Nhieu Cach Giai Gon de DH Ly 2014

7/21/2019 Nhieu Cach Giai Gon de DH Ly 2014

1/18

GV: L ThThoTHPT Nhn ChnhThanh Xun - H Ni

1

([email protected])

HNG DN GII THI TUYN SINH I HC KHI A, A1 NM 2014Mn thi : VT LM : 319

THI GM 50 CU (TCU 1 N CU 50) DNH CHO TT C TH SINH.Cho bit: hng s Plng h=6,625.10-34J.s; ln in tch nguyn t e = 1,6.10-19C; tc nh sng trong chn khng c = 3.108 m/s1uc

2= 931,5 MeV.

Chng S cu Ghi chChng 1: Dao ng c 10 cuChng 2: Sng c v sng m 07 cu

Chng 3: Dng in xoay chiu 12 cuChng 4: Dao ng v sng in t 04 cuChng 5: Sng nh sng 07 cuChng 6: Lng tnh sng 04 cuChng 7: Vt l ht nhn 06 cu

Tng 50 cu

Chng 1: Dao ng c(10 cu)

Cu 1 (C44): Mt cht im dao ng iu ha vi phng trnh x 6 tcos (x tnh bng cm, t tnh bng s). Pht biu no sau y

ng? A. Tc cc i ca cht im l 18,8 cm/s.B. Chu k ca dao ng l 0,5 s.C. Gia tc ca cht im c ln cc i l 113 cm/s2.D. Tn s ca dao ng l 2 Hz.Gii : Tc cc i ca cht im: vmax =.A =.6=18,8(cm/s) Chn A

Cu 2 (C43): Mt vt dao ng iu ha vi phng trnh )cos(5 tx (cm). Qung ng vt i c trong mt chu k l:A. 10 cm B. 5 cm C. 15 cm D. 20 cm

Gii: s = 4A= 20 (cm) Chn D

Cu 3 (C12): Mt vt khi lng 50g dao ng iu ha vi bin 4cm v tn s gc 3 rad/s. ng nng cc i ca vt l:A. 7,2 J. B. 3,6.10

-4J. C. 7,2.10

-4J. D. 3,6 J.

Gii: 22max 2

1AmWW

)(10.6,34

J

Chn B

Cu 4 (C7): Mt vt nh dao ng iu ha trn qu o thng di 14 cm vi chu k 1s. T thi im vt qua v tr c li 3,5 cmtheo chiu dng n khi gia tc ca vt t gi tr cc tiu ln th hai, vt c tc trung bnh l:A. 28 cm/s. B. 28,0 cm/s. C. 27,0 cm/s. D. 26,7 cm/s.

Gii: Gia tc a= -2x => amin ti v tr x =A (bin dng).+ Vt qua v tr amin ln th nht ng vi gc qut 1 =/3, Vt qua v tr amin ln th hai th qutthm 2 => thi gian: t= T/6+T=7T/6 = 7/6(s)+ Qung ng: s = A/2+4A = 4,5 A= 31,5 (cm)+ Tc trung bnh: v =s/t= 27 (cm/s) Chn C

Cu 5( C35): Con lc l xo dao ng iu ha theo phng ngang vi tn s gc . Vt nh ca con lc c khi lng 100g. Ti thiim t=0, vt nh qua v tr cn bng theo chiu dng. Ti thi im t =0,95s, vn tc v v li x ca vt tho mn v =-x ln th 5Ly 2=10. cng ca l xo l:A. 85 N/m B. 37 N/m C. 20 N/m D. 25 N/m

Gii 1: + PT dao ng ca vt c dng: x = Acos(t -2

).

+ Ti thi im t = 0,95s /v /= 22 xA = /-x/ /x /=2

2A.

x(M1

M2

5,23/ O 3,573/


2/18


2

+ Trong mt chu k vt qua v tr c v = -x hai ln Ln th 1,3,5 c x =2

2A, ln 2,4:

2

2Ax vt qua v tr c

v = -x ln 5 ti thi im t = t1 +2T , vi t1 l khong thi gian vt i t VTCB n bin A v quay li

v tr x =2

2A

+ t1 =8

3T. Do t = T

T2

8

3

8

19T= 0,95 T= 0,4 s

+ T =

2 = 5 =

m

k k = 252m = 25 N/m Chn D

Gii 2: Chn t = 0 lc vt qua VTCB theo chiu (+) => x = Acos(t-/2) (1*) => v = x=-Asin(t-/2) (2*)+ Ti thi im t = 0,95s c v =-x (3*). T (2*) v (3*) => ti t: x = A. sin(t-/2) (4*)

+ T (1*) v (4*) => tan(t-/2)=1 =>T

2t-/2 =/4+ k => t =

28

3 Tk

T (vi kN*)

+ Ln 1 ng vi k=0, ln 2 ng vi k=1.ln 5 => k=4 => t= sTTTT

4,095,08

19

24

8

3

+ 2

2

2

2

4,0

1,0.44

2

T

m

kk

m

T )/(25 mN

Chn D

Cu 6(C22): Mt con lc l xo treo vo mt im c nh, dao ng iu ha theo phng thng ng vi chu k 1,2 s. Trong mchu k, nu t s ca thi gian l xo dn vi thi gian l xo nn bng 2 th thi gian m lc n hi ngc chiu lc ko v l

