ôn tập củng cố kiến thức vật lý 9-nguyễn thị ngọc mai

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530.076 NGUYEN THI NGOC MAI o CUNG CO KIEN THUC TAI LIEU ON THI VAO L(3P 10 - VIETTHEO CHUAN KIEN THQC, KT N A N G N G U Y i N THI NGOC MAI 5^0,076 ON TAP, CUNG CO KIEN THLTC VAT i t TAI LIEU O N THI V A O L O P 10 VIET THEO C H U A N KIEN THQC, KI N A N G (Tdi ban Idn thii hai) THU VIENTINHBINHTHUAN if?\/L NHA XUAT BAN GIAO DUG VIET NAM A3 noi ddu Nham dap ung nhu cau on tap kien thuc va ren luyen ki nang lam bai cung nhu giup hpc sinh tu tin thi vao lop 10 chuyen hoac khong chuyen, chung toi bien soan bp sach on thi vao lop 10 On tap, cung co kien thifc Idp 9. Bp sach nay gom c6 nam cuon : Toan, Ngu van, Tieng Anh, Vat li, Hoa hpc. Pham vi kien thuc cua bp sach tap trung vao chuong trinh va chuan kien thuc, ki nang lop 9 do Bp Giao due va Dao tao ban hanh. Cuon On tap, cung cokien thCfc Vat li 9 gom c6 hai phan : Phan mot. On tap va cung co kien thiifc A - Vat li 6, 7, 8 B-Vatli9 Phan hai. Gidi thieu mot so de thi tuyen sinh vao Idp 10 A - D e bai B - Huong dan giai Ngoai kien thuc trpng tarn va nhung bai tap de cung co kien thuc, cuon sach con gioi thieu mot so de thi vao lop 10 kem voi huong dan each giai, qua do khoi gpi su sang tao cua cae em khi on tap va lam bai. Hi vpng cac em se su dung cuon sach nay mot each sang tao de dat dupe ket qua cao trong ki thi sap toi. Mac du da rat co gkng trong qua trinh bien soan, nhung cung kho tranh khoi nhung so suat, ehung toi mong nhan dupe sy dong gop y kien tii phia ban dpe de Ian tai ban sau, sach dupe hoan ehinh hon. Mpi y kien dong gop xin gui ve : Phong Khai thac - Thj tri/dng Cong ty co phan Oau tiTva Phat trien Giao due PhiTcfng Nam 231 Nguyin Van CiT, Quan 5, TP. Ho Chi Minh hoac qua email: khaithacbanthao@yahoo.com TAC GIA 3 ON T A P V A CUNG CO K I E N T H U G i A - ON T A P V A CCING C6 KIE'N T H U C V A T L( 6 , 7 , 8 I - C d HOC 1. Dan vi do do ddi trong he thong do lUdng hdp phap cua nU6c ta la met (m). Ngoai ra, ngUdi ta con dung ddn v i k m , dm, cm, mm,... 2. Dan vi do the tich thUdng dung la met khoi (m^), l i t (0- Ngoai ra, ngUdi ta con dung ddn v i dm^, cm^, cc,... 11 3. = 1 dm^ ; Dan vi do khoi litOng trong he thong do lufdng hdp phap ciia nifdc ta la kilogam (kg), ta (ta), tan (t). Ngoai ra, ngifdi ta con dung ddn v i g, lang,... 1 tan = 1000 kg ; 4. 1 m l = 1 cm^ = 1 cc. 1 ta = 100 kg ; 1 lang = 100 g. C a c loai lufc a) Trong lUc : P Trong lUc la lUc hut ciia Trai Dat, c6 phiTdng t h i n g diing va c6 chieu hadng ve phia Trai Dat. b) Luc ddn hoi: F + Vat chim xuohg k h i : F^ hay hay d^ > d; d^ < d; + V a t Id \\ing t r o n g c h a t long k h i : P = FA T r o n g Ivldng v a k h o i hay = d; Ixidng H e t h i i c giijfa t r o n g liJdng va kho'i Ivfdng : P = lO.m (vdi m t i n h b a n g kg) Vi du : V a t c6 k h o i l i i d n g 100 g t h i t r o n g lUdng la 1 N . Chu y : K h o i l i f d n g m k h o n g t h a y doi theo v i t r i dat v a t , v i k h o i lUdng c h i l i f d n g c h a t c h i l a t r o n g v a t . Con t r o n g l i i d n g l a lUc h u t ciia T r a i D a t len v a t do n e n t r o n g l i l d n g ciia v a t p h u thuoc vao v i t r i cua v a t t r e n T r a i D a t . 