Ngoài kiến thức trọng tâm và những bài tập để củng cố kiến thức, cuốn sách còn giới thiệu một số đề thi vào lớp 10 kèm với hướng dẫn cách giải, qua đó khơi gợi sự sáng tạo của các em khi ôn tập và làm bài. Mời các bạn cùng tham khảo nội dung chi tiết.
GIOI THIEU MOT SO DE TUYEN SINH VAO LOP 10 A-O^ BAI I - D E T H I T U Y E N S I N H V A O L d P 10 K H O N G CHUYEN DE Bai Phat bieu d i n h l u a t J u n - L e n - x d V i e t he thijtc, t e n va ddn v i cac d a i liJdng c6 t r o n g he t h i i c Bai M o t day dkn dong chat, chieu d a i I, t i e t d i e n S c6 d i e n trd R = 60 Q, dvfdc mSc vao mach dien c6 h i e u dien the 12 V N e u gap doi day d a n l a i t h i cifdng dong dien chay qua day d a n luc l a bao n h i e u ? Bai Dac d i e m ciia m a t can ? N e u h a i b i e u h i e n thiYdng t h a y ciia t a t can t h i ? KhSc phuc t a t can t h i l a l a m cho mit c6 t h e n h i n ro v a t d gan h a y d x a m a t ? K i n h can l a l o a i t h a u k i n h gi ? Bai Xac d i n h t e n cac cUc tvT P va Q ciia ong day k h i k h o a K dong va cho b i e t dau A ciia k i m n a m c h a m t r o n g h i n h ve la cue ttf gi ? ( H i n h 2.1) I *^ * — = ^ f-ij^h 2.1 Bai GiiJa h a i d i e m A, B c6 h i e u d i e n t h e U k h o n g doi, ngUdi t a mSc song song h a i dien t r d = 20 Q va R2 = 30 Q CUdng dong d i e n chay t r o n g m a c h c h i n h la 1,5 A Ve sd m a c h d i e n va t i n h d i e n trcl tUdng dUdng ciia doan m a c h A B 71 T i n h cong suat d i e n cua doan m a c h A B va n h i e t lUdng toa r a ciia dien t r d Ri t r o n g 30 p h i i t Muo'n cong suat d i e n cua t o a n m a c h A B bSng 81 W t h i p h a i cat bdt dien t r d R2 m o t doan c6 d i e n t r d b a n g bao n h i e u O m ? DE Bai V i noi dong d i e n m a n g n a n g lUdng ? N e u v i d u m i n h hoa Bai P h a t bieu d i n h l u a t O m , g h i cong thiic va ddn v i cac dai liidng t r o n g cong thtfc Bai P h a t b i e u q u y tac n & m t a y p h a i Ap dung : H i n h 2.2, k h i k h o a K dong, /////////// h a y xac d i n h ctfc cua 6'ng day t r o n g h i n h ve sau v a cho b i e t h i e n tUdng gi xay r a do'i v d i t h a n h n a m c h a m t h S n g k h i no difdc dUa den g a n ong day ? (Hoc s i n h ve l a i h i n h vao giay t h i va g h i c h i i Hmh 2.2 day d i i cac c h i t i e t ) Bai - Co t h e tao r a d o n g d i e n cam t i n g xoay chieu bSng each nao ? - N e u cac tac d u n g ciia dong d i e n xoay chieu Bai G i i i a h a i d i e m M va N ciia m a c h d i e n c6 h i e u d i e n the l u o n k h o n g doi, c6 mfic noi t i e p d i e n t r d = 30 Q va R2 = 20 Q Cudng dong dien qua mach la 72 A T i n h d i e n trci tUdng dUdng va h i e u d i e n t h e giiJa h a i dau doan mach T i n h n h i e t lifdng toa r a ciia doan m a c h t r o n g t h d i g i a n gid r a J M a c t h e m d i e n trci R3 sdng song vdi R^ T i n h dien t r d R3 cho cong suat ciia d o n g d i e n qua d i e n t r d R2 b a n g I a n cong suat ciia dong dien qua h a i d i e n trci R^va R3 72 DE Bai V i de tai truyen dien nang di xa ngifdi ta phai dung may bien tbe ? Bai Tinb dien trd cua mot day nbom dai 200 m, tiet dien 1,4 mm^ biet dien trd suat ciia nbom la 2,8.10"^ fl.m Bai So sanh sU gio'ng va khac ve cau tao va boat dong ciia dinamo d xe dap va may phat dien xoay chieu cong ngbiep ? Bai He tbiic giQa so' boi giac va tieu