Boxmath- Tuyen Tap Vat Li

8/3/2019 Boxmath- Tuyen Tap Vat Li

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BIN SON: H HONG VIT

Tuyn tp 165 cu trc nghim hay v kh

1 Trn mt si dy n di 120 cm c sng dng. Cc im c bin dao ng 3.5mm nmcch nhau on 15cm. Tm bin cc i. Dao ng ny tng ng vi ha m no?A. Bc 4 B. Bc 3 C. Bc 1 D. Bc 2

Hng Dn

Bin 3,5mm chnh l bin bng (bin cc i):

= 2

= 15cm = = 30 = L

2

= 8 = Ha m bc 8.

Bin 3,5mm khng phi l bin cc i = khong cch t im n nt l:d = 7, 5cm =

2= 30 = = 60

Phng trnh bin :

3, 5 = Abng.sin(

2 d

) = Abng =7

2

2 =L

2

= 4 = Ha m bc 42 Ln Lt t cc in p xoay chiu u1 = U

2(cos(100 t + 1)), u2 = U

2(cos(120 t +

2)); u1 = U

2(cos(110 t+3)) vo hai u on mch gm in tr thun R, cun cm thunc t cm L v t in C mc ni tip th cng dng in trong mch c biu thc tngng l i1 = I

2(cos(100 t)); I2 = I

2(cos(120 t +

2

3)); i3 = I

2(cos(110 t +

23

)). Sosnh I v I ta c:A. I = I B. I < I C.I > I D. .I = I

2

Hng Dn

2 trng hp u u c cng U v I = L.1 1C.1

= L.2 1C.2

= LC = 11.2

=cng hng =

1.2 = 109, 5.

C 3 trng hp u c cng in p ch khc nhau tn s (tng ng ngun c in p khngi ch thay i tn s) = 1 < < 3 < 2 trong 3 lch gn vi cng hng nht = I>I

3 Cho mch in xoay chiu gm on AM ni tip vi on MB.on AM l hp kn ( Xcha 2 trong 3 phn t R,L,C); on mch MB l t in c: C =

20

F.t hiu in th xoay

chiu f = 50 Hz vo hai u on mch AB th thy hiu in th gia 2 trong 3 im bt kA,M,B u l120V.Tnh cng sut bn trong hp X?A. PX = 24, 94 W B. P X = 12, 45 C.PX = 21, 49 D. PX = 25, 32

Hng Dn

V gin vecto ra ta thy tam gic ABM l tam gic u c BM vung gc vi i = AM = 6

PX = ui cos AM = 120.120

ZC.

3

2= 24, 94 W

4 Xt nguyn t Hidro trng thi c bn c r = ro = 5, 3.1011(m).Tnh cng dng in

do chuyn ng ca e trn qu o K gy ra:A. 0.05mA B. 0.95mA C.1.05mA D. 1.55mA

http.//boxmath.vn/ TUYN TP CC BI TP VT L BOXMATH 1


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BIN SON: H HONG VIT

Hng Dn

Phng trnh II Niuton cho chuyn ng trn, lc tnh in ng vai tr lc hng tm.

k.q2

r20= m.r0.

2 ==> =q

r0.

k

m.r0Cng dng in.

I = qT

= q.2

5 1 ngi ng cch 1 ci loa khong 20cm, truc loa, nghe c m mc cng khong60dB. Tnh cng sut pht m ca loa. Cho rng loa c dng 1 hnh nn c na gc nh l30o. Cho bit cng chun l 1012 (

W

m2)

A. 0, 0336 W B. 0, 2336 W C. 0, 3216 W D. 5, 421 W

Hng Dn

Cng m ti v tr ngi ng: I = Io.10L

10 = 106W

m2

Gi R = 20m l khong cch t loa n ngi = Din tch chm cu l: S = 2RhV 1 na gc m ca chm cu l 30o nn h = R(1 cos30o) = Cng sut pht m: P = IS =2I R2(1 cos30o) = 0, 0336 W

6 Ngun sng O c tn s 10Hz,v = 0, 4m/s. Trn 1 phng truyn c 2 im, PQ cchnhau15cm. Bit bin l 1 cm. Khi P c ly cc i th ly ca Q l my?

A. x = 0 B. x = 1 C.x = 2 D. x = 3

Hng Dn

=2df

v=

20, 15.10

0, 4= 7, 5 = (2.3 + 1 )

2= PQ vung pha vi nhau ... khi P c li cc

i = Q c li x = 0

7 1 sng c lan truyn trn mt ng thnh t M n N (M N =17

4) tai 1 thi im no

tc dao ng ca im M l: 2f A .Khi tc dao ng ca im N l: ?A. vN = 0 B. vN = 1 C.vN = 2 D. vN = 3

Hng Dn

dMN =17

4= dao ng ca phn t sng ti M v N vung pha nhau. (khong cch gia hai

im dao ng vung pha bng l phn t bc sng) = vM = 2f A = vmax = vN = 0

8 Mt sng c hc c bc sng lamda, tn s f v c bin l A khng i khi truyn i

trong mt mi trng. Sng truyn t im M n im N cch nhau7

3. Vo mt thi im

no tc dao ng ca M l 2f A th tc dao ng ti N l?A. vN = A.f B. vN = 2A.f C.vN = 0 D. vN = 3A.f

Hng Dn



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BIN SON: H HONG VIT

Ta c phng trnh sng ti M :uM = Acos(2f t)vM = A.2.f.sin(2f t) A.2.f = A.2.f.sin(2f t)

sin(2f t) =

1

2f t = 2

+ 2k

Phng trnh sng ti N :

uN = Acos(2f t +143

)

vN = A.2.f.sin(2f t +14

3)

vN = A.2.f.sin(2

+2

3+ 4)

vN = A.2.f.sin( 6

)

= vN = A.f9 Trn mt nc c 2 ngun kt hp S1,S2 dao ng theo phng trnh ln lt

u1 = acos(50t +

2)cm,u2 = acos(50t)cm. vn tc truyn song 1m/s. hai im P, Q

thuc h vn giao thoa,vi P S1 P S2 = 5cm, QS1 QS2 = 7cm.Hi P,Q nm trn ng cci hay cc tiu ?A. P cc i, Q cc tiu B. P cc tiu, Q cc iC. P, Q thuc cc tiu D. P,Q thuc cc i

Hng Dn

Hai ngun vung pha c = vT = 4(cm)Vi P:S1 P S2 = 5cm = (1 + 1

4) = cc i

Vi Q:QS1 QS2 = 7cm = (1 + 34

) = cc tiu

10 Ti hai im A v B trn mt nc cch nhau 16 cm c hai ngun pht sng kt hp daong theo phng trnh u1 = a cos (30t); u2 = a cos (30t +

2). Tc truyn sng trn mt

nc 30 cm/s. Gi E, F l hai im trn on AB sao cho AE = FB = 2 cm. Tm s cc tiu

trn on EF.A. 28 B. 12 C. 13 D. 21

Hng Dnd1 d2 = (M ).

2M = (2k + 1)

= d1 d2 = 2k + 0, 5

= (16 4) 2k + 0, 5 (16 4) = 12



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BIN SON: H HONG VIT

11 Ti mt nc nm ngang, c hai ngun kt hp A v B dao ng theo phng thng ngvi phng trnh ln lt l uA = a1.sin(40t +

6) cm,uB = a2sin(40t +

2)cm. Hai ngun

tc ng ln mt nc ti hai im A v B cch nhau 18cm. Bit tc truyn sng trn mtnc v = 120 cm/s. Gi C v D l hai im thuc mt nc sao cho ABCD l hnh vung. Sim dao ng vi bin cc tiu trn on CD l ?

A. 2 B. 12 C. 13 D. 21

Hng Dn

d1 d2 = (M ). 2

M = (2k + 1)

=

2

6

= d1 d2 = 6k + 2

= AD BD 6k + 2 AC BC = 1, 5 k 0, 9 = 2

12 Hai ngun kt hp A v B dao ng trn mt nc theo cc phng trnhu1 = 2cos(100t +

2)cm; u2 = 2cos(100t)cm . Khi trn mt nc, to ra mt h

thng vn giao thoa. Quan st cho thy, vn bc k i qua im P c hiu s PA PB = 5 cmv vn bc (k + 1),cng loi vi vn k i qua im P c hiu s PA PB = 9cm. Tm tc truyn sng trn mt nc, cc vn ni trn l vn cc i hay cc tiu?A. v = 200cm/s B. v = 130cm/s C. v = 100cm/s D. v = 230cm/s

Hng Dn

9 = (k + 1) = k + k = 5 = = 4 = v = 200cm/s13 Trong th nghim giao thoa sng trn mt nc, hai ngun AB cch nhau 14,5 cm dao

ng ngc pha. im M trn AB gn trung im I ca AB nht, cch I l 0,5 cm lun daong cc i. S im dao ng cc i trn ng elp thuc mt nc nhn A, B lm tiuim l:A. 18 im B. 30 im C. 28 im D. 14 im

Hng Dn

Vi 2 ngun ngc pha, th ti I l cc tiu, m M l im gn I nht t cc i. Vy khongvn i = 2.0, 5 = 1cm V tr cc i s l: x = (0, 5 + k).i = 0, 5 + k Mt khc: 0 x 14, 5 = 0, 5 k 14 = c 14 gi tr k (v k nguyn)= 28 im cc i (ct na trn elip 14 im, ct na di 14 im).

14 Trn mt nc nm ngang c hai ngun sng kt hp cng pha A, B cch nhau 6,5 cm,bc sng = 1cm. X t im M c MA = 7,5 cm, MB = 10 cm. S im dao ng vi bin

cc tiu trn on MB l:A. 6 im B. 7 im C. 8 im D. 9 im

Hng Dn



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BIN SON: H HONG VIT

Gi N l 1 im bt k thuc MB. Vi 2 ngun ngc pha, N s l cc tiu nu: d d = kMt khc:AB d d |MA MB| = 6, 5 k 2, 5 = 6, 5 k 2, 5 = c 8gi tr k = C 8 im cc tiu trn MB .

15 Trong giao thoa sng trn mt nc, hai ngun A, B cch nhau 14,5 cm dao gn ngcpha. im M trn AB gn trung im I ca AB nht, cch I l 0,5 cm lun dao ng cc i.S im dao ng cc i trn ng elip thuc mt nc nhn A, B lm tiu im l?A. 18 im B. 30 im C. 28 im D. 14 im

Hng Dn

V AB ngc pha nn I dao dng cc tiu, im dao ng cc i gn I nht s cch I:

4= = 21cm Xt iu kin: AB k AB = 7, 25 k 7, 25 = c 14 ng cc i = trnelip C 28 im dao ng cc i ( 1 ng cc i ct elip ny ti 2 im)

16 Trn b mt cht lng c 2 ngun pht sng kt hp S1, S2 d cng pha, S1S2 = 40 cm.Bit sng do mi ngun pht ra c tn s: f = 10hz,v = 2 (

m

s). Xt M nm trn ng thng

vung gc vi S1S2 ti S1. on S1M c gi tr ln nht l bao nhiu ti M c d vi bin cc i?A. 30 B. 15 C. 20 D. 13

Hng Dn

=v

f

= 20cm im M s nm trn ng cc i th nht k t trung im AB: =

MB =

MA + 20 = M B2 = MA2 + 40M A + 400 Li c MB2 = MA2 + AB2 = 40MA + 400 = AB2 = M A = 30cm

17 cho giao thoa 2 ngun sng kt hp ng pha S1 v S2 trn b mt cht lng bit 2 imdao ng cc i trn on thng S1 v S1 cch nhau 1 cm . hai im M v N trn mt chtlng M cch S1 8 cm ,cch S2 l 11cm .N cch S1 l 14cm ,S2 l 10cm s im dao ng cc itrn MNA. 18 im B. 4 im C. 28 im D. 14 im

Hng Dn

2 im dao ng cc i trn S1S2 cch nhau 1cm = = 2cmXt bt phng trnh sau: MS1MS2 k NS1N S2 = 3 2k 4 = 1, 5 k 2 Vy c 4 im cc i ng vi k = 1, 0, 1, 2

18 Chiu 1 bc x in t c bc sng 0, 25(m)vo ca tt t bo quang in c cng thot3, 559(eV).Hiu in th gia anot v catot l 1, 25V to ra in trng u trong khong khnggian ca 2 cc.Vn tc ca e quang in khi n anot l v tha mn:

A. 0m/s v 0, 97.106m/s B. 0, 66.106m/s v 0, 97.106m/sC. 0, 71.106m/s v 0, 97.106m/s D. 0m/s v 0, 71.106m/s

Hng Dn



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BIN SON: H HONG VIT

Ta c:hc

= Ao + Wd1 = Wd1 = 2, 2556.1019(J)

Trng hp cc e bc ra vi vn tc cc i:p dng nh l ng nng:

1

2mv2

2 Wd1 = eUAK = v2 = 0.97.106m/si vi cc e bc ra vi vn tc u bng 0 v n Anot: 1

2mv2

2 = eUAK = v2 = 0.66.106m/sVy 0, 66.106

v2

0, 97.10

6

19 Cho dy AB c nh c th thay i l bng rng rc. f = 20 Hz, khi thay i l ta thy gia2 ln c sng dng lin tip th l ln lt l 90 v 100cm. Tm V?A. v = 200cm/s B. v = 130cm/s C. v = 100cm/s D. v = 400cm/s

Hng Dn

Gia 2 ln c sng dng lin tip (m chiu di dy ln th 2 ln hn chiu di dy ln th 1)m 2 u dy c nh nn khi c sng dng th chiu di dy lun = s nguyn ln b sng .

