Tổng hợp công thức lớp 12

7/28/2019 Tng hp cng thc lp 12

1/16

CHNG 0. C HC VT RN1. NG HC VT RN* Lin h gia tc gc v tc di

v = R

* Gia tc tip tuyn

an= R

2

* Gia tc php tuyna

t= R

* Gia tc ton phn

a = at

2+ a

n

2

* Tc gc trung bnh

=

t

* Gia tc gc trung bnh

=

t

* Gia tc gc tc thi = t+

0

* Gc quay c trong thi gian t

=1

2t

2+

0t

* Cng thc c lp vi thi gian

2

0

2= 2.

* Gc quay c trong giy th n

n= (n 0,5)+0

* Chu k quay u

T = 1f=2

* Tc quay

=n

30

n: svng quay/pht

2. MOMEN QUN TNH* Cng thc tng qut

I = (mR2)

* Momen qun tnh ca vnh trn ng chtI = mR

2

* Momen qun tnh ca a trn ng cht

I =1

2mR

2

* Momen qun tnh ca thanh quay quanh 1 uI =

1

3ml

2

* Momen qun tnh ca thanh quay quanh trc gia

I =1

12ml

2

* Momen qun tnh ca hnh cu ng cht

I =2

5mR

2

CNG THC LP 12

gocriengtrenban.com6 1


2/16

* Momen qun tnh ca mt cu ng cht

I =2

3mR

2

* Cng thc nh l Huyghens

IM= I

O+md

2

d: khong cch MO

3. NG LC HC VT R N* Momen lc

MF/O = F.dd: cnh tayn

* Phng trnh ng lc hc vt rn

=M

I

=M

FM

can

I

* o hm ca momen ng lng

M =dL

dt4. NNG LNG* Momen ng lng

L = I

* Bo ton momen ng lng cho 2 a

I11 + I22 = (I1 + I2 )* ng nng quay

Wd=

1

2I

2

* ng nng va quay va tnh tin

W =1

2I

2+1

2mv

2

* Lin h gia ng nng quay v momen ng lng

Wd=

L2

2I

CNG THC LP 12

2


3/16


4/16

* Lc ko vF

kv= kx

* Lc n hiF

dh= k | l

0+ x |

* Lc n hi ln nhtFdhmax

= k(l0+ A)

* Lc n hi nh nhtFdhmin

= k | l0

A |

* Chiu di l xo v tr cn bng

lVTCB

=lmax

+ lmin

2

* Bin con lc

A = lmax

lmin

2

* L xo ghp song song

k//= k

1+ k

2

1

T2=

1

T12

+1

T22

* L xo ghp ni tip1

k//

=1

k1

+1

k2

T

2

= T12

+T22

k1l

1= k

2l

2

* Ghp hai vt nng m = m1 + m2

T2= T

1

2+T

2

2

* L xo b ct k1

k2

=

l2

l1

* iu kin vt m khng trt trn vt MkA

(m +M) g

: hsma st gia m v Mk:cng ca l xo (N/m)

3. CON L C N* Tn s gc, chu k con lc n

=gl

T = 2 lg

* Lin h gia n v di v n v gcs = l v = l '

v

2=

2gl(coscos0 )

* Tc cc iv0= l

0

* Lin h chu k v chiu di dy

T1

2

T22

=

l1

l2

* Lc cng dyT = mg(3cos 2cos

0)

CNG THC LP 12

4


5/16

* Lc cng dy ln nhtTmax =

mg(3 2cos0 )

* Lc cng dy nh nhtT = mgcos

0

* Lc ko vFkv = mgsin

* Thay i nhit

T2

2

T1

2= 1+t

: hsndi (K-1)t:bin thin nhit

* Thay i caoT2

T1

=

R + h

R

R: bn knh trit

* Chu tc dng ca lc ngoi

T2

2

T12=

g

g '

g ' = g a g ' = g

2+ a

2

a: gia tc ngoi (m/s2)

* Lc in

a = q Em

E: cngin trng (V/m)

* Chnh lch thi gian ca ng h

t= t.(1T

2

T1

)

t: thi gian xem xtT1: chu kng

t > 0:ng hchy nhanh

* Chu k con lc b vng inh

T = (l1

g+

l2

g)

