1983-11-10 — Page 23

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頁三第張六第日六初月十年亥癸曆夏

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1984 中學會考試題預習專欄

物理 (A)

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PHYSICS (8)

H.K. LO

MILE & DALE PHESS LTD.

Answers to Exercise 4

1. Form the given graph:

(a)In stage V

the

track takes the longest time, it takes 40 seconds.

(b)Stage IV and stage V

show that the truck travel the greatest distance (400 m). (c)Stage I and stage II

show that the truck travel in uniform speed.

Stage IV obtains

the grestest speed

(20 mg-

(14) Since the slope or

the graph represents.

the acceleration, therefore stage III shows the struck travels with

greatest accele- ration for its slope. is the greatest.

(iii)Stage IV shows that

truck obtains the greatest kinetic energy for its

speed is the

greatest in this stage.

(e)Total distance

travelled

=total area under the

graph

=-(10) (30)+(10)(20)+

(10+20) (10)

~(20)(20)+3(20.) (40).

-1300 m

(Ans.)

As shown in the fig.,

the angle between the 50 cm aru and the

vertical 190 when the piece is in equilibrium.

Let

d-density of the

metal rod in kgm

A cross-sectional

area of the rud in

2

Weight of the 30 cm

arm AC

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

-0.3dAg

Weight of the 50 cm

arm BC.

»afo.5)(A)(g)

-0.5dAgra.

Moment of the weights

of AC and BC about the vertical line

through P are eugal in magnitude.

H

0.3dAg(G2H

1H)=0. 5dAg(Go

(where f and

Go are

the centres of gravity

of the 30 cm arm and

50 cm arm

respectively)

30 H=5G,K

G1H=ACsine

-0.15sine

G&K=3BCcos@

-0.25cose

3(0.15sine)="

5(0.25cos◊)

sine 5(0.25)4

cose 3(0.15)

25

tang -2.7778

0=70 12

(Ans.)

Let the tension in the string be 1, weight of the block be W, volume of the block be Vand the upthrust on the

block in water be

中央

(i) In fig.

T-W

=8

In fig.

B =T+W

2+8 10 N B=(P){g}(v)

(Ang.)

where P=density

郭日僑華

在交通報告。 今年代,卅分,新侂簡報】第一線」,卅分,新聞五時,新閎簡報

交通報告」繼續「八十」交通報告,繼續「畏光。

十時,新聞簡報-

SEX RECHE **•

「賽作品選」,卅分,新卅分,新習簡報,交通

經濟行報導,城市故[交通報告,{ A - BE

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;經濟行情報導,繼接AU匦躪彄」,卅分,

十一時,新聞簡報 繼續「A L B E A T

一時,新出報告, ja

(b) Let the mass

liquired be M

Volume of metal required

M

8000

Total volume of the

combined body

wh. B.

四期星

推

日十月一十年三八九一曆公年二十七國民華中

國際金融股市行情

一九八三年十一月九日(當地時間) 倫敦股市《金融時報指數)

$ 0,20 (719.90) ,五種主要港股被(買盤及資

匯雙

H-K.$ 7.15 7.35

今樂府

「香港第一台

中古 香港第二台「機一鍵悠夜空中」,

「發光第一線」,卅分至5時,新聞商報,

大時,新聞簡報, 同

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七時,新聞簡報

六時,而且與第二

四及IC三兆赫

E

*七時,新問天地,

八華南海域天氣報告)

八時,交通報告,

「磴濱快訊」,一八十

冊分,新聞簡報,「英 繼續「是光第一線」,維籟「輕談淺唱不夜天

「年代」,卅分,新餓麼

張疑難」,繼續「光

報,交班報告。一

九時,新聞簡報,

八時,新簡訊,一般「輕談淺唱不箋」

Plastic beads

suspended in water

An electric pump Postic

Tiny pin

& bag

Hip X

Water from the pian pricks

8000+10

(in m

Mass of block A

-0.8 kg

Total mass, of the

where

pricks on the rubber

combined body

T

tension in the cord

tube

-Funnel

weight of the

copper block.

0.089 g

(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=

M 8000

M+0.8

2286 kg.

(Ans

As shown in the fig. pressure at A(P)

atmospheric pressui -760 mmllg

P-760 mmilg.

Let the pressure at

B be PR

of water

1000 kgm

A

(Ans.)

10-1000(10)(v)

V=1x10-3

(iii)Let the density

of block A be f

(p)(g)(v)=w

(/^-)(10)(10 ̃3)=8

800 kgm

Relative density of

block

·A·

800 1000 =0.8

(Ans

V_~(40–10)A = 30A

where is the cross- sectional area of the tube

Apply Boyles law

PVP_V

AMA

•760(40A)=P. (304)

A mulig

But

5040.

weight of the

iron block

Bouyant for force

exerted on the

copper block

Bouvant force

exerted on the

iron block

Consider the cop block: TW

B

According to the Archimedes' Principle, Rouyant force =

+Water

Which organ in the body does the electric pump represent?

