Công thức lớp 10

8/14/2019 Cng thc lp 10

1/15

CNG THC VT L LP 10

Ch :-Ti liu ny khng bao gm tt c cc cng thc vt l c s

dng trong chng trnh lp 10, m ch mc thng k li mcy nht c th.-Cc cng thc ny c son theo chng trnh Vt l lp 10Nng cao. Nhng cng thc no chng trnh Chun khng dnghoc t dng sc ghi ch bng k hiu

-Mi thc mc, ng gp vui lng gi v a ch http://gocriengtrenban.wordpress.com

Cng thc lp 10 1/15

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8/14/2019 Cng thc lp 10

2/15

CHNG: NG HC CHT IM

STT NI DUNG GHI CH

ACc khi nim c bn:

1 Ta :

x = OM

2 Khong cch:

d= x2 x

1

3 Qung ng:

s = x

x0

Chuyn ng 1 chiu

4 Vn tc trung bnh:

v =s

t

=s1+ s

2

t1+ t

2

5 Gia tc:

a = v v0

t

B Chuyn ng thng u:

6 Phng trnh chuyn ng:

x= vt+ x0

Bt u ty, cc cng thcu ly t0 = 0.Nu t0 0, thay t bng (t - t0)Chiu ca v l chiu chuynng.

Du ca v ph thuc vochiu dng chn.

7 Khi 2 xe gp nhau:

x1= x

2

C Chuyn ng thng bin i u:

8 Phng trnh chuyn ng:

x=

1

2at

2

+ vt+ x0

Nhanh dn u: a, v cngdu.Chm dn u: a, v trnh du.

Cng thc lp 10 2/15

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8/14/2019 Cng thc lp 10

3/15

STT NI DUNG GHI CH

9 Phng trnh vn tc:

v = at+ v0

Thc cht l cng thc tnhvn tc tc thi.

10 Cng thc c lp vi thi gian:

v2 v

0

2= 2as

Khi khng c thi gian th nhn cng thc ny.

11 Vn tc trung bnh:

v =v + v

0

2

Ch p dng cho chuyn ngthng bin i u.

D S

r

i t

do:

12 Ta ri t do:

y =1

2gt

2+ v

0t

Chn gc ta ti v tr th,chiu dng hng xung.

13 Qung ng ri t do sau t giy u tin:

s =1

2

gt2

Cho vt th ri khng vn tcu.

14 Qung ng vt ri trong 1 giy cui:

s= h st1

t: thi gian vt ri chm t.st-1: qung ng vt ritrong gian (t - 1).

15 Vn tc chm t:

vcd = 2gh

Cho vt th ri khng vn tcu.Nu c vn tc u th pdng cng thc khng phthuc vo thi gian.

16 Thi gian chm t:Thay y = h vo phng trnh ta

17 cao cc i:

hmax

=v0

2

2g+ h

Chn gc ta ti mt t,chiu dng hng ln.h: cao nm vt.

E Chuyn ng trong h quy chiu quntnh (chuyn ng tng i)

Cng thc lp 10 3/15


8/14/2019 Cng thc lp 10

4/15

STT NI DUNG GHI CH

18 Cng thc Galilei:

v13 = v12 + v23 Cng thc vn tc tng i.

F Chuyn ng trn u:

19 Lin h gia n v di v n v gc:

= R.s

= R.v

n v di: mn v gc: radrad = 1800n v rad ca mt gc c gitr bng di cung trn cbn knh R = 1 m b chn bigc .

20 Gia tc hng tm:

aht=

v2

R= R

2

Ch c trong chuyn ng

cong.

21 Lc hng tm:

Fht= m

v2

R

Ch c trong chuyn ngcong.

22 Chu k:

T =2R

v=

2

23 Tn s:

f =1

T

24 @ Chuyn ng trn bin i u:

a = aht + att 2 vect gia tc vung gc vinhau.att: gia tc tip tuyn.Qu o hnh xon c chkhng cn trn na.

Cng thc lp 10 4/15


8/14/2019 Cng thc lp 10

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CHNG: NG LC HC CHT IM

STT NI DUNG GHI CH

A Tng hp, phn tch lc:1 Phn tch lc:

Fx = Fcos

Fy = Fsin

: gc hp gia lc F vphng Ox.

