有教佳華頁三第張七第二日八廿月一十年午戊圈夏
1979
WAH KIU YAT PO
報日僑華
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數學
Mathematics
solution to exercise
Let
(n−2-) (n−1)
374
1-
rove:
日七十月二十年八七九一膳公年七十六民華中
三期星
(0.5m)(4200)(20)
to the length of the wire if
the tension is kept constant.
The doameter of the third sphere is (b) The ratio of
volume of the
sphere. to
210000 420000m
1302000
210000 420000
ird,
sphere.
21.728
255 250
Solutions
2abSiúX=2)
ilgrai
Let
Cos(180
xy Cose
jolution:
Let 100% be the number of article and sy be the cast
rice of each article. Total receipt mn articles
sold at marked price =sy(1+40%){ 100x=5x-20x)
-$105%
2.6 kg of ice should be:
where Lis the or
added.
of the wire and
the
final length of the
the absolute zer
craut
According
The fractio increased
280A
385A
100
BD.Cose
cinx AC.BD.Sine AC.BD.Cosh
Total receipt on articles sold at all cost pnice
where A is the
sectional
the
=Tany.
=10xy
266.67
(Ans)
Total receipt on selling all
=$105xy+#10x5
=$1 5xy
Answer: The absolute ze
temperature is -266,67°C.
Solution:
(a)ühen 164
the diameter
The profit
(ii) Let the unknowa temp
ature be 9. C
In figu
straight
115xy-100%
x100%
100xу
line, LBAC is greate
280
-15%
266
345 266.67+0
17
Sinua
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ADARB
Usb Fare
similar to
each
Area of APB
Areu of
quad.AKE.
2044-43 49.
物理
PHYSICS (12)
Suggested answers to
Exercis
Let the final temperature
of
system - be
the x-intercept y-inter
intercept representa
61
An
The unkaova temperat-
ure is 61.
In figure
3. (a) Since
* 2(0.9) 1
·(2)
(2)
- (0.9)
0.81T
Froof:
(a)ZA1B=90°
cle)
Given
iP=ZATB=ZFKB=96*
ZPBA-LAPU;ZKF3= ZAU (Zin al
segment)
ZUM=Z) AB;Z KBI=
rd Alba (3*Z of a) A ̧AUF,AFB, TKB
are similar to
other (AL
of AAIR
area of ALB
AB*
of ATKB
PE
Ratio of ared,
similar A)
Area of AAPB
area, ofAAPB
Area of APKB.
'B'
stheerom)
Aren ofAJIB
AB2
Aren of quad Ai‚Ð
ZAB
Area ofàllhe
of quad HKB.
AD=9Cos20 cm
=8.457cm
By sin formicła
Sin20”. Sinzaug
3$iu20°
LAUQ=30°52 ZAQ0=180 -20 ~30 52
129 8
·AU
129
n129′′8" -
Sin20
4.586 cm
OD=(8.157-4.536)can
=3_921cm
The distance of
from: BC
bition:
(a) Let d en he thre
deameter of
Sphere.
)))a
+9
Heat liberated by atean
0.5 x 2268000
the reciprocal of the focal
0.5 x 4200 x (100-t).
1344000 - 2100t
Bu
04
Heat absorbed by
0.5 x 336000
0.5 x 4200
168000 + 2100€
1344000 — 2100t
1680904 2100t
4200t - 1176000
t = 280.
Hence, the final temperature
is 280 C according to the
above calculation. It is
et the correct answer
assume that the
iteas are all melted,
herefore, the final
temperature must be 100 c
and part of the steam' is
-condensed.
(b) Let the mass of ice
required be m kg
fstal heat liberated by steam
0.5 x 2368000
+0.5 x 4200 x
1302000J
(100~20)
25cm (Ass.)
(b) The linear magnifica-
tion of the glana is √9
Since it is used as
magnifying glass, the image
se formed must be virtual.
Real-is-positive
ht of the block ia
air
of the black
of the black
O.SIT
object distance
image distance
focal length 25.
2.197
-25
-50
Ans: The distance between
the book and the lens is
16.67 cm.
4. Frequencie
wax attacked
255
5
250 Hz.
the fork with
Total heat aba
ice
The frequency
a stretched
* (0,5 + *) x 336000
vire în înversely proportional
(Ans.
5. Let the tensions f
the wire in figure 3
and figure 4 be T and To
respectively.
Also, let the frequency
of the tuning fork be f and
the mass per unit length
the wire be
Apply
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