu

Trong :1

LC l tn s gc ring, 2T LC l chu k ring,

1

2f

LC l tn s ring

00 0

QI Q

LC v 0 00 0

Q I LU I

C C C

* Nng lng in trng )(cos222

1

2

1 220

22 t

C

Q

C

qquCuW

C

* Nng lng t trng )(sin22

1 2202 tC

QLiW

L

* Nng lng in t W = WC+WL =2

20

2

1

2 oLI

C

Q

Ch : Mch dao ng c tn s gc , tn sfv chu k T th nng lng in trng bin thin vi tn s gc 2, tn s 2v chu k T/22. Sng in tTc lan truyn trong khng gian v = c = 3.10-8m/sMy pht hoc my thu sng in t s dng mch dao ng LC th tn s sng in t pht hoc thu bng tn s ri ng mch.

Bc sng ca sng in t 2v

v LC

f


9


10/19

Lu: Mch dao ng c L bin i t L Min LMax v C bin i t CMin CMax th bc sng ca sng in t pht (thu) Min tng ng vi LMin v CMin

Max tng ng vi LMax v CMaxLu :1. Bc sng thay i do ghp t: T C1 ghp song song vi t C2 2 = 22

21

T C1 ghp ni tip vi t C2 22

21

2

111

2. iu kin thu c sng in t l xy ra hin tng cng hng fT= fP

3.Nng lng mch b xung bng nng lng hao ph do ta nhit in trR l W = I2.R.t

4. Trong mt chu k c 4 thi im WC= WL ; cc thi im ny cc u nhau mt khong thi gian bng T/4.

CHNG V. IN XOAY CHIU

1. Biu thc in p tc thi v dng in tc thi: u = U0cos(t + u) v i = I0cos(t + i)Vi = u i l lch pha ca u so vi i, c

2 2

2. Dng in xoay chiu i = I0cos(2ft + i)* Mi giy i chiu 2fln* Nu pha ban u i = 0 hoc i = th ch giy u tin i chiu 2f-1 ln.

3. Cng thc tnh khong thi gian n hunh quang sng trong mt chu kKhi t hiu in th u = U0cos(t + u) vo hai u bng n, bit n ch sng ln khi u U1.4

t

Vi 1

0

osU

cU

, (0 < < /2)

4. Dng in xoay chiu trong on mch R,L,C

* on mch ch c in tr thun R: uR cng pha vi i, ( = u i = 0)U

IR

v 00U

IR

Lu: in tr R cho dng in khng i i qua v cU

IR

* on mch ch c cun thun cm L: uL nhanh pha /2 so vii , ( = u i = /2)

L

UI

Z

v 00L

UI

Z

vi ZL = L l cm khng

Lu: Cun thun cm L cho dng in khng i i qua hon ton (khng cn tr).* on mch ch c t in C: uCchm pha /2 so vii , ( = u i = -/2)

C

UI

Z v 00

C

UI

Z vi

1CZ

C l dung khng

Lu: T in C khng cho dng in khng i i qua (cn tr hon ton).

* on mch RLC khng phn nhnh 2 2 2 2 2 20 0 0 0( ) ( ) ( )L C R L C R L C Z R Z Z U U U U U U U U

;sin ; osL C L C Z Z Z Z R

tg cR Z Z

vi2 2

+ Khi ZL > ZC hay

1

LC > 0 th u nhanh pha so vi i

+ Khi ZL < ZC hay1

LC < 0 th u chm pha so vii

+ Khi ZL = ZC hay1

LC = 0 th u cng pha vi i.

Lc MaxU

I = gi l hin tng cng hng dng inR


10


11/19

5. Cng sut to nhit trn on mch RLC: P = UIcos = I2R.6. Hiu in th u = U1 + U0cos(t + ) c coi gm mt hiu in th khng i U1 v mt hiu in th xoay chiuU0cos(t + ) ng thi t vo on mch.7. Tn s dng in do my pht in xoay chiu mt pha c P cp cc, rto quay vi vn tc n vng/pht pht ra:

60

pnf Hz

T thng gi qua khung dy ca my pht in = NBScos(t +) = 0cos(t + )Vi 0 = NBS l t thng cc i, N l s vng dy, B l cm ng t ca t trng, S l din tch ca vng dy, = 2fSut in ng trong khung dy: e = NSBsin(t + ) = E0sin(t + )Vi E0 = NSB l sut in ng cc i.