Ta c : Gi s b sng ( mi b sng c l = /2) l n th s b sng ca ln th 2 khi dy cchiu di l 100cm l n+1 . = 90n

=100

n + 1=

2= n = 9T gii ra nc = 20cm =

V = .f = 400cm/s

20 Ht nhn 92, 234U phng x alpha, ngay sau khi sinh ra ht a bay vo t trngu c B = 0.5T, theo phng vung gc vi cc ng sc t, bit khi lng cc htU = 233.9904T h = 229.9737, a = 4.0015.1u = 1.66.1027 = 931, 5MeV/C2

A. 5, 27m B. 2, 37m C. 1, 27m D. 1, 07m

Hng DnK1 + K2 = 14, 1588

229, 9737.K1 4, 0015.K2 = 0= KHe = 13, 92MEV = m.v

2

2=> v =

13, 92.1, 6.1013.24, 0015.1, 66

6, 7.1014

Ta c: q.v.B =m.v2

r=> r =

m.v

q.B 4, 0015.1, 66.10

27.

6, 7.1014

2.1, 66.1019.0, 5= 1, 07m

21 Mt si dy cng gia 2 im c nh cch nhau 75 cm.Ngi ta to song dng trn dy.2tn s gn nhau nht cng to ra song dng trn dy l 150 hz v 200 hz.Tn s nh nht to

ra sng dng trn dy l?A. fmin = 22Hz B. f min = 50Hz C. f min = 100Hz D. f min = 25Hz

Hng Dn

f =kv

2l=

kv

1, 5

(k=1)= fmin = v

1, 5

k1v = 225

k2v = 300=

k1 = 3

k2 = 4

= fmin = 50Hz



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BIN SON: H HONG VIT

22 Bi 1: Mt con lc n dao ng nh ti ni c gia tc trng trng 9, 8(m/s2) vi dydi 1(m) qu cu con lc c khi lng 80(g).Cho con lc dao ng vi bin gc 0,15(rad)trong mi trng c lc cn tc dng th n ch dao ng 200(s) th ngng hn.Duy tr daong bng cch dng mt h thng ln dy ct sao cho n chy c trong 1 tun l vi bin gc 0,15(rad). Bit 80% nng lng c dng thng lc ma st do h thng cc bnh rng

ca.cng cn thit ln dy ct l?A. 133, 5J B. 266, 1J C. 103, 5J D. 117, 2J

Hng Dn

Do trong dao ng iu ha chu k dao ng ca vt l 1 hng s, nn trong dao ng tt dn thi lng ny cng khng i:T = 2

lg

= 2s

Mt khc nng lng gim trong 1 chu k cng khng i. T y ta c nng lng gim trong 1sbt k l bng nhau, v bng:

W1s =W0

200=

0, 5.m.g.l20

200= 8, 82.103J

Cng cn thit ln dy ct gm cng thng l cn v cng thng lc ma st bnh rngCng thng lc cn: W1t = W1s.7.24.60.60V: 80% nng lng c dng thng lc ma st do h thng cc bnh rng ca, nn cng cnthit ln dy ct l: W1t.5 = 133, 5J

23 Mt on mch khng phn nhnh gm 1 in tr thun R = 80,mt cun dy c intr thun r = 20 , t cm L=0,318 H v mt t in c in dung C = 15, 9F,c tn s fthay i c.Vi gi tr no ca f th in p gia 2 bn t t gi tr cc i:

A. 71Hz B. 71Hz C. 61Hz D. 55Hz

Hng Dn

2 =2LC R2C2

2L2C2= f = 61(Hz)

24 t vo 2 u dy thun cm c t cm 0, 3/(H) mt in p xoay chiu.Bit gi trtc thi ca in p v cng dng in ti thi im t1 l: 60

6(V)v

2(A), ti thi im

t2 l 60

2(V) v

6(A). Tn s ca dng in l:A. 60Hz B. 50Hz C. 100Hz D. 40Hz

Hng Dn

V u, i lch pha nhau 1 gc

2nn ta c h thc:

UoIo

=

u21 u22i22 i21

= ZL = ZL = 60 = w = 200 = f = 100

25 Hai con lc ging nhau c cng T = 0,2 s. bit A2 = 3.A1. BAit rng lc u 2 vt gpnhau v tr cn bng v chuyn ng ngc chiu nhau.Khong thi gian gia 2 ln vt nnggp nhau lin tip l?

A. 0, 02s B. 0, 04s C. 0, 03s D. 0, 01s

Hng Dn

Khi 2 vt dao ng vi cng chu k m ban u li gp nhau ti v tr cn bng th c sau



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BIN SON: H HONG VIT

T

2= 0, 01s hai vt li gp nhau ti v tr cn bng, khong thi gian ny khng ph thuc vo t

l bin 2 vt (Cn ch rng 2 vt ny c cng v tr cn bng)

26 Cho on mch in xoay chiu gm 2 phn t X v Y mc ni tip. Khi t vo hai uon mch in p xoay chiu c gi tr hiu dng l U th in p hiu dng gia 2 u phn

t X l 3U, gia 2 u phn t Y l 2U. hai phn t X v Y tng ng l?A. T in v in tr thun B. Cun dy thun cm v in tr thunC. T in v cun day thun cm D. T in v cun dy khng thun cm

Hng Dn

p n A, B loi v: Nu mch c R, C hoc R, L th: U2 = U2X + U2Y = UX ; UY iu ny khngtha mn p n C loi v: Nu mch ch c L, C th: UX UY |= (2

3)U iu ny khng tha mn

p n D tha mn (v hnh s gii thch c t l gia cc i lng hon ton tha mn27 Cho dng in gm R ni tip L ni tip C( vi t C c th thay i c), hai u t

C c mt vn k o tr s in p i qua t. in p hiu dng 2 u mch khng i, tns ca dng in, in tr v cm khng ca cn dy khng i. Khi C = C1 = 10(F) vC = C2 = 20(F) ngi ta thy vn k cho kt qu o nh nhau. Tm C gi tr ca vn kt ln nht. Bit L thun cm?

Hng Dn

Khi thay i C nhng P khng i chng t = I khng i = Z khng i= ZL = ZC1 + ZC2

2

Khi thay i C UC max th ta c:UC = I.ZC =U

R2 + Z2LZ2C

2ZLZC

+ 1

=U

y

Nh vy UC max th y min, theo tnh cht tam thc bc 2 th1

ZC=

ZLR2 + Z2L

= C

28 t vo 2 u dy thun cm c t cm0, 3

(H) mt in p xoay chiu.Bit gi tr tc

thi ca in p v cng dng in ti thi im t1 l 606(V) v 2(A), ti thi im t2l 60

2(V) v

6(A). Tn s ca dng in l:

A. 60Hz B. 50Hz C. 100Hz D. 40Hz

Hng Dn

Do cun dy ch cha cun thun cm L nn ta thy lc no u v i cng vung pha vi nhau.Do vyta c: ti thi im t bt k nu in p tc thi l u v i th:(

u

U0)2 + (

i

I0)2 = 1

Thay s ta c:

(60

6

U0)2 + (

2

I0)2 = 1

(60

2

U0)2 + (

6

I0)2 = 1



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BIN SON: H HONG VIT

=

U0 = 120

2(V)

I0 = 2

2(A)= ZL = 2f L = U0I0 = 60 = f =

60

2L= 100(Hz)

29 Mt con lc l xo gm vt M v l xo c cng k dao ng iu ha trn mt phngnm ngang, nhn vi bin l A1. ng lc vt M ang v tr bin th mt vt m c khilng bng vi vt M chuyn ng theo phng ngang vi vn tc vo bng vn tc cc i ca

M, n va chm vi M.Bit va chm gia 2 vt l hon ton n hi xuyn tm, sau va chmvt M tip tc dao ng iu ha vi bin A2.T s bin dao ng ca vt M trc v sauva chm l :

A.A1A2

=22

B.A1A2

=

3

2

C.A1A2

=2

3D.

A1A2

=1

2

Hng Dn

Lc vt M bin th M ang c 1 Wtmax = 0, 5.k.A21 v ng lc ny vt m n v truyn cho

M1 : Wdmax = W = 0, 5.k.A21 T : = Ws = k.A21 = 0, 5.k.(

2A2)

2 = A1A2

=2

2

30 Mt con lc l xo, vt c khi lng m dao ng cng bc di tc dng ca ngoi lcbin thin iu ha vi tn s f.Khi f=f1 dao ng cng bc khi n nh c bin l A1, khif=f2 (f1


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BIN SON: H HONG VIT

32 cho hai ngun kt hp t cch nhau 2m dao ng cng pha di chuyn trn on AB .ngita thy c 5 v tr m c to cc di bit tc m trong khng kh 350(

m

s) tn s c gi tr

tha mn nm trong khong no?

Hng Dn

Di chuyn thy 5 v tr m to nht = trong on AB c 5 bng = 52

2f 437.5HZMt khc ta cng lu ch c 5 cc i tng 5 b sng v t b sng th 1 v th 5 c th cnc khong cch ti ngun 1 on 0, 1 = Vt dng li A1A

= 2, 5 = Vt dng li vtr cch O 1 khong 0,02 (m) Qung ng vt i c:

1

2.K(x21 x22) = Fc.S Vi x1 = 0, 1; x2 =

0, 02 = s = 48(cm)

56 Mt on mch xoay chiu gm in tr R, cun dy thun cm L v mt hp X mc nitip. Hp X gm 2 trong 3 phn t RX ; LX ; CX . t vo hai u on mch mt hiu in th

xoay chiu c chu k dao ng Tnlc ZL =

3R. Vo thi im no thy URL t cc i,sau thi gian

T

12th hiu in th gia hai u hp X l UX t cc i. Hp X cha nhng

phn t no?

Hng Dn

C tanuRLi

=

3 =

3 = uRL nhanh pha hn i 1 gc /3

C uRL nhanh pha hn uX 1 gc

6= Trong mch X c tnh cm khng = X phi c L

Nu X c C =

UX

nhanh pha hn uRL

=

loiVy X cha LX , RX

57 Nu tc quay ca roto tng thm 60 vng trong mt pht th tn s ca dng in domy pht ra tng t 50Hz n 60Hz v sut in ng hiu dng do my pht ra thay i 40Vso vi ban u. Hi nu tip tc tng tc ca roto thm 60 vng/pht na th sut inng hiu dng khi do my pht ra l bao nhiu?

Hng Dn

Nu roto quay tng 60 v/1ph = 1v/1sf

f=

6

5= n +

1

n n = 5

Do my pht ra thay i 40V so vi ban u nn ta c:NBScosA =

40

w w =2

Tip tc tng tc ca roto thm 60 vng/pht Vy n = 7 = f = 70Hz; w = 140 = =140.

2

= 280

58 Trong th nghim I-Yng v giao thoa nh sng cho khong cch gia hai khe l 1mm t

hai khe n mn l 1m.Ta chiu vo hai khe ng thi hai bc x 1 = 0, 5m v 2.Trn brng L = 3mm ngi ta quan st c t c 9 cc i ca c hai bc x trong c 3 cc itrng nhau hai trong s trng nhau hai u 2 bng?



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BIN SON: H HONG VIT

Hng Dn

1C 9 C + 3 C trng nhau Trn L c tt c 12 C S C h 1

L

2.i1

2+1 = 7 s cc i h 2 l: 127 = 5

L

2.i2

2+1 = 5 2 = a.i2D

= 0, 75

29 cc i trong c 3 cc i vn trng = 6 cc i khng trng = xt dfrac12 vng thc 3 cc i khng trngKhong cch 2 cc i trng nhau : i = dfracL2 = 1, 5mm (GT ni 2 vn ngoi l vn trng).

Khong cn i1 ca vn 1 l : i1 =1.D

a= 0, 5mm = k1 = i

i1= 3 vn 1 trng bin l vn bc

3 (k1=3) = vn 2 trng bin l vn bc 2.K vn trng k1.1 = k2.2 = 2 = 0, 75m

59 Mt cun dy mc ni tip vi t in c in dung thay i c ri mc vo ngun in

xoay chiu u = U0cos(t) . Thy i C cng sut ta nhit tren cun dy cc i th khi , in p hiu dng gia 2 bn t l?