A. 0,2 s B. 0,1 s C. 0,3 s D. 0,4 s

Gii: + Ta c: tdn+tnn=T m tdn =2 tnn => gc qut M2OM3 = 2/3+ Lc ko v (lc hi phc) lun hng v v tr cn bng, khi l xo dn th lc n hi hng ln, khi lxo nn th lc n hi hng xung => Khi vt dao ng t v trA/2 -A -A/2 v t OAO (VTCB)(ng vi gc qut trn cung M2M3 v cung M4AM1) lc hi phc v lc n hi cng chiu.+ Lc n hi v lc ko v ngc chiu ng vi gc qut trn cung M1M2 v M3M4 => =2./6=/3=> t=T/6 = 0,2s p n A

Cu 7(C1): Mt con lc l xo gm l xo nh v vt nh khi lng 100g ang dao ng iu ha theo phng ngang, mc tnh th

nng ti v tr cn bng. T thi im t1 = 0 n t2 =48

s, ng nng ca con lc tng t 0,096 J n gi tr cc i ri gim v 0,064

J. thi im t2, th nng ca con lc bng 0,064 J. Bin dao ng ca con lc lA. 5,7 cm. B. 7,0 cm. C. 8,0 cm. D. 3,6 cm.

Gii 1: + T t1 n t2 ng nng ca con lc tng t 0,096 J n gi tr cc i ri gim v 0,064 J => thi im t1 x1 0 hocx1>0, v1 W = 0,128(J) .

+ thi im t1:

24

3

128,0

096,01 12

2

1

2

22

1 Ax

A

x

A

xA

W

W

+ Xt x1 0 (trng hp x1>0, v1 t=T/12+T/8= 5T/24 =/48 => T =/10 => =20 (rad/s)+ W=(1/2) m2A2 => 0,128 =(0,1/2).202. A2 => A= 8.10-2(m) = 8(cm) Chn C

Gii 2: Ti thi im t2 W = Wt C nng ca h W = W + Wt = 0,128 J

Ti t1 = 0 Wt1 = WW1 = 0,032J =4

W/x1/=

2

A. Gi s ti t1 x1=-A/2

Th ti t2 =48

(s) c x2 =

2

2A. Thi gian vt i t x1 = -

2

An gc ta ri n x2 =

2

2A

x

M

A

5,23/

O

M3 -A/2

M4

0

6/

x

M2

5,23/O

M1/6 /4

-A/2 2/A

M0

M1

5,23/O

M2;4

/4

M0

M1;

5,23/O

M2;4

/4


3/18


3

t =12

T+

8

T=

24

5T= t2t1 =

48

T =

10

1(s) Tn s gc ca dao ng =

T

2= 20 rad/s

W =2

2

maxmv=

2

22Am

A =2

2

m

W=

400.1,0

128,0.2= 0,08 m = 8 cm Chn C

Cu 8(C36): Mt con lc n dao ng iu ha vi bin gc 0,1 rad; tn s gc 10 rad/s v pha ban u 0,79 rad. Phng trnh

dao ng ca con lc l

A. 0 1 20 t 0 79 rad, cos( , )( ) B. 0 1 10t 0 79 rad , cos( , )( )

C. 0 1 20 t 0 79 rad, cos( , )( ) D. 0 1 10t 0 79 rad , cos( , )( )

Cu 9(C16): Mt vt dao ng cng bc di tc dng ca mt ngoi lc bin thin iu ha vi tn s f. Chu k dao ng ca v

l A.1

2 f. B.

2

f

. C. 2f. D.

1

f.

Gii: Chu k dao ng ca vt bng chu k ca kc cng bc => Tvt =Tcb = 1/f Chn D

Cu 10(C40): Cho hai dao ng iu ha cng phng vi cc phng trnh ln lt l 1 1x A t 0 35 cmcos( , )( ) v

2 2x A t 1 57 cmcos( , )( ) . Dao ng tng hp ca hai dao ng ny c phng trnh l x 20 t cmcos( )( ) . Gi tr

cc i ca (A1 + A2) gn gi tr no nht sau y?A. 25 cm B. 20 cm C. 40 cm D. 35 cm

Gii 1: T gin vc t, p dng nh l hm s sin ta c:

)35cos(55sin2)20sin()90sin(70sin 0021

0

2

0

1

0

AAAAA

=> 00max21 70sin/)55sin.2()( AAA )(87,34 cm

Chn A (Khi )35;1)35cos(00

Gii 2:21

2

2121

2

21

2

2

2

1

0

21

2

2

2

1

2 684,2202)684,0(2)(70cos2 AAAAAAAAAAAAAAAA

=>4

)(684,2

20)( 221222121

AAAAAA (V )2 2121 AAAA

=> 1,316 1600)(2

21 AA => max21 )( AA )(87,34 cm Chn AHt chng 1

020

070

0902A

1A

A


4/18


4

Chng 2: Sng c(7cu)

Cu 11(C49): Mt sng c truyn trn mt si dy rt di vi tc 1m/s v chu k 0,5s. Sng c ny c bc sng lA. 150 cm B. 100 cm C. 50 cm D. 25 cm

Gii: Bc sng: =v.T = 0,5 (m) =50(cm) Chn C

Cu 12(C6): c lng su ca mt ging cn nc, mt ngi dng ng h bm giy, gh st tai vo ming ging v th mhn ri t do t ming ging; sau 3 s th ngi nghe thy ting hn p vo y ging. Gi s tc truyn m trong khngkh l 330 m/s, ly g = 9,9 m/s2. su c lng ca ging l

A. 43 m. B. 45 m. C. 39 m. D. 41 m.

Gii1: Ta c )2(876,247,358'019806609,9)3(2

1

2

1

2

11

2

1 ttttvgt

h

=> )876,23(330h )(41m Chn D

Gii 2: v

h

g

htts

2)(3 21 )(41mh Chn D

Cu 13(C5): Mt sng c truyn dc theo mt si dy n hi rt di vi bin 6 mm. Ti mt thi im, hai phn t trn dy cnglch khi v tr cn bng 3 mm, chuyn ng ngc chiu v cch nhau mt khong ngn nht l 8 cm (tnh theo phng truyn sng)

Gi l t s ca tc dao ng cc i ca mt phn t trn dy vi tc truyn sng. gn gi tr no nht sau y?

