8 de Thi Hsg Thcs Kem Huong Dan Giai

THI HSG VT L LP 9 S 1 ( Thi gian 150 pht ) Bi 1 : Cho mch in MN nh hnh v di y, hiu in th hai u mch in khng i UMN = 7V; cc in tr R1 = 3 v R2 = 6 . AB l mt dy dn in c chiu di 1,5m tit din khng i S = 0,1mm2, in tr sut = 4.10-7 m ; in tr ca ampe k A v cc dy ni khng ng k : M UMN N a/ Tnh in tr ca dy dn AB ? R1 R2 b/ Dch chuyn con chy c sao cho AC = 1/2 BC. D Tnh cng dng in qua ampe k ? A c/ Xc nh v tr con chy C Ia = 1/3A ? A C B Bi 2 Mt vt sng AB t cch mn chn mt khong L = 90 cm. Trong khong gia vt sng v mn chn t mt thu knh hi t c tiu c f sao cho trc chnh ca thu knh vung gc vi vt AB v mn. Khong cch gia hai v tr t thu knh cho nh r nt trn mn chn l = 30 cm. Tnh tiu c ca thu knh hi t ? Bi 3 Mt bnh thng nhau c ba nhnh ng nc ; ngi ta vo nhnh (1) ct thu ngn c cao h ( c tm mng rt mng ngn khng cho TN chm vo nc ) v vo nhnh (2) ct du c cao bng 2,5.h . a/ Mc cht lng trong nhnh no cao nht ? Thp nht ? Gii thch ? b/ Tnh chnh lch ( tnh t mt thong ) ca mc cht lng mi nhnh theo h ? c/ Cho dHg = 136000 N/m2 , dH2O = 10000 N/m2 , ddu = 8000 N/m2 v h = 8 cm. Hy tnh chnh lch mc nc nhnh (2) v nhnh (3) ? Bi 4 S bin thin nhit ca khi nc ng trong mt ca nhm c cho th di y 0 C 2 O 170 175 Q( kJ ) (

Tnh khi lng nc v khi lng ca nhm ? Cho bit nhit dung ring ca nc C1 = 4200J/kg.K ; ca nhm C2 = 880 J/kg.K v nhit nng chy ca nc l = 3,4.105 J/kg ? c l lam - a )

HNG DN GII S 1 - HSG L LP 9Bi 1 a/ i 0,1mm2 = 1. 10-7 m2 . p dng cng thc tnh in tr R = . b/ Khi AC =BC 2

l ; thay s v tnh RAB = 6 S

RAC =

1 .RAB RAC = 2 v c RCB = RAB - RAC = 4 3

Xt mch cu MN ta c

c/ t RAC = x ( K : 0 x 6 ) ta c RCB = ( 6 - x ) 1

R1 R 3 = 2 = nn mch cu l cn bng. Vy IA = 0 R AC RCB 2

* in tr mch ngoi gm ( R1 // RAC ) ni tip ( R2 // RCB ) l R = * Cng dng in trong mch chnh : I =U =? R

3.x 6.( 6 x) + =? 3 + x 6 + (6 x )

* p dng cng thc tnh HT ca mch // c : UAD = RAD . I =

* Ta c cng dng in qua R1 ; R2 ln lt l : I1 =

3.x .I = ? 3 +x 6.( 6 x ) .I = ? V UDB = RDB . I = 12 x U AD U DB

R1

=?

v I2 =

R2

=?

+ Nu cc dng ca ampe k gn vo D th : I1 = Ia + I2 Ia = I1 - I2 = ? (1) Thay Ia = 1/3A vo (1) Phng trnh bc 2 theo x, gii PT ny c x = 3 ( loi gi tr -18) + Nu cc dng ca ampe k gn vo C th : Ia = I2 - I1 = ? (2) Thay Ia = 1/3A vo (2) Phng trnh bc 2 khc theo x, gii PT ny c x = 1,2 ( loi 25,8 v > 6) AC R AC = * nh v tr im C ta lp t s = ? AC = 0,3m CB RCB Bi 2 HD : Xem li phn l thuyt v TK hi t ( phn s dng mn chn ) v t gii Theo bi ta c = d1 - d2 =L + L2 4.L. f L L2 4.L. f = 2 2 L2 4.L. f

Bi 3 HD: a/ V p sut cht lng ph thuc vo cao v trng lng ring ca cht lng hn na trong bnh thng nhau p sut cht lng gy ra cc nhnh lun bng nhau mt khc ta c dHg = 136000 N/m2 > dH2O = 10000 N/m2 > ddu = 8000 N/m2 nn h(thu ngn) < h( nc ) < h (du ) b/ Quan st hnh v : (1) (2) (3)? ? 2,5h

2 = L2 - 4.L.f f = 20 cm

? hh M N h E

H2O Xt ti cc im M , N , E trong hnh v, ta c : PM = h . d1 (1) PN = 2,5h . d2 + h. d3 (2) PE = h. d3 (3) . Trong d1; d2 ; d3 ln lt l trng lng ring ca TN, du v nc. cao h v h nh hnh v . h.( d 1 d 3 ) h.d 1 h.d 1 + Ta c : PM = PE h = h1,3 = h - h = - h = d3 d3 d3 + Ta cng c PM = PN h = ( h.d1 - 2,5h.d2 ) : d3 h1,2 = ( 2,5h + h ) - h = h.d 1 2,5h.d 2 h.d 3 d3 2

+ Ta cng tnh c h2,3 = ( 2,5h + h ) - h = ? c/ p dng bng s tnh h v h chnh lch mc nc nhnh (3) & (2) l h - h = ? Bi 4 HD : Lu 170 KJ l nhit lng cung cp nc nng chy hon ton O0C, lc ny nhit ca nhm khng i. S : m H O = 0,5 kg ; m Al = 0,45 kg2

S 2

THI HSG VT L LP 9

( Thi gian 150 pht ) Bi 1 Mt cc nc c khi lng 200g nhit - 100C : a/ cc nc chuyn hon ton sang th hi 1000C th cn mt nhit lng l bao nhiu kJ ? Cho nhit dung ring ca nc v nc l C1 = 4200J/kg.K ; C2 = 1800 J/kg.K. Nhit nng chy ca nc l = 3,4.105 J/kg ; nhit ho hi ca nc l L = 2,3.106 J/kg. b/ Nu b cc nc trn vo ca nhm ng nc 200C th khi c cn bng nhit, ngi ta thy c 50g nc cn st li cha tan ht. Tnh khi lng nc ng trong ca nhm lc u bit ca nhm c khi lng 100g v nhit dung ring ca nhm l C3 = 880 J/kg.K ? ( Trong c hai cu u b qua s mt nhit vi mi trng ngoi ) Bi 2 : Mt khi g hnh hp ch nht c din tch y l S = 150 cm2 cao h = 30cm, khi g c th ni trong h nc su H = 0,8m sao cho khi g thng ng. Bit trng lng ring ca g bng 2/3 trng lng ring ca nc v d H 2O = 10 000 N/m3. B qua s thay i mc nc ca h, hy : a) Tnh chiu cao phn chm trong nc ca khi g ? b) Tnh cng ca lc nhc khi g ra khi nc H theo phng thng ng ? c) Tnh cng ca lc nhn chm khi g n y h theo phng thng ng ? Bi 3 : Cho 3 in tr c gi tr nh nhau bng R0, c mc vi nhau theo nhng cch khc nhau v ln lt ni vo mt ngun in khng i xc nh lun mc ni tip vi mt in tr r . Khi 3 in tr trn mc ni tip th cng dng in qua mi in tr bng 0,2A, khi 3 in tr trn mc song song th cng dng in qua mi in tr cng bng 0,2A. a/ Xc nh cng dng in qua mi in tr R0 trong nhng trng hp cn li ? b/ Trong cc cch mc trn, cch mc no tiu th in nng t nht ? Nhiu nht ? c/ Cn t nht bao nhiu in tr R0 v mc chng nh th no vo ngun in khng i c in tr r ni trn cng dng in qua mi in tr R0 u bng 0,1A ? Bi 4 Mt chm sng song song vi trc chnh ti thu knh hi t c tiu c f = 20 cm. Pha sau thu knh ngi ta t mt gng phng ti I v vung gc vi trc chnh ca TK, gng quay mt phn x v pha TK v cch TK mt khong 15 cm. Trong khong gia TK v gng ngi ta quan st c mt im rt sng : a/ Gii thch v v ng truyn ca cc tia sng ( khng v tia sng phn x qua thu knh ) ? Tnh khong cch t im sng ti TK ? b/ C nh TK v quay gng quanh im I n v tr mt phn x hp vi trc chnh mt gc 450. V ng truyn ca cc tia sng v xc nh v tr ca im sng quan st c lc ny ?