Hng Dn

Khi thay i C Pcdmax th xy ra cng hng = ZL = ZC = UL = UC = 2U0 =Urmax =

U02

= Ucd =

U2r + U2L =

4U20 +

U202

=3U0

2

60 Mt con lc n gm mt si dy nh.khng dn v mt vt nh c khi lng m = 100gdao ng iu ho mt ni g = 10m/s2 vi bin gc bng 0.05rad.Nng lng ca dao

ng iu ho bng 5.104J.Chiu di ca dy treo bng?

Hng Dn

W = mgl(1 cos0) l = Wmg(1 cos0) =

5.104

0, 1.10.(1 cos(0, 05)) = 0, 4m = 40cm

61 Hai ngun kt hp A v B cch nhau L = 21cm dao ng cng pha vi tn s 100Hz.Vntc truyn sng bng 4m/s.Bao A,B bng mt vng trn (C) tm O nm ti trung im AB,bnknh ln hn 10cm.Tnh s vn li (dao ng bin cc i) ct na vng trn (C) nm v

mt pha AB?

Hng Dn

=v

f=

400

100= 4cm .2 ngun dao ng cng pha nn ti O dao ng vi bin cc i. Khong

cch 2 cc i lin tip l

4= 1cm > 0, 5cm Kt hp v hnh thy c 11 im tho mn

62 Dao ng in t trong mch LC l dao ng iu ho.Khi in p gia 2 u cuncm bng 1, 2mV th cng dng trong mch l 1, 8mA.Cn khi in p gia 2 u cun

cm bng 0,9mV th cng dng in trong mch bng 2, 4mA.Bit t cm cun dyL = 5mH.in dung ca t bng?

Hng Dn



18/48

BIN SON: H HONG VIT

0, 5Li12 + 0, 5Cu1

2 = 0, 5Li22 + 0, 5Cu2

2 = C = i12 i22

u22 u12 = 20F63 Mt con lc l xo dao ng iu ha trn mt phng ngang vi chu k T = 2. Khi con lc

n v tr bin dng th mt vt c khi lng m chuyn ng cng phng ngc chiu nva chm n hi xuyn tm vi con lc. Tc chuyn ng ca m trc v sau va chm l2(cm/s) v 1(cm/s). Gia tc ca con lc lc l -2(cm/s2).

Hi sau khi va chm con lc i c qung ng bao nhiu th i chiu chuyn ng.Hng Dn

Va chm n hi xuyn tm nn ng lng v ng nng bo ton, c:mv = m1v1 + mv

= 2m = m1v1 m(v = 2; v = 1) = m1v1 = 3m(1) v

1

2mv2 =

1

2mv21 +

1

2mv2 = m1v21 = 3m(2)

T (1) v (2) c v1 = 1(cm/s) Li c gia tc con lc ti bin dng a = 2A = 2(cm/s2) = A = 2(cm)(T = 2 = =1)

Ti v tr x = A c vn tc v1 nn c A1 =

x2 + v21

2= 5(cm)

Vt i c qung ng S = A + A1 = 2 +

5(cm) (ti bin kia) th i chiu chuyn ng

64 Ti hai im A, B cng pha cch nhau 20cm l 2 ngun sng trn mt nc dao ng vitn s f = 15Hz v bin d bng 5cm. Vn tc truyn sng mt nc l v = 0, 3m/s. Bin dao ng ca nc ti cc im M, N nm trn ng AB vi AM = 5cm, AN= 10cm l?A. AM = 0, AN = 10 cmB. AM = 0, AN = 5 cm

C. AM = AN = 10 cmD. AM = AN = 5 cm

Hng Dn

Vi 2 ngun cng pha,im dao ng vi bin cc i tha mn: d1 d2 = k.im dao ng vi bin cc tiu tha mn: d1 d2 = (k + 0.5)= v / f = 0.3 / 15 = 0.02 m = 2 cm

vi im M ta c hiu ng i:d1 d2 = 5 (20 5) = 10 = 5.2 M dao ng vi bin cci = A(M) = 5 cmvi im N ta c hiu ng i: d1 d2 = 10 (20 10) = 0 = 0.2 N dao ng vi bin cci = A(N) = 5 cm



19/48

BIN SON: H HONG VIT

65 Ti hai im S1, S2 cch nhau 5cm trn mt nc t hai ngun kt hp pht sng ngangcng tn s f = 50Hz v cng pha. Tc truyn sng trong nc l 25cm/s. Coi bin sngkhng i khi truyn i. Hai im M, N nm trn mt nc vi S1M = 14, 75cm,S2M = 12, 5cmv S1N = 11cm,S2N = 14cm. Kt lun no ng:A. M dao ng bin cc i, N cc tiu

B. M dao ng bin cc tiu, N cc iC. M, N dao ng bin cc iD. M, N dao ng vi bin cc tiu

Hng Dn

=v

f= 0.5cm

Hiu ng i vi im M:d1 d2 = 14, 75 12, 5 = 2, 25 = (4 + 0, 5).0, 5 =M dao dng vi bin cc tiu Vi im N ta c hiu ng i: d1

d2 = 11

14 =

3 =

6.0, 5 =

N dao dng vi bin cc i

66 Cho 2 ngun kt hp S1,S2 cng pha cch nhau 20cm, = 2cm. Trung im ca S1S2 lO. Gi M l im nm trn ng trung trc ca S1S2 v gn O nht dao ng cng pha viS1. Tm OM?

Hng Dn

OS1 =20

2pha dao ng ti 1 im trong min giao thoa: (d1 + d2)

cng pha:(d1 + d2)= k2, d1 = d2 = d1 = 2k

d1 l khong cch t S1 in im thuc trung trc ca S1S2 (khc O ) = d1 > OS1 = k > 5v l im gn nht:k=6. OM l cnh gc vung ca tam gic OMS1 vung ti O = OM =

d21 OS21 =

122 102 = 411 6, 63cm

67 Mt vt nh khi lng m t trn mt tm vn nm ngang h s ma st ngh gia vtv tm vn l 0,2. Cho tm vn dao ng iu ho theo phng ngang vi tn s 2hz . vtkhng b trt trn tm vn trong qu trnh dao ng th bin dao ng ca tm vn phi

tho mn iu kin no ?

Hng Dn

Gia tc cc i ca vt m khng b trt l amax = k.g = 0, 2.10 = 2m/s2 y cng l giatc cc i ca tm vn trong qu trnh dao ng:= amax = 2 = w2.Amax Amax = 2

w2=

2

(4)2= 0, 0125m = 1, 25cm

Vy iu kin l A 1, 25cm68 T mt my pht in ngi ta mun chuyn ti ni tiu th mt cng sut in l 196KW

vi hiu sut truyn ti l 98%. Bit bin tr ca ng dy dn l 40, h s cng sut bng1. Cn phi a ln ng dy ti ti ni t my pht in mt in p bng bao nhiu?

Hng Dn



20/48

BIN SON: H HONG VIT

Cng thc tnh hiu sut : H = (1 R.PU2.cos()2

) = 0, 98 = U

69 Trong qu trnh truyn ti in nng i xa, gi thit cng sut tiu th nhn c khngi, in p v dng in lun cng pha. Ban u gim in th trn ng dy bng 15%in p ni tiu th. gim cng sut hao ph trn ng dy i 100 ln cn tng in p ca

ngun in ln bao nhiu???

Hng Dn

U1 = U2 + .U = 1, 15U2 = U2 = U11, 115

= P2 = U2.I = I = 1, 115.P2U1

Cng sut hao ph do ta nhit lc ban u :P = R.(1, 115.P2)

2

U21 gim hao ph 100 = U1 tng ln 10 ln

70 Mt in tr mc vo ngun in xoay chiu th cng sut ca in tr l P. Hi khi mc

in tr ni tip vi mt it l tng ri mc vo ngun in trn th cng sut to nhit trnin tr l bao nhiu?

Hng Dn

khi mc diot th dng b chn: T2

= P = P2

71 Mt mch dao ng LC l tng c tn s dao ng ring f0 = 90MHz. Mch ny ni vimt anten thu sng in t. Gi s 2 sng in t c cng nng lng nhng c cc tn s

tng ng f1 = 92MH z, f 2 = 95MHz truyn vo cng anten. Gi bin dao ng ca mchng vi 2 tn s l I1, I2 th I1 ln hn hay nh hn I2

Hng Dn

Vic thu sng theo nguyn tc cng hng. do vy cc tn s m Angten thu c gi tr gn bng vitn s ring th sng r nht (bin mnh nht) = f1 gn f0 hn nn I1 > I2

72 Cho mch in xoay chiu ABgm: on mch AM ch cha C v on mch MB ch chacun dy mc ni tip. Bit UAM =

2UMB, uAB nhanh pha 300 so vi uAM. Nh vy uMB

nhanh pha so vi dng in 1 gc l:?

Hng Dn

Ly UC =

2 => UMB = 1

V gin ra cUC

sin=

UMBsin30

= sin =

2

2Vic thu sng theo nguyn tc cng hng. do vy cc tn s m Angten thu c gi tr gn bngvi tn s ring th sng r nht (bin mnh nht) = f1 gn f0 hn nn I1 ln hnI2 = 135( > 60) = UMB nhanh pha hn I 1 gc 135 60 = 75



21/48

BIN SON: H HONG VIT

73 Nu ni cun dy 2 u on mch gm cun cm L mc ni tip in tr R ,R=1(m) vo2 cc ca ngun in mt chiu c S ko i v in tr trong r th trong mch c dong inkhng i I. Dng ngun ny np in cho mt t in c in dung C = 2.106F khi intch trn t t gi tr cc i th ngt t in khi ngun pi ni vi cun cm L thnh mchdao ng th trong mch c d in t vi chu k T = .106 v I0 = 8I. Gi tr ca r=?

Hng Dn

Ta c w =2pi

T= 2.106;Q0 =

I0w

U0 =Q0C

=I0

wC=

I04

(ch ny thay s vo thi) M I0 = 8I = U0 = 2I Ta c U0 = E = I(r + R) = 2I = r = 1

74 Mt hc sinh lm mt nhn ch s in tr . Hc sinh ny s dng 2 mch in l AB vCD trong AB cha cun cm c 36a(H) v CD cha ci t in c 4a(F) . Sau s dng

ngun in c cng thc l u = Uocost. K n gn ci in tr vo mch AB v s dngngun in trn v cui cng gn ci in tr vo mch CD th thu c kt qu l gc hp bigia u AB v u CD l 1 gc

2. in tr c gi tr l?

Hng Dn

Xt mch AB tana =ZLR

=36a

R.Xt mch CD tanb =

ZCR

=1

4aR= tana.tanb = 1 =

R = 3 .

75 Con lc l xo treo thng ng k = 10, m = 0, 01kg.a vt ln v tr cn bng 8cm ri

bung tay.Tc dng ca lc cn bng 0, 01N.Li ln nht vt t c sau khi qua v tr cnbng?

Hng Dn

A1 A2 = 2FcanK

=2.0, 01

10= 0, 002 = A2 = 0, 08 0, 002 = 0, 078(m)

77 Mt con lc gm vt nng c khi lng 200g v mt l xo c cng 20N/m. Ti thiim t, vn tc v gia tc ca vt nng ln lt l 20cm/s v 2

3m/s2. Bin dao ng ca

vt nng l bao nhiu?

Hng Dn

Cng thc lin h gia vn tc v gia tc ti thi im t ty :v22 + a2 = A2.4

C 2 = k/m = 20/0, 2 = 100; thay vo (0, 20)2.100+(2

3)2 = A2.1002 = A = 0, 04m = 4cm

78 Cho mch in AB gm 1 t in c in dung C, mt in tr hot ng R v 1 cuncm c in tr thun r v c t cm L (theo th t ) mc ni tip vi L=rRC. tvo 2 u on mch mt in p xoay chiu th in p gia 2 u cun cm c biu thc

u = 100cos(.t +

12 ) (V). Vo thi im in p gia 2 u cun cm bng 80V th inp gia 2 u mch AM(AM gm C v R) l 30V. Biu thc in p gia 2 u on mch AM l?

Hng Dn



22/48

BIN SON: H HONG VIT

T L = rRC Zcd vung pha ZAM ta c:

ucd = 80 = 100cost

uAM = 30 = xsint x = 50

Vy uAM = 50cos(.t 512

)

79 Cho mch in AB gm 1 cun cm c in tr hot ng r mc ni tip vi 1 hp kn X

cha 2 trong 3 phn t: in tr hot ng R, cun cm thun L v t in C. t vo 2 uAB mt in p xoay chiu c gi tr hiu dng 130V th in p hiu dng gia 2 u cuncm v 2 u hp X ln lt l 78V v 104V. Hp X phi cha?