(ii) (1)What structural

Water

feature of the arterioles in a nephron is being. demonstrated by the use of clip X2:

(2)What is the

importance of

this feature?

weight (iii)(1) Why do the

of equal volume of liquid displaced.

Volume of copper block mass of copper block

density of copper 0.0894. T

879d

where d is

of water

20.089 8.9d

=0.01. g

≈0.089 g.

plastic beads

not pass through the pin pricks?" (2)State two

substances represented by the plastic beads.

density (iv) Explain why the

0.01 g

=0.079 g.....(1). Similarly, consider the iron block T-W

2

Let m be the mass of the iron block Volume of the iron block

"m' 7.80

(iv) Let the upthrust of

block A in liquid

From fig. (c)

0.5+8 =8

pressure at B

atmospheric pressure +pressure due to

liquid column

750 mmlig + pressure. due to liquid columa

-Pressure due to

liquid column

mg

=mg-;Z;

-(1-78)mg

-0.8718. (2)

By equation (1) and

(2)

0.079g 0.8718mg

x be By.

T+B

3040

760.

Ans.)

760

mfig

(v) Let the density of

liquid X be P

7.5-(.)(10)(103)

=750 kg

Relative density

liquid X

750

1000

(Ans.)

=(13600) (10) (76°) Nm

On the other hand, pressure due to liquid column *(1200)(10)(b)

.(15600)(10)(260)

=(1200)(10)(h)

-13600x760°

3x1200 -2871.11 m

Depth of the lake =2871.11 m+10cm

-2871.11m+0,1m

22871.21 m

A force diagram constructed as following:

(Ans.)

生物

0.079-

0.8718

0.0906 kg

CAD

明德出版社梁偉宏 提供資料|

BIOLOGY (8)

W..

LEUNG

MILL & DALE PRESS LTD

Unit 3: Water and Organisms

5. The diagram below

represents a model which is used to demonstrate the process of

ultrafiltration in bumin nephron,

funnel cannot truly represent the

kidney tubule.

(v) (1)Name the fluid represented by the water draining out from the pin pricka.. (2)Give two

substances which are present in this fluid but absent in urine. (3)Explain their

absence in urine. The following graph shows the data obtained when six potato cores of equal weight were placed in six different concentrations salt solution. +4

0.5

1.5

1⁄4 concentration of salt 50.

hat caused the

potato cores to

have different

weight in different

concentrations of

salt solution?

(ii) What is the

concentration of

the cell sap of the potato cells?

Explain your answer (iii)What was happening

to the cells immersed in the salt solution

(1)at concentration

(2) at concentration

Y?

(3)at concentration

(iv) (1)Name the

phenomenon observed in the cell immersed in

THI

和醇

·希德豐

H.K. 11.10 11.40

H.K.$ 12.70 12.90

3.15

H-K-6 3.00 HKS 2.75 2.90

·紐約股市C這個斯工業指數>

+ 2.131217.07 東京股市C逍續斯工業指數) 2-22.16 (9297.10)

倫敦金市(每盎斯)

U.S.1.65 (380,60)

Jet US 0.75 (383,00)

美冗開市價格:法蘭克聯 2.6835 馬克

倫敦

東京 巴黎

2.4825 美元

236.20日元 M.

8.1460

the salt solution at concentration

(2)Draw a labelled

diagram to show how the cell. would appear under a microscope.

Ansvers

(1): Heart.

(1) The greater diameter

of the aferent arteriole than the efferent arteriole is being demonstrated.

(2) The feature leads to

a high blood pressure inside the

(iii)

glomerulus, which forces water and dissolved substances through the capillary wall into the Bowman's capsule.

(1) They are too large

to pass through the pin pricks..

(2) Plasma proteins and

red blood cells. (iv)The kidneu tubule

has convolutions, sem) permeability and ability of reabsorption, while the funnel has not any.

(1) Glomerular filtrate. (2) Glucose and amino

acids.

(3) They have been

tubules

reabsorbed by the wall of the into the surrounding capillaries.

The gain or loss of water in potato cells.

(ii) 0.85% concentration

of salt solution, At this concentration, the potato cores did not gain or lose. water by osmosis. This means that the concentration of

the cell sap of the potato cells is equal to that of the surrounding-

solution.

(ii))A₺ X: the cells

became turgid

because they gained water from the surrounding hypotonic. solution by

བན

(1) Plusnic) (2)

remained no change because they did not

gain or lose water in an

isotoni 91

solution. the cells became plasmolysed because they lost water in the

hypertonic solution."

Fell Wall

·Placna membrane

Cytoplasm