8. Dng in xoay chiu ba pha : )cos(01 tIi ; )3

2cos(02

tIi & )3

2cos(03

tIi

My pht mc hnh sao: Ud = 3 UpMy pht mc hnh tam gic: Ud = UpTi tiu th mc hnh sao: Id = IpTi tiu th mc hnh tam gic: Id = 3 IpLu: my pht v ti tiu th thng chn cch mc tng ng vi nhau.

9. Cng thc my bin th: 1 1 2 12 2 1 2

U E I N

U E I N

10. Cng sut hao ph trong qu trnh truyn ti in nng:2

2 2

os

PP R

U c

Thng xt: cos = 1 khi 2

2

PP R

U

Trong : P l cng sut cn truyn ti ti ni tiu th;U l in p ni cung cp; cosl h s cng sutca dy ti inl

RS

l in tr tng cng ca dy ti in (lu: dn in bng 2 dy)

gim th trn ng dy ti in: U = IR

Hiu sut ti in: .100%P P

HP

11. on mch RLC c R thay i:* Khi R 0 th I Imax* Khi R th UR URmax

* Khi R = ZL - ZC th P Pmax =CL

ZZ

U

2

2

* Nu cun dy c in tr thun r th: PRmaxR =22 )(

CLZZr v PmaxR = ZL-ZC-r

12. on mch RLC c L thay i:

* Khi2

1L

C th IMax URmax; PMax cn ULCMin Lu: L v C mc lin tip nhau

* Khi2 2

CL

C

R ZZ

Z

th

2 2

axC

LM

U R ZU

R

* Vi L = L1 hoc L = L2 th UL c cng gi tr th ULmax khi1 2

1 2

1 2

21 1 1 1( )

2L L L

L LL

Z Z Z L L

* Khi2 24

2C C

L

Z R Z Z

th ax 2 2

2 R

4RLM

C C

UU

R Z Z

Lu: R v L mc lin tip nhau


11


12/19

13. on mch RLC c C thay i:

* Khi2

1C

L th IMax URmax; PMax cn ULCMin Lu: L v C mc lin tip nhau

* Khi2 2

LC

L

R ZZ

Z

th

2 2

axL

CM

U R ZU

R

* Khi C = C1 hoc C = C2 th UC c cng gi tr th UCmax khi1 2

1 21 1 1 1( )2 2

C C C

C CC

Z Z Z

* Khi 2 242

L L

CZ R Z Z th ax 2 22 R4

RCM

L L

UUR Z Z

Lu: R v C mc lin tip nhau

14. Mch RLC c thay i:

* Khi 1

LC th IMax URmax; PMax cn ULCMin Lu: L v C mc lin tip nhau

* Khi2

1 1

2

C L R

C

thax 2 2

2 .

4LM

U LU

R LC R C

* Khi21

2

L R

L C th

)1(4

..222

maxCRLCR

LUUC

* Vi =1 hoc =2 th I hoc P hoc UR c cng mt gi tr th Imaxhoc Pmax hoc URmax khi 1 2 hay 1 2f f f 15. Hai on mch R1L1C1 v R2L2C2 cng u hoc cng i c pha lch nhau

Vi 1 11

1

L CZ Ztg

R

v 2 2

22

L CZ Z

tgR

(gi s 1 > 2) C 1 2 = 1 21 21

tg tgtg

tg tg

Trng hp c bit = /2 (vung pha nhau) th tg1tg2 = -1.

CHNG VI. SNG NH SNG1. Hin tng tn sc nh sng.* /n: L hin tng nh sng b tch thnh nhiu mu khc nhau khi i qua mt phn cch ca hai mi trng trong sut.* nh sng n sc l nh sng khng b tn sc

nh sng n sc c tn s xc nh, ch c mt mu.

Bc sng ca nh sng n scfv , truyn trong chn khng

fc0

.

* Chit sut ca mi trng trong sut ph thuc vo mu sc nh sng. i vi nh sng mu l nh nht, mu tm l ln nht.* nh sng trng l tp hp ca v s nh sng n sc c mu bin thin lin tc t n tm.

Bc sng ca nh sng trng: 0,40 m 0,76 m.2. Hin tng giao thoa nh sng (ch xt giao thoa nh sng trong th nghim Ing).* /n: L s tng hp ca hai hay nhiu sng nh sng kt hp trong khng gian trong xut hin nhng vch sng v nhng vcti xen k nhau.

Cc vch sng (vn sng) v cc vch ti (vn ti) gi l vn giao thoa.

