TRNG THPT S 1 B TRCH

luyn thi H-C 2009-2010

GV Trng nh Den

THI TH I HC LN 12. Nm hc 2009-2010

Mn. Vt L. Thi gian. 90pht

(S cu trc nghim. 50 cu)

Cu 1: Gi (( v (( ln lt l hai bc sng ng vi cc vch H( v vch lam H( ca dy Ban-me , (1 l bc sng di nht ca dy Pa-sen trong quang ph ca Hir. Biu thc lin h gia ((, ((, (1 l

A. (1 = (( - ((.B.

C.

D.

Cu 2: mt nhit nht nh, nu mt m hi c kh nng pht ra hai bc x c bc sng tng ng (1 v (2 ((1 < (2) th n cng c kh nng hp th

A. hai nh sng n sc c bc sng (1 v (2.

B. mi nh sng n sc c bc sng trong khong t(1 n (2.

C. mi nh sng n sc c bc sng nh hn (1.D. mi nh sng n sc c bc sng ln hn (2.

Cu 3: Pht biu no sau y l sai khi ni v hin tng quang pht quang?

A. Khi c chiu bng tia t ngoi, cht fluorexin pht ra nh sng hunh quang mu lc.

B. Hunh quang v ln quang u l hin tng quan pht quang.

C. Chiu chm tia hng ngoi vo mt cht pht quang, cht hp th v c th pht ra nh sng .

D. Bc sng ca nh sng pht quang bao gi cng ln hn bc sng m cht pht quang hp th.

Cu 4: i vi s lan truyn trong khng gian th pht biu no sau y l sai?

A. Sng in t mang nng lng di dng cc phtn.

B. Trong sng in t, in trng v t trng lun bin thin cng chu k.

C. Sng in t l s lan truyn ca in t trng bin thin theo thi gian.

D. Trong sng in t, in trng v t trng lun bin thin lch pha nhau ())eq \s\don1(\f(,2)).Cu 5: t hiu in th ( khng i) vo hai u on mch RLC khng phn nhnh. t cm v in tr gi khng i. iu chnh C hiu in th hiu dng trn t t cc i. Khi ta c biu thc :

A.

B. C.

D.

Cu 6: Mt vt dao ng iu ho dc theo trc Ox (O l v tr cn bng) c phng trnh: . Xc nh qung ng i c k t lc dao ng n thi im .

A. 22,5 cm B. 23,5 cm C. 24,5 cm D. 25,5 cmCu 7: Cho on mch xoay chiu gm cun dy c t cm L = ())eq \s\don1(\f(1,)) H, in tr R = 50( mc ni tip vi mt t in c in dung thay i c. Ban u in dung ca t in l C = , t vo hai u on mch mt in p xoay chiu tn s khng i f = 50Hz, gim dn gi tr in dung ca t in th lch pha gia in p hai u cun dy vi in p hai u on mch

A. ban u bng ())eq \s\don1(\f(,4)) v sau tng dn.B. ban u bng ())eq \s\don1(\f(,2)) v sau gim dn.

C. ban bng ())eq \s\don1(\f(,2)) v sau khng i.D. ban u bng ())eq \s\don1(\f(,2)) v sau tng dn.Cu 8: Hiu in th hiu dng gia ant v catt ca mt ng Rnghen l U = 12 kV. Coi vn tc ban u ca chm lectrn (lectron) pht ra t catt bng khng. Bit hng s Plng h = 6,625.10-34J.s, in tch nguyn t bng 1,6.10-19C. Bc sng nh nht ca tia Rnghen do ng ny c th pht ra l

A. 10,35nm.B. 73,1966pm.C. 0,73m.D. 1,35.1010m.

Cu 9: Mt vt dao ng iu ha vi chu k T, trn mt on thng, gia hai im bin M v N. Chn chiu dng t M n N, gc ta ti v tr cn bng O, mc thi gian t = 0 l lc vt i qua trung im I ca on MO theo chiu dng. Gia tc ca vt bng khng ln th nht vo thi im

A. t = eq \s\don1(\f(T,6)).B. t = eq \s\don1(\f(T,3)).C. t = eq \s\don1(\f(T,12)).D. t = eq \s\don1(\f(T,4)) .

Cu 10: Con lc l xo treo thng ng, dao ng iu ha vi phng trnh x = 2cos20t (cm). Chiu di t nhin ca l xo l l0 = 30cm, ly g = 10m/s2. Chiu di nh nht v ln nht ca l xo trong qu trnh dao ng ln lt l