Hng Dn

Thy ucd vung pha uX = uX thuc gc phn t th 2 vy X phi cha C v R .

80 Con lc n c dy treo di l = 1m, khi lng m = 20g.Ko hn bi khi v tr cn bngcho dy treo lch mt gc = 30 so vi phng thng ng ri th nh cho chuyn ng .Gc

nh nht hp bi gia tc tip tuyn v gia tc ton phn l?

Hng Dn

at = gsin()

an = 2gl(cos cos30) a =

a2n + a

2t

Gi l gc hp bi at v a = cos() = ata

=sin

4(cos cos30)2 + sin2

= cos =sin

4cos2 + 3 43cos + sin2 =sin

3(cos2 + 4 43cos =sin

(3cos 2)2

= cos = 13cos 2sin

=1

23cos

sin

= min khi cosmax khi = 30

81 Mt con lc l xo gm mt vt nh khi lng 0,02 kg v l xo c cng k = 1N/m. Vtnh c t trn mt gi c nh nm ngang dc theo trc l xo. H s ma st trt giagi d v vt nh l 0,1. Ban u gi vt v tr l xo b nn 10cm ri bung nh con lc

dao ng tt dn. Ly g = 10m/s2. Tc ln nht vt nh t c trong qu trnh dao ng l?

Hng Dn

= 5

2(rad/s) Ta c:kx mg = kA Thay s: = A = 0, 08m = vmax = A = 40

2(cm)

82 c ni tip R vi cun cm L c r ri mc vo ngun xoay chiu. Dng vnk c Rrt ln o U hai u cun cm, in tr v c on mch ta c cc gi tr tng ng l100V, 100V, 173, 2V. Suy ra h s cng sut ca cun cm l?

Hng Dn

Tam gic AOB cn ti A, dng nh l hm cos trong tam gic AOB = gc O1 = 30 =gc A2 = 60 = H s cng sut ca cun cm (coi nh mch L, r) : cos = 0, 5



23/48

BIN SON: H HONG VIT

83 Cho mch in xoay chiu RLC ( cun dy thun cm). t vo hai u mch mt in pkhng i nhng tn s thay i. khi f = f1 th U lmax, khi f = f2 th U cmax, cng sut mchc i khi tn s f lin h nh th no vi f1 v f2?

Hng Dn

UC max khi:1 =

1

LC R

2

2L2

UL max khi :2 =

2

2LC R2C2 Cng sut tiu th cc i khi: 2 =

1

LC

Bin i: 21 =2L R2C

2L2C

V 22 =2

C.(2L R2C); 4 = 21

22 = =

12 = f =

f1f2

84 Cho mch in xoay chiu AB cha R, L,C ni tip, on AM c in tr thun v cundy thun cm 2R = ZL, on MB c t C in dung c th thay i c. t hai u mchvo hiu in th xoaychiu u = U0cost(V), cU0 v khg i. Thay i C = C0 cng sutmch t gi tr cc i, khi mc thm t C1 vo mch MB cng sut ton mch gim mtna, tip tc mc thm t C2 vo mch MB cng sut ca mch tng gp i. T C2 c thnhn gi tr no sau y?

Hng Dn

Lc u do cng hng nn: ZC = ZL = 2R . cng sut on mch gim 1 na tc l sau khighp thm C1 th dung khng ca b t phi tha mn :|ZC ZL| = R nn xy ra 2 trng hp: T H1 : ZC > ZL nn lp t C1 ni tip vi C0 ta c: ZC = 3R = 3/2ZC0 lc .Vy cngsut li tng 2 ln th lc li c: ZC = 2R .Tc phi mc t C2 song song vi Co v C1 khi :ZC2 = 6R = 3ZC0 = C2 =

1

3C0 ...

Tng t cho: ZC < ZL tc lc :ZC = R = ZC2 = R =ZC0

2= C2 = 2C0

85 Mt con lc l xo ngang gm l xo c cng k = 100N/m v vt m = 100g, dao ngtrn mt phng ngang, h s ma st gia vt v mt ngang l = 0, 02. Ko vt lch khiVTCB mt on 10cm ri th nh cho vt dao ng. Qung ng vt i c t khi bt udao ng n khi dng hn l?

Hng Dn

S =k.A2

2..mg= 25(m)

86 Con lc l xo gm vt m = 100g v k=100N/m .dao ng nm ngang.ko ra khi vtcb 1on 3cm. ti t=0 truyn cho v = 30

3(cm/s) theo chiu ra xa vtcb vt bt u d...h

.tnh t ngn nht t khi vt bt u d. n khi l xo b nn Max ?

Hng Dn

T=0,2s.Khi l xo nn max tc l vt li x=-6cm



24/48

BIN SON: H HONG VIT

Vt i t li 3cm = 6cm, quay 1 gc: 3

htT

6.T 6cm = -6cm ht thi gian T

2

= thi gian ngn nht l :T2

+T

6=

2T

3=

2

15(s)

87 Mt con lc l xo, vt m dao ng cng bc khi tn s ngoi lc l f=f1 v vt dao ngn nh th bin o c A1, khi tn s ngoi lc f = f2(f1 < f2 < 2f1)th khi n nh

bin o c A2 = A1. cng lo xo c th c gi tr no?

Hng Dn

Nu gi l tn s goc ca dao ng ring. l tn s gc ca ngoi lc cng bc th.Coi lc mst l hng s v khng ph thuc vo vn tc th:Bin ca dao ng cng bc c tnh theocng thc: A =

F0m|2 2| Vy khi A1=A2 th:

21 +

22 = 2

2 = 42f21 + 42f22 = 2.K

m=

K = 22.m.(f21 + f22 )

88 Mt con lc ng h c coi nh mt con lc n c chu k dao ng T=2s; vt nng ckhi lng m=1kg. Bin dao ng ban u l: = 5. Do chu tc dng ca lc cn khngiF = 0, 001N n dao ng tt dn. Thi gian ng h chy n khi dng li l bao nhiu ?

Hng Dn

Ta c, gim c nng trong 1 chu k bng cng ca lc ma st sinh ra:

E = mgl 02

2 mgl

02

2= Fms.4S0 = 1

2mgl(0

2 02) =1

2mgl(0 0).(0 + 0)) =

1

2

mgl.20

Fms.4S0 = Fms40l

0 =

4Fc

mg

= 4.1040

S ng: N =0

0= 12500

Thi gian dng li hn l: t=N.T=12500.2= 250000s

89 Di tc dung ca mt lc c dng F = 0, 8sin5t(N) mt vt c khi lng 400 dao ngu ha bin dao ng ca vt l: 8cmA. 8cm B. 20cm C. 12cm D. 32cm

Hng Dn

ta c lc ko v : F = Fhl = ma = m2xF = 0, 8sin5t(N) = F0 = 0, 8 m Fmax = F0 = m2A = A = F0

m2vy A=0,08m = 8cm Coi lc l l lc phc hi em F = kx = m2.Asin5t, t y so snh vi bi ri suyra kt qu thi.

90 cho con lc n c chiu di l gia tc trng trng g ang dao ng v chu nh hng calc cn mi trng bng

1

500ln trng lng tc dng ln vt. Hi s ln con lc qua v tr cn

bng n khi con lc n dng hn ?

Hng Dn

Nng lng ban u ca con lc: W = mg2l

.S20



25/48

BIN SON: H HONG VIT

bin thin nng lng trong 1 chu k: W = Ams 4.Fms.S0Sau 1 chu k nng lng cn li l: W1 =

mg

2l.S21

Ta c: W = W W1 = mg2l

.(S20 S21) =mg

2l.(S0 + S1).(S0 S1) mg

2l.S0.S

Vy:4FmsS0 =mg

2lS0S = 8lFms = mgS = 8l mg

500= mgS = 8l

500= SS dao

ng thc hin c ti khi dng li l: n =S0

S s ln i qua VTCB l 2nCng ca lc ma st: Ams = Fms.S, trong S l qung ng di chuyn ca vt, Trong 1 chu kcoi S 4S0, do ta s c: Ams = Fms.4S0 = W

91 Mt on mch khng phn nhnh c dng in sm pha hn hiu in th mt gc nhhn

2:

A. Trong on mch khng th c cun cmB. H s cng sut ca on mch bng khngC. Nu tng tn s ca dng in ln mt lng nh th cc i hiu dng qua on mch gimD. Nu tng tn s ca dng in ln mt lng nh th cc i hiu dng qua on mch tng

Hng Dn

Do u ang tr pha hn i mt gc nh hn

2, chng t mch c RLC v ZC > ZL

Khi f tng w tng,ZL tng,ZC gim. ta thy rng mt s ln hn th gim i, cn s b hn thtng ln nh vy khong cch sai khc gia ZL v ZC gim, hay ZL ZC gim Z gim I tngCn ch trong cu ny lc u: ZC > ZL, cn trong cu 1 lc u ZC = ZL nn khi ZL v ZCthay i th dn n ZL ZC tng Z tng

92 Trong on mch RLC ni tip ang xy ra cng hng. Tng dn tn s dng in v ginguyn cc thng s khc ca mch, kt lun no sau y l ng:A. H s cng sut ca on mch gim B. Cng hiu dng ca dng in tngC. Hiu in th hiu dng trn t tng D. Hiu in th hiu dng trn cun cm gim

Hng Dn

Nu tng tn s dng in th ng ngha vi vic tg W. Khi tng W th ZL tng cn ZC s gim ZC > ZL = Z tng. mt khc R khng i.Ta bit l cos =

R

Z=

chn A

93 Mt con lc n c chiu di 1m dao ng iu ha ti ni c gia tc trng trng l10(m/s2). Gc ln nht v dy treo hp vi phng thng ng gc a0 = 0.1rad. Ti v tr dytreo hp vi phng thng ng gc a = 0.01rad th gia tc ca con lc c ln lA. 0, 1 B. 0, 0989 C. 0, 14 D. 0, 17

Hng Dn

Gia tc ton phn ca vt atp = a2tt + a

2ht

vi att; aht ln lt l gia tc tip tuyn v gia tc hng tm ca vtTa c: aht = v2

R= v

2

l; att =

2x

Tnh v : v =

2gl(cos cos0)Thay s c v=0,3145 m/s tnh c aht = 0, 0989



26/48

BIN SON: H HONG VIT

att = 2x = 2.l. = 0, 1

vy a ton phn atp =

a2tt + a2ht =

0, 12 + 0, 09892 = 0, 14

94 Mt con lc n c chiu di 0, 992m, qu cu nh 25(g). Cho n dao ng ti ni c giatc trng trng 9, 8m/s2 vi bin gc 40, trong mi trng c lc cn tc dng. Bit conlc n ch dao ng c 50(s) th dng hn. Xc nh hao ht c nng trung bnh sau 1co nng trung bnh sau 1 chu k :A. 22J B. 23J C. 20J D. 24J

Hng Dn

T = 2

lg

= 2(s) W = mgl(1 cos 0) = 5, 92.104(J) S chu k thc hin = 502

= 25(chuk)

Vy trung bnh 1 chu k gim:5, 92.104

25 23, 7.106

95 Mt con lc n dao ng tt dn, c sau mi chu k dao ng th c nng ca con lc lib gim 0,01 ln. Ban u bin gc ca con lc l 900. Hi sau bao nhiu thi gian th bin gc ca con lc ch cn 300. Bit chu k con lc l:A. T = 0, 5s B. T = 100s C. T = 50s D. T = 200

Hng Dn

gim c nng l :W = mghcos

6=

3

2W Cho gim c nng trong mi chu k l 0,01 ln

= c nng cn li sau n chu k l :W1 = 0.99n.W = W W1 = W n = 200 = t =n.T = 100s

96 Mt on mch xoay chiu gm R thun ni tip vi cun dy thun cm L thay ic.in p hiu dng hai u L c o bng mt vn k.Gi l gc lch pha gia in phai u on mch v cng dng in trong mch.Khi L = L1 th vn k ch V1 khi L = L2th vn k ch V2.Bit V1 = 2V2 v 1 + 2 = 2 .Tnh t s

P2P1

Hng Dn

Sin(1) =

V1

V ; Sin(2) = cos(1) =

V2

V ==>

V21

V2 +

V22

V2 = 1 = V2

= V

2

1 + V

2

2 = 5V

2

2 = V2

R2 =5V22 V2 = 4V22= P2

P1=

U2.cos(2)2

R

U2.cos(1)2

R

=cos(2)

2

cos(1)2= cotan((2)

2) =V2R2V22

= 4

97 Con lc l xo gm vt nng m = 100g v l xo nh c cng k=100N/m. Tc dng mtngoi lc cng bc bin thin iu ha bin F0 v tn s f1=6Hz th bin dao ng A1.Nu gi nguyn bin F0 m tng tn s ngoi lc n f2=7Hz th bin dao ng n nhl A2. So snh A1 v A2 :

A. .A1 = A2 B. A1 > A2 C. A2 > A1 D. Cha iu kin kt lun

Hng Dn

Ta c tn s dao ng ring ca h l fo = 5Hz;So snh ta thy fo < f1 < f2;T c th kt lun



27/48

BIN SON: H HONG VIT

l A1 > A2( tn s ca lc cng bc cng xa tn s dao ng ring th bin dao ng cng gim ( ch sosnh c v 1 pha))

98 Mt ng h qu lc m giy c chu k bng 2s, qu lc c coi nh mt con lc n viy treo v vt nng lm bng ng c khi lng ring l: 8900 kg/m3 . Gi s ng h treotrong chn khng. a ng h ra ngoi khng kh th chu k dao ng ca n bng bao nhiu?Bit khi lng ring ca khng kh trong kh quyn l: 1,3 kg/m3 . B qua nh hung ca lccn khng kh n chu k dao ng ca con lc. Chn p n ng.A. .2, 00024s B. 2, 00035s C. 2, 00012s D. 2, 00015s

Hng Dn

T

T=

Dkk2D

= 7, 3.105 = T = 2, 00015

99 Mc mt ti thun tr 3 pha i xng hnh tam gic vo 3 dy pha ca mng in xoaychiu 3 pha, ton ti tiu th cng sut l 600 W. Nu t mt dy pha ton ti tiu th cngsut l bao nhiu?