* Hiu ng i ca nh sng (hiu quang trnh) d = d2-d1 =D

ax

Trong : a = S1S2 l khong cch gia hai khe sngD = OI l khong cch t hai khe sng S1, S2 n mn quan st S1M = d1; S2M = d2x = OM l (to ) khong cch t vn trung tm n im M ta xt

* V tr (to ) vn sng: d = k x = ka

D vi kZ

k = 0: Vn sng trung tm; k = 1: Vn sng bc (th) 1; k = 2: Vn sng bc (th) 2

* V tr (to ) vn ti: d = (k + 0,5) x= (k+0,5)a

D vi kZ

k = 0, k = -1: Vn ti th (bc) nht; k = 1, k = -2: Vn ti th (bc) hai; k = 2, k = -3: Vn ti th (bc) ba

* Khong vn i: L khong cch gia hai vn sng hoc hai vn ti lin tip: i=a

D

S1

DS2

d1

d2I O

xM

a


12


13/19

* Nu th nghim c tin hnh trong mi trng trong sut c chit sut n th bc sng v khong vn: n=n

v in =

n

i

a

Dn

* Khi ngun sng S di chuyn theo phng song song vi S1S2 th h vn di chuyn ngc chiu v khong vn i vn khng i.di ca h vn l:

0

1

Dx d

D=

Trong :D l khong cch t 2 khe ti mn; D1 l khong cch t ngun sng ti 2 khe; d l dch chuyn ca ngun sng * Khi trn ng truyn ca nh sng t khe S 1 (hoc S2) c t mt bn mng dy e, chit sut n th h vn s dch chuyn v pS1 (hoc S2) mt on:

0

( 1)n eDx

a

-=

* Xc nh s vn sng, vn ti trong vng giao thoa (trng giao thoa) c b rng L (i xng qua vn trung tm)+ S vn sng (l s l): NS = 2

i

L

2+1

+ S vn ti (l s chn): Nt = 2

5,02i

L

Trong [x] l phn nguyn ca x. V d: [6] = 6; [5,05] = 5; [7,99] = 7* Xc nh s vn sng, vn ti gia hai im M, N c to x 1, x2 (gi s x1 < x2)

+ Vn sng: x1 < ki < x2+ Vn ti: x1 < (k+0,5)i < x2

S gi tr k Z l s vn sng (vn ti) cn tmLu: M v N cng pha vi vn trung tm th x1 v x2 cng du v M v N khc pha vi vn trung tm th x1 v x2 khc du.* Xc nh khong vn i trong khong c b rng L. Bit trong khong L c n vn sng.

+ Nu 2 u l hai vn sng th:1

Li

n=

-

+ Nu 2 u l hai vn ti th: Lin

=

+ Nu mt u l vn sng cn mt u l vn ti th:0,5

Li

n=

-

* S trng nhau ca cc bc x 1, 2 ... (khong vn tng ng l i1, i2 ...)+ Trng nhau ca vn sng: xs = k1i1 = k2i2 = ... k11 = k22 = ...+ Trng nhau ca vn ti: xt = (k1 + 0,5)i1 = (k2 + 0,5)i2 = ... (k1 + 0,5)1 = (k2 + 0,5)2 = ...

Lu: V tr c mu cng mu vi vn sng trung tm l v tr trng nhau ca tt c cc vn sng ca cc bc x.* Trong hin tng giao thoa nh sng trng (0,4 m 0,76 m)

- B rng quang ph bc k: x = ka

D( - t)vi v t l bc sng nh sng v tm

- Xc nh s vn sng, s vn ti v cc bc x tng ng ti mt v tr xc nh ( bit x)+ Vn sng: x = k

a

D vi kZ Vi 0,4 m 0,76 m cc gi tr ca k

+ Vn ti: x= (k+0,5)a

D vi kZ Vi 0,4 m 0,76 m cc gi tr ca k

- Khong cch di nht v ngn nht gia vn sng v vn ti cng bc k:[k ( 0,5) ]Min t

Dx k

a

ax [k ( 0,5) ]M tD

x ka

Khi vn sng v vn ti nm khc pha i vi vn trung tm.

ax [k ( 0,5) ]M tD

x ka

Khi vn sng v vn ti nm cng pha i vi vn trung tm.