6. K h o i Ixidng r i e n g : D Cong t h i i c t i n h k h o i l i i d n g r i e n g : D = Trong do : D : k h o i lUdng r i e n g (kg/m^) ; m : kho'i l i i d n g (kg) ; V : t h e t i c h (m^). 7. Trong lifoTng rieng : d Cong thiic t i n h t r o n g l i i d n g r i e n g : d = p Trong do : d : t r o n g lUdng r i e n g (N/m ) ; P : t r o n g liTdng (N) ; V : the t i c h (m^). Cong thxic t i n h t r o n g lUdng r i e n g theo k h o i lUdng r i e n g : d = TO cong thijfc : d = l O . D , t a suy r a : D = 8. May lO.D ^ . cor doTn g i a n a) Mat phang nghieng (Hinh 1.1) • r \, - Bo qua m a s a t : „ = - r Trong do : F la lUc tac d u n g (N) ; Hinh 1.1 P la t r o n g lUdng v a t (N) ; h la do cao cua m a t p h a n g n g h i e n g (m) ; I la chieu d a i cua m a t p h a n g n g h i e n g ( m ) . - Co ma sat (hao p h i ) t h i h i e u suat H cua m a t ph&ng n g h i e n g la : Ph .100% b) Don bay (Hinh 1.2) 0 : d i e m tUa ; O O O i , O2 : d i e m d a t lUc ; F i , F2 : cac lUc tac d u n g . 0 0 1 = / i ; 0 0 2 = ^2 D i e u k i e n can bSng ciia don bay : = -y- Hinh 1.2 7 9. c) Rong roc : Rong roc la m o t b a n h xe q u a y difdc q u a n h m o t t r u e , v a n h b a n h xe c6 r a n h de d a t day keo. LTng d u n g : Gin + Tae d u n g : D o i h i i d n g eiia lUe tae d u n g ; F = P. + R o n g roe q u a y diJcJe q u a n h mot t r u e c6' d i n h . + R o n g roe c6 d i n h ( H i n h 1.3) - t r e n d i n h eot ed de keo cd, cong n h a n xay dUng d u n g dua gach v i i a len eao,... Rong roc q u a y diidc q u a n h + R o n g roc dong ( H i n h 1.4) - I Hinh 1.3 /////////// ® m o t t r u e d i dong, d i chuyen ciing vdi vat. + Tae d u n g : T h a y doi do Idn ciia Ivtc tae d u n g (giam lUc keo). F = | ; s - = 2h P a l a n g : G o m m o t hoac n h i e u cap r o n g roc. D u n g p a l a n g cho phep g i a m Ixic keo, dong t h d i l a m doi h u d n g ciia lUc nay. C i l d u n g m o t cap r o n g roc (mot r o n g roc eo' d i n h , m o t r o n g roc dong) t h i Idi 2 Ian ve lUc ( H i n h 1.4a). F = P ; 2n s = 2 n h (vdi n l a so' cap cua r o n g roc) Chuyen dong deu va chuyen dong khong deu a) Van toe trong chuyen dong deu s + Cong t h i i e t i n h v a n toe : v = — t Trong do : v : v a n toe (km/h ; m/s) ; s : q u a n g dUdng d i ditde ( k m , m) ; b) Van t : t h d i g i a n d i het q u a n g d i f d n g (h, s). toe trung binh trong chuyen dong khong deu ^ X s - Cong t h i l c t i n h v a n toe t r u n g b i n h t r e n m o t q u a n g dUdng : v^^jj = — . - Cong t h i i c t i n h v a n toe t r u n g b i n h t r e n ca q u a n g d i f d n g chuyen dong : _ + S2 + + Sn t l + t2 + + t„ 8 10. Ap s u a t . A p s u a t c h a t l o n g . B i n h t h o n g n h a u a) Ap suat F Cong t h i i c t i n h ap s u a t : p = Trong do : p : ap suat (N/m^ ; Pa) ; F : ap l u c ( N ) ; S : dien t i c h m a t b i ep (m^). b) Ap suat chat long - Cong thijtc t i n h ap s u a t c h a t l o n g : p = d . h Trong do : p : ap s u a t chat l o n g (N/m^^; d : t r o n g lifdng r i e n g c h a t l o n g (N/m^) ; h : do cao cot c h a t l o n g (m). (h diidc t i n h tvf d i e m t i n h ap suat den m a t t h o a n g c h a t long). c) Binh thong nhau : T r o n g b i n h t h o n g n h a u chiia c i i n g m o t c h a t l o n g d i i n g yen, cac miic chat l o n g d cac n h a n h l u o n l u o n cl c u n g m o t do cao. d) Nguyen F tdc hoat dong cua may thuy lite : ^ *2 Trong - S ©2 do: F j la lUc tac d u n g l e n p i t t o n g c6 dien t i c h S j ; F2 la lUc tac d u n g l e n p i t t o n g c6 d i e n t i c h S2. 