1983-10-11 — Page 24

頁四第張六第日六初月九年亥癸膰夏

WAH KIU YAT PO

報日僑

·二期星

1984

中學會考試題預習專欄

Ans The maximum reight

reached by the 10 kg block is 0.4+2-2.4 above the floor.

ams*

- f

Ans.)

(b) As shown in the

figure below.

fr

物理

·f3.

(B)

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ail: (<720 kgm3) water (100kg m

PHYSICS (4)

H.K. LO

MILI & DALE PRESS. LTD.

Answers to Exercise

-(a)

ams"

15kg

Joka

As shown in the

figure above,

・ams"2

Let T-tension in the

string,

a acceleration.

of the system. Apply Newton's 2nd law to the block

separately,

15(10)-x-150(2)

(1)+(2): 50=25a

(2)

mg

(Ans.)

a-2 into

T-10(10)=10(2)

T=120 N(Ans.

(c) Apply the formula

2

v=u2às

Let v be the

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

initial velocity is

日

02_2(a)(2)

2(2)(2)

#8

-2,8284 ms

(d) Let the time required

for the 15 kg block

to reach the floor

be t (seconds).

Apply the formula

0

inital vel

0 me

t

Substitute s-

2=1 (2)t2

t=√2

Let the velocity of thepile driver as·

it strikes the pile

ms-1 be v

By conservation of energy

nghi

where m-mass of the pile-driver

10 kg:

height fallen 12 m

=2(10)(12).

#240-

V1=15.492 ms

(b) Let the common

velocity required be

V ma

1

Initial momentum of the system =(10)(v) kgms

Final momentum the system =(10+12)v kgms

By conservation linear momentum, 10v, =(10+1:!)v

10

39(15.492)

*7.042 ms.

(c) Let the average

retarding force exerted by the gre on the pile he F. newtons. Hence,

Work done against F -loss in kinetic

energy loss in energy

potenti, fas

Work done against F -F(0.5) -

Loss in kinetic

energy

(10+12)2

(22)(7.042)

-545.5 J

Loss in potential energy

=(10+12) (10) (0-5)

=110 J

F(0.21317 +110)

1.4142 s

30kg

15 kg 10 kg

As shown in the fig. above, let the velocity of the 10 kg block be when it

reaches the level at

̧2 = 2(a)(2)

=2(2)(2)

=8

Hence, the string.

slackens, assume the

block moves bmeters

farther before drops down.

22=2(g)(h)

=2(10)

ams'

20kg

(Ans.)

6001

roką

both horizontal plane

As shown in the fig.

above, let the tensions in the strings be T, and

"the acceleration of

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

600-T-10a

1

• (1)

(2)

-T1=20s.

To÷30a... (3).

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

a=10 ms (Ang. (ii)Substitute a=10

into (1) and (2),

we have

T=600–10(10)

(Ang.)

1

-500 'N

=30(10)

-300 N

(Ans.)

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As shown in the fig and f

above frictional.

are the

forces on the 10 kg, 20 kg and 30 kg blocks respectively f=20 N fo=10_N F-60 N

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

1

T

Apply Newton's 2nd law to the blocks

separately, 600-T1-20=10a |

1

-40=20a

T2-60=30a.

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

600-120-60a".

a8 ms

(2)

(3).

(11)Substitute a'=8 into.

(1),

Leth depth of oil.

A cross-sectional

area of the cube

.01 sq.m.)

Upthrust due to oil on the "cube ={720)(g) (13). #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 =7200A+19,000(0.1–b)A

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

Weight of the cube =900(0.1)(A)(g) =900A BAR

upthrust

ht of the

(1000–2800h)A

h=0.0357 m.

the cube rises

0.0357 -1 cm

0.0357 -0,01 m

=0.0257 m

生物

cube

T'=500 N

(Ans.)

Substitute a=8 into

=900A

(3)

་: 300 N

(Ans.)

(a) V.n.-4 (b)

(Ans.)

M.A: V.R.