Hng Dn

3 ti i xng mc tam gic = Cng sut mi pha : P = Ud2

R= 200W

Khi t 1 dy = RntR//R = Cng sut tiu th l : P = U d2

2R/3=

3

2Ud2

R= 300W

100 Ni 2 cc ca mt my pht in xoay chiu mt pha vo 2 u on mch AB gmgin tr thun R = 30 mc ni tip vi t in.B qua in tr cc cun dy ca my pht.khi roto quay u vi tc i n vng/ pht th I hiu dng trong mch l 1 A. Khi roto quayu vi tc 2n vong/ pht th I hiu dng trong on mch l

6 A. Nu roto quay vi tc

3n vng/ pht th dung khng ca t l:A. 4

5 B. 2

5 C. 16

5 D. 6

5

Hng Dn

Nhn xt E = NBS t l thun vi n (tc vng) v ZC =1

C.

t l nghch (tc vng)

Th1 : E = k.n,ZC = k/n = I2 = k2.n2

R2 +k2

n2

= 1 = R2 + k2

n2= k2.n2

Th2 : E = k.2n, ZC = k/2n = I2 = k2.4n2

R2 +k2

4n2

= 6 = 6R2 + 3k2

2n2= k2.4n2

Kh k t 2 PT trn = k2

n2= 24 = k

n= 12

5 TH3: ZC =

k

3n= 4

5



28/48

BIN SON: H HONG VIT

101 Cho mch in xoay chiu gm cun dy thun cm c t cm L v in tr R ni tipvi mt t in C.t vo hai u on mch in p u = 100

2cos100.t(V) .Khi in p

hiu dng trn t c gi tr gp 1,2 ln in p hai u cun dy .Nu ni tt hai du t inth cng hiu dng khng thay i v c gi tr l: 0,5A.Cm khng ca cun dy c gi trl:

A. 50 B. 160 C. 100 D. 120

Hng Dn

Gi thit cho 2 TH u c cng I = Trong c hai TH chng cng c Z = ZC = 2ZL(1)GT cho TH1 c UC = 1, 2Ud = ZC = 1, 2ZRL = Z2C = 1, 44(Z2L + R2).(2)Mt khc Z = U/I = 200 = 2002 = (ZLZC)2 + R2 th (1) v (2) vo ta c ZC = 240 =ZL = 120

102 Mc vo mch RLC khng phn nhnh mt ngun in xoay chiu c f thay i c khi

f1 = 60Hz th cos = 1 khi f2 = 120Hz th cos = 0.707.Hi khi f3 = 90Hz th h s cngsut bng bao nhiu:A. 0, 872 B. 0.486 C. 0, 625 D. 0, 781

Hng Dn

w1.L =1

w1.C= L.C = 1

w21(1)

tan2 =w2.L 1

w2.CR

= 1 => R = w2.L 1w2.C

tan3 =w3.L 1

w3.CR

=w3.L 1

w3.Cw2.L 1w2.C

=w23.LC 1w22.LC 1

.w2w3

Th (1) vo = tan3 = f23 f21

f22 f21.f2f3

= 3 = cos3 = A2Gi ZL khi f = 60Hz l x th ZL = ZC = x

Khi tn s tng gp i th lc ny ZL = 2x

ZC =

x

2 R =

3x

2

Khi tn s tng 1, 5 th

ZL =3x

2

ZC =2x

3

= tan3 =3x

2 2x

33x

2

=5

9= cos3 = 0, 874

103 Mch in xoay chiu khng phn nhnh gm R, cun dy thun cm v t in C mcni tip Zc = 2ZL vo mt thi im khi hiu in th trn in tr v trn t c gi tr tc

thi tng ng l 40V v 30V th hiu in th gia hai u mch in l?A. 55V B. 85V C. 85V D. 25V

Hng Dn



29/48

BIN SON: H HONG VIT

1

uL = u0Lcos(t +

2)

uC = u0Ccos(t 2

)

=

uCu0C

+uL

u0L

= 0uCu0C

+uLu0L

= 0

ZC = 2ZL = u0L =u0C

2

=

uL =

15 =

u = uR + uL + uC = 55(V)

2ZC = 2ZL = |uC| = 2|uL| = 30V do uL v uC ngc pha = uL = 15 = u = uL+uC+uR =15 + 30 + 40 = 55

104 ng thu tinh di 2m, ct nc trong ng cao 1m. Pht mt sng m vo ng th nghem to nht. Rt nc i 20cm li nghe m to nht ln na. Bit tc truyn m l 340m/s.a/Tm tn s m.b/Tnh s nt v s bng trong trng hp ct nc cao 90cm.

c/Gi nguyn cao ct nc l 50cm. Thay i tn s t 0 n f1 th c 3 ln nghe m tonht. Tnh f1 v tng tn s nghe m to nht ln na.

Hng Dn

a.Khong cch gia 2 mc nc bng

2= 20 =

2= = 40(cm) = f = 340/0, 4 = 850(Hz)

b. Ct nc cao 90(cm) = Phn kh trong ng cao 1, 1(m) 1, 1 = (k + 1/2)

2= (k + 1/2).0, 2 =

k = 5 = 6 bng v 6 ntc. f1 = 3f0

vi f0 = v/4l = 1703 = f1 = 170(Hz)( vi l=0,15m) tg tn s nghe m to nht ln na l f0

105 t mt in p u = Uocos(t) V vo 2 u mt on mch gm cun dy mc ni tipvi t in C c in dung thay i c.Ban u t in c dung khng 100, cun dy c

cm khng 50.Gim in dung mt lng C =103

8

F th tn s dao ng ring ca mch

l 80 ( rad/s).Tn s gc ca dng in trong mch l:A. 40rad/s B. 60rad/s C. 100rad/s D. 50rad/s

Hng Dn

C =1

100.w

L =50

w= C = 1

100.w 10

3

8= Tn s gc dao ng :

1

L.C=

1

50w .(

1

100.w 103

8 )

= 80 = = 40



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BIN SON: H HONG VIT

106 Mt mch in xoay chiu gm cun dy c in tr thun 40, t cm1

3 H, Mt

t in c in dung C thay i c v mt in tr thun 80 mc ni tip. t vo hai umch mt in p xoay chiu c gi tr ln nht 120V, tn s 50Hz, Thay i in dung ca tin n gi tr C0 th in p t vo hai u mch cha cun dy v t in cc tiu. Dngin hiu dng trong mch khi l?

A. 1A B. 0, 7A C. 1, 4A D. 2A

Hng Dn

UrLC =U.

(ZL ZC)2 + r2(ZL ZC)2 + (r + R)2

UrLCmin = (

(ZL ZC)2 + (r + R)2(ZL ZC)2 + r2

)max

(1 +R2 + 2Rr

(ZL

ZC)2 + r2)max = ZL = ZC = Cng hng in = I = U

R + r= 1(A)

107 mt thong ca mt cht lng c hai ngun kt hp A,B cch nhau 10 cm, daong theo phng thng ng vi phng trnh ln lt l uA = 3cos(40t +

6) cm; uB =

4 cos(40t +2

3) cm. Cho bit tc truyn sng l 40 cm/s. Mt ng trn c tm l trung

im ca AB, nm trn mt nc, c bn knh R = 4 cm. S im dao ng vi bin 5 cmc trn ng trn l:A. 30 B. 32 C. 16 D. 15

Hng Dn 32 + 42 = 52 = 2 sng ti ca im c bin bng 5 phi vung pha. lch pha hai sng ti : =

2(d1 d2)

+ 2 1 = 2(d1d2) +

2= = (2k + 1)

2=

d1 d2 = k.2

Xt s im tho K trn ng knh(MN) ca ng trn thuc ng ni 2 ngun MN d1 d2 MN = bn th vo tm k tha mn (m c N s) S im trn ng trn : Nx2(Nu du bng c xy ra th 2 im M v N bn ng tnh vo vic x2 nh) =

32

108 Mt on mch xoay chiu gm R bin tr ,cun dy thun cm v t in C mc nitip.in p hiu dng hai u R c o bng mt vn k ,ng vi hai gi tr ca bin tr lR1 v R2 th vn k chV1 v V2.Bit

V2V1

.Gc lch pha 1 v 2 gia in p u ca mch vi

cc dng i1 v i2 tho mn 1+2 =

2.t N2 = n2+1.H s cng sut mch khi bin tr R1 l :

Hng Dn

cos(1) =V1V

; cos(2) = sin(1) =V2V

d = V21

V2+

V22V2

= 1d = V2 = V21 + V22 d =1

cos(1)2=

V

2

1 + V

2

2V21

= 1 + n2 = N2 = cos(1) = 1N109 Mt vt dao ng vi phng trnh x = 4cos(4t

4).Tnh thi gian chuyn ng i c

qung ng 6cm k t lc bt u?



31/48

BIN SON: H HONG VIT

Hng Dn

T pt dao ng ta c omega = TTa c bin dao ng l 4cm. Trong 1 chu k dao ng l 4A. Vy qung 6 cm.

Vy thi gian t =T

4+

T

8.

110 2 cht im cng xut pht t gc ta v bt u dao ng iu ha theo cng 1chiu trn trc Ox vi bin bng nhau v chu k 3s v 6s. t s tc ca 2 cht im khigp nhau l:A. 2 B. 4 C. 1 D. 3

Hng Dn

Gi s PT 2 cht im : x1 = Acos(2

3.t +pi/2) v x2 = Acos(

3.t +

2)

khi chng gp nhau ==>2

3.t =

3.t + k2 v 2/3.t =

3.t + k2

= t = 6k v t = 2k ==> t=2s

= |v1||v2| =|A.2.f1.sin(2.

3.2 +

2)|

|A.2.f2.sin(2.6

.2 +

2)|

= |v1||v2| =T2T1

= 2

111 Trong th nghim giao thoa vi hai ngun pht sng ging nhau ti A,B trn mt nc.Khong cch hai ngun l AB = 16 cm. Hai sng truyn i c bc sng = 4 cm. Trn ng

thng xx

song song vi AB, cch AB mt khong 8 cm, gi C l giao im ca xx

vi ngtrung trc ca AB. Khong cch ngn nht t C n im dao ng vi bin cc tiu nmtrn xx l:A. 1, 42 cm B. 1, 5 cm C. .2, 15 cm D. 2, 25 cm

Hng Dn

Gi d1 v d2 l khong cch t im M cn tm ti A v B, ta c:

d2 d1 = (M ) 2

= 4k + 2

k = 0

d2 d1 = 2 =

d2 d1 = 282 d21 +

82 d22 = 16

= d1 = 10, 36 = d = 8

d21 82 = 1, 42

112 Hai dao ng iu ha cng phng cng tn s vi bin ln lt l 5 cm v 12 cm.Bin dao ng tng hp ca hai dao ng khng c gi tr no sau y:A. 6cm B. 16cm C. 17cm D. 7cm

Hng Dn

|A1 A2| A |A1 + A2| = A (16, 17, 7)



32/48

BIN SON: H HONG VIT

113 Mt si dy dn hi OM=180cm hai u c nh. khi c kch thch trn dy hnh thnh6 b sng. bin dao ng ti l 3cm. khong cch gia hai im gn nht dao ng ngcpha v c cng bin 1,5cm l?A. 20cm B. 10cm C. 2, 5cm D. 5cm

Hng DnDy 2 u c nh = cng thc tnh bin 1 im trn sng dng a = Abungsin(2.d/)Gi thit a = 1, 5cm; Abung = 3cm = sin( 2.d

) =

6= d =

12(d khong cch ngn nht

n nt)= 2 im ngc pha c bin 1,5cm m ngc pha th cch nhau d = 2/12 = /6 Gi thit cho trn dy c 6 b sng = d = 6

2= 180 = = 30cm = d = 5cm

= 60cm = d = 10cm

114 1 con lc n c chiu di 1, 73m dao ng trn mt chicxe ang ln khng ma st xung1 ci dc nghing 1 gc 300C so vi phng ngang. Tnh chu k dao ng?