11. C o n g ccf h o c . C o n g s u a t a) Cong cd hoc - Cong t h i i c t i n h cong cd hoc : A = F . s Trong do : A : c o n g cd hoc ( J ) ; F : lUc t a c d u n g (N) ; s : q u a n g d U d n g v a t c h u y e n ddi (m). l J = l N . l m = l N . m Chii y : - Cong t h i i c t r e n c h i s\i d u n g k h i h i f d n g cua lUc tac d u n g trCing vdi h u d n g c h u y e n dong ciia v a t . 9 K h i hildng cua luc tac dung vuong goc vdi hudng chuyen dong t h i : A = 0. - K h i hifdng cua lUc tac dung ngUdc vdi hifdng chuyen dong t h i : A = -F.s - AHieu suat cua may cd : H = -r^.100% A Trong do : Ai : cong c6 ich (J) ; A : cong toan phan (J). Chii, y : Cong cd ich la cong can thiet de lam vat dich chuyen. Cong toan phan la tdng cong cd ich va cong hao phi : A = A^ + Ajjp. b) Cong suat - Cong thdc t i n h cong s u a t : 9°= — = F . v Trong do : v : van toe (m/s) ; t : thdi gian thiic hien cong (s) ; A : cong thvfc hien (J) ; P/': cong suat (W) ; F : luc tac dung (N). 1 W = — = i J / s ; 1 kW (kilooat) = 1000 W ; Is 1 M W (megaoat) = 1000000 W - Cach t i n h cong cd hoc thong qua cong suat : Ttf cong thiJc : — ^ A = PAt (J ; W h ; kWh). BAITAP Mot vat xuat phat tiif A chuyen dong deu ve B each A 240 m vdi van toe 10 m/s. Cving liie do, mot vat khae chuyen dong deu t i i B ve A. Sau 15 s hai vat gap nhau. T i n h van toe ciia vat t h i i hai va vi t r i hai vat gap nhau. {DS .• V 2 = 6 m/s ; s A C = 150 m) H a i xe chuyen dong deu tren cung mot diidng thang. Neu di ngUdc chieu t h i sau 15 phut khoang each giiia hai xe giam 25 k m . Neu di ciing chieu t h i sau 15 phiit, khoang each giiia hai xe chi giam 5 k m . Hay t i m van toe cua moi xe. {DS ; v i = 60 km/h ; V 2 = 40 km/h) 3. H a i xe c h u y e n dong t h a n g deu tu: A den B each n h a u 120 k m . Xe 1 d i h e n tuc k h o n g n g h i v d i v a n toe V j = 15 k m / h . Xe 2 k h 6 i h a n h sdm hdn xe 1 l a 1 h n h u n g doc dUdng p h a i n g h i 1,5 h . H o i xe 2 p h a i c6 v a n toe b a n g bao n h i e u de t 6 i B eving m ot luc vdi xe 1 ? (DS : 4. V2 = 16 k m / h ) M o t cano chay xuoi dong song d a i 150 k m . V a n toe cua cano k h i nxidc k h o n g chay l a 25 k m / h , v a n toe ciia dong nxidc chay l a 5 k m / h . T i n h thdi g i a n cano d i h e t doan song do. (DS .• t = 5 h) 5. M o t chiee x u o n g may c h u y e n d o n g t r e n m ot d o n g song. N e u x u o n g ehay x u o i dong tvf A den B t h i m a t 2 h , eon neu x u o n g ehay ngUde dong txi B ve A t h i phai m a t 6 h . T i n h v a n toe eua x u o n g m a y k h i nUde y e n l a n g va v a n toe cua dong nxidc. B i e t k h o a n g each gifla A va B l a 120 k m . (DS : 6. Vx = 40 k m / h ; Vn = 20 k m / h ) T r o n g m ot b i n h t h o n g n h a u chiia t h u y n g a n , ngUdi t a do t h e m vao mot n h a n h a x i t s u n f u r i c v a n h a n h con l a i do t h e m nxidc. K h i cot nifdc t r