買二第張八第 日九十月二十年未巴瑟 WAH KIU YAT PO
報日僑華
L
N
14.
6
1980
中學會考試題預習專欄
730m
二期星
PQ // BY
n
10
(opp sides //gram)
1-0.01
(c) T
w
BY
*9/
= 2.525 x 10
x 10-2
n
CX
(ii)
AY
I - I
R
=
數學
(+A)
明德出版社女長波提供資料
B
(b) (PRB LABC
Mathematics (18)
C. P. Mana
(a) In AADL,
LRPB
/BAC
(Mill & Dale Press Ltd).
AD
Solution to Exercise 8
H
250 tao450 250m
(3rd of 4)
RPB
AB
are similar
BAC
Section B
(b) In ABDL,
1.
(c) RP = PR
250
PB AC
BD
I
13cm
Tem
(c) In AABD,
AB
- AÐ2+BD2_
2AD-BD • cos(90°-61°)
2502+116.62-21250x
2
tan65° 116.6m
But,
Hence,
AQ PB (proved)
= V
1 volt
RQ BY (given)
2
a
PQ = AB
= 0.014
(opp sides //gram)
FL
* 2.50
RP AY
V - IR
-
AY AQ
11 -
CX (given)
CX/gram
AYXC is a
(opp sides equal & //)
corrs RQ/BY) Teorr s BY/CX)
[PER ACB
(corr sides, similar As)
日五月二年〇八九一公年九十六國民華中育教僑華
0.01 x 2.5
As shown in the above figure, in order to measure voltage up to 1.0 volt, a high resistance R must be connected in series with the meter.
The potential difference
the system is
V
aeross
R
(+9)
10 x 10-3 A
(1) (1)
(1 - 12)
*
(12) (x)
11
.". (1-1) (pq) + (y) (y)
(1-1)(2)(1)(6)
10
=
Ia A
≈ 0.8333A
.. the reading of the
aameter is 0.83334
(4) The reading of the
voltmeter is
12(4)
= 3.333 V
4.(a) Let V be the potential
difference of the
electric source.
The energy required të
boil the amount of
water by A
i.e.
E
BA AC
а
116,6c0829°
12.5 x 0.01
V
x 10
R
A
The energy required to boil the same amount of water by B by 8
(a) In AABD,
132
J
82+72-2x8x7x cos/BAD
.*. cos/BAD 2x8x7
- -0.5
/BAD 120°
*.* ABCD is a cyclic quad.
.*. [BCD 180°-120°
In ABCD,
10
sin/CBD
sin/CBD
-
60°
13 sin60°
10 sin60°
13
LCBD 41° 46'
=
(b) Area of A ABD
· †x8x7• si¤120° aq.cm.
24.25 sq.co.
(c) Let r om be the radius
of the circle
>
To prove:
Q
25105
(a) If PR QS,
ABCD is a //gram.
(b) If SR - BD, then AC 1 BD.
Proof:
(a)
•
•
AB = DC
1.01
AB
158.4m
15. Given: AB- CD, AP - PB,
DQ - QC, QS LBD, PR LBD.
1980
= 97.50
2.(a)
中學會考試題預習專欄
fin
volts)!
20
物理
(+^)
15
明德出版社魯榮家提供資料
Physics (18)
10
W. K. Lo
(Mill & Dale Press Ltd)
5
Suggested solutions to
Exercise Nine
1.(a)(i)
(Given)
12.
7cm
...
2r
A
13
sin120°
r = 7.506 cm.
8cm
AP PB
DQ QC
.*. DQ - PB
QSL BD
PR 1. BD (Given)
•*. LQSD - (PRB - 90°
SQ * PR (Given)
*. ADSQ = ABRP (R.H.S.)
[SDQ /PBR
(a) Area of ABCD =
8x7 sq. cm.|
(b)
(b)
Area of AASP
- 56 sq.cm.
- Area of AQCR
*
*x(7-y) sq.ca.
Area of APBQ
-
Area of ADSR
†•y(8-x) sq.cm.
Area of PQRS
-56-2{}x(7-y)+ty(8-x)]
(56-7x-8y+2xy) #q.cm.
Area of PQRS
and y
- 26 sq.cz.
1.3
**
26 56-7x-8(x)+
動
26 =
2
2x(x)
56-7x-4x+x2
-x-11x+30 = 0
(x-5)(x-6) ⭑0
x = 5 or 6
(c) When x = 6,
y =
1(6)
PS .