Hng Dn

T = 2

l

g

g = g.cos30o = g

3

2

T = 2

3

g

3

2

= 22

g=

2

2

= 2

2(s)

115 mt con lc l xo dao ng iu ha theo phng ngang vi nng lng dao ng 1 J vlc n hi cc i l 10 N.gi Q l u c nh ca l xo .khoang thi gian gia 2 ln lin tipQ chu tc dng ca lc ko 5 cn 3 N l 0,1s.tnh qung ng ln nht m vt i c trong 0,4s.

Hng Dn

kA2 = 2J; kA = 10(N) = A = 0, 2m = 20cm. dng gin vct ta c 2 ln lin tip lc kol 5

3 ng vs gc quay 600, i xng qua trc nm ngang. = T

6= 0, 1s = T = 0, 6s

Vy 0, 4s = 2T /3 = T/2 + T /6. Trong

T

2s th qung ng lun = 2A. vy qung ng max khi trong

T

6s vn tc ln nht.

= cng gn VTCB vn tc cng ln = gc quay T6

= 600, i xng qua trc thng ng.= S = 2A + A/2 + A/2 = 3A = 60cm



33/48

BIN SON: H HONG VIT

117 Cho on mch RLC mc ni tip, vi cun cm thun c t cm L thay i c.in p xoay chiu gia hai u on mch lun n nh. Cho L thay i. Khi L = L1 th inp hiu dng gia hai bn t in c gi tr ln nht, in p hiu dng gia hai u in trR bng 220V. Khi L = L2 th in p hiu dng gia hai u cun cm c gi tr ln nht vbng 275V, in p hiu dng gia hai u in tr bng 132V. Lc ny in p hiu dng gia

hai bn t in l ?A. 96V B. 451V C. 457V D. 99V

Hng Dn

cho ULmax = 275V th khng nn cho UR = 132V L thay i Ucmax = cng hng in = U = UR = 220V Khi L thay i ULmax = uRC vung gc u = 1

U2R=

1

U2RC+

1

U2= URC = 165 = UC =

U2RC U2R = 99V

118 Cho on mch in AB khng phn nhnh mc theo th t :mt cun cm ,mt t inc in dung Cthay i c ,mt in tr thun R = 50 .Gia A,B c mt in p xoay chiulun n nh u = 160

2sin.t (V).Cho C thay i .Khi dung khng ca t in bng 40 th

in p gia hai u cun cm lch pha

3so vi in p gia hai u mch MB(mch MB cha

C v R)v cng sut tiu th ca mch AB ln nht Pmax.Gi tr ca Pmax bng?A. 328, 00W B. 840, 50W C. 672, 50W D. 537, 92W

Hng Dn

C thay Pmax = cng hng in = ZL = ZC = 40 = tan(RC) = ZC/R =4/5 = RC = 38, 660 Ud lch pha so vi URC 1 gc

3= cun dy khng thun cm m c r = rL = 6038, 66 =

21, 340 = tan(rL) = ZLr

= r = 102, 38 = Pmax = U2

R + r= 168W

119 Cho mch in AB gm mt t in c in dung C; mt in tr hot ng R v mtcun cm c in tr thun r v c t cm L ( theo th t ) mc ni tip vi L = rRC.t vo hai u on mch mt in p xoay chiu th in p gia hai u cun cm c biu

thc u = 100cos(t +

12 )(v). Vo thi im in p gia hai u cun cm bng 80V th inp gia hai u mch AM( AM gm C v R) l 30V. Biu thc ca in p gia hai u onmch AM l:A. u = 50cos(t 5

12)V B. u = 50cos(t

4)V

C. u = 200cos(t 4

)V D. u = 200cos(t 512

)V

Hng Dn

T L = rRC suy ra in p tc thi UAM vung pha vi UMB suy ra AM =

12

2=

512

U0AM =

1 (UOMB80

)2.30 = 50(V)



34/48

BIN SON: H HONG VIT

120 Con lc n gm qu cu c khi lng m mang in tch q, dy treo nh, khng dn,khng dn in. Khi khng c in trng, con lc dao bng b vi chu k 2s, khi c in trngtheo phng thng ng con lc dao ng b vi chu k

3s, bit ln lc in trng lun

b hn trng lc tc dng vo qu cu. o chiu in trng con lc dao ng b vi chu k:A. 1s B.

2s C.

6 D. 2

3

Hng Dn

T1 = 2.

l

g

T2 = 2.

lg +

F

m

T3 = 2.

l

g Fm

T2 < T1 = g > g = g = g +F

m

=

T1T2

=

1 +

F

mg= F

mg=

1

4T1T3

=

1 F

mg= 4

T23=

3

4

= T3 = 43

121 Con lc l xo nm ngang gm vt c khi lng 200g cng k=100N/m, h s ma stgia vt v mp ngang l 0,1. Ban u vt c gi v tr l xo dn 10cm, ri th nh daong tt dn,ly g = 10m/s2. Trong khong thi gian k t lc th cho n khi tc ca vtbt u gim th gim th nng ca con lc l:A. 32mJ B. 20mJ C. 50mJ D. 48mJ

Hng DnWt =

1

2kl0

2 12

kl2 mgSkl = mg(l0 l)

= l = 0, 002(m) = S = l0 l = 0, 098(m) = Wt = 48(mJ)

122 in p hiu dng gia hai cc ca mt trm pht in cn tng ln bao nhiu ln gim cng sut hao ph trn ng dy ti in 100 ln, vi iu kin cng sut truyn n titiu th khng i. Bit rng khi cha tng in p gim in th trn ng dy ti inbng 15% in p hiu dng gia hai cc ca trm pht in. Coi cng dng in trongmch lun cng pha vi in p t ln ng dy.A. 8,515 ln B. 6,25 ln C. 10 ln D. 8,25 ln

Hng Dn

U1 = U1 + UT1; U1 = 0, 15U1

U2 = U2 + UT2php

php =

RI22

RI12 =

U2I2

U1I1 =

1

100

pt = UT1I1 = UT2I2



35/48

BIN SON: H HONG VIT

=

I2I1

=U2U1

=1

10

UT2 = 10UT1

= U2U1

=U2 + UT2

U1=

U110

+ 10UT1

U1=

0, 15

10+ 0, 85.10 = 8, 515

123 Hai mch dao ng l tng L1C1; L2C2 chu k dao ng mch L1C1 gp 2 ln chu k daong mch L2C2, in tch cc i ca trn t ca cc mch u bng Q0. Ti mt thi im

no ln in tch trn t u bng nhau th t s ln cng dng in gia mchth nht v mch th 2 l bao nhiu?

Hng Dn

W = WC + WL Q02

2C=

q2

2C+

Li2

2= i =

Q0

2 q2 = i1i2

=12

= T2T1

= 2

124 Mt vng dy c din tch S=100 cm2 v in tr R = 0, 45 , quay u vi tn s f=50Hz trong mt t trng u c cm ng t B=0,1T xung quanh mt trc nm trong mtphng vng dy v vung gc vi cc ng sc t. Nhit lng ta ra trong vng dy khi n

quay c 1000 vng l:A. 1, 39J B. 0, 35J C. 7J D. 0, 7J

Hng DnQ = RI

2t =E2

2Rt

= BS = 0, 1.100.104.2.50= t = N(vng)

100(vng)2(s)

=50(s)

125 Ni hai cc ca my pht in xoay chiu mt pha vo hai u on mch AB gm

R = 60 in tr thun , mt cun dy thun cm v mt t in mc ni tip.B qua intr ca cun y ca my pht. Khi roto ca my quay u vi tc n vng/pht th cng dng in hiu dng trong on mch l 1A v dng in tc thi trong mch chm pha

4so vi in p tc thi gia hai u on mch. Khi roto ca my quay u vi tc 2n

vng/pht th dng in trong mch cng pha vi in p tc thi hai u on mch. Cng dng in hiu dng trong mch khi l:

Hng Dn

Hiu in th 2 u mch U=k.n1 (k h s t l, n1 s vng quay/P)

TH1 : n1 = n = I = k.nZ

=

4= ZL ZC = R = Z = R2

I1 = 1 =kn

R

2= kn

R=

2

TH2 : n1=2n = i ng pha u = cng hng = I = UR

=2kn

R= 2

2



36/48

BIN SON: H HONG VIT

126 Cho mch in AB theo th t gm ampe k, in tr hot ng R, cun thun cm L,t in C, ni hai u gia cun cm v t in mt vn k,ng thi ni hai u t in mtkha K. Khi mc mch vo hiu in th mt chiu khng i:K m, vn k ch 100v , K ng,vn k ch 25V . Khi mc vo mch hiu in th xoay chiu, K m hay ng, vn k u ch50v. Bit s ch ca ampe k l nh nhau khi K ng. H s cng sut ca mch khi mc vo

hiu in th xoay chiu l:

Hng Dn

Dng 1 chiu :k m: khng c dng qua mch = s ch von k ch hiu in th mch U=100Vk ng : t ni tt s ch von k ch hiu in th 2 u r cun dy = U r = 25V = UR =75V = R = 3r v I = 100/(4r)Dng xoay chiu:k m mch c R,L,r,C, k ng mch c R,L,rGi thit cho k ng hay m vn k ch nh nhau [B]( nu ch cho bng 50V th khng giiquyt c u, do vy theo ti ngh cn thm gi thit l cho i bng nhau na nh) = ZLr =ZLrC = ZC = 2ZL k ng I bng nhau = 100/4r = 50/ZL2 + r2 = ZL2 = 3r2 = ZL = 3.rtan() =

ZC ZL4r

=

3.r

4r=

3

2= cos()

127 Trong on mch c 2 fn t X v Y mc ni tip>Hiu in th xoay chiu t vo Xnhanh pha

2vi h..t xoay chiu vo fn t Y v cng fa vi dng in trong mch.Xc nh

cc fn t X,YA. X l in tr,Y k cun dy thun cm B. l t in,X k in trC. X l in tr,Y l cun dy t cm c in tr thun r khc 0 D. X l t in,Y l cun dy thun

Hng Dn

X l in tr,Y k cun dy thun cm

128 t vo hai u on mch in RLC ni tip mt hiu in th xoay chiu c gi trhiu dng khng i th hiu in th hiu dng trn cc phn t R,L,C u bng nhau v bng

20V. Khi t b ni tt th hiu in th hiu dng 2 u in tr bng?A. UR = 10

2 B. UR = 20

2 C. UR = 30

2 D. UR = 40

2

Hng Dn

U2 = U2R + (UL UC)2 = 202 + (20 20)2

I =URR

=ULZL

=UCZC

=U = 20(V)R = ZL = ZC

Khi ni ttURU

=R

R2

+ Z

2

L

=1

2

= UR = 10

2



37/48

BIN SON: H HONG VIT

129 on mch xoay chiu AB ch gm cun thun cm L ni tip vi bin tr R. Hiu inth hai u mch l UAB n nh, tn s f. Ta thy c 2 gi tr ca bin tr l R1 v R2 lm lch pha tng ng ca UAB so vi dng in qua mch ln lt l :1 v 2.Cho bit 1 + 2 =

2.

t cm L ca cun dy c xc nh bng biu thc:

A. L = R1.R22..f

B. L = R12

+ R22

2..fC. L = lR1 R2l

2..fD. L = R1 + R2

2..f

Hng Dn

tan 1. tan 2 = 1 = L2.(2f)2 = R1.R2 = L =

R1.R2

2..f

130 cho mch RLC ni tip ,in p xoay chiu ca mch l U = Uocos100t .Hiu in thhiu dng qua t bng 1,2 ln hiu in th hiu dng qua cun dy . Khi ni tt t in ,cng hiu dng qua mch ko i v bng 0,5 A .Tm UL

Hng Dn

I khng i nn UC = 2UL Theo gi thit ta c2UL = 1, 2

U2L + U

2r = 2, 56U2L = 1.44U2r = Ur =

4UL3

= U2L + (4UL

3)2 =

U202

= UL = 3U05

2

131 TRn mt cht lng c hai ngun pht sng S1, S2 dao ng vi pt u1 = u2 = 5cos(20t) ,

cch nhau on S1S2 = 10cm . Tc truyn sng trn mt nc l 30cm/s. Trn ng thngi qua S1S2 c bao nhiu im dao ng cng pha vi trung m O ca S1S2

Hng Dn

Trn ng ni 2 ngun c giao thoa tng t sng dng = c v s im ng pha vi trungim nh. Tm s im cc i ng pha vi trung im O = 2 im

