1982-09-29 — Page 24

頁四第六第日三十月八年戌壬醫夏

報日僑。

WAH KIU

YAT PO-

三期星

日九廿月九年二八九一艇公年一十七國民中

育教僑港

1983

中學會考試題預習專欄

Taking moment about F (AC)

100g(AB)

Substitute into (1)

8kg.

ladder is

58

Normal reaction

1708.8N making

489.9N

block 30N

H.K.LO

物理

明德出版社盛國雄提供資料(b)

PHYSICS (2)

MILL & DALE PRESS LTD.

(Ans.)

(b)T 64N

True weight of the

Substitute into (1) 64-8(10) * 88

acting on the

20033

with the

(Ans.).

vertical

(1)As shown in the force

diagram below

(i)If the force is

applied to the 10kg block,

2m

(Ans.).

be the

Suggested solutions to

Exercise one

(Take g = 10 ms

F(OB)

100g(AB)

.༠•.༤

As shown in the force

diagram above, take

moment about A....

where F is the

horizontal force

required.

OA = fx100m

OB * 5-4 sim

OA -OB

24

OB

4.899 m

#4899N

(ADS.)

10kg

where Ry is the force between the blockad the reaction of 5kg block on 10kg block

Let the acceleration of the system be

a s

B1

10a

5a

(1) (2)

also.

F15N

solving R from and (2) we liave

R 5N

(1)

(ADS.)

(ii)Similarly, let the

force between the blocks be Rg and the acceleration of the system be a

15-Rg

xxx

5a!

Rg 10a

from which

ዩ።

-2

ADS. The elevator moves

downwards with acceleration 2ms-2 (Or the elevator has an upward decelerat ion 2ms-2)

(c)If the elevator moves upwards with uniform: speed, then a

From

0 + 3 = L

mg

80N

*

(Ane.)

Hence, the reading of the balance is 80N.T

(d)If the elevator moves

downwards with uniform speed, hence, we have also

A

0*

T80N

upper end from the smooth wall the resultant force acting on the lower end of the ladder. the vertical component of R. the horizontal Component of R. frictional force between the ladder and the floor.

weight of the

ladder.

(ii)Since the ladder

in equilibrium,

therefore,

•N1-f

N2-W - 0

0

(1)

(2)

Taking moment about B Wx3-N1x8 3W8N1 From (3)

3x100(10)

(Ans.)

N1

=600N

Therefore the reading of the spring balance is 80N.

(e)If the cable wire

breaks, then the elevator will fall

10N

(Ane.)

down with acceleration equal to the gravitat ional acceleration g.

Hence

mg

-mg

0(10)(4.899)

(ii)The minimum force

F should tangent

min

to the wheel as shown in the diagram below.

min

100% is tangent to

wheel at point C

which is one of the

extremes of diameter AOC of the circular wheel.

1983

中學會考試題預習專欄

生

物

明德出版社梁永華 提供資料

BIOLOGY (2)

W.W. LEUNG

Take the upward direction as positive. Let the tension in the spring of the spring balance be T, the acceleration of the elevator be a and the mass of the block be

weight of the b

T-ng ma

(a)when a=2,5ms

T 100N-

MILE & DALE PRESS LTD.

Unit 1: Nutrition

3. After a green plant

has been placed in

dark for 2 days,

leaves are picked and

for 2 hours

at 20°C. as in the experiment illustrated. belov.

Leaf A

Water

kept in light

Leaf B

kept in dark

Leaf

(1)

5% glucose solution

kept in dark

At the end of 24 hours the leaves are tested for starch. The results show that starch is present in leaves A and C but a absent in B.

(a)What

is the purpose of placing the plant in dark for 2 days before the experiment? (b)State

a major force that causes the water and glucose solution to rise in the and enter the mesophyll of the

leaves.

tiole

(c)Explain why atarch

present in leaf A. (d)What conslusiona

be drawn from the results for

can

(1) leaf A and leaf B (2) leaf B and leaf C? (e)Further experiments

show that leaf C would not produce starch in the absence of oxygen. How would you explain

this observation

The diagram below, showa a transverse section through the blade of a dicotyledonous leaf.

3

Therefore, the reading of the spring balance will then be ON.

(1)The force diagram is

shown below:

(a)(1) Name the structure

(2) What would happen

if part A is missing?

