hour and 24 hours
boil; the
日四十月十年二八九一厩公斤一十七國民華中
(b) Let the Common
育教健華真三第張七第
日八廿月八年戌壬夏
WAH KIU YAT PO
報日僑華
四期星
1983
a-acceleration of
the system.
中學會考試題預習專欄
生
物
(四)
8
明德出版社梁永華 提供資料
BIOLOGY (4)
W W. LEUNG
· MILL & DALE PRESS LTD.
Unit 1: Nutrition
7. The diagram below
shows the arrangement of teeth in the upper jav.
The dental formula of
adult man is
(a)Which numbered tooth/
teeth is/are
(1) the canine?
(2) the incisor?
the premolar?
(4) the molar?
(b)Is this the dental
pattern of a child or
an adult?
Give á reason for your
answer.
(c)Why does a sheep not
need canine teeth?
8. Bean seeds were soaked:
in water (1 bean
3
aced/cm3 water) for 24 hours and were then removed. The water. remained, liquid X
was used in this
experiment in which 3 sets of apparatus were prepared as shown below.
tarch solution
Set A
Liquid X + Starch
•Distilled wate
Set B
Boiled liquid x + Starch solution
Distilled water
Set C
after the beginning of the experiment,
solutions taken from
the visking tubings
and beakers were
tested for glucose
and starch. The
results are shown in
the following table,
starch
24 hours
glucose
hours
starch
двозия
Liquid in
嘻
T.
A
-
*.
Visking tubing B
-.
C : ..
-
+
A
+: present
+
-
8.
Beaker:
-
0
ent
(a)Suggest the reagents you should use and state the results you would expect for
testing (1) glucose
and (2) starch.
(b)Explain why starch was
absent in the liquid
(c)
surrounding the
Visking tubing after
the experiment.
If the distilled
water in eet A was
replaced with glucose. solution, would glucose be found in the Visking tubing 24 hours later? Explain your answer by using one of the results.
(d) A hypothesis to
explain the results ia that an enzyme was present in liquid X. (1) Which sets of the
results strongly support this hypothesis?
(2) What is the
possible function
of this enzyme
in the seeds?
(e) Design two further
experiment to test
the hypothesis in
(iv) and describe the
results if it is correct.
Answers 7. (0)
6 and 11
7, 8, 9 and 10
4, 5, 12 and
(4) 1, 2, 3, 14,
15 and 16
(b) This is the dental
pattern of an
of
adult because the presence of 6.
molars in the
upper jaw.
(c) Canines are used
to pierce prey and for tearing flesh, Sheep is a herbivore and therefore does not need canine teeth,
8. (e) (1) Glucose: add
Fehling's. solution to the sample and.
solution would become brick red if it contains glucose.
(2) Starch; add
few drops of iodine
solution to
the sample; a blue black coloration.
would be.
formed if
starch is present.
Starch molecules are too large to pass through the semipermeable Visking tubing. (c) Glucose would be
found in the
tubing 24 hours. later. The results obtained in set B show that glucose molecules are small enough to diffuse through the semi ipermeable Visking tubing.
Sets B and C. The enzyme may convert the starch storage to glucose for
the respira- tion of the developing
embryo
Control
Experiment 1 Set B
Experiment 2 Set B, at
room
temperature
Test
Set B+several drops of HC1-
Set B, at 2°C
If the hypothesis in (iv) is correct (ie, an enzyme is present in liquid X), glucose would be present in both controls and
absent in both
tests after the
experiments.
1983
中學會考試題預習專欄
物理
(四)
明德出版社盧國雄 提供資料
PHYSICS (4)
H.K. LO
MILL & DALE PRESS LTD.
Answers to Exercise
()
G
15kg
As shown in the figure above,
Let T-tension in the
string.
Apply Newton's 2nd law.
to the blocks
separately, 15(10)-T-158.
T-10(10)-10a.
