1978-11-15 — Page 20

莫四第張五第日五十月十年午戊匿

WAH KIU YAT PO

郭日僑華

學等填

處籌統務事樂音 會樂音千萬韻樂 穎|小|分|開|四明 辦|學中|始月1年

阿農

而榮糖

三期星

[模會驗該更努力向學。 福局之口頭 尤。該校校監特別強調,各同學有此 一年庭教育當局所爭取之中七二個,已得教育赏 KEREXEKBERƑSEK - SAKE 年謝會樹成積及港大,中大入學成材之影, |J行畢業典弪•校籃謝中神父報告校務出: (轉)香港天主教慈湖中學,於昨八十元 校監謝銮中神父畢業透露 將辦中七二班制

天主教慈幼中學

日五十月一十年八七九一层公年七十六國民華中 育教僑、

三週系樂音大中

會奏演琴風辦舉

樑樂大唱

師生學推會時子

大辦 午八時在崇基學院猋拜荩举行風举 1、將於十一月二千九日(星期三)下

• (RE) BEENKEEEE

--KEE (3) 溯名流,文教界人士,慕幼會其他贏校校長, 堅毅地站立爲惹我中華靑年而努力。

·最後由陳夫人頒獎投過,該日與會話有官 百强在向 業同學訓中推出 越中串

初

三會 紅雙著人之作品。

牛島酒店(口及蔵利利交界繡開 ),並有導車於當晚七時在中寡在后再頭及九

演安門米十元(學生及藝術中心會員語

试泌食節目將包括巴赫 - 莫札特,

· · KOTKECIK

的結果說明了苦一

·收的實這床 咸酸有黑卡换打 一個英語敎學来

主動

日本幼兒啓

觀光

1979

water is denser than oil.

(1.e. mercury level at A

(b) Let the mercury rises x

metres.

中學會考試題預習專欄

will fall to X and mercury

at B will rise to Y)

明德出版社★榮家提供資料

物理

(六

PHYSICS (6)

ggested solutions to

ercise three

d = density of water

Pressure at X due to

water column

Pressure at Z due to oil

2

column and mercury column

B ̧ = f (x) (v)

7.5 - (√) (10) (103)

- 750 kgm

Relative density

liquid X

750 1000

- 0.75 (Ans.)

(b) Let the mass required be

(1200)(10)(h)

(13600)(10)(760

(1200)(10)h

13600 × 760

x 1200

287111mm

"epth of the lake

2871.11m + 10cm

- 2871.1lm + 0.1m

-2871.21

(Ana,)

Cose-

[POQ

2x37°36 55°12

shown in the figure, the angle between the 30cm-arm and the vertical is 0 when the piece is in equilibrium Let de density of the

metal rod in kgm

cross-sectional

area of the rod

2.

in .

Weight of the 30ensam AC

(d)(0.3)(A)(g)

- 0.3dág

Weight of the 50cm-arm BC

(d)(0.5)(4)(x)

0.50Ag

Moments of the weights of AC and BC about the vertical line through P are equal in magnitude,

0.3dAg(G,H) - 0.5dAg(GK) (where G1 and Gig are the centres of gravity of the 30cm-arm and the 50cm-arm

ctively)

3G H - 5G2K

JACsino

M

Apply

Volume of metal required

M

1979

Pressure density g

height

8000

中學會考試題預習專欄

14. Solution. Let hm be the

height of the hill.

In A ACF

AF-hCot4

In AADE

AE=hCot10°)

Total volume of the combined

(0.17

+

x)gd

明德出版社文長波提供資料

In A BAF

body

M

8000

(in m3)

0.8kg

數學

(六)

Mathematics

-(0.17-x)(0,7d)g+2x(13,6d)g

0.17 + x

(0.17-x)(0.7) +

0.002

Length of the oil column

- 0.17 - 0.002

0.168m

Mass of oil required (volume of oil column x (density of oil): (0.168_x0.00001)

(0.7x1000)

- 0.001176 kg

- 1.176 x 10kg (Ans.)

3. Let the tension in the

string be. I, weight

block be W, volume or the

block be Y and the upthrust

on the block in water be B

figure(a) (1) In figure (8)

8N

In figure (b)

0.15 ino

#BC©oso

figure (b)

Mass of block A Total mass of the combined

body

= (M + 0,8) kg.

