Tài liệu ôn tập và củng cố kiến thức Vật lí 9 - Ôn thi tuyển sinh vào lớp 10: Phần 1

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:



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-

Luc do vat CO tinh dan hoi h i bien dang tac dung vao vat khac goi la lUc
dan hoi.

-

Do bien dang cua vat dan hoi la hieu gifla chieu dai k h i bien dang va
chieu dai tU nhien ciia vat : Al = I - IQ.

-

Dac diem cua lUc dan hoi : Do bien dang dan hoi ciia vat cang Idn t h i lUc
dan hoi cang Idn.

Chu y : K h i 16 xo hi nen hay bi keo dan t h i no se tac dung lUc dan hoi len cac
vat tiep xuc hay gan vdi hai dau cua no.
Yi d u : Neu ta moc qua nang vao mot chiec 16 xo dang dUdc treo tren moc,
t h i 16 xo dan dai ra, Ivic nay 16 xo sinh ra lUc dan hoi keo qua nang len va
lilc dan hoi keo moc treo xuo'ng.
5


5.

c) Life ma sat:
-

F^g

LUc m a sat t r U d t , lUc m a sat I a n x u a t h i e n k h i mot v a t trUdt hay Ian t r e n
be m a t cua v a t k h a c va can t r d c h u y e n dong ciia v a t .

-

L u c m a sat n g h i x u a t h i e n k h i mot v a t c h i u tac d u n g cua liic nhUng v i n
k h o n g c h u y e n dong.

-

hue m a sat c6 the c6 i c h hoac c6 h a i .

d) Luc day

Ac-si-met:

- Do Idn ciia hJc day A c - s i - m e t : F ^ = d . V
Trong

do :

FA : l i i c day A c - s i - m e t (ddn v i l a N ) ;
d : t r o n g l i l d n g r i e n g cua c h a t l o n g (ddn v i la N/m'^);
V : t h e t i c h p h a n c h a t l o n g h i v a t c h i e m cho (ddn v i la m^).
-

K h i v a t n o i t r e n m a t c h a t l o n g t h i V la the t i c h cua p h a n v a t c h i m t r o n g
c h a t long.

-

S i i n o i : D i e u k i e n de v a t n o i , c h i m :

P <

+ V a t noi len k h i :

P >

+ 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


V
1.

e) Tdc dung

sink

li : D o n g d i e n qua cd t h e ngUdi c6 t h e gay co giat, t i m

n g t t n g dap, n g a t t h d , t h a n k i n h b i te Uet,...
L / n g d u n g : T r o n g Y hoc, ngUdi t a d u n g dong dien t h i c h hdp de t r i benh, v a t h
t r i Ueu,...

BAI TAP
G i a i t h i c h t a i sao c a n h q u a t d i e n sau m o t t h d i g i a n boat dong l a i c6 n h i e u b u i
b a m vao, dac b i e t d mep c a n h q u a t ?

Cong viec m a d i e n d u a t r e n tac d u n g nao cua dong dien ? C h u o n g dien hoat

3.

T r o n g b o n g den day toe, bo p h a n nao t h U d n g l a m bSng v o n f a m ? T a i sao ?

2.

dong dUa t r e n tac d u n g nao cvia dong d i e n ?
4.

Co t h e t h a y n a m c h a m d i e n t r o n g chuong d i e n b a n g n a m c h a m v i n h cUu difdc
k h o n g ? T a i sao ?

5.

H a i qua cau A va B gan v d i gia dd
b a n g n h u a ( H i n h 1.11). K h i l a m cho
q u a cAu A n h i e m d i e n , h a i l a n h o m
gan v d i no xoe r a . No'i qua cau A v d i
qua cau B b a n g m o t t h a n h k i m l o a i .
Co h i e n tUdng g i xay r a v d i :
a) H a i l a n h o m d qua cau B ?
b) H a i l a n h o m d q u a cau A ?

6.