o n g n h a n h t h i i h a i cao 72 c m t h i t h a y mvte t h u y n g a n d h a i n h a n h n g a n g n h a u . T i m do cao ciia eot a x i t s u n f u r i c . B i e t t r o n g liidng r i e n g eiia a x i t s u n f u r i c va nude I a n lUdt la d j = 18000 N / m ^ v a dg = 10000 N / m ^ . (DS : hA = 40 em) 7. M o t cue nxidc da eo t h e t i c h V = 360 cm"^ n o i t r e n m a t nxidc. a) T i n h t h e t i c h V ciia p h a n 16 r a k h o i m a t nUdc. B i e t k h o i liJdng r i e n g ciia nxidc da l a 0,92 g/cm^. (DS : V = 28,8 m^) b) So s a n h t h e t i c h cua cue nxidc da va p h a n t h e t i c h nUdc do cue nxidc da t a n ra hoan toan. 8. M o t k h o i go h i n h hop c h i i n h a t c6 t i e t d i e n S = 40 em^, do cao h = 10 c m , kho'i lifdng m = 160g. a) T h a k h o i go vao nxidc. T i m chieu cao cua p h a n go n o i t r e n m a t nxidc ? B i e t k h o i lUdng r i e n g ciia nxidc l a DQ = 1000 kg/m^. (DS .• x = 6 cm) b) Bay gid k h o i go difdc k h o e t m o t 16 h i n h t r u d gii3a eo t i e t d i e n AS = 4 em", sau A h v a difdc l a p day c h i e6 kh6'i lu:dng r i e n g l a = 11300 k g / m ^ . K h i t h a k h o i g6 vao nUde t h i ngUdi t a t h a y mxic nxidc b a n g v d i m a t t r e n cua kho'i go. T i m do sau A h ? (DS : A h = 5,5 em) 11 9. M o t t a u t h u y b i c h i m , nu6c t r a n vao t a t ca cac k h o a n g r o n g cua t a u . De dUa t a u l e n m a t nU6c, n g i f d i t a g&n vao t a u m ot so' phao va bdm day khong k h i vao phao de phao n o i l e n m a t nudc va keo t a u l e n theo. Cho biet the t i c h ciia m o i phao k h i b d m day k h o n g k h i l a YQ = 10 m^, t r o n g l i i d n g t a u la P = 10^ N , t r o n g lUdng r i e n g cua nxidc l a d = 10000 N / m , bo qua the t i c h cua t a u va t r o n g l i f d n g cua phao. a) H o i p h a i can to'i t h i e u bao n h i e u phao de d\ia t a u n d i l e n m a t nUdc ? (f)S .• n = 10 phao) b) Cho b i e t ap suat nxidc t a i n d i t a u c h i m (do k h i quyen va Idp nxidc tii do len den m a t niJdc gay ra) l a p = 3.10^ N / m ^ , ap suat k h i quyen t r e n m a t nifdc l a po = 10^ N / m ^ . T i m do sau cua nxidc ndi t a u c h i m . (DS .- h = 20 m) 10. Tu: ben A doc theo m ot bd song, m ot chiec t h u y e n va m o t chiec be cilng bat dku c h u y e n dong. T h u y e n c h u y e n dong ngUdc dong nudc con be dxiOc t h a t r o i theo d o n g niJdc. K h i t h u y e n c h u y e n dong dUdc 30 p h u t den v i t r i B t h i t h u y e n q u a y l a i v a c h u y e n d o n g x u o i d o n g nUdc. K h i den v i t r i C, t h u y e n dudi k i p chiec be. Cho b i e t v a n toe cua t h u y e n do'i v d i dong nudc l a k h o n g ddi, v a n toe ciia d o n g nifde l a V j . a) T i m t h d i g i a n tii luc t h u y e n q u a y l a i t a i B eho den luc t h u y e n dudi k i p chiec be. {DS .