(b)(1) What is the most.

significant difference between. upper epidermal cells and mesophyll cells? (2) How does this

feature of upper epidermis contribute to plotosynthetic efficiency?.

(c) What is the advantage, for the palisade. cells to arrange at right angles to the leaf surface? leaf surfac

(d) (1) Name the

structure B and

C

(2) What are their respective functions? (3) How does the

distribution of B and C contribute to photosynthetic efficiency?⠀

(e) State two structural

differences between the cells labelled D and E.

(Ana.)

Ans. The direction of the force acting

on the upper end of the ladder is perpendicular to the vertical wall with magnitude 600N.

(iii)The frictional force

fis horizontal.

From (1).

N1 = 600N (Aas. ) (iv)The_ resultant R

(1)Let v` and

velocities of the

block A after the

bullet emerges and the bullet as it emerges from block A respectively.

By conservation of. energy

1(2)

2 ms

(2)(c)(0,2)

(Ang.)

(ii)By conservation of

linear momentum

The initial momentum

(0.01)(2400) 24 kgms-1

The final momentum

A

(2) v +(0,01) (v1). (2)(2)+0.011

4+0.011

24-440.01v1

- 2000 ms

(Ans

(b)Let v be the velocity

of the block B and the bullet after the bullet hit the block

B.

Since the bullet is. embedded in block B, hence, by conservation of momentum. (0.01)(2000).

(0.01 +0.99}v

20m87

√6002

1600

3.8N

(Ans.)

-0.375

By conservation of energy

20°33 (Ang)

Ans, The resultant at

the end of the

(f) How does the presence of stomata in the lower rather thair upper epidermis contribute to photosynthesis?

(g) (1) What is F?

(2) Give a function

ANSWERS:

of it.

(a) The purpose is to

destarch the leaves so that distinct results can be obtained if starch is formed during the experiment.

(b) The transpiration

pull of the leaves. (c) Photosynthesis has

taken place in leaf

(d) (1) Light is required

for photosyn thesis.

(2) Leaf can convert.

glucose into starch.

The conversion

does not require light,

(e) It seems that energy

is required for the conversion of glucose. into starch. Leaf C could not produce starch in the absence of oxygen because no respiration occured to produce the energy energy

(a) (1) Cuticle.

(2) If cuticle is

absent, the leaf would be subjected to fungal and bacteria

(0.01+0.99)(20)2 (0.0140.00)gh

20m

infection,

mechanical.

(Ans.)

damage, and excessive loss

of water by evaporation..

(b) (1) The absence of

chloroplasta in the upper epidermis.

(2) Sun light may

pass through the upper epidermis into the mesophyll cells for photo- synthesis.

(c) This feature permite

light to penetrate deep into the

photosynthetic tissue without passing through very many cell walls; so

avoiding loss of light by absorption and reflection.

(d) (1) B: xylem

C: Phloen (2) B: 1. For the

transporta- tion of

water and minerals from root to leaf.

2. For the

mechanical support for the softer leaf tissues."

C: For the

transportation of food substances from where they are manufactured

in the leaf to growing points and food stored.

areas.

The bouyant force B

at the mid- the part of the

rod submerged in water.

(ii)W Weight of the rod

1,2g 12N

weight at the end

of the rod to be determined.

The volume of the rod.

1.2 0.5xd

(where d is the

density of water)

Volume of rod. submerged in vater

The bouyant force weight of water displaced by the rød

(density of water

(volume of the rod

submerged)xg

dx x10

201

(Ans.)

(iii)Taking moment about

the hinge (Baine)(2.5). (Wsinë)(3)+(waino) +(weine)(6) 2.5x20-12x3+6w

v-2.333N ('Ans.)

(3) Xylem vessels are closer to palisade cella than phloem sieve tubes go that water be easily absorbed by -palisade cells:

for photosyn="25 thesis.

(e) The structural

can

differences between the cells D and E are

as follows:

Cell D

Cell E

absent

present

Thickness of

uneven

chloroplasts

cell wall

(f) The presence of

stomata in the lower rather than appër epidermis enables the plant to lose leas water by evaporation when undergoing photosyn- thesis.

(g) (1) F; Air space/

Inter-cellular

space

(2) To facilitate

gaseoua exchange in mesophy11.