Hinh 1.11

L a y m o t v a t da n h i e m d i e n a m dUa l a i gan m o t qua cau treo t r e n mot sdi td
m a n h . H a y cho b i e t t r o n g cac t r U d n g hdp sau, qua cau c6 b i n h i e m dien
k h o n g ? N e u c6 t h i n h i e m d i e n l o a i g i ?
a) Q u a cau b i h u t l a i g a n v a t n h i e m d i e n ?
b) Q u a cau b i day r a x a v a t n h i e m d i e n ?
Ve sd do m a c h d i e n mac n o l t i e p gom : M o t n g u o n dien gom h a i p i n , mot khoa
K dong, h a i b o n g den. M o t ampe ke de do cvfdng do dong dien cac den, mot
v o n ke de do h i e u d i e n t h e cua m a c h dien. D i i n g m i i i t e n b i e u dien chieu
dong d i e n t r o n g m a c h ?
a) V e sd do m a c h d i e n gom n g u o n d i e n l a h a i p i n no'i t i e p , h a i bong den mac
song song, m o t a m p e k e A do cUdng do dong d i e n m a c h c h i n h , cac ampe ke A i


va A2 do ciicJng do dong dien qua D j va D2, mot cong tic dong ci mach chinh
(c6 ghi chieu dong dien, ciJc (+), cvJc (-) cua nguon dien, chot (+), ch6't(-) cua
ampe ke).
b) Ampe ke A chi I = 1 A, ampe ke Ax chi I j = 0,5 A, ampe ke A2 c.hi I2
bao nhieu ?
c) Neu thao mot trong hai den t h i bong den con l a i c6 sang khong ?
{DS : b) I2 = 0,5 A ; c) Van sang binh thifdng)
9.

Ve sd do mach dien c6 nguon dien gom ba pin mic noi tiep, bong den, cong
tkc dong, ampe ke do cifdng do dong dien chay qua bong den, von ke do hieu
dien the gifla hai dau bong den. Dung miii ten xac dinh chieu dong dien trong
mach. Den c6 ghi 6 V. Nguon dien sii dung phai c6 hieu dien the la bao nhieu

de den sang binh thifdng ?
(DS : 6 V)

10. Cho nguon dien gom ba pin mfic noi tiep, hai bong den mSc no'i tiep, mot cong
tSc dong, day dan, mot ampe ke do ciJdng do dong dien trong mach dien, mot
von ke do hieu dien the gifla hai dau bong den 2.
a) Ve sd do mach dien. Dung mui ten bieu dien chieu dong dien trong mach.
b) Cho biet den 1, den 2 sang binh thifdng k h i cUdng do dong dien qua no Ian
liJdt la 0,15 A va 0,3 A. De khong den nao bi hong t h i cUdng do dong dien
trong mach dien 16n nhat la bao nhieu ?
(SS.-b) 0,15 A)
11. Cho nguon dien gom hai pin mkc noi tiep, hai bong den mac no'i tiep, mot von
ke, mot ampe ke, mot cong tic va day din.
a) Ve sd do mach dien dung k i hieu da hoc, ampe ke do cu:dng do dong dien
chay qua bong den 2 ; von ke do hieu dien the giiia hai dau bong den 1.
Dung mui ten bieu dien chieu dong dien trong mach dien.
b) Hieu dien the cua mach dien la 6 V, von ke chi 3 V. Hieu dien the giiJa hai
dau den 2 la bao nhieu ?
{DS .• b) 3 V)
12. Co nam nguon dien loai : 1,5 V ; 3 V ; 6 V ; 9 V ; 12 V va hai bong den giong
nhau deu c6 ghi 3 V. C^n mic no'i tiep hai bong den nay vao mot trong nam
nguon dien tren. Diing nguon dien nao la phu hdp nhat ? V i sao ?
(DS : Nguon dien 6 V)

21


B - ON TAP VA CONG cd KIEN THUC VAT LI 9
I - D I E N HOC
1.

D i e n trd

cua day dan

a) D i e n t r d ciia day d&n b i e u t h i miic do can t r d dong dien nhieu hay i t cua
day d a n .
b) Cong thvfc xac d i n h dien t r d day d a n : R =

U
I

Trong do :
R l a d i e n trcl (Q) ; U l a h i e u d i e n t h e (V) ; I l a cifdng do dong dien (A).
1 k i l o o m (kQ) = 1000 Q ;
1 megaom ( M Q ) = 1000000 n .
c) K i h i e u :
d) Cach xac d i n h d i e n t r d b a n g von ke, ampe ke :
T h i e t lap m a n g d i e n n h u H i n h 1.12.
-

M a c a m p e ke no'i t i e p v d i R de do ciJdng do

5)

dong dien I R qua R .
-

M a c von ke song song v d i R , de do h i e u d i e n
the

-

H'mh 1.12

giua h a i dau R .