A. 0,105. B. 0,179. C. 0,079. D. 0,314.

Gii: Ta c: )(248.22

3

2cm

d

. Tc sng: v=f =24.f (cm/s);

Tc dao ng cc i ca mt phn t trn dy l: vd =A=2f. (0,6) (cm/s)

Suy ra: T s: 2024

6,0.2

f

f

v

vd

0,157 Chn B

Lu : C th tnh : Hai phn t gn nhau nht c li 2

Achuyn ng ngc chiu nhau cch nhau d =

3

= 8 cm = 24 cm.

Cu 14(C9): Trong mt th nghim giao thoa sng nc, hai ngun S1 v S2 cch nhau 16 cm, dao ng theo phng vung gc vmt nc, cng bin , cng pha, cng tn s 80 Hz. Tc truyn sng trn mt nc l 40 cm/s. mt nc, gi d l ng trung

trc ca on S1S2. Trn d, im M cch S1 10 cm; im N dao ng cng pha vi M v gn M nht s cch M mt on c gi trgn gi tr no nht sau y? A. 7,8 mm. B. 6,8 mm. C. 9,8 mm. D. 8,8 mm.

Gii 1: Bc sng: )(5,0 cmf

v ;

TH1: N nm ngoi on OM lch pha dao ng ca 2 im M so vi N trn trung trc d ca AB:

+ N dao ng cng pha vi M khi: kddkddk 1212 )(2

+ Hai im M1 v M2 gn M nht dao ng cng pha vi M ng vi )(5,105,01012 cmkdd

+ Khong cch MN: )(8,081085,10 2222 cmMN )(8mm

TH2: N nm trong on OM lch pha dao ng ca 2 im N so vi M trn trung trc d ca AB:

)(2 21 dd

+ N dao ng cng pha vi M khi: kddkddk 2121 )(2

+ Hai im M1 v M2 gn M nht dao ng cng pha vi M ng vi )(5,95,01012 cmkdd

+ Khong cch MN: )(88,085,9810 2222 cmMN )(8,8 mm

Vy (MN)min = 8 (mm) => Gi tr gn nht l 7,8 (mm) Chn A

u(mm

6

5,23/O 3/2 A/2

M

N

3

O

M

N

10

8cm 8cmS1

d1

d2

S1O

M

NS2

10

8cm 8cm

d1

d2

)(2 12 dd


5/18


5

Gii 2: Bc sng: )(5,0 cmf

v ; + Gi s )cos(21 tAuu SS .

+ PT sng ti 1 im trn trung trc: )2

cos(2

dtAu

+ Nhn xt:

40

.2M

d=> M, N dao ng cng pha vi ngun.

+ Cc im dao ng cng pha vi ngun tha mn:

2.

2

k

d

=> kkd 5,0 168 k+ dM =10 cm => k =20 => c 2 v tr ca N gn M dao ng cng pha vi M ng vi k =19 v k =21

+ TH 1: k =19 => dN = 9,5 (cm) => )(88,085,9810 2222 cmMN )(8,8 mm

+ TH 2: k =21 => dN = 10,5 (cm) => )(8,081085,10 2222 cmMN )(8mm

Vy (MN)min = 8 (mm) => Gi tr gn nht l 7,8 (mm) Chn A

Cu 15(C33): Trn mt si dy n hi ang c sng dng n nh vi khong cch gia hai nt sng lin tip l 6 cm. Trn dy cnhng phn t sng dao ng vi tn s 5 Hz v bin ln nht l 3 cm. Gi N l v tr ca mt nt sng; C v D l hai phn t trndy hai bn ca N v c v tr cn bng cch N ln lt l 10,5 cm v 7 cm. Ti thi im t1, phn t C c li 1,5 cm v ang

hng v v tr cn bng. Vo thi im2 1

79t t s

40 , phn t D c li l

A. -0,75 cm B. 1,50 cm C. -1,50 cm D. 0,75 cm

Gii : + Khong cch gia hai nt sng lin tip = /2=6 (cm) => =12(cm)

+ Bin ca mt im cch nt mt khong l d l :

daA

2sin2

=> )(5,112

7.2sin3);(25,1

12

5,10.2sin3 cmAcmA DC

+ lch pha dao ng ca phn t C thi im t1 v thi im t1 + s l: 75,11840

79

.2 f+ Li ca C thi im t2 l )(25,1 cm , tc l ang bin (+).

+ V C v D nm hai bn b sng lin k nn chng lun dao ng ngc pha. Do , khi C bin dng th D ang bin m

Vy li ca D l DD Ax )(5,1 cm Chn CCu 16(C11): Trong mi trng ng hng v khng hp th m, c 3 im thng hng theo ng th t A; B; C vi AB = 100 mAC = 250 m. Khi t ti A mt ngun im pht m cng sut P th mc cng m ti B l 100 dB. B ngun m ti A, t ti Bmt ngun im pht m cng sut 2P th mc cng m ti A v C l

A. 103 dB v 99,5 dB B. 100 dB v 96,5 dB.

C. 103 dB v 96,5 dB. D. 100 dB v 99,5 dB.

Gii:

+ Lc u t ngun m ti A:

2

00 4lg10lg10100

B

B

RI

P

I

I

=>

10

2

0

104

BRI

P

+ Khi t ngun m 2P ti B:

)10.2lg(.10

4

2lg10

10

2

0 A

ARI

PL

)(103dB

2

10

22

0

2

0 5,1

10.2lg.10

5,1.4

2lg10

4

2lg10

AC

CRI

P

RI

PL

)(488,99 dB Chn A

M0

25,15,1

M

A

B C

100m 150mB C

A


6/18


6

Cu 17(C38): Trong m nhc, khong cch gia hai nt nhc trong mt qung c tnh bngcung vna cung (nc). Mi qungtm c chia thnh 12 nc. Hai nt nhc cch nhau na cung th hai m (cao, thp) tng ng vi hai nt nhc ny c tn s tha

mn12 12

c tf 2f . Tp hp tt c cc m trong mt qung tm gi l mtgam (m giai). Xt mtgam vi khong cch t nt n

cc nt tip theo R, Mi, Fa, Sol, La, Si, tng ng l 2 nc, 4 nc, 5 nc, 7 nc, 9 nc, 11 nc, 12 nc. Tronggam ny, nu m ng vnt La c tn s 440 Hz th m ng vi nt Sol c tn s l

A. 330 Hz B. 392 Hz C. 494 Hz D. 415 Hz

Gii1: + V hai nt nhc cch nhau na cung th hai m (cao, thp) tng ng vi hai nt nhc ny c tn s tha mn

1212 .2tc

ff Hai tn s lin tip cch nhau na cung c t s bng 2(1/12) =1,059. Nt SON v nt LA cch nhau 2nc nn ta c

22

059,1

400

059,1

Ls

ff )(392 Hz Chn B

Gii2: + Ta c: 121212121212 )1(12

)2(

1212

)1( 44404)2.(2)2.(2.2 LSSLtnccncctncc fffffffff )(392Hz

Chn B

Gii 3: Ta c: 1212 )2(12

)3(

1212

)1(

12

)2(

1212

)1( 2.2.2.2;.2.2.2.2 tnccncctnccncctncc ffffffff1212

).( 2 tN

ncNc ff

=>12 2Ntc ff Vi N l s na cung cch nhau gia fc( cao) v ft (thp)

Ht chng 2


7/18

GV: L ThThoTHPT Nhn ChnhThanh X

Cu 18(C48): in p u 141 2 100 tcos A. 141 V B. 200 V

Cu 19(C47): Dng in c cng i 2

trn in trl A. 12 kJ B. 24 kJGii: Nhit lng ta ra trn in tr: 2IQ

Cu 20(C24): t in p 0u U 100cos

l 0i I 100 t Acos . Gi tr ca

Gii: on mch xoay chiu ch c t in th i

Cu 21(C26): Mt on mch in xoay chiubng R. lch pha ca in p gia hai u o

A.4

. B. 0.

Gii: lch pha gia u v i: tanR

ZL

Cu 22(C2): t in p xoay chiu n nh v cm thun c cm khng ZL v 3ZL = 2ZC. in p gia hai u on mch MB nh hnh v

A. 173V. B. 86 V.

Gii 1: Ta c T = 2.10-2s = 100 rad/s; u

Ta c: LxMBxCAN uuuuuu ; (1*);

+ T (1*) => xMBxCAN uuuuu 33;222

+ T (2*) v (3*) =>4

5

32

uuu MBAN

x

+ Hiu in th hiu dng: 7410xU

un - H Ni

7

Chng 3: in xoay chiu (12 cu)

(V) c gi tr hiu dng bng C. 100 V D. 282 V

2 100 tcos (A) chy qua in trthun 100 . Tron

C. 4243 J D. 8485 J

)(10.1230.100.2 32 Jt )(12 kJ Chn A

V4

vo hai u on mch ch c t in th c

bng A.3

4

. B.

2

. C.

3

4

.

nhanh pha /2 so vi u => =/4+/2 = 3/4 Chn

gm in trthun R mc ni tip vi mt cun cm thn mch vi cng dng in trong mch bng

C.2

D.3

.

1 4/ Chn A

o hai u on mch AB mc ni tip (hnh v). Bit t th biu din s ph thuc vo thi gian ca in p gi . ip p hiu dng gia hai im M v N l

C. 122 V. D. 102 V.

N = 200cos100t (V) uMB = 100cos(100t +3

) (V)

ZL = 2ZC => LC uu 32 (2*)

Lu3 (3*) .

cos37205

)3

100cos(300)100cos(0

tt

)(023,6 V Chn B

g 30 giy, nhit lng ta ra

ng dng in trong mch

D.2

.

un c cm khng vi gi tr

in c dung khng ZC, cun a hai u on mch AN v

)100( t


8/18


Cu 22(C2): t in p xoay chiu n nh vcm thun c cm khng ZL v 3ZL = 2ZC. in p gia hai u on mch MB nh hnh v

A. 173V. B. 86 V.

Gii 2: Ta c T = 2.10-2s = 100 rad/s

+ uAN = 200cos100t (V) uMB = 100cos(100t

+ T 3ZL = 2ZC UC = 1,5UL

+ V gin vc t nh hnh v: OC= UMB = 5

=> tam gic OCD vung gc ti C ( =/2) =>

+ Do UMN = UX =22

LMB UU = 50(

Gii 3: Ta c T = 2.10-2s = 100 rad/s; u

+ V ZC = 1,5ZL => LC UU

5,1

+ Ta c: 5,15,1 MBLXMB UUUU

XAN UUU

+ T (*) v (**) =>ANMBX UUU

5,15,2+ UAN = 100 2 (V) ; UMB = 50 2 (V) gc t

Nn XU .5.1.22.1002.50.5,15,222222

Gii 4: +Ta c T = 2.10-2s = 100 rad/s;

+ V gin vc t nh hnh v: OC= UMB = 5gc COD =/3=> tam gic OCD vung gc ti

+V ZC = 1,5ZL => LC UU

5,1+ Ta c: tanMBL UU (*) v 5,2 LU

+ Hiu in th hiu dng: UMN = UX=cos

MBU

un - H Ni

8

o hai u on mch AB mc ni tip (hnh v). Bit t th biu din s ph thuc vo thi gian ca in p gi . ip p hiu dng gia hai im M v N l

C. 122 V. D. 102 V.

3

) (V)

2 (V), OD= UAN = 100 2 (V) =2OD; gc COD =/3

L + UC = 50 6 (V) UL = 20 6 (V)

22 )620() = 86,02V Chn B

N = 200cos100t (V) uMB = 100cos(100t +3

) (V)

(*)5,1 LX U

.