ctJ cua k i n h liip ? Do boi giac cua k i n h lup la lOx Vay tieu cvi ciia kinh lup la bao nhieu ? Bai Lam the nao de tron hai anh sang c6 mau khac ? Sau k b i tron, mau ciia anh sang thu diidc c6 phai la mot bai mau ban dau kbong ? Bai Giiia diem A, B ciia macb dien c6 hieu dien the 12 V kbong doi c6 mfic noi tiep dien trd Rj = 20 Q va R2 = 30 Q Tinh : Dien trd tUdng dufdng cua doan macb AB Cvfdng dong dien qua moi dien trd Mac them R3 = 60 Q song song vdi R2 Tinh cong suat cua doan macb A B luc 73 DE4 Bai D i e n t r cua day d a n b i e u t h i dieu gi ? Cho b i e t cong thijtc, dcfn v i Ve so m a c h d i e n t r o n g c6 sit d u n g v o n ke v a ampe ke de xac d i n h dien t r d cua m o t day d i n Bai H i e n t U d n g k h i i c x a a n h sang l a gi ? So s a n h goc t d i v a goc k h i i c xa k h i t i a sang t r u y e n t ^ k h o n g k h i sang nvf6c Bai De l a m m o t b i e n trci c6 d i e n trci I d n n h a t l a 30 Q b a n g day d a n n i k e l i n c6 dien trd s u a t l a p = 0,4.10"^ Q m va c6 t i e t d i e n deu l a 0,5 m m ^ T i n h chieu dai cvia day d a n ? Bai NgUdi t a d u n g m o t m a y b i e n t h e cuon sd cap c6 500 vong, cuon t h i i cap c6 50000 v o n g de t a i d i e n n a n g c6 cong s u a t l a 1000 k W tii n h a may den n d i t i e u t h u H i e u d i e n t h e d a t vao h a i d a u cuon thijt cap l a 200000 V va dien trd t r e n dUdng day t a i d i e n l a 20 Q T i n h h i e u d i e n t h e ci h a i d a u cuon sd cap T i n h cong s u a t hao p h i toa n h i e t t r e n difdng day t a i d i e n Bai D a t m o t v a t A B c6 d a n g mvii t e n cao c m v u o n g goc v i t r u e c h i n h cua mot t h a u k i n h h o i t u , each t h a u k i n h 30 c m , t h a u k i n h c6 t i e u cU 10 cm D u n g a n h A ' B ' cua v a t A B ? ( k h o n g can d i i n g t i le) Xac d i n h v i t r i va chieu cao cua a n h M u o n cho a n h cao gap l a n v a t t h i p h a i d i c h u y e n v a t theo c h i e u nao ? T a i ? 74 ^ BE5 Bail Neu quy tSc nam tay phai O hinh 2.3a diing mui ten chi chieu cua dUdng siic tit tai cac diem A, B, C Hay xac dinh cac txi c\ic cua nam cham Van dung quy t i c xac dinh chieu ciia dUdng siic t\X t a i cac diem A, B d h i n h 2.3b B A b) Hinh 2.3 Bai Mot mach dien c6 hieu dien the U khong doi, c6 mac song song dien t r d Rj, R2 Q R Chiing minh rhng cung mot thdi gian t : ^ = H2 ^1 Bai Kinh lup la dung cu dung de lam gi ? K i n h lup la loai thau k i n h gi ? Tieu cU cvia kinh lup c6 dac diem gi ? Bai Hay cho biet : V i phai sii dung tiet kiem dien nang ? Bai K h i nhin xuo'ng suoi, ta thay hinh nhu suo'i can hdn NhUng k h i ta xuong t h i suoi sau hdn Hay giai thich hien tiidng hxidc Bale Mot bep dien loai (220 V - 880 W) difdc s^ dung v6i hieu dien the 220 V de dun soi 1,5 l i t nxidc txi nhiet ban dau 28°C 1, Tinh thdi gian dun soi nvfdc Tinh dien trd cua bep dien Biet nhiet dung rieng ciia va bo qua moi hao phi nxXdc la 4200 J/(kg.K) 75 DE Bai Dien trd cua day dan phu thuoc gi ? Viet cong thiic bieu dien sU phu thuoc ? Ghi ro ddn vi cac dai lifdng c6 cong thiic ? Bai De lam mot bien t r d c6 dien trd Idn nhat la 30 Q, bang day dan Nikelin c6 dien t r d suat p = 0,4.10"^Qm va day c6 chieu dai toi da la 37,5 m Can chon day CO tiet dien phu hdp la bao nhieu ? Bai Mo ta cau tao ciia nam cham dien Co the lam tang luc tijt