Tinh

, t a xac d i n h dUdc gia t r i R p h a i t i m .

e) S u p h u thuoc ciia d i e n t r d vao cac y e u to'ciia day d i n :
D i e n t r d ciia day d a n t i le t h u a n v d i chieu d a i / ciia day dan, t i le n g h i c h vdi
t i e t d i e n S cua day d a n v a p h u thuoc vao v a t l i e u l a m day d a n .
Cong thijtc b i e u d i e n s\l p h u thuoc ciia d i e n t r d : R =

.

Trong do :
I l a c h i e u d a i day d a n (m) ; S la t i e t d i e n ciia day d a n (m^) ; p la dien t r d

j

sua't ciia cha't l a m day d a n (Qm).
T\i cong t h i i c t r e n suy ra : I =

R.S
R •

22

Co t h e sii d u n g cong t h i i c sau de t i n h S :
9
d^
S = 3,14.r2 = 3 , 1 4 . ^

Trong

do :

r : b a n k i n h day d i n ;

d : dtfdng k i n h day d^n.

2. B i e n t r d
a) Bien trd
B i e n t r d l a d i e n t r d c6 t h e t h a y doi difdc t r i so' va c6 t h e
sii d u n g de d i e u c h i n h cUdng do dong d i e n t r o n g m a c h .
b) Y nghia

cdc so'ghi

tren bien trd

T r e n b i e n t r d c6 g h i so' 6m v a so' ampe, h a i so' nay cho b i e t gia t r i d i e n t r d I d n
n h a t cua b i e n t r d va cifdng do dong d i e n I d n n h a t dUdc phep qua b i e n t r d .
3.

D i n h l u a t 6m

a) CUdng do dong dien chay qua day d a n t i le t h u a n v d i h i e u d i e n t h e d a t vao
h a i dau day d a n v a t i le n g h i c h v d i d i e n t r d ciia day d a n .
b) He t h i i c b i e u d i l n d i n h l u a t : I = ^ ( A )
K
Trong do :
I l a cUdng do dong d i e n , ddn v i l a ampe (A) ;
U l a h i e u d i e n t h e , ddn v i l a vdn (V) ;
R l a dien t r d cua day d i n , ddn v i l a 6m (Q).
c) V a n d u n g d i n h l u a t O m cho doan m a c h mac n o i t i e p ( H i n h 1.13)
l = I i = I , = ... = I„

R.

U = U i + U2 + ... + Un

- r

Hinh 1.13

Rtd = R i + R2 + ••• + Rn
d) V a n d u n g d i n h l u a t O m cho doan m a c h mSc song song ( H i n h 1.14)
I = I l + l 2 + ... + In

R,

U = U i = U2 = ... = U „
h
R

Ri

R,

R.

R,

B

Hinh 1.14
23


e) Van dung dinh luat Om cho doan mach mkc hon hdp (Hinh 1.15).

R3

->—
Hmh 1.15
I A B - I AC + I C B
UAB

= UAC

+

UCB

hay:

hay:

Ii =
UAB

12 + 13
= U I + U2;

UAB

= U I +

U3

R2.R3
AB -

AC + I^CB

'•

RAB-I^1

+

R2

+

R3

4. D i e n n a n g . C o n g c u a dong d i e n . C o n g s u a t
a) Dien

nang

Dong dien c6 nang lifdng v i no c6 the thuc hien cong va cung cap nhiet lifclng.
Nang lUdng cua dong dien dUdc goi la dien nang.

Dong dien qua quat dien lam canh quat dien quay (thUc hien cong).

-

Dong dien qua bep dien lam bep dien nong len (cung cap nhiet lifdng).

-

b) Cong ciia dong dien
-

Cong cua dong dien san ra trong mot doan mach la so' do lifdng dien nang
chuyen hoa thanh cac dang nang liidng khac trong doan mach do.

-

Cong ihtic :
2.

A =

= Ult = T R t =

R

Trong do :
.-Pla cong suat (W) ;

t la thdi gian dong dien chay qua (s) ;

A la cong do dong dien san ra (J) ; U la hieu dien the (V) ;
I la cUdng do dong dien (A) ;

R la dien trd (Q).

1 k W h = 3600000 J .
Tren thiic te, liJcing dien nang sii dung difdc do bang cong td dien.
Moi so' dem ciia cong td dien cho biet ludng dien nang da sii dung la :
1 kilooat gid (kWh).
24