A. 28,5cm v 33cm.B. 31cm v 36cm.C. 30,5cm v 34,5cm.D. 32cm v 34cm.

Cu 11: Cho on mch RLC ni tip, t vo mch HT: u =100sint(V). Bit uRL sm pha hn dng in qua mch 1 gc /6rad; uC v u lch pha 1 gc /6rad. Hiu in th hiu dng gia hai bn t l

A. 100 (V)B. 200 (V)C. 100(V)D. 200/(V)Cu 12: Chu k ca m c gi tr no sau y m tai con ngi khng th nghe c?

A. T = 6,25.10-5s.B. T = 6,25.10-4s.C. T = 6,25.10-3s.D. T = 625.10-3s.Cu 13: Thc hin giao thoa sng c trn mt cht lng vi 2 ngun kt hp S1 v S2 pht ra 2 sng c cng bin 1cm v cng pha,bc sng = 20cm th ti im M cch S1 mt on 50 cm v cch S2 mt on 10 cm s c bin

A. 1,5 cm B. 2 cm C. 3 cm D. 2,5 cm.Cu 14: phn r thnh vi chu k bn r 4,47.109 nm. Mt khi c pht hin cha 46,97mg v 2,315mg . Gi s khi khi mi hnh thnh khng cha nguyn t ch v tt c lng ch c mt trong u l sn phm phn r ca . Tui ca khi hin nay l bao nhiu?

A. 2,6.109 nm.B. 2,5.106 nm.C. 3,57.108 nm.D. 3,4.107 nm.

Cu 15: Ln lt t vo hai u mt on mch RLC mc ni tip cc in p u1, u2, u3 c cng gi tr hiu dng nhng tn s khc nhau, th cng dng in trong mch tng ng l i1 = I0cos100(t, , i3 = Ieq \l(\r(,2))cos(110(t ())eq \s\don1(\f(,3))). H thc no sau y l ng?

A. I > eq \s\don1(\f(I0,)).B. I eq \s\don1(\f(I0,)).C. I < eq \s\don1(\f(I0,)).D. I = eq \s\don1(\f(I0,)).Cu 16: Mt con lc n gm mt vt nh c treo vo u di ca mt si dy khng dn, u trn ca si dy c buc c nh. B qua ma st v lc cn ca khng kh. Ko con lc lch khi phng thng ng mt gc 0,1 rad ri th nh. T s gia ln gia tc ca vt ti v tr cn bng v ln gia tc ti v tr bin bng

A. 0,1.B. 0.C. 10.D. 5,73.

Cu 17: Mt con lc l xo dao ng iu ho theo phng thng ng, ti v tr cn bng l xo gin 4(cm). B qua mi ma st, ly g= Kch thch cho con lc dao ng iu ho theo phng thng ng th thy thi gian l xo b nn trong mt chu k bng 0,1(s). Bin dao ng ca vt l:

A. B.4(cm). C.6(cm). D.8(cm). Cu 18: Trong th nghim giao thoa nh sng vi khe Ing (Y-ng), khong cch gia hai khe l 2mm. Chiu sng hai khe bng nh sng hn hp gm hai nh sng n sc c bc sng 500 nm v 660 nm th thu c h vn giao thoa trn mn. Khong cch nh nht gia hai vn sng cng mu vi vn trung tm l 9,9mm, khong cch t mt phng cha hai khe n mn quan st l

A. 1,5m.B. 1m.C. 2m.D. 1,2m.Cu 19: It l mt ng v phng x. Sau 12,3 ngy th s phn r cn li 24% s phn r ban u, hng s phn r ca l

A. 2,45.10-6 s-1.B. 3,14.10-6 s-1.C. 1,34.10-6 s-1.D. 4,25.10-6 s-1.

Cu 20: Mt cht im dao ng iu ha. Khi i qua v tr cn bng, tc ca cht im l 40cm/s, ti v tr bin gia tc c ln 200cm/s2. Bin dao ng ca cht im l

A. 0,1m.B. 8cm.C. 5cm.D. 0,8m.

Cu 21: Mch dao ng LC l tng dao ng vi chu k ring T = 10-4s, in p cc i gia hai bn t in U0 = 10V, cng dng in cc i qua cun dy l I0 = 0,02A. in dung ca t in v h s t cm ca cun dy ln lt l

A. C = 7,9.10-3F v L = 3,2.10-8H.B. C = 3,2F v L = 0,79mH.

C. C = 3,2.10-8F v L = 7,9.10-3H.D. C = 0,2F v L = 0,1mH.

Cu 22: Pht biu no sau y l sai khi ni v nng lng ca dao ng in t trong mch dao ng LC l tng?

