1973-01-18 — Page 16

報日僑華

日八十月一年三七九一群公年二十六國民華中宵救備華

0.17.

四期

WAH KIU YAT PO

日五十月二十年于壬歷鉴

莫四第張四第

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「初級現代溫稗」及「自由

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1973英文中學會考試題預習專欄

WAA* MM**

生物科(十三)

"Answers

week

Ans.

Biology (13)

the questions of last!

12 13.

14. B

45% 0

6. E

16. B

17. A

8.

18. D

19, 0 120.0

9. A 10.07

II. Ans.**

(a) Osmosis is the process of motion of water molecules through a semi-permeable membrane from a pure solvent or a dilute solution into a higher concentrated solut-

on. The mixing condition continu- es until the solutions between the Semi-permeable membrane have the same concentration. Then the dif- fusion of water ceases

A comparision of Asmosis and dif- fusion:

(2)

Osmosis

06curs be tween two

solution of different cono entrat- ion.

(11) results in the format

ion of osmotic pressure,

(iii) Only the

Diffusion

happens, be- tween gases, solutes or solute with) solvent. of different con- centrations. junly a physical process withi ino pressûre.

formed.

The molecules water mols of the solute cules pass or solvents through the pass through▾ semi-permeable the membrane membrane from if present,- low concentrat from higher -ion to high concentration concentration. to low:con- centration., takes place takes place. mostly on mostly on plant cells intracellular and root hairs and extracell of the plants. -ular body.

fluids 01/ animals. (b) Object: To show that osmosis takes place in living tisbues of & potato,

Hv

Procedures Two halves of a potato with the base peeled off were made to stand in a dish of distilled water. A cavity was scooped out in each half of the potato. Into the cavity

801 of one concentrated sugary

was put. The other one was first boiled to kill the living tissues, then an equal volume of the sugary solution was also poured into

the cavity.

At the beginning of experiment.

Distilled

At the end of experiment,

TII. Ans.

(a) Irritability is the power of response to the external stimuli

in the surrounding environment “ found in living organisms

In plants, they respond ti light, water, gravitation and Plysical contact in the form of tropisms.

(b) Planta respond to common exter -nal stimuli in the form of tropic movemtns as they grow unequally on the parts of the plant body. The bending curvatures in plant growth are due to unequal growth. rates on different sides of the plant body and the result of un- equal distribution of auxins, For example:

(1) In geotropic response--- the main roots of the plant are positively geotropic; they grow downwards towards the centre of the earth while stems grow away from the centre of the earth and are said to be negatively go- 10.

(2) In phototropic response-- The Shoot is positively phototrop ic i.e. it grows towards the source of light. Roots are negat- ively phototropic and grow away from the direction of light ill-

umination

(3) In hydrotropic response-

roots are stimulated by the water or soil moisture, and grow towards the source of water or moisture. (4) In contact response

Roots grow away from solid objects such as stones, while stems tend to make contact with objects. This 18 remarkably seen in climbing planta such as pea and vine. There -fore roots are negatively triguo- tropic and stems are positively thigmotropic

(c) Object:

To show the positive- ly phototropic in the shoots of broad bean seedlings.

Procedure: A pot of normal grow Lng broad bean seedlings was put into a light-proof box which has a small hole at one side. The seed

lings were allowed to grow for few days. Y

Observation: After a few days, the shoots of the seedlings curved towards the direction of light illumination. The shoots of plant. are therefore positively phototro pic...

Light-proof box

Light

Ikeminat

Shoot of the Seedling

At the end of experiment. over Shoots show positively photo-

Result: The level of the sugary solution in the living potato was found to rise up, finally the sugary solution had overflown. No change of the level of sugary solution had been detected in the cavity of the

e other. Conclusion: From the above obser vations we know that the tissues. of the living potato had been acting as a semi-permeable membrare which allowed water in the dish to pass into the sugary solution as a result of osmosis. No change was shown in the other half of potato because the living tissues of the potato were killed by boiling, and the tissues were no longer semi permeable, thus the process of Osmosis could not take place.