(**)5,1 LXC UU

o biMB

U

vAN

U

l3

XU3

cos.2.100.50

)(02,86 V

AN = 200cos100t (V) uMB = 100cos(100t +3

) (V)

2 (V), OD= UAN = 100 2 (V) =2OD; ( =/2)

)3/tan(MB =>5,2

3tan

tan

)3/tan(5,2

)7,34cos(

2500

)(86V Chn B

in c dung khng ZC, cun a hai u on mch AN v

07,34

LU

XU

ANU

D

MBU

3/

C

U

C

LU

XU

O

ANU

D

MBU

3/

CU

C

UL+UC

C

L

XU

O

ANU

C

D

MBU

3/


9/18


9

Cu 23(C30): t in p u U 2 t Vcos (vi U v khng i) vo hai u on mch mc ni tip gm n si t cghi 220V100W, cun cm thun c t cm L v t in c in dung C. Khi n sng ng cng sut nh mc. Nu ni thai bn t in th n ch sng vi cng sut bng 50W. Trong hai trng hp, coi in trca n nh nhau, b qua t cm ca

n. Dung khng ca t in khng th l gi tr no trong cc gi tr sau? A. 345 . B. 484 . C. 475 . D. 274 .

Gii: in trca n l: )(484100

22022

m

m

P

UR .

+ Ni tt hai bn t, cng sut gim mt na nn ta c:2222

22

2

22

2

)()(5,0)(

5,0CLL

CLL

ZZRZRZZR

RU

ZR

RU

=> 0,5R

2+0,5ZL

2= R

2+ZL

2-2 ZL ZC+ZC

2=> 0,5ZL

2-2 ZL ZC +ZC

2+ 0,5R

2=0 => =ZC

20,5ZC2-0,25R

2= 0,5ZC

2-58564

K 0CZ => 23968,342CZ Chn D

Cu 24(C4): t in p u = 180 2 cos t (V) (vi khng i) vo hai u on mch AB (hnh v). R l in tr thun, tin c in dung C, cun cm thun c t cm L thay i c. in p hiu dng hai u on mch MB v ln gc lch

pha ca cng dng in so vi in p u khi L=L1 l U v1, cn khi L = L2 th tng ng l 8 U v2. Bit1 +2 = 900. Gi

tr U bngA. 135V. B. 180V. C. 90 V. D. 60 V.

Gii 1: T bi v gin vc t nh hnh v, ta c: 222 )8(180 UU => )(60VU Chn D

Gii 2: V 021 90 nn )(sin:1180

8

1801sinsin

2

2

2

2

2

2

1

2

U

UUU MB

=> )(60VU Chn D

Gii 3:8

1tan

8cossin;sin 1121

ABAB U

U

U

U

180.3

1

3

1sin

sin

11cot 1

1

2

2Ug

)(60V Chn D

Gii 4: Do ln0

21 90 nn 1tan.tan 21 => )8(2 UUCL

1. 21 R

ZZ

R

ZZ LCCL . t x= CL ZZ 1 ;

2LC ZZy =>2

Rxy (1) +8

18

180(*);

180

22

22

222

221

xRy

yRxU

yR

yUU

xR

xU CLCL (2)

+ T (1) v (2) => 0784224 RxRx

22

Rx Thay vo (*) ta c U = 60 V Chn D

RC

A L BM

)( 180ABU

1RU

2RU

)8(2 UUMB

(1 UUMB

1

2.

I

2LU

)8(2 UUCL

)(1 UUCL

ABU

ABU

2

1

1CU

I

1LU

)8(2 UUCL

)(1 UUCL

ABU

ABU

2

1

2CU


10/18


10

Cu 24(C4): t in p u = 180 2 cos t (V) (vi khng i) vo hai u on mch AB (hnh v). R l in tr thun, tin c in dung C, cun cm thun c t cm L thay i c. in p hiu dng hai u on mch MB v ln gc lch

pha ca cng dng in so vi in p u khi L=L1 l U v1, cn khi L = L2 th tng ng l 8 U v2. Bit1 +2 = 900. Gi

tr U bngA. 135V. B. 180V. C. 90 V. D. 60 V.

