1984-10-09 — Page 20

頁四第張五第 日五十月九年子甲歷夏

WAH KIU YAT PO

報日僑華

1000

菲律賓披

第一般西隔线

100

100

100

1001

WRX/00

KO 00

台灣母

8:00

N

100

#TO

第一印度勞理

100

澳門紙

12:00

南非蘭特

100

100

1000

#9-00

南韓紙

SAKE

1.00

R-00

100

路上法郎

100

#O

西德馬克

100

1000

五|荷商

XHE

100

100

"KHO

CO/EEX

1985

中學會考試題預習專欄

(四)

明德出版社盧雄國·提供資料

物理

PHYSICS (4)

H.K. LO

MILL & DÁLE PRESS. LTD.

Answers to Exercise 2

1.

(a)

linky Tams"

As shown in the figure above,

Let T-tension in the

string,

a acceleration

of the system.

Apply Newton's 2nd law to the block

separately,

15(10)-Y-15a... (1) T-10(10)=10a. (1)+(2): 50=25a

a=ms

2

(Ans.)

(b) Substitute a-2 înto

(2).

T-10(10)=10(2)

(c)

Apply

2

T=120N(Ans.)

the formula

v ̄=u+2as⠀

Let 1

be the

velocity of the 15kg. block when it strikes the floor. The

initial velocity is

ms

-1

2=2(a)(2)

=2(2)(2)

=8

-2.8284 ms

(d) Let the time required

for the 15 kg block

(e)

to reach the floor

be t (seconds).

Apply the formula

Buvo toat2

inital velocity

sat

Substitute s-2,

2 = 12 (2) + 2

=1.4142

isleg

išką тока

(Ans.)

As shown in the fig. above, let the velocity of the 10kg

when it block be g

reaches the level at

-2(2)(2)

=8

Hence, the string slackens, assume the block moves h meters father before drops down.

变透时徧置成,令年七月正式生效,十月六日 與挪威簽響該項協議十分有利澳門,以後澳門 協調政務司高贵榜拇上在簽約儀式後表示,今次 玉挪威的精品會增長,此協議是去年十二月 20 【特訊)综合外電報

澳門經濟

美國

行正式簽約儀式。

擴展 美

河普鍵徑設微

二期星

健仍擴經

全然展

處。拒態上較,產生車汽美句下月九

5. 減略期同年無

健全校仍有一年光錢意根感展,

582-4*K (二編雖然失業人數近3

則有九

有百分之一定點八,以此推 九路 百八十五馁,較上年減少 一是句只有八發業5而上年8

「六萬快,爲一九八年八

• ABHALLAKEK+Z 丹以被提供生產华堂,但

同 道:一一》九月下何美國, Capacitidi

同時通貨股仍然偏向

RØDHEN 12 将棋向多元化。母公司設於新加坡,爲東南亞 尚上市公司,經營油脂,食品及清潔劑用品,業

〔一九八四华十月八日)

香港金融行情表

其市場推廣及管理有所貢献。南帕公司是本港15 陳君加入南和公司必能對

香港南順公司委任

九九金衙司对〔阳市(一九四

圖;陳福深照: 南酪香港有限公司日

陳福深商務經理

不致暴跌爲準,否則上罕勢將上場。 ↓,以倍形而論,服息可能性不大,但仍須以美元 再度微減,妨做經濟發臊最重要减免乃息率上升, 可見,雖不及八四年上半年興盛,仍可使失業率 皆,因而一九八五年仍能有百分之三,四成長率 健全,不同之處只是游費需求已改由基未開支代 搔濟陷入衰退。一般相信枧濟掏展主要動力仍然

·公司總經運或執行董事。

驗,在未加入南阳公司

前委任陳福深先生塔商務

營業、市場推廣及廣告經

+陳君飛任本港數人液告事

經理。陳君具有十七年的8

35888

(se) 1700

(220121

(24)79100

#KO TO

KOLO

1-00

v,2 = 2(g)(h)

8

b=2(10)

0.5 m

Ans: The maximum height

reached by the 10 kg block is 0.4+2-2,4 m above the floor.

(a) Let the velocity of

the pile driver as it strikes the pile

be v ms

ï

1

By conservation of energy,

1

mgh=nv, 2

where mass of the

pile-driver: -10 kg

and h height fallen

=12 m.