Since the combined body floats

in water with the upper sur- face just covered by water, therefore the density of the

combined body is equal to

the density of water

1000

M+ 0.8

8000

1.6

0,2286 kg (Ans,)

Section B

11, Solution.

(a) Let x, y, z be the vol-

ume of the portion A, B

and C respectively.

x+y

27y = x+y 11:26

216x

189x9X+z

1:y - 1:189

xyz - 1:26:189

The ratio of the vel-

ume

of the portion A, B and Cis 1;26:189

(b) Let a, b and c be the

surface area of the pertien A, B and C respectively.

a

a+b

Cos/EAF

hCot10

BC t4

[EAF-66°38

The bearing of the path

is Nob 38

15. Given: /PAQ=120° , ABC

equilateral triangle

To prove:

an

PB CQ-BC (b) PB:CQ=AP2;AQ2 Proof: (a) * ABC is an equilateral triangle (given)

[ABC=/ACB-/BAC=60°

AB-BC-CA

/APB+/PAB-LABC-60° (Ext

•ZPAB=120° (given) of ▲)

LAPB +[CAQ=60°

APB-CAQ (subat,)

Similarly, (PAB=/CQA

A APB

ZABP4/CAQ (3rd [

(3rd)

are similar

QAC (Equiangular A.)

PB AB

(Cerr sides, Similar

PB.CQ=AC.AB

(b)

AC

... PB-CQ

BC

1:8.

Area of A APB Area of A

AQ

areas,

Area or

25c080

3(0.15sino) – 5(0.25coso)

Bine

cost

0.25) 3(0.15)

tanė - 25 - 2,7778

(Ans.)

(a) As shown in the figure, AB is the original mercury level. If water and oil are

right poured into left and

Limbs respectively, the mer- cury in the limb which con

tains oil will rise because

(11) B

where

(0,2)(10) + 8 10N (Ans,) (f)(8)(V)

density of water 1000 kgm

➡ 1000(10)(V)

· 10-3 -3 (Ams.) (iii) Let the density of block A be f

(P

(3)(19) (103) 8

800 kgm

Relative density

block A

800-

1000

0.8 (Ana.)

(iv) Let the upthrust of

block A in liquid X be From figure (c)

(0,05)(10)+B ̧.➡8

Bx

7.5N (ADS.)

(v) Let the density

liquid X be S

As shown in the figure Pressure at A (P)

atmospheric pressure 760 mmEg-

760 mmHg

et the pressure at B be 1

40A

(((40–10)A = 30A

Where A in the cross-

sectional area of the tube Apply Boyles: Law

PV

A A

·P.V.

P. (30A)

mHg

760(40A)

3040

pressure at B

Atmospheric pressure

pressure due to

column

760mmHg → pressure due to

liquid column

pressure due to liquid

column 304 0

760 NH8

760

• (13600)(10) (76°) Nm‍ On the other hand,

pressure due to liqui

UMIA

·366 = a+b+c

36a w a+Ba+c 27a-C

aic - 1:27 1

:bic - 1:8:27

The ratio of the sur...

ce

area of the portion

A, B and C is 1:8:27

12. Solution. Let 8X be the

size of the instalment.

• [8500(1+12%)-x} (1+12% )-X=0

8500x1.12-1.12X-X=0

2.12X-8500x1.122

X

8300x1.122 5038-1

12

The annual payment is equal to $5030

13. Solution. Let /POQ-20° and

OPwr

ON 1 PQ

[PON=/NOQ=d

In APON, ON-rCore

Since (OAB-45/

LOAB 45

AN ON- rCese Area of square ABCD

=(2r€ose.)

Ces

Area of

Area of A ACQ

PB:CQAPAQ'

(Ratio of

Similar

(Same

altitude)

16. Given: TP, TQ are

fangents of the circle PN=NQ; TH=HQ ·

To prove: Q, N, R, H ́are' cencyclic

Proof: Join QR, NH

FN-NQ; QH-HT (given). NH//PT (mid-point theorem) ANHR RFT (Alt / _NH//PT)

PT is the tangent of the circle

RPT/PQR ( in alt. seg) .e." /PQR➡/NHR

N, Q, H, Rare concyclic (converse, /s in the same seg)

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