HNG DN GII S 2 - HSG L LP 9Bi 1 S : a) 615,6 kJ ( Tham kho bi tng t trong ti liu ny ) 3

b/ m = 629g . Ch l do nc khng tan ht nn nhit cui cng ca h thng l 00C v ch c 150g nc tan thnh nc. Bi 2 HD : a) Gi chiu cao phn khi g chm trong nc l x (cm) th : (h-x) + Trng lng khi g : P = dg . Vg = dg . S . h ( dg l trng lng ring ca g ) x + Lc y Acsimet tc dng vo khi g : FA = dn . S . x ; H khi g ni nn ta c : P = FA x = 20cm b) Khi khi g c nhc ra khi nc mt on y ( cm ) so vi lc u th lc Acsimet gim i mt lng FA = dn . S.( x - y ) lc nhc khi g s tng thm v bng : F = P - FA = dg.S.h - dn.S.x + dn.S.y = dn.S.y v lc ny s tng u t lc y = 0 n khi y = x , v th gi tr trung bnh ca lc t khi nhc khi g n khi khi g va ra khi mt nc l F/2 . Khi cng phi thc hin l A =1 1 .F.x = .dn.S.x2 = ? (J) 2 2

c) Cng l lun nh cu b song cn lu nhng iu sau : + Khi khi g c nhn chm thm mt on y th lc Acsimet tng ln v lc tc dng lc ny s l F = FA - P v cng c gi tr bng dn.S.y.Khi khi g chm hon ton, lc tc dng l F = dn.S.( h x ); thay s v tnh c F = 15N. + Cng phi thc hin gm hai phn : - Cng A1 dng nhn chm khi g va vn ti mt nc : A1 =1 .F.( h - x ) 2

- Cng A2 nhn chm khi g n y h ( lc FA lc ny khng i ) A2 = F .s (vi s = H - h ) S : 8,25J Bi 3 HD : a/ Xc nh cc cch mc cn li gm : cch mc 1 : (( R0 // R0 ) nt R0 ) nt r cch mc 2 : (( R0 nt R0 ) // R0 ) nt r Theo bi ta ln lt c cng dng in trong mch chnh khi mc ni tip : Int = (1) Cng dng in trong mch chnh khi mc song song :I SS =U = 0,2A r + 3R0

r + 3 R0 =3 r = R0 . em gi tr ny ca r thay vo (1) U = R0 Ly (2) chia cho (1), ta c : r+ 3 0,8.R0 + Cch mc 1 : Ta c (( R0 // R0 ) nt R0 ) nt r (( R1 // R2 ) nt R3 ) nt r t R1 = R2 = R3 = R0 0,8.R0 U = = 0,32 A I R0 2,5.R0 Dng in qua R3 : I3 = . Do R1 = R2 nn I1 = I2 = 3 = 0,16 A r + R0 + 2 2 0,8.R0 U = = 0,48 A 2.R0 .R0 5.R0 + Cch mc 2 : Cng dng in trong mch chnh I = . r+ 3 . R0 3

U = 3.0,2 = 0,6 A R0 (2) r+ 3

4

Hiu in th gia hai u mch ni tip gm 2 in tr R0 : U1 = I. dng in qua mch ni tip ny l I1 = l

2.R0 .R0 = 0,32.R0 cng 3.R0

0,32 .R0 U1 = = 0,16 A CD qua in tr cn li 2.R0 2.R0

I2 = 0,32A. b/ Ta nhn thy U khng i cng sut tiu th mch ngoi P = U.I s nh nht khi I trong mch chnh nh nht cch mc 1 s tiu th cng sut nh nht v cch mc 2 s tiu th cng sut ln nht. c/ Gi s mch in gm n dy song song, mi dy c m in tr ging nhau v bng R0 ( vi m ; n N) Cng dng in trong mch chnh ( Hv ) I + I = U 0,8 = m m ( B sung vo hv cho y ) r + .R0 1 + n n

cng dng in qua mi in tr R0 l 0,1A ta phi c :I = 0,8 = 0,1.n m 1+ n

m + n = 8 . Ta c cc trng hp saum n S in tr R0 1 7 7

2 3 4 5 6 7 6 5 4 3 2 1 1 15 16 15 12 7 2 Theo bng trn ta cn t nht 7 in tr R0 v c 2 cch mc chng : a/ 7 dy //, mi dy 1 in tr. b/ 1 dy gm 7 in tr mc ni tip. Bi 4 HD : Xem bi gii tng t trong ti liu v t gii a/ Khong cch t im sng ti gng = 10 cm ( OA1 = OF - 2.FI ) b/ V nh ca im sng qua h TK - gng lun v tr i xng vi F qua gng, mt khc do gng quay quanh I nn di IF khng i A1 di chuyn trn mt cung trn tm I bn knh IF v n im A2. Khi gng quay mt gc 450 th A1IA2 = 2.450 = 900 ( do t/c i xng ) Khong cch t A2 ti thu knh bng IO v bng 15 cm S 3

THI HSG VT L LP 9

( Thi gian 150 pht ) Bi 1 Hai bn kim loi ng cht, tit din u v bng nhau, cng chiu di = 20cm nhng c trng lng ring khc nhau : d1 = 1,25.d2 . Hai bn c hn dnh vi nhau mt u v c treo bng si dy mnh ( Hv ) /////////// thanh nm ngang, ngi ta thc hin 2 cch sau :

1) Ct mt phn ca bn th nht v em t ln chnh gia ca phn cn li. Tnh chiu di phn b ct ? 2) Ct b mt phn ca bn th nht. Tnh phn b ct i ? Bi 2 5