Gii 5: UMB = 22

2

)(

)(

CL

CLAB

ZZR

ZZU

; 1, 2 l ln

+ tan1 =R

ZZ LC 1; tan2 =

R

ZZ CL 2 m 1 +2 = 900 R2 = (ZC- ZL1)( ZL2- ZC)

+ U =2

1

2

2

1

)(

)(

CL

CLAB

ZZR

ZZU

, 8 U =

2

2

2

2

2

)(

)(

CL

CLAB

ZZR

ZZU

8 (ZL1- ZC)2

[R2

+ (ZL2ZC)2] = (ZL2- ZC)

2[R

2+ (ZL1ZC)

2]; 8 (ZL1- ZC)

2[R

2+

2

`1

4

)( CL ZZ

R

] =

2

`1

4

)( CL ZZ

R

[R

2+ (ZL1ZC)

2]

8 (ZL1- ZC)2

[1 +2

`1

2

)( CL ZZ

R

] =

2

`1

2

)( CL ZZ

R

[R

2+ (ZL1ZC)

2] => 8 (ZL1- ZC)

2= R

2;

U =2

1

2

2

1

)(

)(

CL

CLAB

ZZR

ZZU

=

8

8180

22 R

R

R

= 60V Chn D

Cu 25(39): t in p xoay chiu c gi tr hiu dng U = 200 V v tn s khng thay i vo hai u on mch AB (hnh v)

Cun cm thun c t cm L xc nh; R = 200 ; t in c in dung C thay i c. iu chnh in dung C in p hiudng gia hai u on mch MB t gi tr cc tiu l U1 v gi tr cc i l U2 = 400 V. Gi tr ca U1 lA. 173 V B. 80 V C. 111 V D. 200 V

(Lu:Bi ton gii theo cch khng sdng dliu R =200)

Gii: + in p hiu dng cc i gia 2 u on mch MB l: (AD CT c chng minh di)

LL

MB

ZZR

URUU

)4(

2

22max2 )400(2

).4(

..2

22

U

ZZR

RU

LL

=> RZL 5,1

+ in p hiu dng cc tiu gia 2 u on mch MB l:22

min1

L

MB

ZR

URUU

222 5,1

.200

RR

R)(111 V

Chn C

* Chng minh cng thc (Cu 39 - m 319)

UMB =22

22

)( CL

C

ZZR

ZRU

=

22

22)(

C

CL

ZR

ZZR

U

=Y

U(1*)

Y = )''

(;2)(

222

222

22

22

v

uvvu

v

uy

ZR

ZZZZR

ZR

ZZR

C

CCLL

C

CL

R

CA L BM

BA

L R C

NM

BA

L X C

NM

I

2LU

)(1 UUCL

ABU

2

1

1CU

ABU


11/18


11

Y =

222

22222

)(

2)2()2)((

C

CCCLLCLC

ZR

ZZZZZRZZZR

222

32223222

)(

2422)2222(

C

CCLLCCCCLCL

ZR

ZZZZZRZZZZRZRZ

=

222

22

)(

)(2

C

CLCL

ZR

RZZZZ

Y = 0 khi ZC =

2

LZ

2

4 22 RZZZ LLC

Khi UMB =

LL

MB

ZZR

URUU

222max

4

2

(Chng minh: Bin i y tm maxMBU khi bit CZ tng ng

+ Ta c:2

4

2

4 2222 RZZRZZZZZ LLLL

LCL

+222222

222222

222

2

222

2

22

22

4424

4424

2

4

2

4

)(

LLLL

LLLL

LL

LL

C

CL

ZRZRZZR

ZRZRZZR

RZZR

RZZR

ZR

ZZRy

LLLL

LLL

LLL

LLL

LLL

ZZR

ZZR

ZZRZR

ZZRZR

ZRZZR

ZRZZRy

22

22

2222

2222

2222

2222

4

4

)4(4

)4(4

44

44

+ Trc cn thc mu s =>

2

222

4

4

R

ZZRy LL

=>

R

ZZRy

LL

2

4 22 (2*)

Thay 2(*) vo (1*) =>

LL

MB

ZZR

URUU

222max

4

2(pcm) )

* Tm minMBU +22

22

)( CL

C

MB

ZZR

ZRUU

- Xt 2 trng hp:

+ TH1: minMBCL UZZ khi ZC =0=> 22minmin11L

MB

ZR

URUUU

)

11

(

2

2U

MS

U

R

Z

U

L

(hoc => ngay UMBmin


12/18


12

( Hoc TH2:

22

2

222

22

22

2

22

22

2

min122

2

)(

)(

)(

LCCLL

C

LCL

CMB

CLZR

R

ZZZZR

ZR

ZR

R

ZZR

ZR

U

UUZZ

00

2

0

2)( 2222222242222224

MS

ZZRZZ

MS

ZRZZRZRRZZRZZRR CLLCCCLLLCCL(vi mi C tha mn

CL ZZ )

Vy:22

min1

L

MB

ZR

URUU

(pcm) )


13/18


13

Cu 26(C41): t in p u = U 2 2 ftcos (f thay i c, U t l thun vi f) vo hai u on mch AB gm on mch AMmc ni tip vi on mch MB. on mch AM gm in trthun R mc ni tip vi t in c in dung C, on mch MB chc cun cm thun c t cm L. Bit 2L > R2C. Khi f = 60 Hz hoc f = 90 Hz th cng dng in hiu dng trong mch ccng gi tr. Khi f = 30 Hz hoc f = 120 Hz th in p hiu dng hai u t in c cng gi tr. Khi f = f1 th in p hai u onmch MB lch pha mt gc 1350 so vi in p hai u on mch AM. Gi tr ca f1 bng.