132 Trong hin tng giao thoa song mt nc, hai ngun kt hp ngc pha.AB=20cm ,

bc sng do 2 ngung phat ra 10cm . Mt dim M trn mt nc cch A mt khong L c AMvung gc vi AB .Tm gi tr ln nht ca L quan st c cc i giao thoa?A. 17, 5cm B. 37, 5cm C. 12, 5cm D. 42, 5cm

Hng Dn

2 Ngun ngc pha nn :Ta c M dao ng cc i th MB MA = (k + 1/2) hay AB2 + L2 L = (k + 1/2)C th nhn xt ngay L t GTNN th ng vung gc vi AB phi giao vi hypepol gn vi

ng trung trc nht. ng vi k = 0202

+ L2

L =1

2 = 5 = L = 37.5cm



38/48

BIN SON: H HONG VIT

133 Cho 3 linh kin gm in tr thun R = 60 cun cm thun L v t in C. Lnlt t in p xoay chiu c gi tr hiu dng U v 2 u on mch ni tip RL hocRC th biu thc cng dng in trong mch ln lt l i1 =

2cos(100.t

12(A) v

i2 =

2cos(100.t +7

12(A) Nu t in p trn vo 2 u mch RLC ni tip th dng in

trong mch c biu thc?A. i = 22cos(100.t +

3(A) B. i = 2cos(100.t +

3(A) C. i =

2

2cos(100.t +

4(A) D. i = 2cos(100.t +

4(A)

Hng Dn

2 cch mc u c : I0 = const, U0 = const = Z = const = ZL = ZC = ZL

R=

ZCR

= tan1 =tan(2) = 1 = 2 = u i1 = (u i2) = u =

1

2(i1 + i2) =

4= tan1 =

tan(

4 +

12) = 3 =ZLR = ZL = 603 = U = 120(V)

Cch mc 3: Cng hng u = i =

4= I0 = 120

60

2 = 2

2 = i = 2

2Cos(100t +

4)

134 Cho mch in xaoy chiu RLC ni tip, cun dy thun cm. Bit L = CR2 t vo 2u on mch in p xaoy chiu n nh, mch c cng h s cng sut vi 2 gi tr ca tns gc 1 = 50 v 2 = 200(rad/s) H s cng sut ca on mch bng:A.

1

2B.

213

C.312

D.1

2

Hng Dn

L = CR2; 2 =1

LC=

1

C2.R2

=

12 = 100 = L = R100 ; C =1

100.R= cos = R

R2 + (ZL ZC)2= cos =

213

135 Trn dy cng AB vi hai u dy A, B c nh , c ngun pht sng cch B mton SB = 5 (cho bit trn dy c sng dng ). Tm im M gn S nht thuc on SB

m sng tng hp c bin b A=2a , c dao ng tr pha hn dao ng pht ra t S mt gc 2 ?

Hng Dn

Pt sng ti M:2a.cos(2.d

+

2)cos(t

2)

Yu cu bi ton:2.d

+

2= k2 = 1

4 k 5.25

V M gn S nht nn ta chn k = 1 = d = 34



39/48

BIN SON: H HONG VIT

136 Mt ngun phng x nhn to c chu k bn r l 8 gi, c phng x ban u bng128 ln phng x an ton cho php. Hi phi sau thi gian ti thiu bao nhiu c th lmvic an ton vi ngun phng x ny?A. 56 gi B. 48 gi C. 32 gi D. 64 gi

Hng DnSau khi phng x gim i 128 ln th an ton nn: 2 tT = 128 = t = 8.7 = 56

137 Mt nhc c pht ra m c bn c tn s f1 = 420 Hz. Mt ngi ch nghe c m caonht c tn s l 18000 Hz, tm tn s ln nht m nhc c ny c th pht ra ngi nghe c.

Hng Dn

fnghe < 18000 = nf1 < 18000 = n < 18000420 = n < 42, 857;[nmax] = 42 = fmax =420.42 = 17640(Hz)

138 Mt vt dao ng iu ho vi phng trnh x = 10 cos

t +

3

cm. Khong thi gian

k t lc vt bt u dao ng n khi vt i c qung ng 50 cm l?A. t =

7

3(s) B. t = 2, 4(s) C. t =

4

3(s) D. t = 1, 5(s)

Hng Dn

T2 T1 = nT + t n

= S = n4A + s; s = s + s = 50 = 1.4.10 + 10 = n = 1 t

t = 0 x = A2

= 5(cm); v < 0 s = 10 = x = A2

x = A2

= t = T12

+T

12=

T

6=

T2 T1 = T + T6

=7T

6=

7

3(s)

139 Mt vt dao ng iu ho vi phng trnh x = A cos

2t

T

cm. Khong thi gian

ngn nht k t lc vt bt u dao ng n thi im m ng nng bng 3 ln th nng u

tin l?A. tmin =

T

4B. tmin =

T

8C. tmin =

T

6D. tmin =

T

12

Hng Dn

p dng cng thc x = An + 1

vi n l Wd = nWt = x = A2

Lc u vt bin +A nn v tr vt n u tin l x = +A

2

T n tmin = T6

s



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BIN SON: H HONG VIT

140 Mt vt dao ng iu ho vi bin A v chu k T. Trong khong thi gian t =3T

4,

qung ng nh nht m vt i c l:A. 4A A2 B. 2A + A2 C. 2A A2 D. A + A2

Hng Dn

Bi gii tng qut cho dng ton ny: Nu t < T/2Smax = 2Asin(

2

) v Smin = 2A(1 cos(2 )) = .t Nu t > T/2 (phn tch t = n.T/2 + t1(n > 1, t1 < t2)

Smax = n.2A + 2Asin(

2) v Smin = n.2A + 2A(1 cos(

2))

p dng vo bi trnt = 3T /4 = T /2 + T /4

Smax = 2A + 2Asin(.T/4

2

) = 2A + A

2 v Smin = 2A + 2A(1

cos(.T/4

2

)) = 2A + 2A(1

2

2

) =

4A A2

141 t vo 2 u AB mt mch in xoay chiu c in p hiu dng UAB=120V c R nitip cun cm khng thun v t C bin thin .Gia R v cun dy l im M. Gia cun dyv t l im N. Cho R = 20, r = 10, ZL = 20 .Ngi ta thy khi C=Cm th in p hiudng gia M v B t gi tr cc tiu U1min. Gi tr l :A. 60V B. 50V C. 40V D. 30V

Hng Dn UMB = IABZMB =

UAB(20 + 10)2 + (20 ZC)2

102 + (20 ZC)2

=UAB

302 + (20 ZC)2102 + (20 ZC)2

=120

1 +800

102 + (20 ZC)2UMBmin = ZC = 20 UMBmin = 40V

142 Cho mt hp kn bn trong cha 2 trong 3 thit b R,L,C mc ni tip. t mt hiu

in th khng i U = 100V vo 2 u on mch th thy dng in chy qua mch l 1A .Khi t vo on mch hiu in th xoay chiu c biu thc u = 200cos(100t

4V th

thy dng in hiu dng qua mch c gi tr khng i. Cc thit b trong hp kn l:A. Cun dy khng thun cm c R = 100B. Cun dy thun cm c ZL = 100C. in tr thun R = 100 v cun dy thun cm c L =

1

H

D. in tr thun R = 100 v t in c in dung C =104

F

Hng Dn

C dng 1 chiu i qua = mch ch c R,L hoc ch c RGT cho 2 trong 3 = Mch c R,L = Cu (C)



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BIN SON: H HONG VIT

143 Mt on mch xoay chiu gm in tr R, cun dy thun cm L v mt hp X mc nitip. Hp X cha 2 trong 3 phn t RX ; LX ; CX . t vo 2 u on mch mt hiu in thxoay chiu c chu k dao ng T , lc ZL = sqrt3.R . Vo thi im no thy URL tcc i, sau thi gian T

12th thy hiu in th 2 u hp X l Ux t cc i. Hp X cha:

A. RX; LX B. RX ; CX C. LX ; CX D. khng xc nh c

Hng Dn

=2

T.

T

12=

6= RL X =

3 X = X = 6 = A

144 t mt ngun in xoay chiu c hiu in th hiu dng U v tn s f vo hai u caon mch gm RLC mc ni tip, trong cun dy thun cm.Ni hai u t vi mt ampe kth thy n ch 1A ng thi dng in tc thi chy qua ampe k chm pha

6so vi hiu in

th tc thi gia hai u on mch.Nu thay ampe k bng mt vn k th thy n ch 167, 3V

,ng thi hiu in th tc thi gia hai u vn k chm pha mt gc

4 so vi hiu in thtc thi gia hai u on mch. Hiu in th hiu dng ca ngun in xoay chiu l ?A. 175V B. 150V C. 100V D. 125V

Hng Dn

Nhn vo gin ta c:UR = ULC = UC UL(1)UL = tan(30).UR =

3UR3

(2)

T 1 v 2 UR = UC1+

3

3

= 106, 06V = UL = 61, 23V = U =

(UL UC)2 + U2R = 150V

145 Trong th nghim Ing v giao thoa nh sng khe hp S pht ra ng thi ba bc x1; 2; 3. Trn mn trong khong gia hai vn sng lin tip c mu ging mu vn sng trungtm, nu hai vn sng ca hai bc x trng nhau ta ch tnh mt vn sng th s vn sng quanst c trn on l l bao nhiu?

Hng Dn Gi N1; N2; N3 l s vn sng ca bc x 1; 2; 3N123 l s vn trng ca ba bc x 1; 2; 3N12; N13; N23 l s vn trng ca tng cp bc x s vn sng :N1 + N2 + N3 (N12 N123) (N23 N123) (N13 N123) 2N123

146 t vo 2 u mch RLC mc ni tip mt hiu in th xoay chiu c gi tr hiu dngkhng i th in p hiu dng trn cc phn t R, L, C u bng nhau v bng 20V. Khi nitt t in th in p hiu dng 2 u in tr R l?

A. 10V B. 102V C. 20V D. 302V

Hng Dn

t vo 2 u mch RLC mc ni tip mt hiu in th xoay chiu c gi tr hiu dng khng



42/48

BIN SON: H HONG VIT

i th in p hiu dng trn cc phn t R, L, C u bng nhau v bng 20V = Cng hng= UAB = 20VKhi ni tt t in th in p hiu dng 2 u in tr R = UR = 10sqrt2

147 Mch RLC ni tip. Khi t vo mch hiu in th xoay chiu c tn s gc th mchc tnh cm khng. Cho bin i th ta thy tn ti mt gi tr ca khi cng hiudng c tr s ln nht l Imax v 2 gi tr1 v 2 vi 1 2 = 100 cng dng introng mch bng nhau v tha mn I1 = I2 = Imax

3, bit 2

2

H. Tm in tr R?

Hng Dn

Z1 = Z2 = 0 = 12 = 12 = 02 = 1LC

I1 =Imax

3= U

R2 + (ZL1 ZC1)=

U

3R= 8R2 = (ZL1 ZC1)2

I2 =Imax

3

=

UR2

+ (ZL2 ZC2)=

U

3R

=

8R2 = (ZL2

ZC2)

2

21 = 200(2 > 1); 12 = 1LC

= 8R = (ZL1 ZC1) = (ZL2 ZC2) = L2L1 =

R =200.2

2

.

8= 200

148 Cng sut m thanh cc i ca mt my nghe nhc gia nh l 10W. Cho rng c truyntrn khong cch 1m, nng lng m b gim 5% so vi ln u do s hp th ca mi trngtruyn m. Bit I0 = 10-12W/m2. Nu m to ht c th mc cng m khong cch 6m l:A. 102dB B. 107dB C. 98dB D. 89dB

Hng Dn

C 1m gim5

100vy cn

95P

100

Sau 6 m cn li l (95

100)6P

I =P

4.d2

L = 10.lgP0(0, 96)

2

4.d2.I0= 102dB

149 Hai im A, B cng phng truyn sng, cch nhau 24cm. Trn on AB c 3 im A1,A2, A3 dao ng cng pha vi A; 3 im B1, B2, B3 dao ng cng pha vi B. Sng truyntheo th t A, A1, B1, A2, B2, A3, B3, B, bit AB1 = 3cm. Bc sng l:A. 6cm B. 3cm C. 7cm D. 9cm

Hng Dn

AB = 3 + 3 = 24 = = 7(cm)



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BIN SON: H HONG VIT

150 Trong th nghim Y-ng, ngun S pht bc x n sc , mn quan st cch mt phng haikhe mt khong khng i D, khong cch gia hai khe S1S2 = a c th thay i (nhng S1 vS2 lun cch u S). Xt im M trn mn, lc u l vn sng bc 4, nu ln lt gim hoctng khong cch S1S2 mt lng x th ti l vn sng bc k v bc 3k. Nu tng khongcch S1S2 thm 2x th ti M l:

A. vn sng bc 7 B. vn sng bc 9 C. vn ti th 9 D. vn sng bc 8

Hng Dn

4D

a= k

D

a a = 3kD

a + a= k

D

a + 2a= k = 2; k = 8 = D

151 Trong th nghim v giao thoa nh sng I-ng. Nu lm th nghim vi nh sng n scc bc sng 1 = 0, 6m th trn mn quan st, ta thy c 6 vn sng lin tip tri di trn brng 9mm. Nu lm th nghim vi nh sng hn tp gm hai bc x c bc sng 1 v 2 th

ngi ta thy: t mt im M trn mn n vn sng trung tm c 3 vn sng cng mu vivn sng trung tm v ti M l mt trong 3 vn . Bit M cch vn trung tm 10, 8mm:Tnh bc sng bc x 2?A. 2 = 0, 4m B. 2 = 0, 2m C. 2 = 0, 32m D. 2 = 0, 75m

Hng Dn

Khong cch 6 vn sng: 5i1 = 9 = i1 = 1, 8mm. T M n O c 3 vn trng :3itrung = 10, 8 =

itrung = 3, 6mm.