cua nam cham dien tac dung len mot vat bang each nao ? Bai Dac diem ciia mat lao Cach khSc phuc mat lao Bai Cho R j , R2 mac song song vdi vao hai diem A, B c6 hieu dien the 12 V khong doi, t h i cifdng dong dien qua R j , R2 Ian lifdt la 0,6 A va 0,4 A T i n h dien t r d R i , R2 T i n h nhiet lUdng toa cua mach 10 phut Mac them bong den (6 V - 3,6 W) no'i tiep vdi doan mach song song tren vao hai diem A, B t h i den c6 sang binh thiidng khong ? Giai thich tai ? DE Bai Phat bieu dinh luat Jun - Len-xd Viet he thvfc cua dinh luat, ten va ddn vi cac dai lifdng he thiic Bai Den CO ghi (220 V - 75 W) cho biet gi ? T i n h cUdng dong dien dinh mvtc ciia den Bai Bo phan chinh ciia may phat dien xoay chieu la gi ? 76 Bai4 Phat bieu quy tic nam tay phai Xac dinh ten cac c\lc t\i cua nam cham dien va cho biet hien tifdng gi xay v6i nam cham vinh ciiu k h i dUa lai gan ohg day Hinh 2.4 mm /////////// B H/n/7 2.4 Bai5 Mot vat sang A B ( t ) dat vuong goc v6i true chinh (A n^m tren true chinh) cua thau kinh hoi t u c6 tieu cU 10 cm va vat each thau k i n h 15 cm Hay ve va t r i n h bay each dUng anh AjB^ cua vat AB (Do cao cua vat va anh khong can dung t i le) Neu tinh chat cua anh A j E j Tinh khoang each txi vat den anh DE Bai Phat bieu dinh luat Om Viet cong thflc, ten goi va ddn vi cac dai lifdng c6 cong thijtc Biet cifdng dong dien chay qua day d i n la A k h i dat vao dau day mot hieu dien the 15 V Neu hieu dien the giiia dau day la V t h i cUdng dong dien qua day la bao nhieu ? Bai Cong dung ciia bien trcl Tren bien trd c6 ghi (100 W - A) cho biet y nghia cac so' ghi tren bien trc! Bai Neu nguyen nhan gay hao phi k h i truyen tai dien nang di xa Cach lam giam hao phi Bai Mot nam cham dien dat cho true cua no nSm doc theo nam cham nhif Hinh 2.5 Dong khoa K, dau tien ta thay nam cham hi day xa - - - ( ) ) ) ) ) ) /////////// Xac dinh cUc \xL d dau Q nam cham Sau CO hien tiJdng gi xay v6i nam cham ? Hmh 2.5 77 Bai M o t m a y b i e n t h e can h a h i e u d i e n t h e xoay chieu t\i 220 V xuo'ng V C u o n sd cap c6 4400 vong T i n h so v o n g d cuon t h i i cap Bai M o t n g i i d i can t h i k h i k h o n g deo k i n h , c h i c6 t h e n h i n t h a y ro dUdc v a t xa n h a t d each m a t 50 em H a y cho biet : NgUdi ay p h a i deo k i n h can l a t h a u k i n h loai nao de khae phuc t a t can t h i ? K i n h can t h i c h hdp v d i n g i f d i ay c6 t i e u cU O F b a n g bao n h i e u ? G i a i t h i c h v i n g i i d i ay p h a i deo l o a i k i n h ? = DE - Bai X e t t r i f d n g hdp doan m a c h c6 d i e n t r d R, h a y c h i i n g to cong suat dien ciia doan m a c h dudc t i n h theo cong t h i i c : I^R = Bai Co n h i J n g each nao de s\i d u n g t i e t k i e m d i e n n a n g ? Bai T r i n h bay cau tao v a n g u y e n tSc b o a t dong cua m a y b i e n t h e Bai H a y n e u m o t v i d u ve n g u o n p h a t r a a n h sang t r f i n g va h a i v i d u ve each tao i r a a n h sang Bai C h i e u a n h sang vao m o t t d giay t r f i n g t a se t h a y t d giay c6 m a u gi ? N e u t h a y b a n g t d giay m a u x a n h t a se t h a y t d giay c6 m a u g i ? Bai M o t n g i f d i d i i n g k i n h l u p c6 t i e u cvf c m de q u a n sat m o t v a t nho V a t dat each k i n h em T i n h so'hoi giac cua k i n h l u p D U n g a n h cua v a t qua k i n h A n h cua v a t qua k i n h l u p l a a n h t h a t hay O O ' V a n h ao ? A n h I d n h d n h a y nho h d n v a t bao n h i e u I a n ? 