2 Tng hp lc khng cng phng (nh lhm cos)

F2

= F12

+ F22

+ 2F1F2 cos

: gc hp gia F1 v F2

B 3 nh lut Newton:

3 nh lut 2 Newton:

a =

Fhl

m

Fhl=

m

a

4 nh lut 3 Newton:

F12 = F21C Cc loi lc:

5 Trng lc:

P=

m

g6 Lc hp dn:

Fhd= G

m1m

2

r2

G : H n g s h p d n(6,67.10-11 Nm2/kg2)

7 Lc hp dn ca tri t:

Fhd= G

mM

(R + h)2

M: Khi lng tri tR: Bn knh tri tm: khi lng ca vth: cao ca vt.Ch : Nu vt nm trong hth khng dng cng thc ny.

Cng thc lp 10 5/15


8/14/2019 Cng thc lp 10

6/15

STT NI DUNG GHI CH

8 Lc n hi ca l xo:

Fdh= kl

k: cng ca l xo (N/m)

9 Lc ma st:Fms= N

: h s ma st (khng cn v)H s ma st ny ty thucvo chuyn ng trt hayln.

10 Lc hng tm:

Fht= m

v2

R= m

2R

Ch xut hin trong chuynng trn.

11 Lc qun tnh:

Fqt = m

a

Ch xut hin trong h quychiu phi qun tnh.

D Cc cng thc khc:

12 Gia tc trng trng:

g =GM

(R + h)2

13 Pt chuyn ng nm ngang:

y =g

2v0

2x2 + h

Qu o ng parabol, nh v tr nm.

14 @ Pt chuyn ng nm xin:

y =

g

2v0

2cos

x2 + (tan)x

15 Tm xa:

L =v0

2sin2

g

Chuyn ng nm ngang choy = 0

Cng thc lp 10 6/15


8/14/2019 Cng thc lp 10

7/15

STT NI DUNG GHI CH

16 @ Tm cao:

H =v0

2sin

2

2g

Chuyn ng nm ngangkhng c tm cao

17 cao cc i:

hmax

=v0

2

2g+ h

Chn gc ta ti mt t,chiu dng hng ln.h: cao nm vt.

18 @ Vn tc chuyn ng nm xin:

v2

= v2

x + vy2

vx: vn tc theo phngngang (khng i)

vy: vn tc theo phng thngng (ri t do)

19 @ H vt: Dy khng co gin

T1= T

2= T

a1= a

2= a

Cng thc lp 10 7/15


8/14/2019 Cng thc lp 10

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TNH HC VT RN

STT NI DUNG GHI CH

A Mmen lc1 nh ngha:

MF/O

= F.d

d: cnh tay n ca lc

2 Quy tc momen lc:

Mdongho =

Mnguocdongho

3 Mmen ngy lc:

M = 2F.d

B Quy tc hp lc song song:

4 Song song cng chiu:

F= F1 + F2

F1

F2

=d2

d1

(chia trong) F1d1 = F2d2

5 Song song ngc chiu:

F= F1 F2

F1

F2

=

d2

d1

(chia ngoai) F1d1 = F2d2

Cng thc lp 10 8/15


8/14/2019 Cng thc lp 10

9/15

CC NH LUT BO TON

STT NI DUNG GHI CH

A ng lng:

1 Cng thc ng lng

p = mv2 nh lut bo ton ng lng:

p = p ' m1v1 + m2v2 = m1v '1+ m2v '2Chn chiu dng v chiu.

3 Xung lng:

Ft=

p

Ch chn chiu dng vchiu.

B Nng lng:

4 Cng:

A = F.s.cos

Ch p dng cho trng hplc khng i v qu othng.

5 Cng sut trung bnh:

P=

A

t

6 @ Cng sut tc thi:

P = Fv = F.v.cos7 ng nng:

Wd=

1

2

mv2

8 Lin h gia ng nng v cng:

Wd=W

d2W

d1= A

F

Cng ca ngoi lc F.

9 Th nng trng trng:

Wt =

mgz

Ch chn gc th nng.

10 Lin h gia th nng trng trng v cng:

Wt =Wt1 Wt2 = Ap

Cng thc lp 10 9/15


8/14/2019 Cng thc lp 10

10/15

STT NI DUNG GHI CH

11 Cng ca trng lc (ri)

AP = mgh

Khi vt i ln th thm du tr-

12 Th nng n hi:

Wt=

1

2kx

2

Chn gc th nng v tr tnhin (khng co gin)

13 Lin h gia th nng n hi v cng:

Wt=W

t1W

t2= A

Fdh

14 C nng:

W =Wt+W

d

15 nh lut bo ton c nng:

Wd1+W

t1=W

d2+W

t2

C nng ch bo ton khikhng c ngoi lc khng th.

