教您華
頁一第張五革
日七十月三年丁胶
WAH KIU YAT PO
育教僑華
英文中學會考試題預習專欄
中文中學會考試題預習專欄
一九六七年
物理科 (廿六)
•
梁海明,
(PAYSTCS31 26)78; M. Leurs
學科
(廿六)
D
蒲仲強。
第廿五次預習試題解答
(1)ABCD
正方形,延長,AB系X及戸西便BXAB,BDBD聨
; DP & BC, XC Ś★ R R Q JA SI_RQ➡OP.
5
(設証如題)
(EOR) 1.41=4P (BP-BD)
Solution
1497(a) Resistore rare in
series and are connected
in parallel with the series
a Suppose the four resiatore are ap chosen. that the instrument con-
三期星
日六廿月四年七六九一麿公年六十五國民嘩
b) The apparatus is shown in the diagram. The
voltage drop acro88 the unknown resistor 19
measured by the voltmeter
and the current flowing
by the ammeter. Ohm's
Law in the form R-E/1
is then applied to de-
termine the resistance.
If an accurate result
is desired, the ammeter
should be of as low a
realatance as possible as 10 15 connectea series in the circuit. The voltagter should be of as high a resistance as possible, as it is connected in parallel in the circuit. voltmetar of low resistance will take too large a current.
1
2. 42mcp (APIDC)
4 <P=41=42=32.5′′ ($60)
5 BXC4 ( 18x=90, 8x=BC)!
6 43=2BXC-¿P = 22.5° ( = } #{ 4 À¥££ ↳ * A **}
7. 44—3—22.5° (πtrÃÃ) ¦
8 CQ=DC=CB (22.5" XDC=CB}"
9.BQ,CBQ=CQB=2/12(180-45)=67.5(等腰の産角三角形の角和
10 <PBQTM $0-67.5"=225*== <P (@ <CBP = 90"}
#1
12.
13.
14.
8Q=QP (=H#{4¶1⁄2 21)
• BRQ = < DRC=«RCD-12— 90-22.5 — 6 7 5° { # } #{ MA$«) BORQ (BKBRQ=LR8Q= 67.5"}
RQ=QP (HDR)
Q.ED.
(重)本題時用足理雖減,但轉折甚多如緊能够将知道鹰家的客
角先行列出,証明不覺困難了。
{2}設0萬 △ABC內任意奌,DEFGAB.BC.CA之中奌 XPQR分
21 CO, AO, 802 È DP. EQ. FR.
設証如題)
(SE) ALPE, ED, DO QP BI
PEL08 QD! ±08 (Añ¿‡‡ ##)
L.PEL QD(同线平行线,代换)
3.PEPQ等口(一組対比等豆平村)
4 # DP EQ & Hí mó̟, gimă
中奌(平行四边形对角线)
5. DP. FR***! @
R
D
VEZDY |)
S.M.N两奌重合(一线拜只有一点
Bp DP EQ FR 1*** M.
3)在ABC中試謡:
(a) (b+c-a) tan = (cra_b) tan & = (a+b-c) Lanç
b) C (cos A+ cos B)=2(a+b) sin
a tan tan in
Stan (S-b) tan & = (s-c) tam &
A
Fa
s = (a+b+c)
5-a~(b+c-a), s-b-±(c+a-b)
λ.₤4b+c-a) tax = (c+a-b) taw = (a+b-c) tan
Q. E. D.
El a
Los A=
24C
sin &= [(5-a) (5-8)
ab
202.
1
nected between B and D
shows no deflection. This
means that there is no p.d.
between B and D. Then. from Ohm's Cas
123
..i (1).
Similarly, the voltage drop from 8 to C met equal the voltage drop from D to C, so
•
112 - 124 Dividing eq. (1) by (2)
If
is unknown.”
•
23
4
(2)
and
We can find T by means of
and T are regladed by 4 In practice single length of resistance wire strectched tightly along a metre or half-metre scale. as shown in the following diagram.
B
COG D.
If the resistance wire is uniform, the re
will be proportional the resistance of CD(F
sistance of AD (r) the length AD while will be proportio Therefore,
To obta
resis
mist
Ad
2. Poud
posi be ob
3 Starta
points not sli
4. Find two po
deflections in
until the opposit
the Length of CD.
AD
#DC
208sible
ine
ethe followi
nat toe
Duinter****
ment
should
yeu various rds C. Dc
#re
SMALL
sections, fork otions are as
small as possible. Then take the mean.
5. Interchange the unknown resistors (r,)
and the standard resistor r2. Repeat the
experiment,
50.[a] Let the internal resistance of
the call be r ohms. The e.n.f be v volts. Then if only a 10 ohms coil is connected, we have 0.1(+10)
· *
.(1) If then a 5 ohms coil is connected in parallel, the reault ant resistance will be
·
1
1
20 ई.
10/3
одов
3.18(r. 10)
(1) -(2)
0.1( r 10) 0.18(
> onus.
put r - 5 ohme into (1)
v0.15 + 10 )
1.5
voŝts.