=2gb

-2(10)(12) =240

v.-15.492 ms

(b) Let the commou

velocity required be

-1

V MS

Initial momentum the system =(10)(v,) kgma

Find momentum of the system

=(10+12)v kgms

By conservation o linear momentum, 10v -(10+12}v_

10

10

(15.492)

=7.042 ms

(c) Let the average

retarding force exerted by the ground on the pile be F newtons

Hence,

Work done against F

-loss in kinetic

energy

+ loss in

potential energy

Work done against F

=F(0,5)

Loss in kinetic

energy

(10+12)v

(22)(7.042)2

=545.5 J

Loss in potential energy

=(10+12) (10) (0.5)

=110 J

F(0.5) (545.5+110)

F=1311. (Ans.)

ams

30kg 20kg! Toky Smooth horizon

As shown in the fig above, let the tensions in the

strings be T1 and T2,

the acceleration of the system be a. Apply Newton's 2nd law to the blocks separately,

600-T-1 10a...(1) =20a... - (2). T2-30a....(3)

(1)+(2)+(3):. 600 (10+20+30) a

-2

a-10 ms (Ang.)

(ii)Substitute

8

a 10. into (1) and (2), we have T=600-10(10)

1

500 N T#30(10)

=300.- N

am

(Ans.)

(Ans.)

600N

fa

As shown in the fig. above, f, fg and f are the frictional forces. on the 10 kg, 20 kg and 30 kg blocks respectively. f=20 N

f2-10 N £=60

Let the acceleration of the system be a', the tensions in the strings be T and

1.

Apply Newton's 2nd law to the blocks. separately,

600-T ' -20=19a'..(1)

T' ̧ -T'¿¬40=2Ua'..(2)

1

T

2-60-30a...

(1)+(2)+(3):

600-120-60à·!.

a'8.m:

(ii)Substitute a

(3)

(Ans.)

8 into.

T500 N

Substitute

(3)

300 N

(Ans.)

(a) VR-4

M.A.

V.R.

2=40%

MA÷x 100%-40%

M.A. 16

(c)

(Ans.)

Let the minimum

=(1000)(g) (v•)

100(10)=9 V-9x10-4

Since the volume the cube

=(0.1)?

-1x10-3

m

fraction of the

cube immersed

9x10

1x10

of

(Ana.)

(b) As shown in the

figure below

below

0.1m.

water "(100 kgm3)

Let hadepth of oil

A cross-sectional

area of the cube

(=0,01 sq.m.). Upthrust due to oil on the cube

=(720){g}{hA)

7200hA

Upthrust due to water on the cube =1000(g)(0.1-h)A =10,000(0.1–h) a Total upthrust

exerted on the cube =7200+10,000(0.1-h)A

=10004-2800hA.. =(1000-2800h)A

Weight of the cube -900(0.1)(A)(g). -

=900A*

Total upthrust weight of the cùbe

(1000–200H)A

h=0.0557 m

the cube rises

0.0357 m-1 cm

-0.0357 -0.01 m

=0.0257x

日九月十年四八九一曆公年三十七國民中 育敎濟經 ̇

澳門與挪威 瑞成语造

署一般紡織

the indicator

after 1 hour.

solution

簽紡織協議 适的協議,Jo Liolog

會,計論世界鍚似問題。 他和週五在拉巴斯以東三百二十哩的克斯

造成的對泄耗減少和錫消費的結構性改變。 過剩和美棠政府雖售庫存與使湯補湯業深受影晌 文件還說,對生產國而長期世界經濟所 會議結東時,與會者在一項文件中說,生產 澳門與

議决家國產生

施實續錫

額配口

罩

#WEIGHT

濟

錫出 插電電量

Positions

a) What was the purpose of hanging the glass tube at different positions?

b) What indicator

solution was used in this experiment? Suggest two environ-

oultions

of during the experiment,

be aware.

d) Explain the results.

recorded at positions (1)x, (ii) and (iii)Z. 10. The diagram below.

shows the arrangement of teeth on the human upper jaw.

A,B,C and D are the four types of teeth in man.