- 3
√62+(7-3) 2 2/13 cm
(corr is As)
AB / DC.
2
(Alt is equal) ABCD is a gram. (opp sides equal & //)
DS - BR
DX = XB
(corr sides = AS)
(diage // gram)
SX = XR
i.e. XR
But, XB
*
(subtraction)
180 *BD
.. BR XR - +BD
PA //AX (mid-pt theorem) RPB = XAB
Tcorrs PR / AX)
90°
But (RPB+/PBB
...
/XAB+/PBR
i... i.e.
16. Given: AQ
sum of A)
·
90* 90o
LAXB
M
ACL BD
BQ
S
BY, BQ / CX,
RB-PQ
- RP.BC
Y
CM
P
R
18
1
Force
8
=
x 20
x 10 =
V
x 20
R2R
When the resistors are
connected in series, the equivalent resistance
in A
=
3
(b) The graph shows that the potential difference against current is linear, therefore it obeys Olin's Law.
(c) Resistance of the wire
= slope of the linear
graph
= 4.370
As shown to the figure above, (d) u = f f
a current is sent, through the cuil. Among the Tour sides of the coil, AB and C are perpendicular to the field, there is no force acting on 3D and BC. AB and CD are noted on by two equal and opposite forces. These two forces form a couple aud the coil will then rotale about the axis XY. (ii) The factors are
I
1. the dagnitude of the
current flowing through the coil,
2. the strength of the
magnetic field,
3. the unher of turus
of the coil,
$
Rs
f = A
=
4.37 x 10
100
- 4.37 x 10-8 abri m
Note: If two rezistances
R and Hg are connected in parallel, the equivalent resistance it is given by
} =
R. Ro
+
If three resistances R1Rg and R are connected in parallel, the equivalent resistance is given by
Roy + RHz + Rz
aza
LOV
60
1.2A
RA
+
RE 3RA
Let t minutes are required
to boil the some amount of water
Q
38
A
t
ņ
-
- x 1()
A
30 minutes
(b) When the resistors are
connected in parallel, thể equivalent resistance is
28. R. A B
3R.
A
R
Let the time required to
boil the water be t1 minteg
2
x10
R
A
20
栏
minutes
towards with an
5. (a) Deuterium accelerates
acceleration 1.5 x 1010 (ii)
No. of
Mass no. charger
Deuterium
Pruton
2
1
1
1
-particle
&
2
From the above table, Peuterium and proton have the saza number of charges. Proton will move faster in the electric field because they experience the same maguitude of external force; but proton is lighter.
izu "deaden
F
f
Į
Since F
+
deu
17
10e
1
A
Ра
(8-6)2+32
.em
/13 em
Area of PQRS
2/13•/13*sin/SPQ.
- 26
BIN/SPQ
- 1
[SPQ = 90°
13(a) Tax on an income of $1500
#(1500-600)x38%
- #342
(b) Let 8x be the man's
income before taxation.
600+(x-600) (1-38%)
..
1468
x-600 1400
x = 2000
The man's income
before taxation is $2000.
To prove:
BP, AY = CX,
(a) ABPQ and ACKY
are // grams.
(b) as RPB and BAC
are similar.
AY
(c) AX - AQ
AC
RP.BC
Proof:
(a) *.* RB*PQ
RB
=
(given)
RP
BC PQ
PB / CQ
(converae,
equal ratios)
PB - AQ
AQPH is a
(given)
gran
(opp sides equal & //)
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As shown in the above
figure, in order to use it
to measure o
current up to
1.0 ampere, a low resistauce
shunt
must be conuected
in parallel with the ampeter given.
Let the resistance of the meter be R the current s though the meter and the resistance R be I respectively..
If a current 1 sent through,
then
1
£ it
x *
AR
1
anch
n
: là is
6.0
(a) The equivalent
resistance of the brauch P
(4)(8)(8)
Rpq * ( 3 ) ( 8 ) + ( 8 ) ( 8-80
(b) Y
HxL
- en
(519). un
= 2 +4 × 60
The equivalent resistance of E
Pu
p and B
connected
X2
in parallel is
(816)
= 1.50
- 1 x 10 t
0.013
.*. The equivalent resistance
of the external circuit
a 1.5 1.2 -0.3
R
= 2.40
#
1.5'x1010
30
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=
μ
= ZA
deu
2 x 1.5 x 1010
~ 3x1010
በ 5
(iii) similarly
*deu
10
"deu
deu
-
I'
A
deu
=
વસ
deu
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