Raw potato

Sugary solution Boiled

Tropic

Questions for this week?

(a) What is photosynthesis? State the environmental condit- ons necessary for it to occur. (b). How would you prove that chlorophyll(green pigment) is essential for photosynthesis? (a) What do you understand by aerobic and anaerobic respirat- ion?

(b) By what experiment would you show that heat in generated during the process of respirat¬ 1on?,

1978

中文中學會考試題預習專欄

數學科(十三),張正邦•

第十二次預習題解答

求下列各題三角形的面積,

eke b=24:36 C=18:34 A=36948. 解

== b C Sm A

±x 24 36 × 18.54 Sm36 18 12.18 x 18.54 × 0.592 133 681

a=15. 6=14_C=13] s==(a+b+c)

s == (15+14+13)=21

√s(s-a)(s=b)(3−c)

212(21-15)(21−14) (21-13)

21 × 6 × 7× 8 - 84

知R∠ABC的外楼田半径

EN HE R= abc.

= RMA

Ra

26c Sma

abc

44

4K Sm 7521

(*

abc

1x=bcsmA

a==bcSmA)

Sm 75° = Sun (45° +30°)

·Sus

5. Paco 105°è Of G105 =

Cobo Go 45.

3. Sm 78°32′ ( 11° 28′4′′ Smll 28′′

XC 78°32′ =1

at Sm 18 32 Goll 28 +Sm 11 28 0018 32

Sm (7832 +1128). Sm 90°-1

7 Gs (A-30°)== (BCDA+SMA) BE Go (A)-30°) == Csf (1030 | Smpsnos

√ =

8. Zko SmA = GB = √= &

A&B ALB‡. Sm (A+B) & CO (1-B) ≥ 1.

ARB+ Si A=60°, & COOB=√3 = B=45+

Sm (A+B) = Sm (60° +45°)

Su360 Cos 45 + as 60 Si

as (60-45°)

Cos (A-8)= as co

例

tod vt log b loga & loga d

解

*

ich loquib

NART

JB2 Jed

loja d

- loga b = -lgac*

- 3+ logab - = log b

- loga - loga d'

#

-=-=logab-44-load

tol = BEBA 2 log (b+ Jbz1)

BE log

= loja (b+ √B=1)=2&opa(b+\B=1)

求

x

10:0476

/ 5 × 59 × 0.0088

X

94.62 X-√0.0476 x 5/522 3/2019:×0.0088

之值

og x== log 0.0476 + = log 222

+= (coloy 5059 + Cologs soff)

·T. 8111 +0.4693+7.45049

·T.730.89.

/5059 x 0.0088.

14x=0.5381

0638|3

Hot MB log X + log (x+3)=1 B. log X + log (x+3)-1

log (x(x+3)] = log 12

+3x-10=0

X(x+3)=10. (x+5)(x-2)=C 49 X=2h-5.

$4 213*=516-** 213 511-x+4

akak en 213 = (

X=(-x+4) lo%2 516

-X+4

516 (13.

3041X= 10, 8504,

7126 2,3284

tol*. 3G log (1-2x)3—loj (3+x) =6. Bq. log (1-2x)3- log (3-x)3- 6 3 log (1-2x)-3 boy (3-x)=6

log (1-2x)= log(3-x)=2)

leg 1-2x = log 100

14 X-3.051

Go 60° Go 45 + Sm 60 5m 45.

第十三次預習題,

Ex Int=x I

對數如一般通性

(1) loga mm - logam + logan

E logam

:(3) loga mr =

loga! rlogam

logan

4) loga I'm = logam

ད་

x42=x (0.01)

Co.mat)fx59683-

4. 解方程 13-

=14

lg xx-21+x=

5.