(1)+(2); 50-25a
a2m8
(b)substitute a=2 into (2)
10(10)
T-120 N
(c)Apply the formula
.2
Let an
be the velocity 1
of the 15 kg block when it strikes the floor. The initial
velocity is 0 ms
22(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
t+at2
=0'na
velocity
Subst
ute s=2
· 2′′ (2)+
4142
Ans
As shown in the fig. above, let the velocity of the 10 kg block be when it
reaches the level at
▼22=2(a)(2) -2(2)(2)
=8
Hence, the string slackens, assume the block moves b meters. farther before drops down.
▼22=2(g)(h)
28
*2(10)
40.4 m
The maximum height reached by the 10 kg block is 0.4+2% »2,4 m above the floor.
Let the velocity of the pile driver as it strikes the pile
be
By conservation of energy. ngh=
2
where n mass of the pile-driver 10 kg
and h➡height fallen
12 m
velocity required be.
-1
v ms
Initial momentum of the system
=(10)(v) kgns
-1
Final momentum of the system
(10412) kgms
By conservation 01 linear momentum, 10v;
*(10+12)v
-19(15.492)
-7.042 ms
Let the average retarding force exerted by the ground. on the pile be F newtons.
Hence,
Work done against F
loss in kinetic energy+loss in potential energy Work done against =P(0.5)
Loss in kinetic
energy
(10412)v2 −(22)(7.042)2
-545.5 J
Loss in potential energy
»(10+12)(10)(0.5) #110 J
.F(0.5)-(545.5+110)
F-1311 Ana
・30kg: Iz
a ms
6000
Smooth horizontal plane
As shown in the fig. above, let the tensions in the strings be T
the
and
acceleration of the system be a. Apply Newton's 2nd lav to the blocks separately, 600-T =108.
T1-12-208.
To
·(2)
×30a..........(3)
(1)+(2)+(3): 600-(10+20+30)a
a1Ums
2
(ii)Substitute a=10
into (1) and (2), we have P=600-10(10)
8
1
500: N
30(10) 300 N
Ans.
LOOD
fi
As shown in the figs ab above, f, f, and f, are the frictional forces on the 10 kg, 20 kg and 30 kg blocks, respectively. 120N
13-60N
Let the acceleration of
the system be a', the tensions in the strings be T and T'
2
Apply Newton's 2nd law to the blocka separately
·600-T-20=10&* .(1).
T'2-40×20x
T'2~60=30a (1)+(2)+(3): 600-120-60a?
at-8ms (ii)Substitute
(2)
8 into
(1): T-500 N
Substitute (3). T'2-300′′N
8 into
=2(10)(12)
240
V_R2=4
Ans.
Ana
-15.492 ma
(b)
M. V.R.
x100%-40%
-2gh
x100%-40%
(c) Let the minimum
effort required be
Load-1.6
E Load=50g
500N
.500 -1.6
E -312,5 N Aus.
(d) Let the tension in
the string connected to the 50 kg block be T newtons
T
1.6. effort when effort=600 N
Ꭲ 600
1.6.
T-960 N
Hence, the net force acting on the 50 kg block
-960-50g
-960-500
460 N
Apply Newton's 2nd law,
460 = 50(a)
9.2 ma
Ane
The block will
move upwards with accelera tion 9.2 maTM.
Upthrust of water the cube
weight of the cube
-900(0.1) 8
-900(0.1) (10)
9. N
Ang.
(ii)Let the volume of the
cube immersed in water be V
Upthrust on the cube
-(1000)(g) (v');
1000(10) V
Since the volume the cube.
=(0,1)3
1x1
fraction of the cube immersed.
9x10
1x10
As shown in the figure below.
Ans
water
< 100 kg)
Let h depth of oil
A cross-sectional
area of the
cube (-0.01 Bqit Upthrust due to oil on the cube: =(720)(g) (HA)
*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. -7200£+10,000(0.1–b)▲
-1000A-28006A =(1000-2800h)A Weight of the cube
-900(0.1)(A)(g)
-900A
Total upthrust ght of the cube
(1000-2800h)A
-900A h=0.0557 m
the cube rises -0.0357 m-1 cm -0.0357 -0.01 -0.0257 m
Ans