Mt ng thu tinh hnh tr, cha mt lng nc v lng thu ngn c cng khi lng. cao tng cng ca ct cht lng trong ng l H = 94cm. a/ Tnh cao ca mi cht lng trong ng ? b/ Tnh p sut ca cht lng ln y ng bit khi lng ring ca nc v ca thu ngn ln lt l D1 = 1g/cm3 v D2 = 13,6g/cm3 ? Bi 3 Cho mch in sau Cho U = 6V , r = 1 = R1 ; R2 = R3 = 3 U r bit s ch trn A khi K ng bng 9/5 s ch R1 R3 ca A khi K m. Tnh : a/ in tr R4 ? R2 R4 A K b/ Khi K ng, tnh IK ? Bi 4 a) t vt AB trc mt thu knh hi t L c tiu c f nh hnh v . Qua TK ngi ta thy AB cho nh ngc chiu cao gp 2 ln vt. Gi nguyn v tr Tknh L, dch chuyn vt sng dc theo xy li gn Tknh mt on 10cm th nh ca vt AB lc ny vn cao gp 2 ln vt. Hi nh ca AB trong mi trng hp l nh g ? Tnh tiu c f v v hnh minh ho ? B L1 (M) B x y A O A O1 O2 L2 b)Thu knh L c ct ngang qua quang tm thnh hai na tknh L1 & L2 . Phn b ct ca L2 c thay bng mt gng phng (M) c mt phn x quay v L1. Khong cch O1O2 = 2f. V nh ca vt sng AB qua h quang v s lng nh ca AB qua h ? ( Cu a v b c lp nhau )

HNG DN GII S 3 - HSG L LP 9Bi 1 HD : a) Gi x ( cm ) l chiu di phn b ct, do n c t ln chnh gia phn cn li v thanh cn bng nn ta c : P1.x = P2. . Gi S l tit din ca 2 2

///////////- x

mi bn kim loi, ta c d1.S. .

x = 4cm

x = d2.S. . 2 2 d1( - x ) = d2.

P1

P2

b) Gi y (cm) ( K : y < 20 ) l phn phi ct b i, trng lng phn cn li l : P1 = P1. thanh cn bng nn ta c : d1.S.( - y ). y = d2.S. . 2 2

y . Do

( - y )2 =

d2 2 . hay d1

d2 ). 2 d1 Thay s c phng trnh bc 2 theo y: y2 - 40y + 80 = 0. Gii PT c y = 2,11cm . ( loi 37,6 ) Bi 2 HD :a/ + Gi h1 v h2 theo th t l cao ca ct nc v ct thu ngn, ta c H = h1 + h2 = 94 cm 6

y2 - 2 .y + ( 1 -

+ Gi S l din tch y ng, do TNgn v nc c cng khi lng nn S.h1. D1 = S. h2 . D2 D1 h2 D + D2 h1 + h2 H D 2 .H = 1 = = h1. D1 = h2 . D2 h1 = D2 h1 D2 h1 h1 D1 + D2 h2 = H - h1 b/ p sut ca cht lng ln y ng : 10 m1 + 10 m2 10 Sh1 D1 + 10 Sh 2 D2 = = 10 ( D1 .h1 + D2 .h2 ) . Thay h1 v h2 vo, ta tnh c P. P= S S Bi 3 HD : * Khi K m, cch mc l ( R1 nt R3 ) // ( R2 nt R4 ) in tr tng ng ca mch ngoi l U 4(3 + R4 ) R =r+ Cng dng in trong mch chnh : I = 1 + 4(3 + R4 ) . Hiu in th gia 7 + R4 7 + R4 ( R1 + R3 )( R2 + R4 ) ( R1 + R3 ).I U AB .I I4 = = = ( Thay s, I ) hai im A v B l UAB = R1 + R2 + R3 + R4 R2 + R4 R1 + R2 + R3 + R4 =4U 19 + 5 R4

* Khi K ng, cch mc l (R1 // R2 ) nt ( R3 // R4 ) in tr tng ng ca mch ngoi l U 9 + 15 R4 R' = r + Cng dng in trong mch chnh lc ny l : I = 1 + 9 + 15 R4 . Hiu in 12 + 4 R4 12 + 4 R4 R3 .R4 R3 .I ' U .I ' I4 = AB = = ( Thay s, I ) = th gia hai im A v B l UAB = R3 + R 4 R4 R3 + R412U 21 +19 R4

* Theo bi th I4 =

9 .I 4 ; t tnh c R4 = 1 5

b/ Trong khi K ng, thay R4 vo ta tnh c I4 = 1,8A v I = 2,4A UAC = RAC . I = 1,8V U AC = 0,6 A . Ta c I2 = I2 + IK = I4 IK = 1,2A R2 Bi 4 HD :a/ B2 ( Hy b sung hnh v cho y ) B1 F A1 A2 A2 O B2 I F A1

B1 Xt cc cp tam gic ng dng FA1B1 v FOI : (d - f )/f = 2 d = 3f Xt cc cp tam gic ng dng OA1B1 v OA1B1 : d1 = d/2 d1 = 3/2f Khi di n A2B2 , l lun tng t ta c d2 = f/2 . Theo ta c d1 = 10 + d2 f = 10cm 7

b) H cho 3 nh : AB qua L1 cho A1B1 v qua L2 cho nh o A2B2 . AB qua L2 cho nh A3B3 . Khng c nh qua gng (M). Hy t dng cc nh trn !

S 4

THI HSG VT L LP 9( Thi gian 150 pht )

Bi 1 Mt thanh ng cht tit din u c chiu di AB = = 40cm c dng trong chu sao cho OA =1 OB v ABx = 300 . Thanh c gi nguyn v quay c quanh im O ( Hv ). 3

A Ngi ta nc vo chu cho n khi thanh bt u ni O (u B khng cn ta ln y chu ): a) Tm cao ca ct nc cn vo chu ( tnh t y n mt thong ) bit khi lng ring ca thanh AB v ca 300 nc ln lt l : Dt = 1120 kg/m3 v Dn = 1000 kg/m3 ? B x b) Thay nc bng mt cht lng khc, KLR ca cht lng phi th no thc hin c vic trn ? Bi 2 C hai bnh cch nhit, bnh 1 cha m1 = 2kg nc t1 = 200C, bnh 2 cha m2 = 4kg nc nhit t2 = 600C . Ngi ta rt mt lng nc m t bnh 1 sang bnh 2, sau khi cn bng nhit, ngi ta li rt mt lng nc nh vy t bnh 2 sang bnh 1. nhit cn bng bnh 1 lc ny l t1 = 21,950C : 1) Tnh lng nc m v nhit khi c cn bng nhit trong bnh 2 ( t2 ) ? 2) Nu tip tc thc hin nh vy mt ln na, tm nhit khi c cn bng nhit mi bnh lc ny ? Bi 3 Cho mch in nh hnh v. Bit UAB = 18V khng i cho c bi ton, bng n 1 ( 3V - 3W ) Bng n 2 ( 6V - 12W ) . Rb l gi tr ca bin tr V con chy ang v tr C 2 n sng bnh thng : UAB 1) n 1 v n 2 v tr no trong mch ? r 2) Tnh gi tr ton phn ca bin tr v v tr (1) (2) con chy C ? 3) Khi dch chuyn con chy v pha N th sng ca hai n thay i th no ? M Rb C N Bi 4 Hai vt sng A1B1 v A2B2 cao bng nhau v bng h c t vung gc vi trc chnh xy ( A1 & A2 xy ) v hai bn ca mt thu knh (L). nh ca hai vt to bi thu knh cng mt v tr trn xy . Bit OA1 = d1 ; OA2 = d2 : 1) Thu knh trn l thu knh g ? V hnh ? 2) Tnh tiu c ca thu knh v ln ca cc nh theo h ; d1 v d2 ? 3) B A1B1 i, t mt gng phng vung gc vi trc chnh ti I ( I nm cng pha vi A2B2 v OI > OA2 ), gng quay mt phn x v pha thu knh. Xc nh v tr ca I nh ca A2B2 qua Tk v qua h gng - Tk cao bng nhau ?