78 \ D E 10 Bai B i e n trcf la gi ? G i a i t h i c h so' g h i (20 Q - A ) t r e n m o t b i e n t r d Bai N e u h a i dac d i e m ciia t h a u k i n h de n h a n b i e t l a t h a u k i n h p h a n k i N e u t i n h chat a n h cua m o t v a t tao bcii t h a u k i n h p h a n k i Bai N e u cau tao cua m a y a n h V a t k i n h ciia m a y a n h la l o a i t h a u k i n h g i ? A n h cua v a t can c h u p l e n h i e n ci d a u ? m a y a n h t h o n g thucing t h i t i n h c h a t ciia a n h t r e n p h i m n h u t h e nao ? Bai - N h i n mot ngon den day toe qua m o t k i n h loc m a u do, t a t h a y a n h sang m a u gi ? - N h i n ngon den qua k i n h loc m a u l a m , t a t h a y a n h sang m a u gi ? - Chap h a i k i n h loc noi t r e n v i n h a u v a n h i n ngon den, t a t h a y a n h sang m a u s^m Do c6 p h a i la t r o n a n h sang v6i a n h sang l a m h a y k h o n g ? T a i ? Bai V i k h o n g the d u n g dong d i e n k h o n g doi de chay m a y b i e n t h e ? Bai M o t t h a u k i n h h o i t u c6 t i e u c\l c m M o t v a t sang A B dat t r e n t r u e e h i n h , v u o n g goc v6i t r u e c h i n h va each t h a u k i n h 20 c m Ve a n h A ' B ' eua v a t sang A B K h i ddi v a t l a i gan t h a u k i n h m o t doan c m H a y neu cac dac d i e m cua a n h tao bcii t h a u k i n h 79 I I - D E T H I T U Y E N S I N H vAo LOiP 10 C H U Y E N D E 11 De thi tuyen sink Ldp 10 chuyen TP HCM Nam hoc : 2008 - 2009 Bai Mot chiec binh mong h i n h t r u c6 dien tich day S = 10 cm^ Binh diJdc dat tren mat ban nam ngang, binh c6 chiia n i i d c d nhiet t j = 20°C den cao hi = 10 cm so v d i day binh N g U d i ta tha mot qua cau dac bang dong c6 nhiet t2 = lOO^C vao binh Q u a cau chim het vao nxXdc va niidc binh dang cao them cm so vdi luc ban dau.Cho biet kho'i liJdng rieng cua niidc la D i = 1000 kg/m^, ciia dong la Dg = 9000 kg/m^ ; nhiet dung rieng cua nxldc la Ci = 4200 J/(kg.K), ciia dong la Cg = 380 J/(kg.K) Bo qua sU trao doi nhiet vdi binh va vdi moi trufdng xung quanh, s\ bien doi the tich ciia binh, niidc v a qua c a u theo nhiet T i m nhiet ciia niidc binh k h i c6 can bang nhiet T i m ap lUc c u a qua cau len day binh Bai Vat that A B dat vuong gdc t r u e chinh trUdc mot thau k i n h hoi t u , A nam t r e n t r u e chinh Tieu eU ciia thau k i n h la f Ban dau AB d v i t r i A^Bj each quang „ , f , tam O cua thau k i n h mot dean — < OA^ < f va cho anh la A'^B^ Anh A\B\ cao gap n^ lan A B Sau d6 ddi A B den vi t r i cua A'^B^ Anh cua AB luc l a A'2B2 cao gap n g Ian Ve h i n h mo ta hai AB t r U d n g h d p tao anh n e u t r e n Sii dung cac hinh ve tren va cac phep t i n h hinh hoc, lap bieu thiic tinh n2 theo n^ Bai Mot bien trd R diide mSe vao mot nguon dien Hieu dien the U d hai dau nguon dien no'i vdi bien trd thay doi phu thuoc vao gia t r i R Goi cifdng dong dien mach la I Ta k i hieu y = j Do t h i bieu dien sU phu thuoc eua y vao R la mot diidng thang n h i i H i n h 2.6 80 U = I R - I ( y - l ) = I (12 I =^U= 12-1 T h a o R r a k h o i nguon, mach hci nen ciJdng dong d i e n t r o n g m a c h = = > U = 12 V Bai4 = 24 a D i e n t r d den : dm Cudng dong d i e n qua den : U = H i e u d i e n t h e h a i d a u b o n g den : Cong suat t i e u t h u ciia den : = 0,56 A R + r = I R = 13,44 V ^/^^ = = 7,53 W Cong suat t i e u t h u ciia bo den : Rtd+r Cong suat t i e u t h u cua moi den : => Khi U RU2 R — + r n (R + n r r R ( ^=Rtd.4=R = n RU' PA n (1) (R + nvf va n la n g h i c h b i e n = 110%.