V tr vn trng u tin:x1 = x2 = 3, 6mm. = k11 = k22 = 2 =

1, 2

k2 Dng mod 7:f(x) = 1,2

X; 0, 38 f(x) 0, 76 = X = k2 = 2, 3 = 2 = 0, 4m

152 Mt my bin th c s vng cun s cp gp 10 ln cun th cp. Hai u cun s cpmc vo ngun xoay chiu c in p hiu dng U1 = 220V. in tr ca cun s cp l r1xem nh bng 0 v cun th cp r2 = 2. Mch t khp kn; b qua hao ph do dng Fuco v

bc x. Khi hai u cun th cp mc vi in tr R = 20 th in p hiu dng gia hai ucun th cp bng bao nhiu?A. 18V B. 22V C. 20V D. 24V

Hng Dn

e1e2

=N1N2

= U1 I1r1U2 + I1r2

= 10 = U1U2 + I1r2

= 10 = U1U2 +

U2R2

r2

= 10 = U2 = 20V

e1; e2 Sut in ng cun s cp,th cp.



44/48

BIN SON: H HONG VIT

153 hai ngun kt hp S1va S2 ging nhau ,S1S2=8cm,f=10(Hz).vn tc truyn sng 20cm/s.Hai im M v N trn mt nc sao cho S1S2 l trung trc ca MN. Trung im ca S1S2 cchMN 2cm v MS1=10cm.s im cc i trn on MN l :A. 1 B. 2 C. 3 D. 4

Hng DnDng pitago tnh c MS2 = 2

33cm

Xt im M v im I cMS2 M S1

=

2

33 102

0, 74 v IS2 IS1

=6 2

2= 2

Vt M nm gia vn cc i trung tm v vn cc a bc 1 li c vn cc i bc 2 qua I nghal vn cc i bc 1 s ct MN ti 2 im Vy tng cng l 3 im .

154 Trong gi hc thc hnh, hc sinh mc ni tip mt qut in xoay chiu vi in trR = 352 ri mc hai u on mch ny vo in p xoay chiu c gi tr hiu dng 380 V. Bit qut in ny hot ng ch nh mc vi in p nh mc t vo qut l 220V

v khi y th lch pha gia in p hai u qut v cng dng in qua n l , vicos = 0, 8. Hy xc nh cng sut nh mc ca qut in:A. 90W B. 266W C. 80W D. 160W

Hng Dn

U2 = U2R + U2q + 2UR.Uq.cos() = 3802 = U2R +2202 + 2.UR.220.0, 8 = UR = 180V = I =

URR

= 0, 512 = P = U.I.cos() = 90W

155 Trong mt mi trng vt cht n hi c hai ngun kt hp A v B cch nhau 10 cm,cng tn s. Khi ti vng gia hai ngun ngi ta quan st thy xut hin 10 dy dao ngcc i v ct on S1S2 thnh 11 on m hai on gn cc ngun ch di bng mt na ccon cn li. Bit Tc truyn sng trong mi trng l 50cm/s . Tn s dao ng ca haingun l:A. 25Hz B. 30Hz C.15Hz D. 50Hz

Hng Dn

Ta thy, 10 dy cc i s chia S1S2 thnh 9.

2v 2 on gn ngun c bng 1 na cc on

cn li nn tng khong cch 2 on l

4+

4=

2. T = S1S2 = 10.

2= 10cm = =

2cm = f = v

=50

2= 25Hz

156 Mch dao ng LC gm L v hai t C1, C2. Khi dng L v C1 ni tip vi C2 th khungbt c sng in t c tn s l 5,0MHz , nu t C1 b nh thng th khung bt c sngin t c f1=3MHz . Hi khi dng L v C1 th khung bt c sng in t c f2 bng baonhiu?A. 2, 4MHz B. 2, 4MHz C.7, 0MHz D. 2, 0M Hz

Hng Dn

f2 = f12 + f2

2 = f2 =

52 32 = 4M(HZ)



45/48

BIN SON: H HONG VIT

Khi dng L v C1 ni tip vi C2 :f =1

2

L.

C1C2C1 + C2

= f2 = ( 12

)2

(1

LC1+

1

LC2)

Nu t C1 b nh thng:f12 = (1

2)2

(1

LC2)

khi dng L v C1:f2

2 = (1

2)2

(1

LC1)

157 Khi mc mt in p xoay chiu c gi tr hiu dng 220V vo mt hp en X th thydng in trong mch bng 0,25A v sm pha so vi in p t vo l

2. Cng in p ni

trn nu mc vo hp en Y th cng dng in vn l 0,25A nhng cng pha vi in pt vo. Xc nh dng in trong mch khi mc in p trn vo mch cha X v Y mc nitip.A. C gi tr hiu dng l

12

(A) v tr pha

4so vi in p.

B. C gi tr hiu dng l1

42(A) v sm pha

4so vi in p.

C. C gi tr hiu dng l1

4

2(A) v tr pha

4so vi in p.

D. C gi tr hiu dng l1

2(A) v sm pha

4so vi in p.

Hng Dn

Trong 2 TH X,Y c cng U v I = Zx = Zy = UI

=220

0, 25= 880

Khi mc X ni tip Y : uX chm pha i

2, uY ng pha i = u chm pha i hay i nhanh pha hn

u 1 gc 45 .= I = 220

Z=

220

880

2=

1

4

2= A B

158 Trong cc dng c tiu th in nh qut , t lnh, ng c..... ngi ta phi nng caoh s cong sut nhm:A. tng cng sut tiu th B. Gim cng sut tiu thC. Thay i tn s dng in D. Tng hiu sut ca vic s dng in

Hng Dn

P = Pcoich + RI2 = U.I.cos()

tng Hiu sut s dng = gim cng sut ta nhit = gim I P = const, U = const = I = P/U.cos() gim khi cos() tng

159 Cho mch in RLC ni tip, c in tr 90.t vo 2 u on mch in pu = 100

2cos(100t) (V).Thay i L ta thy khi cm khng cun dy bng ZL th hiu in

th gia 2 u RL t gi tr cc i bng 200V.Tnh gi tr ZL?A. 90 B. 120 C.150 D. 180

Hng Dn



46/48

BIN SON: H HONG VIT

URL = I. sqrtR2 + ZL

2 =U.

R2 + ZL2

R2 + (ZL ZC)2=

UR2 + (ZL ZC)2

R2 + ZL2

=U

Y

(URLMAX )= YMIN =

Y(ZL) =2.(ZL ZC).(R2 + ZL2) 2.ZL(R2 + (ZL ZC)2)

R2 + ZL22

Y = 0 = ZC.ZL2 ZC2.ZL ZC.R2 = 0 = ZL = ZC +

4R2

+ ZC2

2> 0

= URLMAX = 2U R4R2 + ZC

2 ZC160 ln lt chiu vo catot ca mt t bo quang in cc chm nh sng n sc c tn s

f, 2f th vn tc ban u cc i ca cc quang electron tng ng l v, 2v. Nu dng chmsng c tn s 3f th vn tc ban u cc i ca cc quang electron l:

Hng Dn

hf A = mv22

(1);2hf A = m(2v)22

= 4mv22

(2)

Chia (2) cho (1) c:2hf Ahf A = 4 =

hf + (hf A)hf A = 4 =

hf

hf A = 3(3)Tng t ta cng c:3hf Ahf A = (

v3v

)2 = 2hfhf A + 1 = (

v3v

)2

Thay (3) vo ta c:(

v3v

)2 = 2.3 + 1 = 7 = v3v

=

7

161 Mt cun dy c in tr thun R = 100

3, L =3

mc ni tip vi mt on mch X

c tng tr ZX ri mc vo in p c xoay chiu c gi tr hiu dng 120 V, tn s 50 Hz ththy dng in qua mch in c cng hiu dng bng 0,3 A v chm pha 300 so vi inp gia hai u mch. Cng sut tiu th trn on mch X bng:A. 9

3W B. 15

3W C.30W D. 40W

Hng Dn

Nu khng c X th I chm pha hn U 60o, khi c X th chm pha hn 30o, chng t X c r hocC hoc rC. Nhng trong p n khng c P = 0 nn X c r hoc rC.ZRL = 200

3. Vy nu X c r th Z khng th l U

I= 400 m phi l

3ZRL = 600 = X c rC.

Bit gc gia ZRL v Z l 30o tnh c ZrC = 200 = Gc gia ZrC v Z l 60o = Gc giaZrC v I l 30o = cos ca X l cos30o =

32

= P = I2ZrCcos = 9

3W

162 Mc vo mch RLC khng phn nhnh mt ngun in xoay chiu c f thay i c khif1 = 60Hz th cos = 1 khi f2 = 120Hz th cos = 0.707.Hi khi f3 = 90Hz th h s cngsut bng bao nhiu:

A. 0, 872 B. 0.486 C.0, 625 D. 0, 781

Hng Dn



47/48

BIN SON: H HONG VIT

f1 = 60 = cng hng 21 =1

LC(1)

f2 = 120 = tan(2) =

L2 1C2

R(2)

f3 = 90 = tan(3) = L3 1

C3 R (3)

Ly(3)

(2)= tan(3)

tan(2)=

23

23 2122 21

163 Ni 2 cc ca mt my pht in xoay chiu mt pha vo 2 u on mch AB gmgin tr thun R = 30 mc ni tip vi t in.B qua in tr cc cun dy ca my pht.khi roto quay u vi tc i n vng/ pht th I hiu dng trong mch l 1A. Khi roto quayu vi tc 2n vong/ pht th I hiu dng trong on mch l

6 A. Nu roto quay vi tc

3n vng/ pht th dung khng ca t l :

A. 4

5 B. 2

5 C.16

5 D. 8

5

Hng Dn

Nhn xt: E = NBS t l thun vi n (tc vng) v ZC =1

C.t l nghch (tc vng)

Th1 : E = k.n,ZC =k

n= I2 = k

2.n2

R2 +k2

n2

= 1 = R2 + k2

n2= k2.n2

Th2 :E = k.2n, ZC =k

2n= I2 = k

2.4n2

R2 +k2

4n2

= 6 = 6R2 + 3k2

2n2= k2.4n2

Kh k t 2 phng trnh trn = k2

n2= 24 = k

n= 12

5

TH3: ZC =k

3n= 4

5

164 Mt mng mng c chit sut n = 1, 42 c chiu bi nh sng c bc sng = 0, 59.106m.Hy xc nh b dy cc tiu ca mng nu do hin tng giao thoa cc tiasng phn chiu c cng cc tiu.Cho bit chm sng ti song song v vung gc vi mn.A. 0, 21m B. 0, 42m C. 0, 36m D. 1, 21m

Hng Dn

Hiu quang l :L2 L1 = 2d

n2 sin2i 2

; Chm sng vung gc vi mn nn i = 0 =L2 L1 = 2dn

2 c cc tiu :L2 L1 = 2dn

2= (2k + 1).

2= dmin =

2n 0, 21m

165 Mt ng Rnghen hot ng vi in p U, bc sng nh nht ca tia Rnghen do ngpht ra l 5.1010m.Nu cng dng in qua ng l 0, 01A. Gi s ton b ng nng caelectron dng t nng i m cc. S electron p vo catt mi giy v nhit lng cungcp cho i m cc mi pht ln lt l:A. 1,25.1016 ht; 1490 J B. 6,25.1016 ht; 1490 JC. 6,25.1016 ht; 2500 J D. 6,25.1026 ht; 2500 J

Hng Dn



48/48

BIN SON: H HONG VIT

I = n.e => n = Ie

= 6, 25.1016; Uak =h.c

.e= 2484, 4v; wd = e.UAk.n = 24, 844

Ton b ng nng dng t nng m cc: Q = 24, 844.60 = 1940.64J

Boxmath- Tuyen Tap Vat Li

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