HNG DN GII S 4 - HSG L LP 9Bi 1

8

HD: a) Gi mc nc vo trong chu thanh bt u ni ( tnh t B theo chiu di thanh ) l x ( cm ) K : x < OB = 30cm, theo hnh v di y th x = BI. A Gi S l tit din ca thanh, thanh chu tc dng ca trng O lng P t ti trung im M ca AB v lc y Acsimet M H F t ti trung im N ca BI. Theo iu kin cn bng ca I n by th : P.MH = F.NK(1) trong P = 10m = 10.Dt.S. N K V F = 10.Dn.S.x . Thay vo (1) (H2O) Dt MH .. x = B E Dn NK Xt cp tam gic ng dng OMH v ONK ta c v NO = OB - NB =M H M O = ; ta tnh c MO = MA - OA =10cm NK N O

60 x . Thay s v bin i c phng trnh bc 2 theo x : x2 - 60x + 896 = 0. 2 1 = 14cm ( cng c th s dng kin thc 2

Gii phng trnh trn v loi nghim x = 32 ( > 30 ) ta c x = 28 cm. T I h IE Bx, trong tam gic IBE vung ti E th IE = IB.sin IBE = 28.sin300 = 28. v na tam gic u )

Dt 20 .. ; t biu Dn 60 x thc ny hy rt ra Dn ?Mc nc ti a ta c th vo chu l x = OB = 30cm, khi minDn = 995,5 kg/m3 .

b) Trong php bin i a v PT bc 2 theo x, ta gp biu thc : x =

Bi 2 1) Vit Pt to nhit v Pt thu nhit mi ln trt t c : + Phng trnh cn bng nhit bnh 2 : m.(t2 - t1 ) = m2.( t2 - t2 ) (1) + Phng trnh cn bng nhit bnh 1 : m.( t2 - t1 ) = ( m1 - m )( t1 - t1 ) (2) m 2 .t 2 m1 (t '1 t1 ) + T (1) & (2) t ' 2 = = ? (3) . Thay (3) vo (2) m = ? S : 590C v m2 100g 2) ti nhit lc ny ca hai bnh, l lun tng t nh trn ta c kt qu l : 58,120C v 23,760C Bi 3 1) C I1m = P1 / U1 = 1A v I2m = P2 / U2 = 2A. V I2m > I1m nn n 1 mch r ( v tr 1) cn n 2 mch chnh ( v tr 2 ) . 2) t I 1 = I1 v I 2 = I2 = I v cng dng in qua phn bin tr MC l Ib + V hai n sng bnh thng nn I1 = 1A ; I = 2A Ib = 1A . Do Ib = I1 = 1A nn U1 RMC = R1 = = 3 I1 R1 .RMC + ( Rb RMC ) + R2 = r + Rb + 1,5 + in tr tng ng ca mch ngoi l : Rt = r + R1 + RMC U AB = 2 Rb = 5,5 . + CD trong mch chnh : I = Rtd Vy C v tr sao cho RMC = 3 hoc RCN = 2,5 .3) Khi dch chuyn con chy C v pha N th in tr tng ng ca mch ngoi gim I ( chnh ) tng 9

n 2 sng mnh ln. Khi RCM tng th UMC cng tng ( do I1 c nh v I tng nn Ib tng ) n 1 cng sng mnh ln. Bi 4 HD : 1) V nh ca c hai vt nm cng mt v tr trn trc chnh xy nn s c mt trong hai vt sng cho nh nm khc pha vi vt thu knh phi l Tk hi t, ta c hnh v sau : ( B sung thm vo hnh v cho y ) B2 (L) B1 x A1 F O A2 B1d1 . f d1 + f d2. f + Xt cc cp tam gic ng dng trong trng hp vt A2B2 cho nh A2B2 c OA2 = f d2 d1 . f d2. f + Theo bi ta c : OA1 = OA2 = f=? d1 + f f d2 h.O 1 ' A h.O 2 ' A Thay f vo mt trng hp trn c OA1 = OA2 ; t : A1B1 = v A2B2 = d1 d2 . 3) V vt A2B2 v thu knh c nh nn nh ca n qua thu knh vn l A2B2 . Bng php v ta hy xc nh v tr t gng OI, ta c cc nhn xt sau : + nh ca A2B2 qua gng l nh o, v tr i xng vi vt qua gng v cao bng A2B2 ( nh A3B3 ) + nh o A3B3 qua thu knh s cho nh tht A4B4, ngc chiu v cao bng nh A2B2 + V A4B4 > A3B3 nn vt o A3B3 phi nm trong khong t f n 2f im I cng thuc khong ny. + V tr t gng l trung im on A2A3, nm cch Tk mt on OI = OA2 + 1/2 A2A3 . * Hnh v : ( b sung cho y ) B2 B2 B3 x A4 F y O A2 F A3 A2

H

B2 F A2 A1 y

2) + Xt cc cp tam gic ng dng trong trng hp vt A1B1 cho nh A1B1 c OA1 =

B4 * Tnh : K Do A4B4 // = A2B2 nn t gic A4B4A2B2 l hnh bnh hnh FA4 = FA2 = f + OA2 = ? OA4 =? Da vo 2 tam gic ng dng OA4B4 v OA3B3 ta tnh c OA3 A2A3 v tr t gng . S 5

THI HSG VT L LP 910

( Thi gian 150 pht ) Bi 1 1) Mt bnh thng nhau gm hai nhnh hnh tr ging nhau cng cha nc. Ngi ta th vo nhnh A mt qu cu bng g nng 20g, qu cu ngp mt phn trong nc th thy mc nc dng ln trong mi nhnh l 2mm. Sau ngi ta ly qu cu bng g ra v vo nhnh A mt lng du 100g. Tnh chnh lch mc cht lng trong hai nhnh ? Cho Dn = 1 g/cm3 ; Dd = 0,8 g/cm3 2) Mt ng thu tinh hnh tr, cha mt lng nc v lng thu ngn c cng khi lng. cao tng cng ca cht lng trong ng l 94cm. a/ Tnh cao ca mi cht lng trong ng ? b/ Tnh p sut ca cht lng ln y ng bit khi lng ring ca nc v ca thu ngn ln lt l D1 = 1g/cm3 v D2 = 13,6g/cm3 ? Bi 2 Thanh AB c th quay quanh bn l gn trn tng thng ng ti u B ( hv ). Bit AB = BC v trng lng ca thanh AB l P = 100 N : 1) Khi thanh nm ngang, tnh sc cng dy T xut hin trn dy AC thanh cn bng ( hnh 1 ) ? C C T Hnh 1 T Hnh 2 A O O B A B P P 2) Khi thanh AB c treo nh hnh 2, bit tam gic ABC u. Tnh lc cng dy T ca AC lc ny ? Bi 3 Mt hp kn cha mt ngun in c hiu in th khng i U = 150V v mt in tr r = 2 . Ngi ta mc vo hai im ly in A v B ca hp mt bng n c cng sut nh mc P = 180W ni tip vi mt bin tr c in tr Rb ( Hv ) A U B 1) n sng bnh thng th phi iu chnh Rb = 18 . Tnh r hiu in th nh mc ca n ? 2) Mc song song vi n mt bng n na ging ht n. Hi Rb c hai n sng bnh thng th phi tng hay gim Rb ? Tnh tng ( gim ) ny ? 3) Vi hp in kn trn, c th thp sng ti a bao nhiu bng n nh n ? Hiu sut s dng in khi l bao nhiu phn trm ? Bi 4 C hai thu knh (L1) & (L2) c b tr song song vi nhau sao cho chng c cng mt trc chnh l ng thng xy . Ngi ta chiu n thu knh (L1) mt chm sng song song v di chuyn thu knh (L2) dc theo trc chnh sao cho chm sng khc x sau khi qua thu knh (L2) vn l chm sng song song. Khi i mt trong hai thu knh trn bng mt TK khc loi c cng tiu c v cng lm nh trn, ngi ta ln lt o c khong cch gia 2 TK hai trng hp ny l 1 = 24 cm v 2 = 8 cm. 1) Cc thu knh (L1) v (L2) c th l cc thu knh g ? v ng truyn ca chm sng qua 2 TK trn ? 2) Trong trng hp c hai TK u l TK hi t v (L1) c tiu c nh hn (L2), ngi ta t mt vt sng AB cao 8 cm vung gc vi trc chnh v cch (L1) mt on d1 = 12 cm. Hy : + Dng nh ca vt sng AB qua hai thu knh ? + Tnh khong cch t nh ca AB qua TK (L2) n (L1) v ln ca nh ny ?