^/^d„ = 6,6 W, g i a i p h i t d n g t r i n h (1) t i m diidc : n = 2,7 K h i ^(j = 90%.^(j„ = 5,4 W, g i a i p h i l d n g t r i n h (1) t i m diidc : n = 5,5 ^ K h i 90%.^,,^ < < 110% t h i 2,7 < n < 5,5 Do n n g u y e n n e n t a c6 n g h i e m : n = ; ; Bai5 T h d i g i a n chuyen dong tvL A den H r o i den B : t j = — + — T h d i gian chuyen dong t h A n g t i i A den B : t2 = = 140 s = 128 s D a t A H = Zi, B H = Zg C H = x xpTf— , ^ AC CB Zi-x T h d i g i a n chuyen 3n dong : t = + = +'1 V2 = ^ + Vj^pTJ V i1^2 V f^ 107 Dat ^ y = v i ^/x^~+7| - =^ ( y + V2xf = v i ( x ^ + / f ) ( v f - v | ) x - v y x + v^/f - y ^ = (1) D i e u k i e n c6 n g h i e m ciia (1) : A ' = viy2 - ( V ? - viKvfzf - y2) > t m i n => Y m i n => Y = h^'^l t = - ^ + ^ ^ = 124 s va A' = O ^ C H = x = Khido: V, - A = y2 >/|(v2 - v i ) §6,4 =54m = V1V2 D E 12 Bai K h i xe k h d i h a n h t h i xe da k h d i h a n h trade m o t k h o a n g t h d i gian At vdi v a n toe V j va d i difdc q u a n g difdng As = 300 m T a i t h d i diem t = 30 s, xe bat dau d i vao doan diTdng x a u va g i a m v a n toe, k h o a n g each xe g i a m D e n thdi d i e m t = 50 s t h i xe bat d a u d i vao doan ditdng x a u , h a i xe chuyen dong vdi eung v a n toe V2 v a k h o a n g each xe k h o n g doi V a y xe da k h d i h a n h sau xe m o t k h o a n g t h d i g i a n la At = 20 s => V j = — = = 15 m/s -^0 At T r o n g t h d i gian A t j = 20 s t i f t h d i d i e m t = 30 s den t h d i diem t = 50 s xe c h u y e n dong vdi v a n toe V2 Xe chuy en dong vdi v a n toe V ] va k h o ; i n g each xe g i a m d i m o t doan : Asj = 300 - 200 = 100 (m) T a CO : As, = v , A t i - V A t , = 15.20 - V2.20 = 100 Vo = 10 m/s D e n t h d i d i e m t = 90 s, xe hht d a u r d i doan du'dng x a u , k h o a n g each h a i xe t a n g V a y xe c huy en dong t r e n doan dUdng x a u t r d n g t h d i gian : At2 = 90 - 30 = 60 (s) => d a i doan dUdng x a u : s = V2.At2 = 600 m 108 Bai Phitdng t r i n h can bang nhiet k h i lUdng nxidc m vao binh : (mc + m'c')(ti - t) = mc(t' - t j ) Tinh difdc : m'c' = — mc Phtfdng t r i n h can bang nhiet k h i lUdng nUdc m t h i i hai vao binh (mc + m'c')(t2 - t) = 2mc(t' - t^,) Tinh dUdc : t2 = = 44,8"C Bai Ve anh : A' a) Hinh 2.20 - Tha'u kinh hoi t u : AOA'B' c/5 AOAB AFJA'B' CO FiOIj A'B' AB OA' OA a A ' B ' _ A ' B ' _ F\A' _ FiO + OA' _ AB Oil FiO +b FjO Tii hai phUdng t r i n h t i n h dUdc t i e u cU tha'u k i n h hoi t u : fi = OFi = a.b b-a - Tha'u kinh phan k i : AOA'B' oo AOAB AFgA'B' CO AF2OI2 A ' B ' _ OA/ _ a AB ~ OA " b A'B' AB A'B' OI2 " F.';A' F:>0-0A' F.^0 F;,O f.,- Tit hai phUdng t r i n h , t i n h difdc tieu ciJ thau kinh phan k i : Suy : £2 = f i = 20 cm Bai4 Mac ampe ke no'i tiep vdi RQ vao nguon, doc so'chi I j ciia ampe ke : U = II.(RO + R A ) Mac ampe ke no'i tiep vdi R vao nguon, doc so'chi I2 ciia ampe ke : U = I2.(R + R A ) C Mac ampe ke noi tiep vdi RQ va R vao nguon, doc so' chi I3 ciia ampe ke u = ig.cRo + R + RA) I i ( R o + RA) = h-i^o Txi (!) va (3) : Ii.