4. NNG LNG CON L C* ng nng con lc

Wd=

1

2

mv2

* Th nng con lc l xo

Wt=

1

2kx

2

* Th nng con lc n

Wt = mgl(1 cos) =1

2m

2l2

2

* C nng tng qutW =W

t+W

d

* C nng con lc l xo

W =1

2m

2A

2=

1

2kA

2

CNG THC LP 12

5


6/16

* C nng con lc n

W =1

2m

2l2

0

2

* Lin h li v nng lngW

t

W= (

x

A)2= (

0

)2=

1 cos

1 cos0

* Lin h vn tc v nng lngW

d

W= (

v

vmax

)2

5. DAO NG T T DN - CNG HNG* Bin con lc gim sau N chu k

A =4NFc

k=

4Nmg

k* S dao ng thc hin cho n khi tt hn

N=A

A=

kA

4F=

kA

4mg

* Qung ng i c cho n lc tt hn

s =kA

2

2F=

kA2

2mg

* Chu k ngoi lc xe chy trn rnh

T =s

v6. T NG HP DAO NG* Cng pha

A = A1 + A2 = Amax

* Ngc qua

A = |A1 - A2| = Amin* Vung pha

A = A1

2+ A

2

2

* Cng bin

A = 2A1cos(

2)

* Bin tng qut

A2= A

1

2+ A

2

2+ 2A

1A

2cos

* Pha tng hp

tan=A

1sin

1+ A

2sin

2

A1cos

1+ A

2cos

2

7. CON L C TRNG PHNG* S ln con lc gp nhau cho n khi gp li

N1T1= N

2T2

CNG THC LP 12

6


7/16

CHNG 2. SNG C HC1. SNG C HC TNG QUT* Bc sng ca sng c hc

=v

f

* lch pha gia 2 im trn phng truyn

= 2d

* Phng trnh lan truyn sng

u = Acos(2(t

T

x

))

* Tc lan truyn sng

v =FT

: khi lng trn mtn vchiu di dyFT: lc cng dy

2. GIAO THOA SNG

* Bc giao thoa

k=d

* Bc giao thoa ca hai bin b rng giao thoa L

k=L

* Phng trnh giao thoa

u = 2A cos(d

2 d

1)

cos(t

(d2 d

1)

)

3. SNG DNG* S b sng dng

k=2l

* Bc sng di nht trn dy 2 u cnh= 2l

* Bc sng di nht trn dy 1 u cnh= 4l

* Bin mt im trong b/4 - 2a; /12 - a

/8 - a2/2; /6 - a3/2

4. SNG M* Cng m

I =P

4R2

* T s cng m

I1

I2

=

R2

2

R1

2

* Mc cng m

L = log(I

I0

) LA L

B= log(

IA

IB

)

* Trung im

dM=dA+ d

B

2 d

M=

dA d

B

2

CNG THC LP 12

7


8/16

CHNG 3. DNG IN XOAY CHIU1. I CNG IN XOAY CHIU* Gi tr hiu dng

I =I0

2 U=

U0

2

* Cng tc thi

i =U

R

R

i

I0

2

+uL

U0L

2

= 1 i

I0

2

+uC

U0C

2

= 1

* in p tc thiu = u

R

+ uL

+ uC

* Cm khngZL = L = 2 fL

* Dung khng

ZC =1

C=

1

2 fC* Tng tr

Z= R2+ Z

L Z

C( )2= (R + r)

2+ Z

L Z

C( )2

* in p ton mch

U= UR2

+ ULUC( )

2

* lch pha

tan=Z

L Z

C

R=

ULU

C

UR

* Cng dng in hiu dngI =

U

Z=

UR

R=

UL

ZL

=

UC

ZC

R :ZL

:ZC=U

R:U

L:U

C

* H s cng sut

cos= R

Z=UR

U

* Cng sut in

P =UIcos= RI2=

U2

Rcos

2

P = (R + r)I2

* Cng hng in: I max, P maxZ

L= Z

C Z= R

UL=U

CU=U

R

=0=

1

LC f = f0 =

1

2 LC

u, i cng phacos= 1 P =UI

0

=

f0

f

=

ZC

ZL

* u1 vung pha vi u2tan

1tan

2= 1

* iu kin URL khng ph thuc vo RZ

C= 2Z

L

CNG THC LP 12

8


9/16

* iu kin URC khng ph thuc vo RZ

L= 2Z

C

2. KHI R THAY I* P max khi ch c R

R = ZL Z

C

Pmax

=

U2

2R cos=

2

2

* P max khi c R, r

R + r = ZL Z

C

Pmax

=U

2

2(R + r) cos=

2

2

* PR max khi c R, r

R = ZL Z

C( )2

+ r2

Pmax =U

2

2(R + r)