HNG DN GII S 5 - HSG L LP 911

Bi 1 (A) (A) (B) (B)

HD : + h = 2 mm = 0,2 cm. Khi ct nc 2 M N nhnh dng ln l 2.h = 0,4 cm + Qu cu ni nn lc y Acsimet m nc tc dng ln qu cu bng trng lng ca qu cu ; gi tit din ca mi nhnh l S, ta c P = FA 10.m = S.2h.dn 10.m = S.2h.10Dn S = 50cm2 + Gi h (cm) l cao ca ct du th md = D.Vd = D.S.h h ? Xt p sut m du v nc ln lt gy ra ti M v N, t s cn bng p sut ny ta c cao h ca ct nc nhnh B . chnh lch mc cht lng hai nhnh l : h - h Bi 2 C C H H T Hnh 1 T Hnh 2 K I A O O B A B P P HD : Trong c hai trng hp, v BH AC. Theo quy tc cn bng ca n by ta c : 1) T . BH = P . OB (1) . V OB =

AB v tam gic ABC vung cn ti B nn BAH = 450 . Trong tam 22 ; thay vo (1) ta c : T.AB. 2 2 = P. 2

gic BAH vung ti H ta c BH = AB. Sin BAH = AB.AB 2

T=?

2) Tng t cu 1 : T.BH = P.IK (2). C BAH vung ti H BH = AB. sinBAH = AB.sin600 =A . 3 B . V OI l ng trung bnh ca ABK IK = 1/2 AK = 1/2 BH ( do AK = BH ) 2 IK = AB . 3 ; thay vo (2) : T . AB . 3 = P . AB . 3 T = ? S : T = 20 2 N v T = 4 2 4

20N Bi 3 HD : 1) Gi I l cng dng in trong mch chnh th U.I = P + ( Rb + r ).I2 ; thay s ta c mt phng trnh bc 2 theo I : 2I2 - 15I + 18 = 0 . Gii PT ny ta c 2 gi tr ca I l I1 = 1,5A v I2 = 6A. + Vi I = I1 = 1,5A Ud = I = 120V d trong trng hp ny l : H =P

; + Lm tt vi I = I2 = 6A Hiu sut s dng in

p 180 = = 20 % nn qu thp loi b nghim I2 = 6A U .I 150 .6

2) Khi mc 2 n // th I = 2.Id = 3A, 2 n sng bnh thng nn Ud = U - ( r + Rb ).I Rb ? gim ca Rb ? ( S : 10 ) 3) Ta nhn thy U = 150V v Ud = 120V nn cc n sng bnh thng, ta khng th mc ni tip t 2 bng n tr ln c m phi mc chng song song. Gi s ta mc // c ti a n n vo 2 im A & B cng dng in trong mch chnh I = n . Id . Ta c U.I = ( r + Rb ).I2 + n . P U. n . Id = ( r + Rb ).n2 .I2d + n . P U.Id = ( r + Rb ).n.Id + P 12

Rb =

U .I d P n.I d2

r 0 n

U .I d P r .I d2

=

150.1,5 180 = 10 n max = 10 khi Rb = 0 2.(1,5) 2

+ Hiu sut s dng in khi bng : H =

Ud = 80 % U

Bi 4 1) Chng ta hc qua 2 loi thu knh, hy xt ht cc trng hp : C hai l TK phn k ; c hai l thu knh hi t ; TK (L1) l TK hi t v TK (L2) l TK phn k ; TK (L1) l phn k cn TK (L2) l hi t. a) S khng thu c chm sng sau cng l chm sng // nu c hai u l thu knh phn k v chm tia khc x sau khi ra khi thu knh phn k khng bao gi l chm sng //. ( loi trng hp ny ) b)Trng hp c hai TK u l TK hi t th ta thy cho chm sng cui cng khc x qua (L2) l chm sng // th cc tia ti TK (L2) phi i qua tiu im ca TK ny, mt khc (L1) cng l TK hi t v trng trc chnh vi (L2) do tiu im nh ca (L1) phi trng vi tiu im vt ca (L2). ( chn trng hp ny ) ng truyn ca cc tia sng c minh ho hnh di : ( B sung hnh v ) (L1) F1 x F1=F2 F2 y (L2)

c) Trng hp TK (L1) l phn k v TK (L2) l hi t :L lun tng t nh trn ta s c tiu im vt ca hai thu knh trn phi trng nhau ( chn trng hp ny ). ng truyn cc tia sngc minh ho nh hnh di : (L2) (L1) x F1 F2 y

Do tnh cht thun nghch ca ng truyn nh sng nn s khng c g khc khi (L1) l TH hi t cn (L2) l phn k. 2) + Dng nh ca vt sng AB trong trng hp c 2 TK u l hi t : (L1) B F1= F2 A F1 O1 B1 B2 (L2) + Ta thy rng vic i thu knh ch c th i c TK phn k bng mt thu knh hi t c cng tiu c ( theo a ). Nn : 13 A2 A1 O2 F2

- T c) ta c : F1O1 + O1O2 = F2O2 = f2 f2 - f1 = 2 = 8 cm - T 2) ta c : O1F1 + F2O = O1O2 f2 + f1 = 1 = 24cm Vy f1 = 8cm v f2 = 16cm + p dng cc cp tam gic ng dng v cc yu t cho ta tnh c khong cch t nh A1B1 n thu knh (L2) ( bng O1O2 - O1A1 ), sau tnh c khong cch O2A2 ri suy ra iu cn tnh ( A2O1 ). S 6

THI HSG VT L LP 9( Thi gian 150 pht )

Bi 1 Mt thanh ng cht tit din u c nhng mt u trong nc, thanh ta vo thnh chu ti im O v quay quanh O sao cho OA =1 .OB. Khi thanh cn bng, mc nc chnh gia thanh. Tnh 2