(Ro + RA) = h-(R + RA) t i m dildc RA theo R Tu: (1) va (2) : (; + R + RA) Thay RA theo R vao, t i n h difdc R Bai R r R, ,>/>2 2 Cudng dong dien qua mach : = R i + R2 Rtd u u u 3R2 Cong suat den I : Cong suat den I I : = R I ^ = R ^ ^ ^ ^ ^ / W = W ?± l !L4 = 2=>f = ^ •''"2 _ S2xq _ l2-d2 /2d? y/\-R,^ /2S1 _ ^1 ^ ^2 _ ( ^ ^ C no Ta (1) va (2) suy r a : ^ = ^ 1,59 ; | = = ^ * 1,26 D E 13 Bai 1 H i n h 2.20a theo d u n g t i le : C h i can ve dUcing d i ciia h a i t r o n g ba t i a sang qua t h a u k i n h AOAB' cr> AOAB A F O r CO AFAB A'F O A'B' OA' AB OA A'B' or FO AB AB FA OA' ^ OF OA " O A - OF A'B' OA' AB OA B' Hinh 2.20a A ' = 150 cm = K h i d i c h u y e n v a t A B r a xa, t i a t i B I song song t r u e c h i n h k h o n g doi T a i v i t r i m i c i i a v a t , t a ve t h e m t i a t i B O t r u y e n t h i n g de c6 v i t r i m d i cua a n h TvC h i n h ve, so s a n h a n h m d i v i a n h c i i , t a suy r a t i n h c h a t a n h A ' B ' m i : - l a a n h t h a t sau t h a u k i n h - l a i gan t h a u k i n h so vdi a n h c u - nho h d n a n h cu Bai 1 D i e n trcl den : R4 = — = 12 Q M a c h : [(R3 n t R4)//R2] n t R j T i n h dUdc : Rtd = Q ; I^h = 1,9 A D e n sang h i n h t h u d n g : I = Do U = I2R2 Suy r a : U Rtd = 0,5 A U A B - I c h R l = dch - l34)R2, t i n h dUdC leh = A U A B = (Ich - 1.34)^2 = ( R + R4)l34> t i n h dUdc R3 = Q Ill Bai Ba d i e n t r d mac no'i t i e p : I nhiJ n h a u ^/^1 = I2RI;.^4 = I2R2;,#3=:I2R3 R2 Ri y/'i ' Rj //\ = Ba dien t r d mac song song ; U n h u n h a u Goi cong suat t i e u t h u d i e n ciia cac d i e n t r d R^, R2, R3 I a n l\Jdt la ^ , ^^5, J/'Q T a CO : ^ A = Ri = I"Ri =Ri 100 3U^ 2Ri Rq U2 U2 Rl R Theo t r e n : U R l + R o + Rq N2 = RlJ ( I u ^ lOORi 3^1 J ,^/> = 15 W = 3//'4 = W = 7,5 W 9U^ ^j = 9U^ lOORi lOO.r, = 60Q R2 = - R l = 20 Q ; R3 = R i = 120 Q Bai G o i n h i e t ciia he t h o n g k h i co can b a n g n h i e t I a n I , I I , I I I l a t j , t , t (mQ.Cg + m c ) ( t - t ^ ) = mc(t - t ) K h i CO can bang n h i e t I a n I I : mo.Co(ti - t g ) = m c ( t - t j ) K h i CO c a n b a n g n h i e t I a n I : (1) (2) V d i t j - t o = 6°C, t - t i = 4''C Tiif (2) - (1), t a t i m difdc : mQ.Co = 4mc hay - " ^ ^ = mc K h i CO can b a n g n h i e t I a n I I I : (mQ.Cy + 2mc) (ts - t2) = mc(t - i^) (3) 90 Tiif (3) - (2), t a t i m dUdc : tg - t = ^ = 2,9°C 112 j Bai5 Goi t h d i g i a n chuyen dong cua h a i ngifdi cho den liic gap n h a u l a t Ta CO BC^ = AC^ + A B ^ ^ (vg if = (V i tf + f : T h a y so'va g i a i phiTdng t r i n h , t a t i n h dUdc : t = 180 s Suy r a : AC = Vj t = 720 m Goi t h d i g i a n ngUdi I I c h u y e n dong t r e n doan dUdng B M la x Ta CO M D ^ = (AD - B M ) " + : ^ [v2(t - x)]- = (vi t - V2 x)^ + f T h a y so', k h a i t r i e n va r u t gon, t a t h u dUdc phUdng t r i n h : I 144x^ - 54tx + 291600 - t - = D i e u k i e n de phUdng t r i n h c6 n g h i e m x : A = (27t)2 - 144.