13


14/19

3. My quang ph. Cc loi quang ph.a. Quang ph lin tc- Khi nim: QP gm nhiu di mu t n tm ni lin nhau mt cch lien tc c gi l QPLT.- iu kin: Cc cht Rn, lng v kh p sut ln khi b nung nng pht ra QPLT. - c im: Khng ph thuc vo bn cht vt m ch ph thuc nhit ca vt. Khi nhit tng dn, QP m rng n vng cbc song ngn.b. QPVPX- Khi nim: Gm cc vch mu ring r, ngn cch nhau bi cc khong ti.- iu kin: Cc cht kh hay hi p sut thp khi b kch thch.- c im: Mi nguyn t ha hc khi b kch thch pht ra cc bc x c bc song xc nh v cho mt QPVPX ring.c. QPVHT- Khi nim: QPLT thiu mt s vch mu do b cht kh hay hi kim loi hp th gi l QPVHT ca kh hay hi .- iu kin: Nhit ca m kh hay hi hp th phi nh hn nhit ca ngun pht QPLT. d. S o vch quang ph.e. Phn tch quang ph.4. Tia hng ngoi v tia t ngoi.a. Tia hng ngoi.- Khi nim: Bc x khng nhn thy c bc song di hn 0,76 m.- iu kin: Mi vt c nhit (d thp).- Tnh cht: +, Tc dng nhit

+, Gy phn ng quang ha, tc dng ln phim nh+, Bin iu song in t cao tn.+, Gy hin tng quang in trong mt s cht bn dn.

a. Tia t ngoi.

- Khi nim: Bc x khng nhn thy c bc sng ngn hn 0,38 m n 10-9 m.- iu kin: Mi vt c nhit trn 20000 C.- Tnh cht: +, Tc dng ln phim nh, lm ion ha mt s chy kh.

+, Kch thch s pht quang ca nhiu cht, c th gy mt s phn ng quang ha v phn ng ho hc.+, B thy tinh, nc hp th mnh. Tia t ngoi c bc sng t 0,18 m n 0,4 m truyn qua c thch anh.+, Hy dit t bo, lm rm da, hi mt, dit khun, dit nm mc.+, C th gy hin tng quang in.

5. Tia X. Thuyt in t v nh song. Thang sng in t.a. Tia X- Khi nim: Bc x bc sng t 10-8 m n 10-11 m.- iu kin: Chm tia Catt c ng nn ln p vo ming kim loi c nguyn t lng ln (Pt hay W).- Tnh cht: +, Kh nng m xuyn.

+, Tc dng mnh ln knh nh, lm ion ha khng kh.

+, Pht quang nhiu cht, c th gy ra hin tng quang in. Hy dit t bo, dit vi khun.b. Thuyt in t v nh sng. H thc n

v

c

c. Thang sng in tMin ST Bc sng (m) Tn s (Hz)Sng v tuyn in 3.104 - 10-4 104 3.1012Tia hng ngoi 10-3 7,6.10-7 3.1011 4.1014nh sng nhn thy 7,6.10-7 - 3,8.10-7 4.1014 - 8.1014Tia t ngoi 3,8.10-7 10-9 8.1014 - 3.1017Tia X 10-8 10-11 3.1016 - 3.1019Tia Gamma < 10-11 >3.1019

CHNG VII. LNG T NH SNG

1. Nng lng mt lng t nh sng (ht phtn) = h.f=

ch.

2. Tia Rnghen (tia X) min =

E

ch.

Trong 220

2 2

mvmvE e U = = + l ng nng ca electron khi p vo i catt (i m cc)

Ul hiu in th gia ant v catt; v l vn tc electron khi p vo i cattv0 l vn tc ca electron khi ri catt (thng v0 = 0) m = 9,1.10

-31 kg l khi lng electron


14


15/19

3. Hin tng quang in

*Cng thc Anhxtanh: = h.f=2

.. 2max0vmAch

viA =0

.

chl cng thot ca kim loi dng lm catt; 0 l gii hn quang in ca kim loi dng lm catt;

v0max l vn tc ban u ca electron quang in khi thot khi catt ; f, l tn s, bc sng ca nh sng kch thch

* dng quang in trit tiu th UAK Uh (Uh < 0), Uh gi l hiu in th hm20 ax

2M

h

mveU =

Lu: Trong mt s bi ton ngi ta ly Uh > 0 th l ln.* Xt vt c lp v in, c in th cc i VMax v khong cch cc i dMax m electron chuyn ng trong in trng cn ccng E c tnh theo cng thc: maxmax dWVe * Vi U l hiu in th gia ant v catt, vA l vn tc cc i ca electron khi p vo ant, vK = v0Max l vn tc ban u cc ca electron khi ri catt th: 2 21 1

2 2A Ke U mv mv= -

* Hiu sut lng t (hiu sut quang in) H = n / NVi n v n0 l s electron quang in bt khi catt v s phtn p vo catt trong cng mt khong thi gian t.