* R = R1; R = R2 cho cng P

P = U

2

R1+ R

2

R1R

2= Z

L Z

C( )2

(R2

0= R

1R

2 P

max=

U2

2 R1R

2

)

3. KHI L THAY I* P, I, UC max: cng hng

L =1

C2

* UL max

ZL=R2 + Z2C

ZC

; uRC v u vung pha

ULmax

=

U R2+ Z

2

C

R= U

2+U

2

L+U

2

R

(U2Lmax

UCU

Lmax U2 = 0 )

* URL max

ZL=

ZC+ Z

2

C+ 4R

2

2

URLmax

=2RU

4R2+ Z

2

C Z

C

* L = L1; L = L2 cho cng P

ZC= Z

L0=

ZL1

+ ZL2

2* L = L1; L = L2 cho cng UL

1

ZL1

+1

ZL2

=2Z

C

R2+ Z

2

C

1

ZL

=

1

2 (

1

ZL1

+

1

ZL2

) L =2L

1L

2

L1+ L

2

(UL max)

4. KHI C THAY I

CNG THC LP 12

9


10/16

* P, I, UL max: cng hng

C=1

L2

* UC max

ZC=R

2+ Z

2

L

ZL

; uLR v u vung pha

UCmax =

U R2+ Z

2

L

R= U

2

+U

2

C +U

2

R

(U2

CmaxU

LU

CmaxU

2= 0 )

* URC max

ZC=

ZL+ Z

2

L+ 4R

2

2

URCmax

=2RU

4R2+ Z

2

L Z

L

* C = C1; C = C2 cho cng P

ZL= Z

C0=ZC1 + ZC2

2

* C = C1; C = C2 cho cng UC

1

ZC1

+1

ZC2

=2Z

L

R2+ Z

2

L

1

ZC

=1

2(1

ZC1

+1

ZC2

)C=C1 +C2

2(UC max)

5. KHI , f THAY I* P, I max: cng hng

=1

LC

* UL max

=1

CL

C

R2

2

ULmax

=

2LU

R 4LC R2C

2

* UC max

=1

L

L

C

R2

2 U

Cmax=

2CU

R 4LC R2C

2

* = 1; = 2 cho cng P

1

2=

0=

1

LC

* = 1; = 2 cho cng UL

1

1

2+

1

2

2= LC C

2

R2

2

* = 1; = 2 cho cng UC

1

1

2+

1

2

2=

L2

L

R

R2

2

6. MY BI N P

CNG THC LP 12

10


11/16

* My l tngU

1

U2

=

N1

N2

=

I2

I1

* My khng l tng

H=Ptt

Ptp=

RI2

2

U1I1

H: hiu sut my bin p

7. TRUY N TI IN NNG I XA* in tr dy

R = l

S

:in trsut (m)

* Cng sut hao ph

P =RP

2

U2cos

2

* Hiu sut truyn ti

H =100%P

P

* gim thU = I.R

8. MY PHT IN XOAY CHI U* Tn s dng in

f =np

60

n: scp ccp: svng quay/pht

* T thng cc i gi qua khung dy

0 = NBSN: svng

B: cmng t(T)S: din tch khung dy (m2)

* Sut in ng cc i sinh raE

0=

0

* H thc c lp gia t thng v sut in ng

0

2

+E

E0

2

= 1

* Mch mc hnh sao

Id = Ip

Ud = 3Up

P = 3UpIp cos

* Mch mc hnh tam gic

Id = 3Ip

Ud =Up

P = 3UdIdcos

CNG THC LP 12

11


12/16

CHNG 4. DAO NG IN T1. DAO NG IN T* Tn s dao ng

=1

LC f =

1

2 LC

* Cng dng in cc i

I0=Q

0 =U

0

C

L* Cng thc c lp vi thi gian

q = Cu

i

I0

2

+u

U0

2

= 1 i

I0

2

+q

Q0

2

= 1

* Nng lng in t trng

WL =1

2Li

2

WC=1

2Cu

2=1

2 qu=1

2

q2

C

W =WL +WC

* Nng lng hao ph

P = RI2= R

I0

2

2

2. SNG IN T* Bc sng in t

= 2c LC

* Ghp tC

//= C

1+C

2

1

Cnt

=1

C1

+1

C2

* T xoay

C : gc quayc

* Lin h gia bc sng v t

1

2

2

=

C1

C2

* Ghp cun dy Lnt= L

1+ L

2

1

L//

=1

L1

+1

L2

* Lin h gia bc sng v L

1

2

2

=

L1

L2

CNG THC LP 12

12


13/16

CHNG 5. SNG NH SNG1. TN SC, KHC X NH SNG* Lin h gia chit sut v bc sng

1

2

=

n2

n1

(f khng i)

