1979-11-27 — Page 24

頁四第張六第 日八初月十年未己曆夏

WAH KIU YAT PO

報日僑華

二期星

HEENKE

PA

SERFR) BERRY CR

日七十月一十年九七九一圈公年八十六國民華中育教低奲

深饿

冠

而排蝦分

之作們的讚。

欣儒

如

散組公以迅 文英開下歲 括囊量學籍外 校

TLENZELS

AHDALE &

分之微而遭落敗的泰芬妮·麥健士”的成績 REK EREZAK-HARNAI

體勇奪冠軍的科娜,高妮;可資入初公開組英語

生、們以

十分流,而他們的演出十分優異,他們的老師功

·評判於賽華說:他們年紀小,但朗誦之水準

「對虛」式的頒獎塲面,那左起爲季軍黃詩

加之詩優

位席名三二首

高準水人各讚盛判評

三位外额同

·Z SETERN

--

一女獨瓿比賽:而

KREEKAR

以下公開税奔散一

第31屇 校際朗誦節特輯

外首

BN

kgf

fig. (b).

1980

中學會考試題預習專欄

(0.2)(10)

(11) B = (f)(g)(V)

明德出版社魯榮家提供資料

density of

物理

Physics (8)

VK Lo

(Mill & Dale Press It

Answers to Exercise

From the given graph:

(a) In stage V, the truck.

takes the longest time', it takes 40 seconds. (b) Stage IV and stuge V show that the truck travel the greatest distance (400m); (c) Stage I and stage II show

that the truck travel in uniform speed.

| (d)(i) Stage IV obtains the

greatest speed (20us (ii) Siuce, the slope of the

graph represents the

therefore

accelCER As the truck

stage

travels

greatest

acceleration for its sløpe

is the greatest.

111) Stage IV shows that

truck obtains the greatest kinetic energy for its speed is the greatest în this stage.

(e) Total distance travelled

total area under the

grxpl area un

4(10)(50)+(10)(20)

+4(10420)(10)+(20) (20) +4(20)(40) 1300 m

Ang.

water

1000 kgm

1000(10)(v)

10

Let the density

A be

(1)(6)(V)

) (10) (10′′

800kgm

Relative density of

block

200 1000 08

ADS.

(iv) Let the upthrust of

block A in liquid be B From fig. (c)

พ. T+Bx (0,05)(10)+B,

7.5N

(v) Let the density

liquid X be f

L(d)(v) (FM(10)(10′′

* 750 kgm

TẤT

Relative density

fsity of Fiquid X

750

pressure due to liquid column

760mmlg + pressure due to

liquid column

pressure due to

colum

3040

760

760 cmllg

(13600)(10)(760) Nu

Bu the other hand,

re due to li

pressure column

liquid

(1200)(10)(h).

(13600)(10)(700)

(1200)(10)h 13000x760

3x1200

2871 11m

Depth of the lake

2871.11 1.0cm. 2871,11m +0 1m 2871,21m

A force diagram is constructed as following

Where T

Mathematics (8)

C.PL Man

之

(MILL & DALE PRESS LTD) Solution tä Exerc

Section A

+5:6-20724 +8.5=25:15

20 24

x=20k;

-24k; 2-15k Uk+24k-15k 20k+24k-15k

As shown in the figure, the angle between the 50cm-arm and the vertical is when the piece is in equilibrium。 Let d density of the

metal rod sin kgm cross-sectional area of the rod in m.

Weight of the 30cm-arm- Al

(d)(0,3)(A)(g)

-0.584g

Weight of the 50cm-arm HC

(8)(0,5)(1)(g)

0.5dAg A

Moment of the weights of AC and RC about the vertical Jine through P are equal in magnitude

0.5dAg(GH) = 0.5dAg(0,K) (where 6, "and Go are the centres gravity of the 0cm-arm and the 59cm-arm respectively)

36 H

A¤sing 0.158in

BC cose =0.25cose

3(0.15síně)

5(0.25cos6) sine -5(0.25) Cose 5(0.15)

tano

0 = 70" 12"

the Lensic string be 7,

Ans

Let the

Weight of the

block be W volume of the block be Y. and the upthrust. on the block in water be B

fig. (a) fig. (1) In fig. (a

1000

0.75

Let the mass required be M.

Ang

Volume of metal reqi red

M 8000

Total volume of the combined

body

8000 + 10

Mass of block A

0.8kg

Totul mass of the

combined body (M+0.8)kg

Since the combined body floats in water with the upper surface just covered by water, therefore the density of the combined body is equal to the density of water

1000

8000

286kg.

shown in the figure Pressure at A (P ̧)

atmospheric pressure 760 mmHg.

=760mmlig

let the pressure at R be

V = 40A

B

(40-10)A 30A

Where A is the cro

sectional area of the tube

Apply Boyles' Law

tension in the cord

weight of the copper block 0.089g7 weight of the iron block Bouyant force exerted on the copper block Bouvant force- exerted on the iron block

Consider the copper block:

T = W - 3

According to the

Archimedes' Principle,

Bouvant force

weight of

5%

equal volare

of Liquid displaced

Volume of copper block mass of copper block

density of copper

10.0897

·8.9d

where d is the density of water

Solution:

Cost of firice

-5334(160gy

Cost of price.

-$22÷(1+10%).

selling price

is $20,

Solution:

Volume of the single

34

(91)

Radius of the sphere

50 cm

(correct to

significant figures

tant.

ine cose cose

sin cose.

ZAXP

=66°

AP/CH/BQ/ PRAR 4

RB HQ- [CHI – ZA BI

/CPB= /APH

ZTICH-TAB

A BCI BAI CR BR AP BY

similar.

Section B.

9

13.

12cm

6cm

2 cm

10

[DAM=53

Slope of

2

ZDMX =

of A

(4) Mc

tan/DCM-

61

[DCN=45"

01g

in 042:

vl

10. Solution:

0.01g

sinëcose-

0.089g 0.01g

0.0792

Similarly, consider iron block

1 - W2

Let be the mass of the iron block

Volume of the iron block.

201

7,8d

(7.8)dg

0.8718

By equating (1) and

0.079g= 0.8718 mg

tar

Solution:

é m

-3/3.

cm

=6 cm.

-4x2

-0.079

0.8718

0.0906 kg:

1980

Solution.

V

中學會考試題預習專欄

700(403)

(301)

3040

mmg

數學

.00).

pressure at B

明德出版社文長波提供資料

atmospheric pressure

ZAPB÷90′′

Let_0X=r_cm, £XOY=0°

Length of AB

e8

Tength of XY =rg

(2)

:6r+24=8r.

r=12

01÷(1274) cm.

10cm.

The rodii of the two concentric circles are 12 cm "and" 16cm"

respectively.

0 cm

-28 39

+16-2x12

x1bcos28°

=63

AY= 7.94 cm