KLR ca cht lm thanh ? Cho KLR ca nc Dn = 1000 kg/m3 Bi 2 Mt khi nc khi lng m1 = 2 kg nhit - 50C : 1) Tnh nhit lng cn cung cp khi nc trn bin thnh hi hon ton 1000C ? Hy v th biu din qu trnh bin thin nhit theo nhit lng c cung cp ? 2) B khi nc ni trn vo mt ca nhm cha nc 500C. Sau khi c cn bng nhit ngi ta thy cn st li 100g nc cha tan ht. Tnh lng nc c trong ca nhm bit ca nhm c khi lng mn = 500g . Cho Cn = 1800 J/kg.K ; Cn = 4200 J/kg.K ; Cnh = 880 J/kg.K ; = 3,4.105 J/kg ; L = 2,3.106 J/kg Bi 3 Cho mch in c s sau. Bit UAB = 12V khng i, R1 = 5 ; R2 = 25 ; R3 = 20 . Nhnh DB c hai in tr ging nhau v bng r, khi hai in tr r mc ni tip vn k V ch gi tr U1, khi hai in tr r mc song song vn k V ch gi tr U2 = 3U1 : R1 C R2 1) Xc nh gi tr ca in tr r ? ( vnk c R = ) 2) Khi nhnh DB ch c mt in tr r, vnk V ch gi tr bao nhiu ? A V B 3) Vnk V ang ch gi tr U1 ( hai in tr r ni tip ). V ch s 0 ch cn : + Hoc chuyn ch mt in tr, l in tr no R3 D r r v chuyn n i u trong mch in ? + Hoc i ch hai in tr cho nhau, l nhng in tr no ? Bi 4 B I D hnh bn c AB v CD l hai gng phng song song v quay mt phn x vo nhau cch nhau 40 cm. t im sng S cch A mt on SA = 10 cm . SI // AB, cho SI = 40 cm a/ Trnh by cch v mt tia sng xut pht t S phn x trn AB M, phn x trn CD ti N v i qua I ? b/ Tnh di cc on AM v CN ? A S C

HNG DN GII S 6 - HSG L LP 9Bi 1 Tham kho bi gii tt trong ti liu ny Bi 2 14

HD : 1) Qu trnh bin thin nhit ca nc : - 50C 00C nng chy ht 00C * th : 100 0C

1000C

ho hi ht 1000C

0 Q( kJ ) -5 18 698 1538 6138 2) Gi mx ( kg ) l khi lng nc tan thnh nc : mx = 2 - 0,1 = 1,9 kg. Do nc khng tan ht nn nhit cui cng ca h thng bng 00C, theo trn th nhit lng nc nhn vo tng n 00C l Q1 = 18000 J + Nhit lng m mx ( kg ) nc nhn vo tan hon ton thnh nc 00C l Qx = .mx = 646 000 J. + Ton b nhit lng ny l do nc trong ca nhm ( c khi lng M ) v ca nhm c khi lng mn cung cp khi chng h nhit t 500C xung 00C. Do : Q = ( M.Cn + mn.Cn ).(50 - 0 ) + Khi c cn bng nhit : Q = Q1 + Qx M = 3,05 kg Bi 3 HD : 1) Do vnk c in tr v cng ln nn ta c cch mc ( R1 nt R2 ) // ( R3 nt 2r ) . Ta tnh c U AB 12 = cng dng in qua in tr R1 l I1 = 0,4A; cng dng in qua R3 l I3 = R3 + 2r 20 + 2r UDC = UAC - UAD = I1.R1 - I3.R3 = 0,4.5 12 .20 4r 200 = (1) 20 + 2r 20 + r

Tt khi hai in tr r mc song song ta c cch mc l ( R1 nt R2 ) // ( R3 nt c: UDC =

r ) ; l lun nh trn, ta 2

2r 400 (2) . Theo bi ta c UDC = 3.UDC , t (1) & (2) mt phng trnh bc 2 theo r; 40 + r

gii PT ny ta c r = 20 ( loi gi tr r = - 100 ). 4V

Phn 2) tnh UAC & UAD ( t gii )R AC R = CB (3) R AD R DB

S :

3) Khi vn k ch s 0 th khi mch cu cn bng v :

+ Chuyn ch mt in tr : tho mn (3), ta nhn thy c th chuyn mt in tr r ln nhnh AC v mc ni tip vi R1. Tht vy, khi c RAC = r + R1 = 25 ; RCB = 25 ; RAD = 20 v RDB = 20 (3) c tho mn. + i ch hai in tr : tho mn (3), c th i ch R1 vi mt in tr r ( l lun v trnh by tt ) Bi 4 B K 15 I D I

M

H

x S A S C y a/ V nh ca I qua CD v nh ca S qua AB; ni cc cc nh ny vi nhau ta s xc nh c M v N. b/ Dng cc cp ng dng & KH = 1/2 SI. S 7

THI HSG VT L LP 9( Thi gian 150 pht )

Bi 1 Mt m in c 2 in tr R1 v R2 . Nu R1 v R2 mc ni tip vi nhau th thi gian un si nc ng trong m l 50 pht. Nu R1 v R2 mc song song vi nhau th thi gian un si nc trong m lc ny l 12 pht. B qua s mt nhit vi mi trng v cc iu kin un nc l nh nhau, hi nu dng ring tng in tr th thi gian un si nc tng ng l bao nhiu ? Cho hiu in th U l khng i . Bi 2 Mt hp kn cha ngun in khng i c hiu in th U v mt in tr thay i r ( Hv ). r A U B Khi s dng hp kn trn thp sng ng thi hai bng n 1 v 2 ging nhau v mt bng n 3, ngi ta nhn thy rng, c 3 bng n sng bnh thng th c th tm c hai cch mc : + Cch mc 1 : ( 1 // 2 ) nt 3 vo hai im A v B. + Cch mc 2 : ( 1 nt 2 ) // 3 vo hai im A v B. a) Cho U = 30V, tnh hiu in th nh mc ca mi n ? b) Vi mt trong hai cch mc trn, cng sut ton phn ca hp l P = 60W. Hy tnh cc gi tr nh mc ca mi bng n v tr s ca in tr r ? c) Nn chn cch mc no trong hai cch trn ? V sao ? Bi 3 1) Mt hp kn c chiu rng a (cm) trong c hai thu knh c t st thnh hp v song song vi nhau ( trng trc chnh ). Chiu ti hp mt chm sng song song c b rng d, chm tia khc x i ra khi hp cng l chm sng song song v c b rng 2d ( Hv ). Hy xc nh loi thu knh trong hp v tiu c ca chng theo a v d ? ( Trc ca TK cng trng vi trc ca 2 chm sng ) d 2d

2) a) Vt tht AB cho nh tht AB nh hnh v. Hy v v trnh by cch v xc nh quang tm, trc chnh v cc tiu im ca thu knh ? b) Gi thu knh c nh, quay vt AB quanh im A B theo chiu ngc vi chiu quay ca kim ng h th nh AB A s th no ? A c) Khi vt AB vung gc vi trc chnh, ngi ta o B 16

c AB = 1,5.AB v AB cch TK mt on d = 30cm. Tnh tiu c ca thu knh ? Bi 4 Mt ngi cao 1,7 m ng trn mt t i din vi mt gng phng hnh ch nht c treo thng ng. Mt ngi cch nh u 16 cm : a) Mp di ca gng cch mt t t nht l bao nhiu mt ngi nhn thy nh chn mnh trong gng ? b) Mp trn ca gng cch mt t nhiu nht l bao nhiu mt ngi thy nh ca nh u mnh trong gng ? c) Tm chiu cao ti thiu ca gng ngi ny nhn thy ton th nh ca mnh trong gng ? d) Khi gng c nh, ngi ny di chuyn ra xa hoc li gn gng th cc kt qu trn th no ? Bi 5 a) Ngi ta rt vo bnh ng khi nc c khi lng m1 = 2 kg mt lng nc m2 = 1 kg nhit t2 = 100C. Khi c cn bng nhit, lng nc tng thm m = 50g. Xc nh nht ban u ca nc ? b) Sau qu trnh trn, ngi ta cho hi nc si vo bnh trong mt thi gian v sau khi c cn bng nhit, nhit ca nc trong bnh l 500C. Tnh lng hi nc si dn vo bnh ? B qua khi lng ca bnh ng v s mt nhit vi mi trng ngoi. Cho Cn = 2000 J/kg.K ; Cn = 4200 J/kg.K ; = 3,4.105 J/kg ; L = 2,3.106 J/kg