• t = 30 p h u t ) b) Cho b i e t k h o a n g each A C l a 6 k m . T i m v a n toe V j cua dong nUde. (DS :vy = 6 k m / h ) II - NHIET HOC 1. N h i e t l i i d n g . P h i f d n g t r i n h c a n b a n g n h i e t a) Nhiet luang N h i e t l i i d n g l a p h a n n h i e t n a n g m a v a t n h ^ n t h e m h a y m a t bdt d i t r o n g qua t r i n h truyen nhiet. Caeh t i n h n h i e t lUdng toa r a h a y t h u vao : Q = m.c.At Trong do : Q : n h i e t l i i d n g (J) ; m : k h d i lUdng (kg) ; e : n h i e t d u n g r i e n g cua chat J / ( k g . K ) ; At : do t a n g , g i a m n h i e t do (°C). b) Phuang trinh can bang nhiet Qtda ra ~ Q t h u vao 12 BAI TAP 1. NgUdi t a t h a dong t h d i 200 g sSt d 15°C v a 450g dong d n h i e t do 25°C vao 150 g nxldc d n h i e t do 80°C. T i n h n h i e t do k h i can b a n g n h i e t . Cho n h i e t d u n g r i e n g ciia sat l a = 460 J / ( k g . K ) ; cua dong l a C2 = 400 J / ( k g . K ) va ciia nUdc la C3 = 4200 J / ( k g . K ) . (DS : t = 62,4°C) 2. M o t n h i e t lUdng ke c6 k h o i lUdng = 100 g, c h i i a m d t l U d n g nUdc c6 kho'i lifdng m2 = 500 g d c i i n g n h i e t do t^ = 15°C. N g i f d i t a t h a vao do h o n hdp bdt n h o m va thiec c6 kho'i lUdng t o n g cong l a m = 150 g da dUdc d u n n o n g t 6 i 100°C. K h i CO can b a n g n h i e t , n h i e t do l a t = 17°C. T i n h k h o i Ivfdng mg ciia n h o m , m4 cua thiec c6 t r o n g h o n hdp. N h i e t d u n g r i e n g ciia chat l a m n h i e t lUdng ke, nifdc, n h o m , thiec I a n lUdt l a : C i = 460 J / ( k g . K ) , C2 = 4200 J / ( k g . K ) , C3 = 900 J / ( k g . K ) , C4 = 230 J / ( k g . K ) . (DS : mg = 25 g ; m4 = 125 g) 3. Co h a i b i n h each n h i e t . B i n h 1 chiia m^ = 2 k g nxidc d n h i e t dp t j = 40°C. B i n h 2 c h i i a m2 = 1 k g nU6c d n h i e t dp t2 = 20°C. T r u t tU b i n h 1 sang b i n h 2 mot l i i d n g nxidc m (kg), de n h i e t dp b i n h 2 p n d i n h , l a i t r u t m p t lUdng nifdc n h i i vay tijf b i n h 2 sang b i n h 1. N h i e t dp can b a n g ci b i n h 1 luc n a y l a 38°C. T i n h liJdng nif6c m da t r u t d m6i I a n va n h i e t dp can b a n g 6 b i n h 2. (DS : m = 0,25 k g ; t,b = 24°C) 4. M p t ngUdi t h d r e n tpi m p t cai r i u t h e p n a n g m ^ = 8 k g b a n g each n u n g n p n g no den n h i e t dp t^ = 400°C r p i t h a vao m p t xp nU6c c h i l a m2 = 4 k g d n h i e t dp t2 = 40°C. K h i l a m nhxi vay t h i cp h i e n tufdng g i xay r a ? H a y g i a i t h i c h . Chd n h i e t d u n g r i e n g cua na6c l a C2 = 4200 J / ( k g . K ) , cua t h e p c^ = 460 J / ( k g . K ) . (DS : Q i > Q 2 n e n k h i nxidc n o n g t d i 100°C t h i m p t p h a n niJ6c b i hpa h d i v i no t i e p tuc diidc c u n g cap n h i e t ) 13 5. M o t a m n h p m c6 k h o i lifdng l a 250 g c h i i a 1 l i t nxidc d 20°C. T i n h n h i e t liJdng c ^ n de d u n soi lUdng nUdc t r e n . B i e t n h i e t d u n g r i e n g ciia n h o m v a nUdc I a n lUdt l a Cj = 880 J/(kg.K), = 4200 J/(kg.K). {DS : Q = 353,6 k J ) 6. M o t t h a u n h o m c6 kho'i liJdng l a 0,5 k g d i i n g 2 k g nxidc c! 20°C. a) T h a vao t h a u nUdc m o t t h o i d o n g c6 kho'i l i f d n g 200 g l a y d 16 r a . Nxidc n o n g d e n 21,2°C. T i m n h i e t do c i i a bep 16. B i e t n h i e t d u n g r i e n g cua n h o m , nude, d o n g I a n l i f d t l a C i = 880 J/(kg.K), cg = 4200 J/(kg.K), C3 = 380 J/(kg.K). B d q u a sU t o a n h i e t r a m o i t r U d n g ngoai. {DS : t = leO.YS^C) b) ThUc r a t r o n g t r U d n g hdp nay, n h i e t lUdng toa r a m o i t r U d n g l a 10% n h i e t