Cng sut ca ngun bc x: P =

chN

t

fN

t

N ... ; Cng dng quang in bo ho: bh

n eqI

t t= =

* Bn knh qu o ca electron khi chuyn ng vi vn tc v trong t trng u B R =sin

.

Be

vm

Xt electron va ri khi catt th v = v0MaxLu : Hin tng quang in xy ra khi c chiu ng thi nhiu bc x th khi tnh cc i lng: Vn tc ban u cc iv0Max, hiu in th hm Uh, in th cc i VMax, u c tnh ng vi bc x c Min (hoc fMax)4. Tin Bohr- Quang ph nguyn t Hir

* Tin Bohr = h.fnm=nm

ch

.

* Bn knh qu o dng th n ca electron trong nguyn t hir:rn = n2r0

Vi r0 =5,3.10-11m l bn knh Bo ( qu o K)

* Nng lng electron trong nguyn t hir:2

13,6( )

nE eV

n= -

* S mc nng lng- Dy Laiman: Nm trong vng t ngoi

ng vi e chuyn t qu o bn ngoi v qu o KLu: Vch di nht LK khi e chuyn t L K

Vch ngn nht K khi e chuyn t K.- Dy Banme: Mt phn nm trong vng t ngoi, mt phn nm trongvng nh sng nhn thy

ng vi e chuyn t qu o bn ngoi v qu o LVng nh sng nhn thy c 4 vch:Vch H ng vi e: M L; Vch lam H ng vi e: N LVch chm Hng vi e: O L; Vch tm H ng vi e: P L

Lu: Vch di nht ML (Vch H)Vch ngn nht L khi e chuyn t L.

- Dy Pasen: Nm trong vng hng ngoi ng vi e chuyn t qu obn ngoi v qu o MLu: Vch di nht NM khi e chuyn t N M. Vch ngn nht M khi e chuyn t M.

Mi lin h gia cc bc sng v tn s ca cc vch quang ph ca nguyn t hir:13 12 23

1 1 1

v f13 = f12 +f23

hfnm Hfnm

nhn phtn pht phtnEn

EmEn > Em

Laiman

K

M

NO

L

P

Banme

Pasen

HH H H

n=1

n=2

n=3

n=4n=5n=6


15


16/19

2. S co di: Khi mt thanh di lo chuyn ng vi vn tc v th di thanh b co li theo phng chuyn ng ca n.

Khng gian l tng i, ph thuc vo h quy chiu. l =2

2

0 1c

vl (l < lo)

Trong lo: chiu di khi thanh ng yn; l: chiu di khi thanh chuynng

3. S chm li ca ng h chuyn ng: ng h gn vi quan st vin chuyn ng chy chm hn ng h gn vi quan

vin ng yn. Thi gian l tng i, ph thuc vo h quy chiu:

2

2

0

1c

v

tt

(t > to)

to: thi gian o trong HQC K chuynng; t: thi gian o trong HQC K ng yn;4. Khi lng tng i tnh: Khi lng l tng i, ph thuc vo h quy chiu. Khi lng ca vt tng khi v tng.

2

2

0

1c

v

mm

mo: khi lng ngh; m: khi lng ca vt ang chuyn ng

5. H thc gia nng lng v khi lng: E = mc2 =

2

2

0

1

.

c

v

cm

Vy, khi vt c khi lng m th n c nng lng E, khi vt c nng lng E th n c khi lng m. Hai i lngny lun t

l vi nhau theo h s t l c

2

. Khi nng lng thay i E th kh

i lng cng thay i m tng ng v ngc lE = mc2

Ch :- Theo thuyt tng i, i vi h kn, mo v Eo tng ng khng nht thit c bo ton nhng NL ton phn W c boton.

220 .2

1. vmcmW

- C hc c in l trng hp ring ca c hc tng i tnh khi tc chuyn ng ca vt rt nh so vi vn tc nh sng

CHNG VIII. S LC V THUYT TNG I HP

1. Thuyt tng i hpa. Hn ch ca c hc c in

Khi xt chuyn ng ca cc vt c vn tc xp x vn tc nh sng th c hc Newton khng cn ng na.Nm 1905 Einstein xy dng thuyt tng i hp.

b. Cc tin EinsteinTin 1 (nguyn l tng i): Cc nh lut vt l (c hc, in t hc . . .) c cng mt dng nh nhau trong mi h quychiu qun tnh.