A. Nng lng in t bin thin tun hon vi tn s gp i tn s dao ng ring ca mch.

B. Nng lng in trng trong t in v nng lng t trng trong cun dy chuyn ha ln nhau.

C. C sau thi gian bng eq \s\don1(\f(1,4)) chu k dao ng, nng lng in trng v nng lng t trng li bng nhau.

D. Nng lng in trng cc i bng nng lng t trng cc i.

Cu 23: t vo hai u mt on mch R, L, C mc ni tip in p xoay chiu u = 200cos100(t (V) th cng dng in trong mch c biu thc i = eq \l(\r(,2))cos100(t (A). in tr thun trong mch l

A. 100(.B. 200(.C. 282,8(.D. 141,4(.Cu 24: Trong th nghim v hin tng quang in ngi ta cho cc quang electron bay vo mt t trng u theo phng vung gc vi ng sc t th bn knh qu ao ln nht ca quang electron s tng khi

A. ch cn gim bc sng nh sng kch thch.

B. tng bc sng nh sng kch thch v gim cng nh sng kch thch.

C. tng cng nh sng kch thch v tng bc sng nh sng kch thch.

D. ch cn tng cng nh sng kch thch.Cu 25: Khi t vo hai u on mch gm cun dy thun cm (cm thun) mc ni tip vi in tr thun mt hiu in th xoay chiu th cm khng ca cun dy bng ln gi tr ca in tr thun. Pha ca dng in trong on mch so vi pha hiu in th gia hai u on mch l

A. chm hn gc .B. nhanh hn gc .C. chm hn gc .D. nhanh hn gc .Cu 26: Qu trnh bin i t thnh ch ch xy ra phng x ( v (-. S ln phn r ( v (- ln lt l

A. 8 v 10.B. 6 v 8.C. 10 v 6.D. 8 v 6.Cu 27: Chn pht biu sai khi ni v s phng x ca ht nhn nguyn t:

A. Ti mt thi im, khi lng cht phng x cng ln th s phn r cng ln.

B. phng x ti mt thi im t l vi s ht nhn phn r tnh n thi im .

C. phng x ph thuc vo bn cht ca cht phng x.

D. Mi phn r l mt phn ng ht nhn ta nng lng.

Cu 28: Ht nhn phng x ng yn, phng ra mt ht ( v bin thnh ht nhn thori (Th). ng nng ca ht ( chim bao nhiu phn trm nng lng phn r?

A. 18,4%.B. 1,7%.C. 81,6%.D. 98,3%.Cu 29: Ln lt mc in tr R, cun dy thun cm c t cm L, t in c in dung C vo in p xoay chiu u = U0cos(t th cng hiu dng ca dng in qua chng ln lt l 4A, 6A, 2A. Nu mc ni tip cc phn t trn vo in p ny th cng hiu dng ca dng in qua mch l

A. 4A.B. 12A.C. 2,4A.D. 6A.

Cu 30: Trong cc tia: (; X; Catt; nh sng , tia no khng cng bn cht vi cc tia cn li?

A. Tia nh sng .B. Tia Catt.C. Tia X.D. Tia (.

Cu 31: Mch dao ng l tng gm t in c in dung v cun dy c t cm . Khi t = 0, cng dng in qua cun dy c ln ln nht l 0,05A. in p gia hai bn t in t cc i l

A. 1 vn ti thi im t = 0,03s.B. 5 vn ti thi im t = 1,57.10-4s.

C. 3 vn ti thi im t = 1,57.10-4s.D. 7 vn ti thi im t = 0,03s.

Cu 32: Mt sng c c bc sng (, tn s f v bin a khng i, lan truyn trn mt ng thng t im M n im N cch M mt on ())eq \s\don1(\f(,3)). Ti mt thi im no , tc dao ng ca M bng 2(fa, lc tc dao ng ca im N bng

A. eq \l(\r(,2))(fa.B. (fa.C. 0.D. eq \l(\r(,3))(fa.

Cu 33: Chn pht biu ng:

A. nh sng n sc l nh sng m sau khi i qua lng knh khng b lch v y ca lng knh.

B. Trong chn khng, tn s ca nh sng v tn s ca nh sng tm l nh nhau.

C. Trong tt c cc mi trng trong sut, nh sng tm truyn i vi tc nh hn nh sng .

D. nh sng n sc c bc sng thay i khi i qua cc mi trng trong sut khc nhau.Cu 34: Chiu ln lt hai bc x c bc sng (1 v (2 ((2 > (1) vo mt tm kim loi th tc ban u cc i ca cc lctrn quang in tng ng l v1 v v2. Nu chiu ng thi c hai bc x trn vo tm kim loi th tc ban u cc i ca cc lctrn quang in l

A. v2.B. v1 + v2.C. v1.D.

Cu 35: Trong qu trnh dao ng iu ha ca con lc l xo th

A. c nng v ng nng bin thin tun hon cng tn s, tn s gp i tn s dao ng.

B. sau mi ln vt i chiu, c 2 thi im ti c nng gp hai ln ng nng.

C. khi ng nng tng, c nng gim v ngc li, khi ng nng gim th c nng tng.

D. c nng ca vt bng ng nng khi vt i chiu chuyn ng.

Cu 36: Trn mt si dy c sng dng, im bng M cch nt gn nht N mt on 10cm, khong thi gian gia hai ln lin tip trung im P ca on MN c cng li vi im M l 0,1 giy. Tc truyn sng trn dy l

A. 400cm/s.B. 200cm/s.C. 100cm/s.D. 300cm/s.

Cu 37: Hai ngun m O1, O2 coi l hai ngun im cch nhau 4m, pht sng kt hp cng tn s 425 Hz, cng bin 1 cm v cng pha ban u bng khng (vn tc truyn m l 340 m/s). S im dao ng vi bin 1cm trong khong gia O1O2 l:

A. 18.B. 9.C. 8.D. 20.Cu 38: Mt con lc n dao ng ti A vi chu k 2s. a con lc ti a im B th n thc hin 100 dao ng ht 201s. Coi nhit bng nhau. Gia tc trng trng t B so vi ti A:

A. Tng 0,1%B. Gim 0,1%C. Tng 1%D. Gim 1%Cu 39: Mt con lc l xo nm ngang, ti v tr cn bng, cp cho vt nng mt vn tc c ln 10cm/s dc theo trc l xo, th sau 0,4s th nng con lc t cc i ln u tin, lc vt cch v tr cn bng

A. 1,25cm.B. 4cm.C. 2,5cm.D. 5cm.

Cu 40: Mt ci ci pht sng m tn s 1000Hz chuyn ng i ra xa mt ngi ng bn ng v pha mt vch , vi tc 15m/s. Ly tc truyn m trong khng kh l 340m/s. Tn s ca m m ngi nghe c khi m phn x li t vch l

A. 956 Hz.B. 958 Hz.C. 1 046 Hz.D. 1 044 Hz.

Cu 41: Trong nguyn t hir , bn knh Bo l r0 = 5,3.10-11m. Sau khi nguyn t hir bc x ra phtn ng vi vch (vch H() th bn knh qu o chuyn ng ca lctrn trong nguyn t gim

A. 13,6m.B. 0,47nm.C. 0,26nm.D. 0,75m.

Cu 42: Mt khung dy dn phng, quay u vi tc gcquanh mt trc c nh trong mt t trng u, c vc t cm ng t vung gc vi trc quay ca khung, sut in ng cm ng trong khung c biu thc . Vo thi im t = 0, vc t php tuyn ca mt phng khung dy hp vi vct cm ng t mt gc bng

A. 1800.B. 1500.C. 450.D. 900.

Cu 43: Hai dao ng iu ha (1) v (2) cng phng, cng tn s v cng bin A = 4cm. Ti mt thi im no , dao ng (1) c li x = 2 eq \l(\r(,3))cm, ang chuyn ng ngc chiu dng, cn dao ng (2) i qua v tr cn bng theo chiu dng. Lc , dao ng tng hp ca hai dao ng trn c li bao nhiu v ang chuyn ng theo hng no?

A. x = 8cm v chuyn ng ngc chiu dng.B. x = 0 v chuyn ng ngc chiu dng.

C. x = 4eq \l(\r(,3))cm v chuyn ng theo chiu dng.D. x = 2eq \l(\r(,3))cm v chuyn ng theo chiu dng.

Cu 44: Cho mach ien xoay chieu nh hnh ve . at vao hai au A, B mot hieu ien the xoay chieu , hieu ien the tc thi gia cac iem Ava M , M va B co dang : . Bieu thc hieu ien the gia A va B co dang :

A.