lUdng c u n g cap cho t h a u nu6c. T i m n h i e t do t h u c sU cua bep 16. {BS 7. : t = 174,74°C) NgUdi t a t r g n l i n h a i chat long c6 n h i e t d u n g r i e n g , kho'i lUdng, n h i e t do b a n dku cua c h i i n g I a n lUdt l a c^, m j , t^ va C2, m 2 , t 2 . T i n h t i so'kho'i lifting cua h a i c h a t l o n g t r d n g cac t r U d n g hdp sau day : a) D o b i e n t h i e n n h i e t do cua chat l o n g t h i i h a i gap doi so v 6 i do bien t h i e n n h i e t do c u a chat l o n g t h t f n h a t s a u k h i can bSng n h i e t . mg Ci b) H i e u n h i e t do b a n d a u ciia h a i chat l o n g so v d i h i e u giuta n h i e t do can bang va n h i e t do d a u ciia chat long t h u n h i e t bSng t i so' ^ . b bcj mg 8. M o t n h i e t lUdng ke b i n g n h o m c6 kho'i lUdng m^ = 100 g chiia m 2 = 400 g nirdc d n h i e t do t j = 10°C. N g i i d i t a t h a vao n h i e t l i i d n g ke m o t t h o i hdp k i m n h o m v a thiec c6 kho'i lUdng m = 200 g dUdc n u n g nong den n h i e t do t 2 = 120°C. N h i e t do can b a n g ciia he tho'ng l a 14°C. T i n h kho'i l i i d n g n h o m va thiec CO t r o n g hdp k i m . Cho n h i e t d u n g r i e n g cua n h o m , nUdc, thiec I a n lifdt l a c i = 900 J/(kg.K), C2 = 4200 J/(kg.K), C3 = 230 J/(kg.K). (DS : mg = 0,031 k g ; m4 = 0,169 kg) 14 9. Co hai binh each nhiet. B i n h 1 chijfa m j = 2 kg nifdc d t^ = 20°C, binh 2 chiia m2 = 4 kg nxldc ci t2 = 60°C. Dau tien ngUdi ta rot mot phan nxidc m t i i binh 1 sang binh 2, sau k h i can hkng nhiet, ngUdi ta l a i rot mot liJdng nildc m n h u the t\i b i n h 2 sang b i n h 1. Nhiet do can bSng d binh 1 luc nay la t'l = 21,95°C. a) Tinh lUdng nxidc m trong moi Ian rot va nhiet do can hkng cua binh 2. (DS .- m = 100 g ; t'g = 59°C) b) Neu tiep tuc thiJc hien Ian hai, t i m nhiet do can bang cua moi binh. {DS : t,h = 23,76''C) 10. Mot bep dau dun 1 l i t niidc dUng trong am nhom c6 khoi liidng la m2 = 300 g t h i sau thdi gian t^ = 10 phut t h i nUdc soi. Neu dvmg bep va am tren de dun soi 2 l i t nxidc trong cung mot dieu kien t h i sau bao lau nUdc soi ? Cho nhiet dung rieng cua nUdc va nhom Ian liidt la Cj = 4200 J/(kg.K), C2 = 880 J/(kg.K). Biet rang nhiet do bep dau cung cap mot each deu dan. (DS ; t = 19,4 phut) III - QUANG H O C 1. Nhan biet anh sang - Ta nhan biet dUde anh sang k h i eo anh sang truyen vao mat ta. Vi du : Ta n h i n thay bong hoa mau do v i c6 anh sang mau do t i i bong hoa den mat ta. - Ta nhin thay mot vat k h i c6 anh sang truyen t\i vat do vao mat ta. - Vat den la vat khong tU phat ra anh sang ciing khong hat lai anh sang ehieu vao no. Sci di ta nhan ra vat den v i no dUdc dat ben canh nhiing vat sang khac. 2. Nguon sang Nguon sang la vat t\i no phat ra anh sang. V i du : M a t Trdi, den dien dang hoat dong,... 