HNG DN GII S 7 - HSG L LP 9Bi 1 HD : * Gi Q (J) l nhit lng m bp cn cung cp cho m un si nc th Q lun khng i trong cc trng hp trn. Nu ta gi t1 ; t2 ; t3 v t4 theo th t l thi gian bp un si nc tng ng vi khi dng R1, R2 ni tip; R1, R2 song song ; ch dng R1 v ch dng R2 th theo nh lut Jun-lenx ta c : U 2 .t 3 U 2 .t 4 U 2 .t1 U 2 .t 2 U 2 .t Q= = = = = R1 .R2 R R1 + R2 R1 R2 (1) R1 + R2 * Ta tnh R1 v R2 theo Q; U ; t1 v t2 : + T (1) R 1 + R2 =U 2 .t1 Q

+ Cng t (1) R1 . R2 =

U 2 .t 2 U 4 .t1 .t 2 .( R1 + R2 ) = Q Q2U 2 .t1 U 4 .t1.t 2 .R + = Q Q2 =

* Theo nh l Vi-et th R1 v R2 phi l nghim s ca phng trnh : R2 0 (1)U4 Thay t1 = 50 pht ; t2 = 12 pht vo PT (1) v gii ta c = 10 . 2 Q2

1 .U 2 0 Q

U 2 .t1 10 .U 2 + U2 (t1 +10 ).U 2 30. Q Q R1 = = = Q 2 2.Q

v R2 = 20.

U2 Q

Q.R1 Q.R2 = 30 pht v t4 = = 20 pht . Vy nu dng ring tng in tr th 2 U U2 thi gian un si nc trong m tng ng l 30ph v 20 ph . Bi 2 17* Ta c t3 =

HD : a) V s mi cch mc v da vo thy : + V 1 v 2 ging nhau nn c I1 = I2 ; U1 = U2 + Theo cch mc 1 ta c I3 = I1 + I2 = 2.I1 = 2.I2 ; theo cch mc 2 th U3 = U1 + U2 = 2U1 = 2U2 . + Ta c UAB = U1 + U3 . Gi I l cng dng in trong mch chnh th : I = I3 U1 + U3 = U rI 1,5U3 = U - rI3 rI3 = U - 1,5U3 (1) + Theo cch mc 2 th UAB = U3 = U - rI ( vi I l cng dng in trong mch chnh ) v I = I1 + I3 U3 = U - r( I1 + I3 ) = U - 1,5.r.I3 (2) ( v theo trn th 2I1 = I3 ) + Thay (2) vo (1), ta c : U3 = U - 1,5( U - 1,5U3 ) U3 = 0,4U = 12V U1 = U2 = U3/2 = 6V b) Ta hy xt tng s cch mc : * S cch mc 1 : Ta c P = U.I = U.I3 I3 = 2A, thay vo (1) ta c r = 6 ; P3 = U3.I3 = 24W ; P1 = P2 = U1.I1 = U1.I3 / 2 = 6W U 1,5U 3 * S cch mc 2 : Ta c P = U.I = U( I1 + I3 ) = U.1,5.I3 I3 = 4/3 A, (2) r = = I3 9 Tng t : P3 = U3I3 = 16W v P1 = P2 = U1. I3 / 2 = 4W. c) chn s cch mc, ta hy tnh hiu sut s dng n trn mi s : U +U 3 U .100 % = 60% ; Vi cch mc 2 : H 1 = 3 . 100 % = 40%. + Vi cch mc 1 : H 1 = 1 U U + Ta chn s cch mc 1 v c hiu sut s dng in cao hn. Bi 3 HD : Tiu din ca thu knh l mt phng vung gc vi trc chnh ti tiu im a) Xc nh quang tm O ( ni A vi A v B vi B ). Ko di AB v BA ct nhau ti M, MO l vt t thu knh, k qua O ng thng xy ( trc chnh ) vung gc vi MO. T B k BI // xy ( I MO ) ni I vi B ct xy ti F b) V TK c nh v im A c nh nn A c nh. Khi B di chuyn ngc chiu kim ng h ra xa thu knh th B di chuyn theo chiu kim ng h ti gn tiu im F. Vy nh AB quay quanh im A theo chiu quay ca kim ng h ti gn tiu im F. c) Bng cch xt cc cp tam gic ng dng v da vo bi ( tnh c d v d ) ta tm c f. d) Bng cch quan st ng truyn ca tia sng (1) ta thy TK cho l TK hi t. Qua O v tt//(1) xc nh tiu din ca TK. T O v mm//(2) ct ng thng tiu din ti I : Tia (2) qua TK phi i qua I. Bi 4 HD : K a) IO l ng trung bnh trong MCC D D b) KH l ng trung bnh trong MDM KO ? M H M c) IK = KO - IO d) Cc kt qu trn khng thay i khi ngi di chuyn v chiu cao ca ngi khng i nn di cc ng TB I trong cc tam gic m ta xt trn khng i. C O C

Bi 5 Tham kho bi tt trong ti liu ny

18

S 8

THI HSG VT L LP 9

( Thi gian 150 pht ) Bi 1 Tm vn OB c khi lng khng ng k, u O t trn im ta, u B c treo bng mt si dy vt qua rng rc c nh R ( Vn quay c quanh O ). Mt ngi c khi lng 60 kg ng trn vn : a) Lc u, ngi ng ti im A sao cho OA =2 OB ( Hnh 1 ) 31 OB ( Hnh 2

b) Tip theo, thay rng rc c nh R bng mt Pa-lng gm mt rng rc c nh R v mt rng rc ng R, ng thi di chuyn v tr ng ca ngi v im I sao cho OI = 2) c) Sau cng, Pa-lng cu b c mc theo cch khc nhng vn c OI =1 OB ( Hnh 3 ) 2

Hi trong mi trng hp a) ; b) ; c) ngi phi tc dng vo dy mt lc F bng bao nhiu tm vn OB nm ngang thng bng ? Tnh lc F do vn tc dng vo im ta O trong mi trng hp ? ( B qua ma st cc rng rc v trng lng ca dy, ca rng rc ) ////////// ///////// /////////

F F

F F

O

A

B

O

I

B

O

I

B

Hnh 1 Hnh 2 Hnh 3 Bi 2 Mt cc cch nhit dung tch 500 cm3, ngi ta b lt vo cc mt cc nc nhit - 80C ri rt nc nhit 350C vo cho y ti ming cc : a) Khi nc nng chy hon ton th mc nc trong cc s th no ( h xung ; nc trn ra ngoi hay vn gi nguyn y ti ming cc ) ? V sao ? b) Khi c cn bng nhit th nhit nc trong cc l 150C. Tnh khi lng nc b vo cc lc u ? Cho Cn = 4200 J/kg.K ; Cn = 2100 J/kg.K v = 336 200 J/kg.K ( b qua s mt nhit vi cc dng c v mi trng ngoi ) Bi 3 Cho mch in nh hnh v, ngun in c hiu in th khng i U = 120V, cc in tr R0 = 20 , R1 = 275 : - Gia hai im A v B ca mch in, mc ni tip in tr R = 1000 vi vn k V th vnk ch 10V - Nu thay in tr R bng in tr Rx ( Rx mc ni tip vi vnk V ) th vn k ch 20V a) Hi in tr ca vn k V l v cng ln hay c gi tr xc nh c ? V sao ? b) Tnh gi tr in tr Rx ? ( b qua in tr ca dy ni ) ( Hnh v bi 3 ) Bi 4 R1 bng n 1( 6V - 6W ) s dng c ngun in C R c hiu in th khng i U = 12V, ngi ta dng thm A V B 19

mt bin tr con chy v mc mch in theo s 1 hoc s 2 nh hnh v ; iu chnh con chy C cho n 1 sng bnh thng : + U a) Mc mch in theo s no th t hao ph in nng hn ? Gii thch ? 1 1 X X C B A C B A