B.

C.

D.

Cu 45: Sau khi c tch ra t ht nhn , tng khi lng ca 2 prtn v 2 ntrn ln hn khi lng ht nhn 4He mt lng l 0,0305u. Nu 1u = 931, nng lng ng vi mi nucln, tch chng ra khi ht nhn 4He l bao nhiu?

A. 7,098875MeV.B. 2,745.1015J.C. 28,3955MeV.D. 0.2745.1016MeV.

Cu 46: on mch xoay chiu AB ch gm cun thun cm L, ni tip vi bin tr R. Hiu in th 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 vi dng in qua mch ln lt l (1 v (2. Cho bit (1 + (2 = ())eq \s\don1(\f(,2)) . t cm L ca cun dy c xc nh bng biu thc:

A. L = 1.R2))eq \s\don1(\f(,eq \l(\l(2(f)))) .B. L = 2,1))

eq \l(\l( ))eq \s\don1(\f(+eq \l(\l( ))Req \l(\o\al(2,2)))),eq \l(\l(2(f)))) .C. L = 1eq \s\don1(\f(eq \l(\l( ))R2)),eq \l(\l(2(f)))) .D. L = eq \s\don1(\f(R1+eq \l(\l( ))R2,eq \l(\l(2(f)))) .

Cu 47: Con lc vt l l mt thanh mnh, ng cht, khi lng m, chiu di , dao ng iu ha (trong mt mt phng thng ng) quanh mt trc c nh nm ngang i qua mt u thanh, ti ni c gia tc trng trng g. Bit momen qun tnh ca thanh i vi trc quay cho l I = m 2 . Dao ng ca con lc ny c chu k l

A. T = 2

B. T =

C. T =

D. T =

Cu 48: Mt rng rc khi lng M , bn knh R , c th quay t do xung quanh trc c nh ca n . Mt si dy qun quanh rng rc v u t do ca dy c gn mt vt khi lng m. Gi cho vt ng yn ri th nh . Khi vt m ri xung c mt on bng h , th tc ca n thi im

A. t l thun vi R.B. t l nghch vi R.C. t l nghich vi R2.D. khng ph thuc R.

Cu 49: ng c khng ng b 3 pha hot ng bng dng xoay chiu tn s 50Hz. Ti trc quay ca rto, mi cun dy to ra t trng c cm ng t cc i B0. thi im t, cm ng t tng hp do 3 cun dy gy ra ti trc quay l th sau 0,01s, cm ng t tng hp ti l

A.

B. .C. .D. B0.

Cu 50: Mt on mch in xoay chiu gm in tr thun R mc ni tip vi t in C. Nu dung khng ZC bng R th cng dng in qua in tr lun

A. tr pha ())eq \s\don1(\f(,4)) so vi in p gia hai bn t in.B. sm pha ())eq \s\don1(\f(,2)) so vi in p hai u on mch.

C. sm pha ())eq \s\don1(\f(,4)) so vi in p hai u on mch.D. tr pha ())eq \s\don1(\f(,2)) so vi in p gia hai bn t in.

1C11B21C31B41C

2A12D22A32B42A

3C1323D33D43D

4D14C24A34C44

5C15A2535B45A

6A16A26D36B46A

7D1727B37D47B

8B18D28D3848A

9C19C29C39C49A

10C20B30B40C50A

Mt mmen lc c ln 30Nm tc dng vo mt bnh xe c mmen qun tnh i vi trc bnh xe l 2kgm2. Nu bnh xe quay nhanh dn u t trng thi ngh th ng nng ca bnh xe thi im t = 10s l

A. E = 22,5 kJ.B. E = 18,3 kJ.C. E = 20,2 kJ.D. E = 24,6 kJ.

t vo hai u mt cun dy in p xoay chiu c biu thc u = 100cos(100(t) (V) th dng in qua cun dy c cng hiu dng bng 2A v sau thi gian 1 gi, nhit lng ta ra trn cun dy l 36.104J. Biu thc cng dng in tc thi qua cun dy l

A. i = 2eq \l(\r(,2))cos(100(t + ())eq \s\don1(\f(,4))) (A).B. i = 2eq \l(\r(,2))cos(100(t + ())eq \s\don1(\f(,3))) (A).

C. i = 2eq \l(\r(,2))cos(100(t - ())eq \s\don1(\f(,3))) (A).D. i = 2eq \l(\r(,2))cos(100(t - ())eq \s\don1(\f(,4))) (A).Mt cht im chuyn ng trn xung quanh mt trc c mmen qun tnh i vi trc l I. Kt lun no sau y l khng ng?