3. Vat sang Vat sang gom nguon sang va nhiing vat hat lai anh sang ehieu vao no. 15 4. D i n h l u a t t r u y e n t h a n g c u a a n h s a n g : Trong moi trildng trong suot va dong t i n h , anh sang truyen di theo dutdng thang. Chii y : Trong moi trifdng trong suot va khong dong tinh, anh sang khong truyen theo diidng thSng. V i du : khong k h i tren sa mac d gan mat dat t h i nong, len cao t h i lanh, mat do khong k h i khong deu, anh sang c6 the truyen theo diJdng cong nen gay ra hien tiidng ao anh. 5. Dvfcfng t r u y e n c u a a n h s a n g dildc bieu dien bang mot diidng thang c6 m i i i ten chi hiidng goi la tia sang. S * M Trong thiJc te', ta khong nhin thay mot tia sang ma chi nhin thay chum sang gom rat nhieu tia sang hdp thanh. Co ba loai chum sang : - Chum sang song song : cac tia sang khong giao nhau tren diJdng truyen cua chiing (Hinh 1.5). K h i nguon sang d rat xa vat nhan sang t h i chum sang t6i dUdc coi la chum sang song song. V i du : Anh sang tii Mat Trdi chieu den Trai Dat diidc coi la chum sang song song. - Chiim sang hoi t u : cac tia sang giao nhau tren dUdng truyen cua chiing (Hinh 1.6). - Chum sang phan k i : cac tia sang loe rong ra tren dvJdng truyen ciia chung (Hinh 1.7). Hinh 1.5 6. Hinh 1.6 Hinh 1.7 tTng d u n g c u a sij^ t r u y e n t h a n g c u a a n h s a n g Giai thich hien hifdng nhat thiJc, nguyet thUc. - Giai thich hien tudng bong to'i va bong nika to'i; - Do quy l u a t chuyen dong ciia T r a i Dat va M a t Trang, nen ngUdi ta c6 the t i n h dUdc mot each chinh xac ndi va ngay, gid xay ra nhat thuc hay nguyet thUc. '. ; ,. . / ' ;'"• V 1. BAITAP Chieu mot tia sang t6i gUdng phSng. Biet goc tdi la 30°. Tinh goc tao bdi t i a phan xa va mat phSng gUOng. (DS : 60°) 2. Chieu mot tia sang vao gUdng phang vdi goc tdi la 60°. Tinh goc giiia t i a tdi va tia phan xa. (DS : 120°) 3. Goc tao bdi tia phan xa va phap tuyen cua mat gUdng t a i diem tdi la 40° t h i goc hdp bdi tia phan xa va tia tdi la bao nhieu ? {DS : 80°) 4. Mot ngiidi cao 1,6 m diing each gifdng phSng mot khoang 3 m. Hoi anh ngUdi do cao bao nhieu va each gUdng mot khoang bao nhieu ? Ve hinh theo t i xich 1 m tUdng L(ng 1 cm. 5. Mot hoc sinh cao 1,5 m diing each gildng phSng mot khoang 80 cm. Hoi anh each hoc sinh do mot khoang bao nhieu ? (DS : 160 cm) 6. Dat mot vat sang A B gan sat trildc ba gvfdng G^, G2, G3 cd eung kich thudc. Giidng G i cho anh ao Idn hdn vat. Gutdng G2 cho anh ao nhd hdn vat. GUdng G3 cho anh ao cao bSng vat. G^, Gg, G3 la giidng gi ? V i sao ? 7. Cho mot diem sang S dat triidc gUdng phang nhif H i n h 1.8. Hay diing hinh ve de xac dinh khoang khong gian can dat mat de cd the n h i n h thay anh S' ciia S. •S Hinh 1.8 8. Hinh 1.9 Xac dinh tren hinh ve anh A'B' cua vat A B ( H i n h 1.9) va vung dat mat de nhin thay anh do. r'Hi; V!ENT!MHBiHHTHUAM 17 9. Cho h a i gvfdng p h A n g song song nam ngang, m a t p h a n xa h u d n g vao n h a u . C h i e u t i a s a n g S I l e n gifdng G j . H a y ve t i e p t i a p h a n xa I a n Ivfdt t r e n g U d n g Gi r o i G2. Co n h a n x e t g i ve p h i l d n g c i i a t i a p h a n xa c u o i c u n g va p h U d n g ciia t i a t d i ? 10. • p h a n g va h a i d i e m S va R. ^ T r o n g H i n h 1.10 c6 ve m o t gUdng s a) D u n g t h u d c ke c6 chia do va e ke, • h a y ve t i a t d i qua d i e m S cho t i a p h a n x a d i q u a R. b) M o t a b a n g I d i each ve cua em. /////////////////////////^^^^ Hinh 1.10 IV - D I E N HOC 1. D o n g d i e n l a d d n g cac d i e n t i c h d i c h e h u y e n ed h u d n g . K h i CO d o n g d i e n t r o n g day d a n k i m l o a i , cac e l e c t r o n t u do d i c h ehuyen cd h i l d n g v d i v a n toe k h o a n g tu: 0,1 m/s t d i 1 m/s. T h e m a k h i ddng cong tac d i e n t h i b d n g den s a n g h a u n h u t i i c t h i , mac d u day d a n ed t h e r a t d a i . Do l a v i k h i d d n g cong tac, cac e l e c t r o n t U do cd san d m d i ehd t r o n g day dan n h a n dUdc t i n h i e u gan n h i i c u n g m o t luc va h a u n h U dong loat ehuyen d o n g cd h u d n g . 