(291600 - 9t^) > Suy r a : t > 144 s hay t ^ j n = 144 s 27t Khinay: x = — = 27 s => B M = V g x = 351 m A D = v j t = 576 m DE 14 Bai 1 Chon he q u y chieu : - A B Chieu dUdng t\i A den B ( H i n h 2.21) ^ ^ „ Hinh2.21 - 100- - Goc toa dp : t a i A - Goc t h d i g i a n : liic h a i xe k h d i h a n h t o i = to2 - + + PhUdng t r i n h c h u y e n dong ciia xe : X j = X Q I + V J (t - tgi) = 40t PhUdng t r i n h chuyen dong ciia xe : X = Xo2 + V ( t - to2) = 100 - 60t X Q A = ; (+) v, X Q B K h i h a i xe gap n h a u : Xi = X2 o 40t = 100 - 60t loot = 100 ^ t ^ (h) V i t r i gap n h a u each A : x = 40.1 = 40 k m V a y h a i xe gap n h a u sau gid c h u y e n dong va each A la 40 k m 113 Do t h i chuyen dong cua hai xe (Hinh 2.22) U(kni) Hai xe each 50 k m trUdc k h i gap vao luc : X2 - x i = 100 - 60t - 40t => 50 = 100 - 100 t o 100 t = 50 2,5 =^ t = 0,5 h Hinh 2.22 Bai Goi t la nhiet k h i binh C can bang nhiet PhUdng t r i n h can bang nhiet : Qtoa = Q t h u 3t = 90 ^ t = 30°C o m.c.(50-t) = m c ( t - ) o - t = t - » Goi la kho'i liidng nifdc ban dau c6 binh A ; t' : nhiet hon hdp Phifdng t r i n h can bang nhiet : Q t ^ j = Q t ^ u 10 m i = 20m m i ( - ) = 20m mi(50 - t') = 50m - 30m c=> m t ' - 30m = 50mi - m j t ' - 50m + mt' ce> m.c.(t' - 30) = (mi - m).c.(50 - t') mi m _ = Bai Ve h i n h (Hinh 2.23) - Goi OB la phap tuyen => OB la tia phan giac cua SOA Ta SOA = AOC + a = 90° + a CO : SOB = 90° + a SOA - Goc hdp bdi tia tdi va mat phang gUdng : - Goc SOM = 90° - SOB = 90° - 90° + a 90° - a Hinh 2.23 114 Ap dung : K h i a = ° ^ MOC = 90° + 20° I = 55° Bai4 TT2 Dien trd den : 192 = -^f^ =— = 20 Q TrUbng hcfp (Hinh 2.24) : V i U i = U2 ; I i = I2 ^ ^1 = ^2 = 7,2 W +^ TrUbng hdp (Hinh 2.25) : V i R ys^B mac no'i tiep vdi R2 : UAB _ RAB _ U2 R2 _ R^B U-U2 R2 ^ R AB = 1^2 Rj + Rg R i + R3 ^ R = 13,3 Q Dien trci den (Hinh 2.26) : R2 = H|m2 = R, = 30 Q ^^dm2 > 4,8 • R2 M V i RjyiN mSc no'i tiep vdi R4 : U4 12 12 Bai R4 ^ RMN U ^ J_ = _±_ = J_ + J_ ^ R4 R4 R4 RMN ^1 = 12 Q ^2 - Tim Ij va I2 : Ta CO dong dien di vao chot M va di chot N Do U = 4RA U4 = 3RA tijtc la : U C N > U D N hay Vc > V Q Nen dong dien di qua ampe ke A2 c6 chieu tu: C sang D U c N = UcD + U D N = R A = I2RA + 3RA ^ I2 = A Xet tai niit D ta CO : I i + I2 = I = I i + - A => I i = A - Tim R, RA : Ta viet phifdng t r i n h hieu dien the : UMN = UMD + UDN = 28 = 2RA + 3RA ^ R A = 5,6 Q TUdng t u ta cung c6 : U ^ N - U M C + U C N = 5R = 5,6 = ^ R = ^ 28 = R + 4.5,6 (vi IR = N C = ^ = m N Hinh 2.27 B T r o n g t h d i g i a n t , nguldi d i c h u y e n dUdc m o t doan dUdng B N = Z = v t T r o n g t h d i gian nay, bong cua d i n h d a u d i c h u y e n difdc m o t doan : /' = B C = B N + N C = ^ V a n toe d i c h u y e n cua bong t o i ciia d i n h d a u : /' ^ = T= 3v = 1,8 m/s Bai4 - K h i K i m d , K2 dong : R2 va R3 b i doan m a c h , d o n g d i e n c h i qua R j So'chi ciia ampe ke la : Ij = ^ = 0,4 A => R j = 15 Q - K h i K j dong, K2 md : R i va R2 b i doan m a c h , dong d i e n c h i qua R3 So'chi ciia ampe k e ' l a : Lj = ^ = , A => R3 = 30 Q K3 - K h i K j va K2 md : R j , R2, Ra mSc no'i t i e p So'chi ciia ampe ke la : I2 = U Ri + R o + R.> = , A =^ R o = 15 Q K h i K i va K o dong : R j , R2, R3 mflc song song So c h i cvia ampe ke la : I4 = — ^ = U 1 Ri Ro + Rq = lA 11' Bai T i n h d U d c c U d n g d o d o n g d i e n d i n h m i i c c i i a m o i d e n : Ii = , A ; l = 0,5A;l3=lA D o I = I2 + I3 n e n k h o n g t h e d i e u c h i n h Cdch de b a d e n s a n g b i n h t h u d n g : M a c RQ s o n g s o n g v d i d e n Uo = U i = V ; ^ = ^ l „ = I + I3 - I j = A = Q ^0 UX = U A B - U I - U = V ; R = ^ Cdch = + I = 1,5 A = 2Q : M a c RQ v a o A v a M UO = U