育教節樂音頁三第張七第 日八廿月二年戌壬夏
WAH KIU YAT PO
報日僑華
二期星
日三廿月三年二八九一层公年一十七國民翰中
第三十四届
香港學校音樂節特輯
小組合奏亞軍:故添男校。
tan60°
PB
45
CB
sh2-2(—)2+2(50)2
+5000
5000
38.73
Height of the flagstaff is 38.73
15. Given:
AHCD is ʼn square ADFC is an equilateral A ZEDC=/ECD-15'
To Prove:
fa) BCFE is a rhombus
AAEB is an
equilateral A
Proof:
80
1982
中學會考試題預習專欄
博物
理
(四)
明德出版社 盧雄國提供資料
Physics (24)
Suggested solutions tó
K. Lo
EL & DALE PRESS LTD.,
Exercise 12
1. Let the length of the
wire be Lm.
(a)
0.3m
Apply the principle o Wheatstone bridge.
L-0,3
* (2)
0.4V
Ans.
(b)
the balance length
is x from A.
Resistance of AB - =(20)
20x z
Resistance of AD
20.2
Ans.
42m
(L-1.2)m
the balance length
(c) the current through
Bin (a) is much greater than the current in (b),
therefore, the temperature (a) is much
in
than in higher
Consequently, the resistance of B in (s) would be
greater gor resistance increases as temperature increases.
(1) Let the current
through the battery be I, the currents
hrough the
voltameter and the
heater be I, and I
respectively, Apply
the formula
m = zit
mass of copper
posited.
8.91x10
time taken
1 hr.
3600 8.
kg
As shown in the figure,
e.c.e. of
copper
3.3×10~7
(0.5x4200x8)
(0.5x420x6)
23108.ĴA
Heat energy given out by the Heater ****81000J
Assuming no heat. losses, or gains. from the
surroundings
23100 - 81000
0-35
the rise
Ans.
temperature of the copper vessel and
its content in 35
(十四)
明德出版社 交長渡提供资料
MATHEMATICS (Syll.2) (24) C. P. Man
MILL & DALE PRESS LTD.
Solution to exercise 10
Section }}°
10. Solution:
(a) het OX=rcm, ¿XOY=0"
length of AB
* (r+4}0 = 8 length of XY
r@ -6
·(1)
(2)
The numbers are
3.5, and 23
Solutionzət
Let x litres, of wine
in the cask at first.
At first
After 1st
operation
After 2nd
operation
wine(in litres) Water(in litrea)
0.
-x
100-19
∙100
Bix
(a) Apply
• (2)
(1) 200-
2002
R
(2)x(1)
XOY
(0.3)(1.2
= (L-9.3)(1.2)
3)(L-1.2) ..
51 =
(0.3)(1.2)
L= 0 (rejected)
or L 1.5
the length of the
wire is 1.5m (b) Substitute the
描
value of Linto (1)
KARFARHOL
Y
1.2-0.
the resistance
of the bulb B
2002
Ans
1300
Ans
PIV
200
200
1A
Ans
400 X
Soun
Apply the principle of Wheatstone bridge.
PX
Where s
R хо
= 1204 and the resist- ance of bulb
0.8m
Son
120
Ала
3.3x10x3600
0.75A
the current I taken from the battery is given by
10
I
3x0.75A 2.25A
(11) The current
through the heater
- I - 1,
Gr+22
GA=(1244) cm
- 16 cm
12
The radii of the two concentric circles are 12 em and 16 cm respectively
(b) XOY
28°39+
-2x12x16 con28 39.
Ans.
63
AY 7.94 cm
Solution:
Let x,y and a be the numbers.
100(x-4)
19% -800x+1600 – 0. (19x-40)(x−40)
40 or
40
19 (rejected)
40 litres of wine in in the cask at first
13. Solution:
(a) ́ ́p and q are the
roots of x2+4x-6)=0
-4; pq=-b
p2+q2 -(n+q)2=2pq
-(-4)-2(-6)
28
=(p+q)(p −pq+q^) mal (28+6)
--136
+q2)+(p2+q3)
28-136 -108.
28(-1 -7808
The required equation is
2 x2+108x-3808=0
- - - -
• (1)
14. Solution:
.(2)
As shown in the figure, resistance
of AB
0.8
* 20
= 1652
Resistance of AD
- 20 + 60
B. measured by the circuit. R* 100-40
R 180KL The current through B
3 120+180. -0.01A
(iii) The power
dissipated by B (0.01) (180) -0.018W
- 2.25A-0.75A
- 1.5 A
Energy given out by the heating coil in the period
(1,5) (10)(3600)
81000J
(iii) Let the
temperature of the copper vessel and its content rises.
Heat energy abosrbed by the copper vessel and its content
Let † - † - - - k Fixmky 5k ; z=11k
5k+3 Tik+3
K+3
25k2+30k+9×11k +3.6k
£k(7k-3)
k=
or 0
(rejected)
Som
Som
Let TP - hm
AP
tan30°
(a) /EDC=LECD=15°
DE EC
(Given)
(sides opp. equal Za)
DF - FC
(sides of
quilateral 4)
EF is the bisector of DC
(1 bisector
locus)
ZFCD - 60
(Lof
LEFC
ZFEC
equilateral A)
(sum of 4)
*90o
30" 180°
RUM
ZFEC-/FCE-75".
FE FC (siden opp.
equal 48)
But DC-FC-CB
FE-PC-CB (Subst)
FE//CB (alt. 28 equal)
BCPE in a rhombus (b) ZENC=LEFC»30" (opp. 48 of
rhombus)
4 EBA-900–30°
-60o
Similarly
ZAEB-180°.
SAED in an
equilateral-A