2. Moi n g u o n d i e n d e u c6 h a i c\ic : Cue dUdng ( k i h i e u dau +), cxlc am ( k i hieu dau - ) . Cac n g u o n d i e n t h i i d n g d u n g l a : p i n , acquy, may p h a t dien,... 3. D o n g d i e n c h a y t r o n g m a c h d i e n k i n bao gom cac t h i e t bi dien dUde no'i l i e n v d i h a i eUe ciia n g u o n dien bang day d i e n . 4. C h a t d a n d i e n l a c h a t cho d d n g d i e n d i qua V i d u : Bac, dong, v a n g , n h o m , sat, t h u y n g a n , t h a n c h i , eae d u n g dich axit, k i e m , muo'i, nutdc t h i i d n g d i i n g . 5. C h a t e a c h d i e n l a c h a t k h d n g cho d d n g d i e n d i qua V i d u : Nude n g u y e n chat, k h o n g k h i k h o , go k h o , chat n h u a , deo, sanh, s i i , thuy tinh,... 6. M a c h d i e n : N g u o n d i e n , v a t t i e u t h u d i e n , day d a n , k h o a nd'i v d i n h a u tao t h a n h mach dien. M a c h d i e n dildc m d t a b a n g sd do va tii sd do cd t h e lap m a c h d i e n t i i d n g l i n g . 18 K i h i e u m o t so' bo p h a n m a c h d i e n : a) Nguon dien + P i n , acquy : + Bo p i n , bo acquy : b) Vat tieu thu dien + Bong den : + C h u o n g dien : + D o n g CO d i e n : c) Cong tdc + Cong the dong : + Cong tac m d : d) Vat tod nhiet (ban la, bep dien,...) 7. C h i e u d o n g d i e n theo quy U6c la chieu t\i cUc dUdng qua day d a n va cac t h i e t b i dien t d i cUc a m ciia nguon d i e n . 8. C a c tac dung cua dong dien a) Tdc dung nhiet : D o n g d i e n qua m o i v a t d a n t h o n g t h i t d n g deu l a m cho v a t d a n nong l e n . U n g d u n g : H o a t dong ciia bep d i e n , b a n l a , den d i e n day toe,... b) Tdc dung phdt sdng : D o n g d i e n qua v a t d a n l a m cho v a t d a n nong len t d i n h i e t do cao t h i p h a t sang. L/ng d u n g : H o a t dong ciia den d i e n , bong den ciia b u t t h i i dien,... - Day toe den b i dot nong m a n h va p h a t sang k h i c6 dong dien chay qua. K h i den sang b i n h t h i t d n g t h i n h i e t do k h o a n g 2500"C, nen day toe bong den diJde l a m bang vonfam, v i vonfam la chat eo n h i e t do nong chay cao (3370"C). - B o n g den eua b u t t h i i dien : D e n sang do v u n g chat k h i d giQa h a i d a u day cua bong den p h a t sang. c) Tdc dung tit : D o n g d i e n c6 tac d u n g t i l v i no c6 t h e l a m q u a y k i m nam cham. l / n g d u n g : C h u o n g d i e n , rdle t\i, q u a t dien,... d) Tdc dung hod hoc : K h i cho dong d i e n qua d u n g dich muo'i dong t h i no t a c h dong r a k h o i d u n g d i c h , tao t h a n h Idp dong b a m t r e n t h o i t h a n noi v6i cue a m . LTng d u n g : M a d i e n de cho'ng g i , l a m dep do d u n g , m a y moc. 19
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