A. Tng ng thi khi lng ca cht im ln hai ln v khong cch t cht im n trc quay ln hai ln th mmen qun tnh tng 8 ln.

B. Tng khong cch t cht im n trc quay ln hai ln th mmen qun tnh tng 4 ln.

C. Tng khi lng ca cht im ln hai ln th mmen qun tnh tng ln hai ln.

D. Tng khong cch t cht im n trc quay ln hai ln th mmen qun tnh tng 2 ln.

Mmen qun tnh ca vt rn i vi trc quay xc nh

A. bng tng mmen qun tnh ca cc phn khc nhau ca vt i vi trc quay .

B. c th dng hay m ty thuc vo chiu quay ca vt.

C. cng ln th mmen lc tc dng ln vt cng ln.

D. khng ph thuc vo v tr trc quay.

Mt a mi c mmen qun tnh i vi trc quay ca n l 1,2 kgm2. a chu mt mmen lc khng i 16Nm, mmen ng lng ca a ti thi im t = 3,3s l

A. 70,4 kgm2/s.B. 52,8 kgm2/s.C. 66,2 kgm2/s.D. 30,6 kgm2/s.

Mt bn trn phng nm ngang bn knh 4m c trc quay c nh i qua tm bn. Mmen qun tnh ca bn i vi trc quay ny l 40kg.m2. Mt ngi khi lng 60kg ng trn bn st trc quay. Bn ang quay u vi tc gc 2rad/s th ngi trn bn i ra mp bn. B qua ma st trc quay, lc cn ca mi trng v xem ngi nh mt cht im. Tc gc ca bn khi ngi ra ti mp bn l

A. 0,16rad/s.B. 0,08rad/s.C. 0,078rad/s.D. 0,314rad/s.

Vt c trc quay c nh, mmen qun tnh i vi trc quay ny l I, quay vi tc gc th mmen qun tnh L v ng nng W lin h vi nhau bi h thc

A.

B.

C.

D.

Mt lc c ln khng i tc dng ln mt vt ti im M cch trc quay ca vt mt on OM = R khng i. Khi gi ca lc ng thi vung gc vi OM v trc quay th n gy ra cho vt gia tc gc (, khi gi ca lc vn vung gc vi trc quay nhng n hp vi OM gc 300 th gia tc gc m lc ny gy ra cho vt l

A. ())eq \s\don1(\f(,2)).B. 3(.C. (eq \s\don1(\f()),2)).D. (.

M

L

R

C

B

A

N

A

B

Trang 1/5

_1302178852.unknown

_1302178896.unknown

_1302179051.unknown

_1302180794.unknown

_1334601445.unknown

_1334601596.unknown

_1334601612.unknown

_1334601548.unknown

_1302808889.unknown

_1303548460.unknown

_1302179079.unknown

_1302179091.unknown

_1302179100.unknown

_1302179103.unknown

_1302179097.unknown

_1302179081.unknown

_1302179074.unknown

_1302179076.unknown

_1302179052.unknown

_1302178944.unknown

_1302179033.unknown

_1302179041.unknown

_1302179044.unknown

_1302179047.unknown

_1302179038.unknown

_1302179020.unknown

_1302179021.unknown

_1302178983.unknown

_1302178932.unknown

_1302178933.unknown

_1302178897.unknown

_1302178863.unknown

_1302178869.unknown

_1302178875.unknown

_1302178866.unknown

_1302178858.unknown

_1302178861.unknown

_1302178853.unknown

_1289545204.unknown

_1302178785.unknown

_1302178812.unknown

_1302178850.unknown

_1302178851.unknown

_1302178849.unknown

_1302178787.unknown

_1289545215.unknown

_1289545221.unknown

_1302178782.unknown

_1289545218.unknown

_1289545211.unknown

_1271875493.unknown

_1279253267.unknown

_1288436706.unknown

_1279253266.unknown

_1243238902.unknown

_1264962265.unknown

_1264962298.unknown

_1271875490.unknown

_1264962196.unknown

_1243238926.unknown

_1243238865.unknown

_1243238899.unknown

_1243238849.unknown

_1243045682.unknown