Is this the dental pattern of a child an adult? Give a reason for your

answer.

b) What is the dental

formula for this

c) pattern of teeth? (1) Indicate by letter the tooth type

which is reduced in sheep. (ii)Why is not needed in

sheep?

d) What is the importance

of the action of teeth, on food in the process of digestion?.

e) Explain how the

following methods can prevent tooth decay. To have a balanced diet.

(i) (ans.)

生物

(四)

(ii) To brush the teeth

after each meal,

effort required be T.KU 202##*(i)To use dental

Load

=1.6

E

Load=50 g

500

-500 N.

1.6.

. E. F=312.5 N (Ans.)

(d) Let the tension in

8

the string connected to the 50 kg block be 7 newtons

T

1.6 effort

when effort=600 N

T 600

T-960 N

Hence, the net force acting on the 50 kg block

=960-50 g

=960–500

460 NA

Apply Newton's 2nd law, 460=50(a)

a9,2 ms

-2

(Ans.) Ans: The block will

move upwards

with accelera- tion 9.2 ms-2

Upthrust of water on the cube

weight of the cube 3

-900(0.1)

=900(0.1) 3 (10)

9 N

(Ang.)

(ii)Let the volume of the

cube immersed in

water be V'.

Upthrust on the cube

BIOLOGY (4)

F. LEUNG

MILL & DALE: PRESS LTD.

Unit 1: Nutrition

9. A student designed an

experiment to investigate the effect of light on the

disclosing agent.

11. After a green plant

had been placed in total darkness for two days, the leaves were picked and treated at 20°C as shown in the diagram. below.

In light

water

release of carbon dioxide by green leaf. The apparatus is shown below.

Leaf A

In darkness

Water

Leaf B

-In darkness:

Light

source

Wire

·Cork -Green leaf.

·Glass Lube ∙Water

·Indicator Solution

He hung the glass tube at different positions before the light. source for one hour. cach time and recorded the results as shown in the following table.

(The colour of indicator solution is orange in air.)

Y

red orange yellow

glucose Leaf C Solution

利亞 - 玻利維亞 -

並且說,美國不應 尼和馬來西亞的礦業

政到湖個穩定爲止, 繼楼實灏出口配額i *E-EX,

協會閉幕會議中决定

八波料能亞克巴

·部長和其他官員,

C.

in which water was a product.

Would leaf C contain

starch if the dish was kept in light? Answers:

9.

To change the

intensity of light. received by leaf,

h) Bicarbonate indicator

solution.

(1) The temperature

should be kept. constant.

(ii)Light from other

sources is not allowed.

Under strong light.

the photosynthesis

occurred to a

greater extent than respiration. The decreased carbon dioxide

concentration turned the solution red. (ii) Under medium light the photosynthetic rate and respiratory rate are balanced, The unchanged carbon dioxide M concentration caused the solution to remsin orange. (iii)Under dim light the photosynthesis

occurred to a

smaller extent than respiration. The increased

dioxide

carbon

concentration turned the solution yellow.

10. a) This is the dental

pattern of an adult because of the presence of 6 molars (D) on the upper jaw.

2 b) 15, c, pq,

(i) B

(ii)Canines are used to

pieroe prey and for. tearing flesh. Sheep is a herbivore and.

therefore need not have canine teeth, d) The food can be broken down into tiny pieces: so that the surface. area for the actions of digestive enzymes can be much increased.

The resistance of teeth to bacterial decay can be increased,

(ii) The food residue

lodged between the. teeth can be removed.

(iii)Dental plaque caused

the

11.

After a few hours leaves were tested for starch. The results showed that starch was present in leaves A and C but absent in B.

a) What is the purpose of

placing the plant in darkness. for 2 days before the experiment?. b) State a major force

that caused the water and glucose solution: to rise and enter the leaves.

Explain why starch was present in leaf A.

d) What conclusions can be drawn from the results for

(i) leaf A and leaf B (ii)leaf B and leaf. C e) State two processes

that occurred in leaf

by poor tooth

brushing habits can

be revealed.

a) To destarch the leaves

so that distinct results can be obtained if starch is formed. A

b) The transpiration pull

of the leaves.

c) Photosynthesis had

taken place in leaf A.

d)

(1) Light is needed for

the formation of starch.

(ii)Glucose can be

converted into starch by leaf.

The conversion does not need light.

Respiration. i)Condensation of

glucose to starch Yes.

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