* nh lut khc x nh sngn1sini = n

2sinr

n1i = n2r

r = rdo r

tim

* Cng thc lng knhA = r

1+ r

2

D = i1+ i

2 A

D = Dtim D

do

i = i ' =D

min+ A

2 r = r ' =

A

2

* Cng thc thu knh

1f= (n 1)( 1

R1

+ 1R

2

)

2. GIAO THOA NH SNG* Hiu quang l

r=

xa

D

* Khong vn giao thoa

i =D

a

* Bc giao thoa

k=x

i=

r

* Mi trng trong sut chit sut n

in=

i

n

* t bn mng trn ng i nh sng

x0= n 1( )

eD

a

* Dch chuyn ngun sng ln xung

x0 =D

D1

d

D1: khong cch tngunn 2 khed:di ca ngun sng

x0:di ca hvn/VSTT

* B rng quang phx

1= x

1d x

1t

* Quang ph chng lnh = x

1d x

2t

* Vn sng trng nhau

k1

1

= k2

2

CNG THC LP 12

13


14/16

CHNG 6. LNG T NH SNG1. HIN TNG QUANG IN* Nng lng ca photon

= hf =hc

=

1,9875.1025

* Cng thot

A =hc

0

* ng nng ban u cc i ca electron

Wd0max

=

1

2mv

0max

2

* Cng thc quang in

= A +Wd0max

* Hiu in th hm/in th cc i

eUh=W

d0max

* Cng dng quang in

I =n

e.e

t

* Cng sut bc x chiu ti

P =np.

t

* Hiu sut quang inH=

ne

np

* ng nng electron p vo anotW

dAW

d0max

= eUAK

* Bn knh electron bay trong t trng u

R =mv

eB

B: cmng t(T)m: khi lng electron

* ng nng electron p vo i catot (ng tia X)W

d=Q +

2. QUANG PH HIDRO* Nng lng electron mc n

En

=

13,6

n2 (eV

)

* Bc sng bc x khi chuyn di

Em En = hfmn =hc

mn

1

31

=

1

32

+

1

21

* Bn knh mc nng lng

r= n2r0

* Vn tc electron mc n

v =

9.109e2

r.n

* Nng lng ion ho t mc n: En

CNG THC LP 12

14


15/16

CHNG 7. VT L HT NHN1. TNG QUT V HT NHN* ht khi

m = Zmp + (A Z)mn mX

* Nng lng lin kt/Nng lng phn r

W = m.c2= m.931(MeV)

* Nng lng lin kt ring

w =W

A

* S mol

n =N

NA

=

m

A

2. PHNG X* Hng s phng x

=ln2

T=

0,693

T

* Lng cn li

N = N02

t/T= N

0e

T

m = m02t/T

= m0eT

n = n02t/T

= n0eT

H = H02t/T

= H0eT

* S ht sinh ra

N = N0 (1 2t/T

)

* T s ht mi v c sau thi gian t

= 2t/T1

* Khi lng cht mi

mY=

AY

AX

m0(1 2

t/T)

* phng x

H = N1 Ci = 3,7.1010Bq1 Bq = 1 phn r/s

* Tc phn r/chiu x

t=N

H

N: sphn r

3. PHN NG HT NHN* Bo ton s khi/in tch

At= A

s

Zt= Z

s

* Nng lng to ra/thu vo

W = mtm

s( )c2= m

s m

t( )c2= W

s W

t

* ng nng - ng lng

p = mv = 2mK

pt = ps* ng lng trong phng x

pY = p

Y: ht nhn con

CNG THC LP 12

15


16/16

PH LCMT S CNG THC THNG GP KHC* Mi quan h gia 2 i lng cng pha, ngcpha

x1

x01

=

x2

x02

* Mi quan h gia 2 i lng cng pha, ngc

phax1

x01

2

+x2

x02

2

= 1

x1,x2: cc gi trtc thix01,x02: cc gi trcci

* Cng thc log, mx

ax

b= x

a+b

(xa)b= x

ab

elnx

= x

CNG THC LP 12

Documents

Tổng hợp công thức lớp 12