16 cao ng nng bng n ln th nng:

h =h0

n +1

Nu th nng bng m lnng nng th thay n = 1/mCh p dng khi lm bi trcnghim hoc kim tra kt qu.

17 Hiu sut:

H =Aci

Atp

Aci: Cng c chAtp: Cng ton phn

C Va chm:

18 Va chm mm:

v ' =m

1v1+ m

2v2

m1+ m

2

ng lng c bo ton.Cc i lng u tnh theogi tri s.

19 @ Va chm n hi:

v '1=

(m1 m

2)v

1+ 2m

2v

2

m1 + m2

ng lng v c nng cbo ton.Cc i lng u tnh theogi tri s.

Cng thc lp 10 10/15


8/14/2019 Cng thc lp 10

11/15

@ C HC CHT LU

STT NI DUNG GHI CH

A p sut

1 nh ngha p sut

p =F

S

F: p lc (N)S: din tch (m2)

2 iu kin cn bng p sut:

pA = pB

pA v pB l p sut theo 2chiu ngc nhau.

3 p sut ca ct cht lng:p = gh

: khi lng ring (kg/m3)

p sut tnh:

p = gh + p0

p0: p sut kh quyn (Pa)

p sut ng:

p=

1

2 v

2

B Cc nh lut:

4 Nguyn l Pascal:

F1

F2

=

S1

S2

=

d1

d2

p sut truyn i nguyn vntrong lng cht lng:d: on dch chuyn ca dintch

5 nh lut Bernoulli1

2v

2+ gh = const

Cng thc lp 10 11/15


8/14/2019 Cng thc lp 10

12/15

CHT KH

STT NI DUNG GHI CH

A Cc qu trnh bin i cht kh:

1 Qu trnh ng nhit:

p1V1= p

2V2

nh lut Boyle-Mariotte

2 Qu trnh ng tch:

p1

T1

=

p2

T2

nh lut Charles

3 Qu trnh ng p:

V1

T1

=

V2

T2

nh lut Gay-Lussac

4 Phng trnh trng thi kh l tng:

p1V1

T1

=

p2V2

T2

5 @ Phng trnh Claperon - Mendeleev:

pV = nRT

n: s molR: hng sNu p tnh bng atm, V tnhbng lt th R = 22,4/273

B Nng lng trong cc qu trnh:

6 Nguyn l 1:

Q = U+ A

Quy c:Q > 0: nhn nhitQ < 0: ta nhit

A > 0: nhn cngA < 0: sinh cng

7 Qu trnh ng p:

A = pV

Bnh xilanh.

8 Qu trnh ng tch:

A = 0

9 Qu trnh truyn nhit:Q = mC(T1 T2 )

C: nhit dung ring ca cht(J/kg.K)

Cng thc lp 10 12/15


8/14/2019 Cng thc lp 10

13/15

STT NI DUNG GHI CH

10 Qu trnh ng nhit:

U = 0

11 Qu trnh chuyn pha:Q = m

: nhit ha hi (nhit halng) (J/kg)

12 Hiu sut ng c nhit:

H =Q

1Q

2

Q1

13 Hiu sut ng c nhit l tng:

H =T1

T2

T1

Hiu sut cao nht ca ng

c nhit

Cng thc lp 10 13/15


8/14/2019 Cng thc lp 10

14/15

CHT RN

STT NI DUNG GHI CH

A Bin dng c

1 bin dng ti:

=

l

l0

2 Sut n hi:

=

F

S

n v: Pa

3 Lc n hi tng qut:

F= ESl

l0

E: sut Young (Pa)

H sn hi:

k=ES

l0

Cn gi l cng ca l xo(N/m)

B Bin dng nhit

6 S n di:

l= l0(1+t)

: h s n di (K-1)

7 S n khi:

V = V0 (1+ t)

: h s n khi (K-1)

Cng thc lp 10 14/15


8/14/2019 Cng thc lp 10

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CHT LNG

STT NI DUNG GHI CH

A Hin tng cng b mt:

1 Lc cng b mt tng qut:

F= l

: sut cng b mt (N/m)

2 Lc cng b mt dc b mt khung dy:

F= 2l

l: chiu di thanh trt

3 Lc cng b mt khi khung dy nhc ln:

F= l

l: chu vi khung dy

B Cc cng thc tnh chu vi:

4 Chu vi hnh ch nht:

l= (a + b).2

a, b: chiu di v rng

5 Chu vi hnh trn:

l= 2R

6 Chu vi hnh xuyn:

l= 2(R1+ R

2)

Cng thc lp 10 15/15


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Công thức lớp 10