Tin 2 (nguyn l v

s bt bin ca tc nh sng): tc nh sng trong chn khng c cng ln bng c trong mi hquy chiu qun tnh, khng ph thuc vo phng truyn v vo tc ca ngun sng hay my thu. c = 299 792 458m/s 3.108m/s (tc ln nht ca vt cht trong t nhin)

CHNG IX. VT L HT NHN

1. Hin tng phng x* S nguyn t cht phng x cn li sau thi gian t N = N0.2 t/T = N0.e-t* S ht nguyn t b phn r bng s ht nhn con c to thnh v bng s ht ( hoc e- hoc e+) c to thnh:

N = No-N = N0(1- e-t)* Khi lng cht phng x cn li sau thi gian t m = m0.2 t/T = m0.e-t

Trong : N0, m0 l s nguyn t, khi lng cht phng x ban u;T l chu k bn r ; = ln2/T = 0,693/T l hng s phng x

v T khng ph thuc vo cc tc ng bn ngoi m ch ph thuc bn cht bn trong ca cht phng x.* Khi lng cht b phng x sau thi gian t m = mo-m = m0(1- e-t)

* Phn trm cht phng x b phn r: t

o

em

m

1 ; Phn trm cht phng x cn li: t

o

em

m


16


17/19

2p

1p

p

* Khi lng cht mi c to thnh sau thi gian t )1()1(. 01111tt

A

o

A

emM

Me

N

NMM

N

Nm

Trong : A, A1 l s khi ca cht phng x ban u v ca cht mi c to thnh,NA = 6,022.10

-23 mol-1 l s Avgar.Lu: Trng hp phng x +, - th A = A1 m1 = m* phng x H: L i lng c trng cho tnh phng x mnh hay yu ca mt lng cht phng x, o bng s phn rtrong 1 giy. H = H0.2

t/T = HN0.e-t = .N

H0 = N0 l phng x ban u. n v: Becren (Bq); 1Bq = 1 phn r/giy hay Curi (Ci); 1 Ci = 3,7.1010 BqLu: Khi tnh phng x H, H0 (Bq) th chu k phng x T phi i ra n v giy(s).

2. H thc Anhxtanh, ht khi, nng lng lin kt* H thc Anhxtanh gia khi lng v nng lng

Vt c khi lng m th c nng lng ngh E = m.c2

Vi c = 3.108 m/s l vn tc nh sng trong chn khng.* ht khi ca ht nhn A

ZX m = m0 m

Trong m0 = Zmp + Nmn = Zmp + (A-Z)mn l khi lng cc nucln. m l khi lng ht nhn X.* Nng lng lin kt E = m.c2 = (m0-m)c2* Nng lng lin kt ring (l nng lng lin kt tnh cho 1 nucln):

A

E

Lu: Nng lng lin kt ring cng ln th ht nhn cng bn vng.

3. Phn ng ht nhn

* Phng trnh phn ng:4

443

332

221

11 X

AZX

AZX

A

ZXAZ

Trong s cc ht ny c th l ht s cp nh nucln, eletrn, phtn ... Trng hp c bit l s phng x: X1 X2 + X3X1 l ht nhn m, X2 l ht nhn con, X3 l ht hoc

* Cc nh lut bo ton+ Bo ton s nucln (s khi): A1 + A2 = A3 + A4+ Bo ton in tch (nguyn t s): Z1 + Z2 = Z3 + Z4+ Bo ton ng lng: 4321 pppp

+ Bo ton nng lng: KX1 + KX2 + E = KX3 +KX4Trong : E l nng lng phn ng ht nhn 21

2X x xK m v= l ng nng chuyn ng ca ht X

Lu: - Khng c nh lut bo ton khi lng trong cc phnng ht nhn.- Mi quan h gia ng lng pX v ng nng KX ca ht X l: 2 2X X X p m K = - Khi tnh vn tc v hay ng nng K thng p dng quy tc hnh bnh hnh

V d: 21 ppp

bit ),( 21 pp

hay cos...2 2122

21

2

1ppppp

hay cos...2)()()( 22112

222

112 vmvmvmvmmv

hay cos...2 22112211 KmKmKmKmmK

Tng t khi bit ),( 11 pp

hoc ),( 22 pp

Trng hp c bit: )( 21 pp

2 2 21 2p p p= +

Tng t khi )( 1 pp

hoc )( 2 pp

v = 0 (p = 0) p1 = p21 1 2 2

2 2 1 1

K v m A

K v m A= = Tng t v1 = 0 hoc v2 = 0.