A. 60 Hz B. 80 Hz C. 50 Hz D. 120 Hz

Gii: Lu : U t l vi f

+ Vi tn s'

1f v'

2f th 21 II 22

2

2

2'

1

2'

11CRLC

(1)

+ Vi tn s 3f v 4f th 21 CC UU 43

43

11

LC

LC(2)

+ Vi tn s 1f th CR UU 221

2 1

CR

(3)

+ Thay (2), (3) vo (1), =>

2'

2

2'

143

2

1

1121

22222

1129600

5

129600

492.9

)180(

1

)120(

1

240.60

21

=> 5/3602

1 => )(5,801 Hzf Chn B

* Chng minh cng thc (Cu 41 - m 319)

* Lu : U t l vi f Gi U l hiu in th ng vi 1 n v

+ U t l vi f Vi tn s '1f v'

2f th 21 II =>2

1

1

2

1

)'

1'(

'

CLR

U

=

2

2

2

2

2

)'

1'(

'

CLR

U

2

/

1

/

1

22/

2

2

/

2

/

2

22/

1 )1

()1

(C

LRC

LR

22/

1

22/

1

22/

222/

2

22/

2

22/

1

1212

CC

LLR

CC

LLR

22/

1

2/

2

2/

222/

1

2/

2

22/

222/

2

2/

1

2/

122/

2

2/

1

22/

1

22

CC

LLR

CC

LLR

C

L

CR

C

L

CR

2)()(

2))(()(

22/

1

2/

2

2/

2

2/

122/

2

2/

122/

1

2/

2

2/

2

2/

1

2/

2

2/

122/

2

2/

1

22

2

2

2'

1

2'

11CRLC

+ U t l vi f Vi tn s 3f v 4f th 43 CC UU 2

33

2

33

)

1

( CLR

ZU C

=

2

424

2

44

)

1

( CLR

ZU C

R2

+ (3L -C3

1

)

2= R

2+ (4L -

C4

1

)

2(3L -

C3

1

) = - (4L -

C4

1

)

CC

L43

43

11)(

LC

143 (pcm)

900

RU

CU

MBU

AMU

045


14/18


15/18


Ch

Cu 30(C42): Trong mch dao ng LC l tnqua cun cm thun bin thin iu ha theo thA.lun ngc pha nhauC. vi cng bin

Cu 31(C34): Mt mch dao ng LC l tndng in cc i trong mch l I0. Dao ng i

A. 0

0

4 QT

I

B. 0

0

QT

2I

Gii :C

Q

2

2

0 =2

2

0LI LC =

2

0

2

0

I

Q; T = 2 L

Cu 32(C31): Mt t in c in dung C tchc t cm L2 th trong mch c dao ng i cun cm thun c t cm L3=(9L1+4L2) th

A. 9 mA. B. 4 mA.

Gii: +0

2

020

1

01

1;

1Q

CLIQ

CLI

+ ;254.4949 111213 LLLLLL

Cu 33(C17): Hai mch dao ng in t LC l

mch l 1i v 2i c biu din nh hnh v.

bng

A.4

C

B.3

C

Gii : T th ta suy ra c phng trnh bi

.81i

Suy ra biu thc in tch tng ng l:2

81q

T ta c: )(2000cos(021 tQqq

un - H Ni

15

ng 4: Dao ng v sng in t(4 cu)

g ang c dao ng in t t do, in tch ca mt bn t i gian

B. lun cng pha nhau

D. vi cng tn s

ang c dao ng in t t do vi in tch cc i cn t t do trong mch c chu k l

C. 0

0

2 QT

I

D.

0

3 QT

I

=>0

02

I

QT

Chn C

in Q0. Nu ni t in vi cun cm thun c t c n t t do vi cng dng in cc i l 20mA ho rong mch c dao ng in t t do vi cng dng

C. 10 mA. D. 5 mA.

12

1

2

2

02

01 44 LLL

L

I

I

5/525 01031

3

03

01 IIL

LmA4 C

tng ang c dao ng in t t do vi cc cng

ng in tch ca hai t in trong hai mch cng m

C.5

C

D.10

C

u din dng in trong mi mch l

2000cos(10.6);)(2

2000cos(10 323

tiAt

200cos(2000

10.6);)(2000cos(

000

10. 3

2

3

qAt

);C 21, qq vung pha )(2

010max21 QQQqq

=> max21 )( qq = ))(/5( C

in v cng dng in

a t in l Q0 v cng

0

L1 hoc vi cun cm thun 10 mA. Nu ni t in v

in cc i l

hn B

dng in tc thi trong ha

thi im c gi tr ln nh

))(A

))(2

At

)(10.5 62

02 C

Chn C


16/18


16

Chng 5: Sng nh sng (7 cu)

Cu 34(C27): Hin tng chm nh sng trng i qua lng knh, b phn tch thnh cc chm sng n sc l hin tngA. phn x ton phn. B. phn x nh sng. C. tn sc nh sng. D. giao thoa nh sng.

Cu 35(C25): Gi n, nt v nv ln lt l chit sut ca mt mi trng trong sut i vi cc nh sng n sc , tm v vng. Spxp no sau y l ng?A. n< nv< nt B. nv >n> nt C. n >nt> nv D. nt >n> nv

Gii: n = c/v = c/f m f khng i >t n < nt Chn ALu : Chiu tng ca bc sng v chit sut ngc nhau.

Cu 36(C23): Trong th nghim Y-ng v giao thoa nh sng, khong cch gia hai khe l 1 mm, khong cch t mt phng cha hakhe n mn quan st l 2 m. Ngun sng n sc c bc sng 0,45 m . Khong vn giao thoa trn mn bng

A. 0,2 mm B. 0,9 mm C. 0,5 mm D. 0,6 mm

Gii: )(10.9 4 ma

Di

)(9,0 mm Chn B

Cu 37(C21): Trong chn khng, bc sng nh sng lc bngA. 546 mm B. 546 m C. 546 pm D. 546 nm

Cu 38(3): Khi ni v tia hng ngoi v tia t ngoi, pht biu no sau y ng?A. Tia hng ngoi v tia t ngoi gy ra hin tng quang in i vi mi kim loi.B. Tn s ca tia hng ngoi nh hn tn s ca tia t ngoi.C. Tia hng ngoi v tia t ngoi u lm ion ha mnh cc cht kh.D. Mt vt b nung nng pht ra tia t ngoi, khi vt khng pht ra tia hng ngoi.