A B - U = V ; R o = ^ I = l2 + l - I i = lA = 6Q UX = U A B - U - U I R^ = ^ = = V ; I ,= II= 0,5A 6a 118 MuC L U C Trang Ldi noi ddu Phdn mot ON T A P VA C U N G CO K I E N THlTC A - O n t a p v a c i i n g c o k i e n thijfc V a t l i 6, 7, I - Ccf hoc I I - N h i e t hoc 12 I I I - Q u a n g hoc 15 I V - D i e n hoc 18 B - O n t a p v a c u n g co k i e n thtfc V a t l i I - D i e n hoc 22 I I - D i e n t i f hoc 46 I I I - Q u a n g hoc 52 Phdn hai G l l T H I E U M O T S O D E T H I T U Y E N S I N H V A O L O P 10 A - De bai I - De t h i t u y e n s i n h vao 16p 10 k h o n g c h u y e n 71 I I - De t h i t u y e n s i n h vao 16p 10 c h u y e n 80 B - Hufofng d a n g i a i I - De t h i t u y e n s i n h vao 16p 10 k h o n g c h u y e n I I - De t h i t u y e n s i n h vao 16p 10 c h u y e n 88 105 TAI Chiu track nhiem xudt ban : C h u t i c h H o i d o n g T h a n h v i e n k i e m T o n g G i a m doc N d O T R A N A I T o n g b i e n t a p k i e m P h o T o n g G i a m doc NCJlivfeN Q U Y T H A O To chikc hdn thuo vd chiu tnich nhiem noi dunii •' Tong G i a m doc C T f l ' f'liui nr va I'hat t r i c n G i a o iliu- Phiftfng Nam I ' H A M VAN P h o T o n g l)ien t a p Bien PHIJNG T H A N H HONG I'HAN X U A N THANH tap Idn dau : HHY^N - D I N H T H I T H A I Bien Q U V N H tap tdi bdn : D I N H T H IT H A I Thie't Q U Y N H ke- Che bdn : N(;uYi:N K I M T O A N Trinh bay bia : NGUYftN M A N H H U N G Cong ty Co phan Dau t i / va Phat trien Giao due Phucng Nam Nha xuat ban Giao due Viet Nam giu" quyen eong bo tae pham ON TAP CUNG CO KIEN THUG VAT L I LIEU O N T H I V A O L O P 10 - V I E T T H E O C H U A N K I E N T H L / C , KT N A N G Ma so: T9L32p3-DTN S(Ydang ky K H X B : 20-2()13/CXB/100-1982/GD In 5.000 cuon, kho 17 x 24cm In tai Cong ty cd phan in Ben Tre In xong va nop liTu chieu thang nam 2013 120 C N G T Y C P H A N DAU TU VA PHAT TRIEN GIAO DgC P H U O N G NAM Dia ch?: 231 Nguyin van CC/, Quan 5, TP Ho Chf Minh Di$n tho^i: (08) 38 357 197 - Fax 38 350 002 Website : http://www.sachhoctro.com.vn MOflBANTIMOpC BO SACH ON THIVAO L(5fP 10 flEMG A N H I Ban doc c6 the mua sach tai cac Cong ty Sach - Thiet bj trUdng hoc d cac dja phUdng hoac cac ciTa hang sach cua Nha xuat ban Giao due Viet Nam : 187 Giang V6 ; 14/3 Nguyin Khanh Toan ; 232 Tay Sdn ; ] -Tai TP Ha Npi : 25 Han Thuyen ; 51 L6 Due ; 45 Hang Chuoi ; • 67B CCfa B I C ; 45 Pho Vpng ; ngo 385 Hoang Quoc Viet ] 78 Pasteur; 247 Hai Phong ; 71 Ly ThUdng Kiet i - Tai TP Da Ning : -Tai TP Ho Chi Minh, 2A Dinh Tien Hoang, quan ; 231 Nguyin Van CU ; 240 Tran Blnh Trpng, quan ; 116 Dinh Tien Hoang, phUdng 1, quan Binh Thanh - Tai TP Can Thd : 162D dUdng thang 2, quan Ninh Kieu J Website : www.nxbgd.vn UMIM ... AF - = ^ ^ T Q — OAi=(ni-l).f TriCdng hdp I I : AO A''oB''o ^ AOA2B2 : => ^ = ST^ A2B2 -OA2-OF OF- = n2 =^ OAg = n2.0A2 OA2 n^OAa-f f _ n2 + Do OA''i = OA2 => n j - = — ^2 => n2 = _(n2+l)f - n... den : 1 92 = -^ f^ =— = 20 Q TrUbng hcfp (Hinh 2. 24) : V i U i = U2 ; I i = I2 ^ ^1 = ^2 = 7 ,2 W +^ TrUbng hdp (Hinh 2. 25) : V i R ys^B mac no''i tiep vdi R2 : UAB _ RAB _ U2 R2 _ R^B U-U2 R2 ^ R... = 90 ° + a CO : SOB = 90 ° + a SOA - Goc hdp bdi tia tdi va mat phang gUdng : - Goc SOM = 90 ° - SOB = 90 ° - 90 ° + a 90 ° - a Hinh 2. 23 114 Ap dung : K h i a = ° ^ MOC = 90 ° + 20 ° I = 55° Bai4 TT2