M

×100%-40%

x100%-40%

M.A.-16

(Ans.)

effort required be F. Load

E·

load=50g

500

-500 N

1.6

E-312.5 N.

(c) Let the minimum

(Ans.)

(d) Let the tension in

the string conuected to the 50 kg block be T newtons.

T effort when

1.6

effort=600 N

1.6

600 .

T=960 N

Hence, the net force

acting on the 50 kg block

=960-50 g

-960-500/

-460 N

Apply Newton's 2nd. law,

460=50(a)

a-9.2 ms

(ans.)

Ans The block will

move upwards

with accelera- tion 9.2 ms-2.

日一十月十年三八九一圈公年二十七國民華中育教僑華

in the dark for 2 days before the experiment?

(b)State a major force

that causes the water and glucose solution: to rise in the petiole

and enter the mesophyll of the leaves.

(c)Explain why starch is

present in leaf A.

(d)What conslusions can be drawn from the results for

(1)Leaf and leaf R (2)leaf B and leaf wa. of B and lea (e)Further experiments

show that leaf C would ot produce starch in the absence of oxygen. How would you explain. this observation?

8. The diagram below

shows a transverse.. section through the blade of a

(Ans.)

甲

CE)

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BIOLOGY (4)

W W. Leung

MIEL & DALE PRESS LTD.

Unit 1: Nutrition

7. After a green plant

has been placed in dark for 2 days, the

are picked and

C for 2 hours

at 20 °C as in the experiment illustrated below.

انها

kept in light

Upthrust of water on the cube.

kept in darkness

weight of the cube

-900(0.1)

=900(0.1)3(10)

=9% N

(Aus.)

(ii)Let the volume of the

cube immersed in

water be V1

Upthrust on the cube cube

-(1000) (g) (V!) 1000(10)V1-9 V1=9x10 ̄”

M

Since the volume: the cube

=(0,1)3 =1x10-3

fraction of the cube immersed.

9x10-4

1x10

Leaf

dicotyledonous leaf.

感

(a)(1)Name the structure

A.

(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 epidermie

contribute to photosynthetic efficiency?

(c)What is the advantage

for the palisade

cells to arrange at right angles to the leaf surface?

(d)(1)Name the

structure B and (2)What are their

respective functions? (3)How does the

distribution of Band C

contribute to photosynthetic efficiency?

(e)State two structural

differences between

the cells labelled D and E.

(f)How does the presence

of stomata in the. lower rather than upper epidermis contribute ta photosynthesis?

(g)(1)What is F?

(2)Give a function

of it.

ANSWERS:

7.

5% glucose solution

kept in darkness

of 24 hours At the end of 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

(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 A.

(a)(1)Tight is required

for photosynthesis. (2)tenf can convert

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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.

8.

(a)(1)Cuticle.

(2)If cuticle is

absent, the leaf would be subjected to fungal and bacterial infection; mechanical damage, and excessive loss of water by evaporation.

(b)(1)The absence of

chloroplasts in the upper epidermis. (2)Sun light may.

pass through the upper epidermis into the

mesophyll cells for photosynthesis,

(c)This feature permits

light to penetrate:

deep into the photo-

synthetic tissue

without passing

through very many

cell walls, so

avoiding loss of

Light by absorption

and reflection,

(d)(1)B: xylem

C. phloem

(2)B:1.For the trans-

portation of water and minerals from root to leaf. 2. For the mechan-

ical support for the softer: leaf tissues. C:For the trans-

portation of food substances from where they are manufactured in the leaf to

grow ing points and food stored are areas.

(3)Xylem vessels are

closer to palisade cells than phloem sieve tubes so that water can be easily absorbed by palisade cells for photosynthesis.

(e)The structural dif-

ferenes between the'.

and “E are

chloroplasts

Thickness of

cell wall

Cell D

present

uneven

Cell E

absent

even

(f)The presence of

stomata in the lower rather than upper epidermis enables the plant to lose less water. by evaporating when undergoing photosyn- thesis.

(g)(1)F:Air space/

(2)TO

Inter-cellular

space

facilitate

gaseous exchange

in mesophy11.

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