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.