10
0.19
ΤΟΩ
0
100
5.2
(2)
Answer: The resistazos of the cell in 5 ohma
and the e.n.f. of the cell is 1.5 v
(b) Since the resistance of a uniform wire 100 om long is 2.5 ohma, the resistance of 70 cm
ohme. long 2,5x70/100 1.75
The e..f. of the Leclanche' cell 18 1.4 volta.
1.4
the current of the wire
175 -0.8 amp.
Answer: The current flowing in the potentiom
meter wire is 0.8 amp.
Topice for revision this week, "Electromagnetic induction.
Wuestions.
51.(a) Describe a simple A.C, gererator and how it:
** la modified by a commutator to provide D.C. (3) Explain the terme electromagnetic induction:
induced electromotive force.
YAR
Draw a diagram of the apparatus you would use to transform 200 vošta A.C. to 15 volte A.C. Explain how the transformer works. b) Draw a diagram of a talephone receiver and
transmitter and explain the principles undab....... Lying their operation.
-End - and then the for
© (Coa A+ cos 8) = C
ab2+ac2¬L+bcrab-b'
Zat
- =
a{b+c*»a^}+b(+α-b3)
"gat
zab
— ab(ark) +c2(a+b)–(a2+ b2) ___ (a + b) [ a b + c 2 (a2= ab + b']
•
Rab
= (a+b) [c= (a+b)" } = (a+b) (c+a_s) (c-a+b)
zab
Zab
(a+b) 2(5-8)=2(5-a) -= 2(a+b), (s-al(s=b)=2(a+b) sin?
90f
道(1)第四題中若∠A小路其蟠果修家因在()想
244 1×2 ¢ Ã ± 44 [ By cos(0) = coral în tarbegan. (a-b)), a 3 1 1 ..
好附铱如下:
24 1⁄2 (A+B) cos
) cost (A
(A-B)
za ark sun At sin 8 2 sen.
Q E D.
siné.
}
God
+C cost (AB) ain Coo
Lot (A~B) sin C
去分母,即得求証
4(a) A ABC & TEA ART BTCD# Y ¿C ̃ BE # DC & AC 14 14 *E #ZA de A (a+b) sin £C = c cot £ ( 4-8)
12. BEI DC
J
・浅内錯角、同社角)
E=43 - •£ (4) b EC=Q
A ABE LE LABE
AE= a+b hæ £1
ab
sin (B+) sin
19. an(8+ £) = cos [ 90° ~ (6+ €)].
(A-B)
*
(5) 7 of 1+ sec 20° — cot. 30°cot 40° 1 AM 33 5
ças 20°+ L——
sor 20°
Show 20
2 an 20er 20+ 2
Zain 20°cos 20".
2 sim 30 cor 10 + sin 20
(sim 40°4 sim 20% sin 20′′ "Aim:40°
cos 70
ZOT 40" cos 30′′.
{sin 400
Cos 30 cos
:40°
Q. 8. D.
a trengs cot 30° cot 40°
(6 Ja ABS #01 /T Z APPOO FUOTF X ¿BAP=X WI
2x + sinRx ----
*****4* 2x+ sin 2x=1, t1⁄2 x y wx, 13
2x+y=%£****.£f (0, 1.57), (279,0). 两真決定之連结成直线典Main2x0*桌坐標,即笃 好求方程式之掘
46
2x+4=
Xt www!
「由图中得交
桌機坐標導
x=0.42
: x=0.42
(24°4'),
并去分母。
# AOP TA=140.
(a+b) sin
(b) A ABC? ZA K 28.
•ccor 1 (A-B).
Q. E D.
2 & CA Z DR CDCB
ALE (a - b) cos = C:
== (A-B). ·
(i) @ CD = c8.1
<D=2CBD = {(180-0)=90-=
¿ABD (90) B = A+B+C C-B
AD=CD-CA=2-b △ABD中依E驢定律
ફ
** M
描绘y Ain2x图解(x取值由0至1:6)並利用之以解 11 2x+2x=-
+1 ~鯛り先
302&p (7–22) – sin 2x = —
Q.E.D.
am (A-8)
QEE
0.6 0.2 0.4 200
0.8 62 16 2.0 242) 3.2 2007 22:55: 45:50. 48945: 91.40 114:35 137*30. 166.25 185:21" 10 0.389 0.717 0932 1.000 0.909 0.676 0.335 -0.058
一时季04 單位(比例尺可斟酌選定團 线绳盡量放大)描出名点恼次传连结得
0.8
1.0 1.2
1.4
16
第廿六次預習試題
(1) 9948£: (a) Xy+xy2+ yʻz+yz*+ Z*x+3x2+3x4%
(b) 4 (x+5) (x+6)(x+10) (X+12)~3x2
(2) X 12+2 + (x-24 ) = ( 2 ) + ( 22 2)
(3)(a) Xx-3x+2 P✯ +x+4x= 18'x*+ 15x-5 $42*4x-78K
* p q z (b) ŝi mn b✯ ✯E (X+ 3)*~~~ ̄ ̄ + (x + 57
整除:
x+3x+13(4x+6x=1),
=4
(5)甲で二人毎日工資本等甲作工者平日研工奴元で作較 甲少8日 得工资40元,若工作的日數互换 总共的工資客得
(6)绘
一旦图解(x取值由花王+4人开利用悉以解方程