* Nng lng phn ng ht nhn E = (M0 - M)c2Trong :

1 20 X XM m m= + l tng khi lng cc ht nhn trc phn ng.

3 4X XM m m= + l tng khi lng cc ht nhn sau phn ng.

Lu: - Nu M0 > M th phn ng to nng lng E di dng ng nng ca cc ht X3, X4 hoc phtn .Cc ht sinh ra c ht khi ln hn nn bn vng hn.

- Nu M0 < M th phn ng thu nng lng E di dng ng nng ca cc ht X1, X2 hoc phtn .Cc ht sinh ra c ht khi nh hn nn km bn vng.


17


18/19

* Trong phn ng ht nhn4

443

332

2

21

11 X

A

ZXA

ZXA

ZX

A

Z

Cc ht nhn X1, X2, X3, X4 c:Nng lng lin kt ring tng ng l 1, 2, 3, 4.Nng lng lin kt tng ng l E1, E2, E3, E4 ht khi tng ng l m1, m2, m3, m4Nng lng ca phn ng ht nhn

E = A33 +A44 - A11 - A22 = E3 + E4 E1 E2 = (m3 + m4 - m1 - m2)c2

* Quy tc dch chuyn ca s phng x

+ Phng x ( 42He ): YAZHeXAZ 4242 So vi ht nhn m, ht nhn con li 2 trong bng tun hon v c s khi gim 4 n v.

+ Phng x - ( 10e- ): YAZeX

AZ 1

01

So vi ht nhn m, ht nhn con tin 1 trong bng tun hon v c cng s khi.Thc cht ca phng x - l mt ht ntrn bin thnh mt ht prtn, mt ht electrn v mt ht ntrin:

n p + e- + Lu: - Bn cht (thc cht) ca tia phng x - l ht electrn (e-)

- Ht ntrin (v) khng mang in, khng khi lng (hoc rt nh) chuyn ng vi vn tc ca nh sng v hunh khng tng tc vi vt cht.+ Phng x + ( 10 e

+ ): YAZeXAZ 1

01

So vi ht nhn m, ht nhn con li 1 trong bng tun hon v c cng s khi.Thc cht ca phng x + l mt ht prtn bin thnh mt ht ntrn, mt ht pzitrn v mt ht ntrin:p n + e+ +

Lu: Bn cht (thc cht) ca tia phng x + l ht pzitrn (e+)+ Phng x (ht phtn)

Ht nhn con sinh ra trng thi kch thch c mc nng lng E1 chuyn xung mc nng lng E2 ng thi phng ra

mt phtn c nng lng = h.f=

ch.= E1 E2

Lu: Trong phng x khng c s bin i ht nhn phng x thng i km theo phng x v .4. Cc hng s v n v thng sdng* S Avgar: NA = 6,022.1023 mol-1* n v nng lng: 1eV = 1,6.10 -19 J; 1MeV = 1,6.10-13 J

* n v khi lng nguyn t (n v Cacbon): 1u = 1,66055.10-27

kg = 931,5 MeV/c2

* in tch nguyn t: e = 1,6.10-19 C* Khi lng prtn: mp = 1,007276u* Khi lng ntrn: mn = 1,008665u* Khi lng electrn: me = 9,1.10-31kg = 0,00055u

TM TT CHNG X. T VI M N V M

1. Ht s cp- Ht s cp l nhng ht c kch thc v khi lng nh hn ht nhn nguyn t. c trng chnh ca cc ht s cp l:+ Khi lng ngh m0, ht c nng lng ngh E0 = m0c2.+ S lng t in tch q ca ht s cp c th l +1, -1, 0 (tnh theo in tch nguyn t e).+ S lng spin s l i lng c trng cho chuyn ng ni ti ca ht s cp.

+ Thi gian sng trung bnh. Ch c 4 ht s cp khng phn r thnh cc ht khc, l prtn, lectron, phtn, ntrin; cli l cc ht khng bn c thi gian sng rt ngn, c t 10-24s n10-6s, tr ntron c thi gian sng l 932s.+ Phn ln cc ht s cp u to thnh cp: ht v phn ht.Phn ht c cng khi lng ngh, cng spin, in tch c cng ln nhng tri du.- Cc ht s cp c phn thnh 4 loi: phtn, leptn, mzn v barion. Mzn v barion c gi chung l harn.C 4 loi tng tc c bn i vi ht s cp l: tng tc hp dn, tng tc in t, tng tc yu, tng tc mnh. - Tt c cc harn u c cu to t ht quac.