R0

+U + U S 1 S 2 b) Bin tr trn c in tr ton phn RAB = 20 . Tnh phn in tr RCB ca bin tr trong mi cch mc trn ? ( b qua in tr ca dy ni ) c) By gi ch s dng ngun in trn v 7 bng n gm : 3 bng n ging nhau loi 1(6V6W) v 4 bng n loi 2(3V-4,5W). V s cch mc 2 mch in tho mn yu cu : + C 7 bng n u sng bnh thng ? Gii thch ? + C mt bng n khng sng ( khng phi do b hng ) v 6 bng n cn li sng bnh thng ? Gii thch ? Bi 5 Mt thu knh hi t (L) c tiu c f = 50cm, quang tm O. Ngi ta t mt gng phng (G) ti im I trn trc chnh sao cho gng hp vi trc chnh ca thu knh mt gc 450 v OI = 40cm, gng quay mt phn x v pha thu knh : a) Mt chm sng song song vi trc chnh ti thu knh, phn x trn gng v cho nh l mt im sng S. V ng i ca cc tia sng v gii thch, tnh khong cch SF ? b) C nh thu knh v chm tia ti, quay gng quanh im I mt gc . im sng S di chuyn th no ? Tnh di qung ng di chuyn ca S theo ?

HNG DN GII S 8 - HSG L LP 9Bi 1 : HD : 1) Ngi ng trn tm vn ko dy mt lc F th dy cng ko ngi mt lc bng F a) + Lc do ngi tc dng vo vn trong trng hp ny cn : P = P F + Tm vn l n by c im ta O, chu tc dng ca 2 lc P t ti A v FB = F t ti B. iu kin cn bngP ' OB 3 3 2 = = P F = .F F = .P = 0,4.10 .60 = 240 N FB OA 2 2 5

+ Lc ko do vn tc dng vo O : F = P F = 600 2. 240 = 120N b)1 .FB . iu kin cn 2 P ' OB P = = 2 P = 2.FB = 4.F P F = 4.F F = = 120 N bng lc ny l FB OI 5 + Ngi ng chnh gia tm vn nn F cn bng vi FB F = FB = 2F = 120 .2 = 240N

+ Pa lng cho ta li 2 ln v lc nn lc F do ngi tc dng vo dy F =

c) + Theo cch mc ca pa lng hnh ny s cho ta li 3 ln v lc. Lc F do ngi tc dng vo dy hng ln trn nn ta c P = P + F . iu kin cn bng lc ny l : 20P ' OB = = 2 P + F = 2.FB FB OI

+ Ngi ng chnh gia tm vn nn F cn bng vi FB F = FB = 3.F = 3.120 = 360N. Bi 2 : HD : a) + Do trng lng ring ca nc nh hn trng lng ring ca nc nn nc ni, mt phn nc nh ln khi ming cc, lc ny tng th tch nc v nc > 500cm3 + Trng lng nc ng bng trng lng phn nc b nc chim ch ( t ming cc tr xung ) Khi nc tan ht th th tch nc lc u ng bng th tch phn nc b nc chim ch, do mc nc trong cc vn gi nguyn nh lc u (y ti ming cc ) b) + Tng khi lng nc v nc bng khi lng ca 500cm3 nc v bng 0,5kg. + Gi m (kg) l khi lng ca cc nc lc u khi lng nc rt vo cc l 0,5 m( kg) + Phng trnh cn bng nhit : ( 0,5 m ). 4200. ( 35 15 ) = m. + 2100.m. [0 ( 8) ] + 4200.m.15 + Gii phng trnh ny ta c m = 0,084kg = 84g. Bi 3 : HD a) C nhiu cch lp lun thy in tr ca vn k c th xc nh c, v d : + Mch in cho l mch kn nn c dng in chy trong mch, gia hai im A v B c HT UAB nn : - Nu on mch ( V nt R ) m RV c gi tr v cng ln th xem nh dng in khng qua V v R UAC = UCB mc d R c thay i gi tr S ch ca V khng thay i + Theo bi th khi thay R bng Rx th s ch ca V tng t 10V ln 20V C dng in qua mch ( V nt R ) Vn k c in tr xc nh. b) Tnh Rx + Khi mc ( V nt R ) . Gi I l cng dng in trong mch chnh v RV l in tr ca vn k th ( Rv + R ). R1 in tr tng - in tr tng ng ca mch [ ( Rv ntR ) // R1 ] l R ' = Rv + R + R1 ng ca ton mch l : Rtm = R + R0 U R' U .U . Mt khc c UAB = Iv . ( Rv + R ) = AB UAB = - Ta c R '+R0 R tm R'

P + F = 2. 3F P = 6F F = 120N

+ Khi thay in tr R bng Rx . t Rx = x , in tr tng ng ca mch [ ( R x ntR v ) // R1 ] = R. L U ' v .( x + Rv ) R' ' .U = Iv .( x + RV ) = lun tng t nh trn ta c PT : . Thay s tnh c R ' '+R0 Rv x = 547,5 . Bi 4 : HD: a) in nng hao ph trn mch in l phn in nng chuyn thnh nhit trn bin tr ( RBC ), nhit nng ny t l thun vi bnh phng cng dng in qua bin tr. s 1 c in tr tng ng ca mch in ln hn nn dng in qua bin tr c cng nh hn ( do U khng i v RCB khng i ) nn cch mc s 1 s t hao ph in nng hn. b) S : S 1 RBC = 6 S 2 RBC = 4,34 c) + Cch mc 7 n u sng bnh thng 21

R' .U = Iv . ( Rv + R ) . Thay s tnh c Rv = 100 . R '+R0

X A X X C

X

X B

X

X

H n 1 H n 2 + Cch mc 6 n sng bnh thng v c mt n khng sng (1) M (1) X X A X X (1) B

X X X (2) N (2) Cch mc ny do mch cu cn bng nn n thuc h (1) mc gia hai im M v N khng sng Bi 5 : HD a) (L) (G) F O S I

+ Theo c im ca thu knh hi t, chm tia sng ti song song vi trc chnh s cho chm tia l hi t ti tiu im. Gng phng (G) t trong khong tiu c OF ( v OI = 40cm < OF = 50cm ) chm tia l s khng tp trung v im F m hi t ti im S i xng vi F qua gng phng (G). + Tnh SF Do tnh i xng nn IF = IS = 10cm . SIF vung ti I nn SF2 = IS2 + IF2 = 102 + 102 = 200 SF = 10 2 cm b) Khi gng (G) quay quanh I mt gc : - Do IF lun khng i nn IS cng lun khng i im S di chuyn trn cung trn tm I bn knh IS = 10cm. - Gng (G) quay gc Gc SIF tng ( Gim ) mt gc 2 . di cung trn m im S di .SI .2 . = chuyn l = cm.180 9

22