Cu 39(C50): Tia XA. mang in tch m nn b lch trong in trng.B. cng bn cht vi sng mC. c tn s nh hn tn s ca tia hng ngoiD. cng bn cht vi tia t ngoi

Cu 40(C20): Trong chn khng, cc bc x c bc sng tng dn theo th t ng lA. nh sng nhn thy; tia t ngoi; tia X; tia gamma; sng v tuyn v tia hng ngoi.B. sng v tuyn; tia hng ngoi; nh sng nhn thy; tia t ngoi; tia X v tia gamma.C. tia gamma; tia X; tia t ngoi; nh sng nhn thy; tia hng ngoi v sng v tuyn.D. tia hng ngoi; nh sng nhn thy; tia t ngoi; tia X; tia gamma v sng v tuyn.

Chng 6: Lng tnh sng (4 cu)

Cu 41(C46): Cng thot lectron ca mt kim loi l 4,14 eV. Gii hn quang in ca kim loi ny lA. 0,6 m B. 0,3 m C. 0,4 m D. 0,2 m

Gii:

)(10.3,010.6,1.14,4

10.3.10.625,6619

834

0 mA

hc m3,0 Chn B

Cu 42(C13): Trong chn khng, mt nh sng c bc sng l 0,60m. Nng lng ca phtn nh sng ny bngA. 4,07 eV. B. 5,14 eV. C. 3,34 eV. D. 2,07 eV.

Gii:Nng lng ca photon: )(10.3125,3 19 Jhc

)(07,2 eV Chn D


17/18


17

Cu 43(C10): Theo mu Bo v nguyn t hir, nu lc tng tc tnh in gia lectron v ht nhn khi lectron chuyn ng trnqu o dng L l F th khi lectron chuyn ng trn qu o dng N, lc ny s l

A.F

16. B.

F

9. C.

F

4. D.

F

25.

Gii: Qu o L c n = 2 r2 = n2r0 = 4r0 =; Qu o N c n = 4 r4 = n

2r0 = 16r0,

Ta c:

16

1

4

1'';

22

4

2

2

4

2

42

2

2

2

r

r

F

F

r

keFF

r

keFF

16

' F

F Chn A

Cu 44(C28): Chm nh sng laze khng c ng dngA. trong truyn tin bng cp quang. B. lm dao m trong y hc .C. lm ngun pht siu m. D. trong u c a CD.

Gii: Ngun siu m l sng c, cn chm laze l sng nh sng l sng i n t khc bn cht nn chm laze khng th lmngun siu m c. Chn C

Chng 7: Vt l ht nhn (6 cu)

Cu 45(C45): S nucln ca ht nhn230

90 Th nhiu hn s nucln ca ht nhn210

84 Po lA. 6 B. 126 C. 20 D. 14

Gii: S nucln = s khi A => S nuclon ca ht nhn 23090 Th nhiu hn ca ht nhn

210

84 Po l 230210 = 20 Chn C

Cu 46(C37): ng v l nhng nguyn t m ht nhn c cng sA. prtn nhng khc s nucln B. nucln nhng khc s ntronC. nucln nhng khc s prtn D. ntron nhng khc s prtn

Cu 47(32): Trong cc ht nhn nguyn t: 4 56 2382 26 92

; ;He Fe U v 23090Th , ht nhn bn vng nht l

A.4

2He . B.230

90Th . C.56

26 Fe . D.238

92U .

Gii: Cc ht nhn bn vng cA

WlKln nht vo c 8,8 MeV/nuclon; l nhng ht nhn nm gia ca bng tun hon ng

vi 50< A


18/18


18

Cu 49(C18): Bn ht vo ht nhn nguyn t nhm ang ng yn gy ra phn ng: 4 27 30 12 13 15 0He Al P n . Bit phn ng

thu nng lng l 2,70 MeV; gi s hai ht to thnh bay ra vi cng (vc t) vn tc v phn ng khng km bc x . Ly khlng ca cc ht tnh theo n v u c gi tr bng s khi ca chng. ng nng ca ht l

A. 2,70 MeV B. 3,10 MeV C. 1,35 MeV D.1,55 MeV

Gii: V cc ht sinh ra c cn g vn tc nn t s ng lng : 1

30

nn

PP

n

P

vm

vm

P

PnP PP 30 (1)

=>

2

2 30 2P P n nm K ( m K ) => nP KK2

3030 nP KK 30 (2)+ Theo L BT ng lng: nP PPP

, v nP PP

(do nP vv

) => nP PPP , Thay (1) vo => nPP 31 (3)

=> 22 31 2n n

m K ( m K ) => KKn 961

4 (4)

+ Thay (4) vo (2) => KKP961

120 (5)

+Theo L BT nng lng => KKKKKKKE nP961

4

961

1207,2 )(1,3 MeV Chn B

Cu 50(C29): Tia

A. c vn tc bng vn tc nh sng trong chn khng.B. l dng cc ht nhn 4

2 He .

C. khng b lch khi i qua in trng v t trng.D. l dng cc ht nhn nguyn t hir.Gii:Tia L dng cc ht nhn ca nguyn t He

4

2 chuyn ng vi tc c2.107m/s, i trong khng kh vi cm, trong

vt rn vi micrmt (T188 SGKCB)Tia b lch trong in trng hoc t trng (SNC T268)

Tia , Xp x tc nh sng; truyn vi mt trong khng kh, vi milimt trong kim loi. (T189)

Tia Bc x in t cn gi l tia . Tia v mt trong tng v v xent mt trong c (T190)

Ht

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Nhieu Cach Giai Gon de DH Ly 2014