C 6 loi quac l u(up), d(down), s(strange), c(charm), b(bottom) v t(top). in tch cc ht quac l 3

e,

2

3

e.

Cc barion l t hp ca ba quac. VD: proton (u,u,d); ntron (u,d,d)Quan nim hin nay v cc ht thc s l s cp gm cc quac, cc leptn v cc ht truyn tng tc l glun, phtn, W ,

v gravitn.


18


19/19

2. H Mt Tri- H Mt Tri gm Mt Tri trung tm h; 8 hnh tinh ln v cc v tinh ca n gm Thu tinh, Kim tinh, Tri t, Ho tinMc tinh, Th tinh, Thin Vng tinh v Hi Vng tinh. Cc hnh tinh ny chuyn ng quanh Mt Tri theo cng mt chv gn nh trong cng mt phng. Mt Tri v cc hnh tinh cn t quay quanh mnh n.Khi lng Mt Tri bng 1,99.1030kg, gp 333000 ln khi lng Tri t. Khong cch t Tri t n Mt Tri xp x 1triu km, bng 1 n v thin vn.- Mt Tri gm quang cu v kh quyn Mt Tri. Mt Tri lun bc x nng lng ra xung quanh. Hng s Mt Tri l H1360W/m2. Cng sut bc x nng lng ca Mt Tri l P = 3,9.1026W. Ngun nng lng ca Mt Tri chnh l cc phn nhit hch. thi k hot ng ca Mt Tri, trn Mt Tri xut hin cc vt en, bng sng nhiu hn lc bnh thng.- Tri t c dng phng cu c bn knh xch o bng 6378km, c khi lng l 5,98.1024kg. Mt Trng l v tinh ca Tt c bn knh 1738km v khi lng l 7,35.1022kg. Gia tc trng trng trn Mt Trng l 1,63m/s2.

Bng so snh cc hnh tinh trong h mt tri vi tri t

Thin th K/c ti mttri (vtv)

Bn knh

(km)

Khi lng sovi tri t

Khi lngring(10

3kg/m

3)

Chu k quay

quanh trcChu k quay

quanh Mt triS v tin

bitThy tinhKim tinh

Tri tHa tinhMc tinhTh tinh

Thin vng tinhHi vng tinh

0,39

0,72

1

1,52

5,2

9,54

19,19

30,07

2.440

6.056

6.375

3.395

71.490

60.270

25.760

25.270

0,055

0,81

1

0,11

318

95

15

17

5,4

5,3

5,5

3,9

1,3

0,7

1,2

1,7

59 ngy

243 ngy

23h56ph

24h37ph

9h50ph

14h14ph

17h14ph

16h11ph

87,9 ngy

224,7 ngy

365,2422 ngy

1,88 nm11,86 nm29,46 nm

84 nm164,80 nm

0

0

1

2

63

34

27

13

3. Sao. Thin h- Sao l mt khi kh nng sng ging nh Mt Tri nhng rt xa Tri t. a s sao trng thi n nh. Ngo i ra c mt ssao c bit nh sao bin quang, sao mi, sao ntron. Khi nhin liu trong sao cn kit, sao tr thnh sao ln, sao ntron hoc en.- Thin h l h thng gm nhiu loi sao v tinh vn.

Ba loi thin h chnh l thin h xon c, thin h elip, v thin h khng nh hnh.Thin H ca chng ta l thin h xon c c ng knh khong 100 ngn nm nh sng, khi lng bng 150 t l

khi lng Mt Tri, l h phng ging nh mt ci a, dy khong 330 nm nh sng, cha vi trm t ngi sao. H Mt Trnm ra Thin H, cch trung tm khong 30.000 nm nh sng v quay vi tc khong 250 km/s.

4. Thuyt Big BangTheo Thuyt Big Bang, v tr c to ra bi mt v n cc ln, mnh cch y khong 14 t nm, hin ang dn n vlong dn. Hai hin tng thin vn quan trng l v tr dn n v bc x nn v tr l minh chng ca thuyt Big Bang.

Tai liu Vt ly n thi: TN - CD - DH

Documents

Ly thuyet Vat ly 12 NC