育教僑華·莫三第張八第日四廿月二年未己曆
WAH KIU YAT PO
郭日倫垂
四期星
日二十月三年九七九一曆公年八十六國民華中
1979
the balance length is
x = from. A.........
of copper
1.6gin30o
1077 kgc
0.54 sq.cm.
(x~12)(x+15) 4
中學會考試題預習專欄
Resistance
AB
(20)
物
理
(牛四)
明德出版社春榮家提供資料:
20x
Resistance of
PHYSICS (24)
Suggested solutions
Exercise 121
1. Let the length of the
ire be b
(2,24+0.54)x(28)
62550 sq.. Length of the
of the squai
The area of the field
420
-0.2 Ans.
62550 m.
250 m
Sölution :
3.3x 10.
0.75A
3600
x= 12 ar -16 (rejected))
The total number
of books produced
altogether in 8ain
(x+x+4)*8
12+12+4)x?
224
15. Given: ABCD is a igram.
DM MC HN NC.
Apply the principle of Wheatstone ridge
(1)
the balance length
0,2m
3(*) apply
(1)
200
200
the resistance.
the bulb B
2002 (âns.)
(11) Apply
PRIV
The current I taken from the
battery is given by
10
10
(a):
31
3 x 0.75A
2.25A (Ans.)
The
urrent through th
394 m
heater
(b)
/BPA
To prove
(a) A ADI! = AXCM (b) ACXD is a #gram (c) Area of ABCD –
of ABYXD
(2)
(0.3)(1.2)
XY
)(3)(1–1.2)
1.2)=(0.3)(1.2).
5L
0
(rejected)
(b)
1.5
the length of the wir
1,5m (Ans.)
(b)(i)
200
200
(Ana
Apply the principle Wheatstone bridge
ubstitute the val
· EX
where S
́into (1)
0.3 1.2-0.3
-0.3-4
(Anm.
120 40
As shown in the figure,
resistance of AB
0.8
162:
20
Resistance of AD
802
60
80
(2)
0.4V (Ans.)
(b)
Χα
1202 and
the resistance of
bulb B measured by
the circuit
R? 100-40
180 (АПВ)
(ii) The current through
1204180
0,01A, (An■,)
(ii) The power dissipated
by B
(0,01) (180)
-0.018W (Ana)
(c) The current through B in (a) is much greater than
the current in (b) there. foret the temperature of B in (a) is much higher than in (b). Consequently, the resistance of B in (a) would he greater for resistance increases as temperature
increases.
(i) Let the current through the battery he. I the
current
through the.
Voter and the heater
be I. and I, respectively.
Apply the formula
נס ...
Z.It
mass of copper deposited.
8.91
2.25A 0.75A
1.5A
Energy given out by the heating coil in the period.
0.894
ZBPA= 63°35'
The inclination of each
of the rope to the
horizontal
Length of each
·894
Proof: (a) DM- MC (Given)
ZIMA = [XMC (vert opp.
[MAD= [MCX (alt. £s.
AD// BC) ^ADM= AXCM (\.S\A,)
--(1,5) (10)(3600)
81000J (Ans.)
(111) Let the temperature of
the copper vessel and its
content rises 0 C
fleat energy absorbed by the copper vessel and its
content
(0.5 x 4200 x 0 )
(0.5
23100 J
420 x 0
Heat energy given out by
the heater
81000 J
Assuming no heat losses, or
gains from the surroundings
23100 - 81000
0 - 35 (And.)
the rise in temperature:
of the copper vessel and its
content la 35 C
數學
(白四)
明德出版社文長波提供資料 Mathematics 24
Solution to Exercise 10
Section B..
11 Solution:
2.8
(BAD≈ 90
BD
120
21.35
1.35c08120°
13.44
BD = 3.67 cm
1.6cm
(n) On the man
Area of ABAC= 4x2,8x1.6
Given: 0
is
AM= MX (corr sides = As
ACXD is
gram.
-(Diags bisect ex
other)
the centre of
circle ABQ
APQ Fis a straight
Similarly ABYC is a gran Area of AXDC= Area of
Area of AABC
Area of A BCY
·· (α) [AQB- £LAPB
(b) Po PB
(c) All// BO
(a) Join 40 and 80.
[AOB= [APB (/s
/s in the
same seg.)
[AQB= ±/A0B (same arc)
~LAQR= 3/APB
{b} _APB= [PQB+[PHQ (EXT /
of A).
·ZPOB= 3/APB (proved)
[PQB= {PBQ= 1/APB
PQ = PB (Sides opp. equal (a)
ARB /AOB (Same are)
/ARB= [PBQ (Subst.)
AR//BQ (Alt /s eq).
14 Solutïon;
(a) The tine to produce
one metal book
60
se.c
The time to produce one plastic hook
AADC
(Diags of// gram)
•BC= CX
AD (opp sides
// gram)
Area of ABCY Area of AXCY (Equal base
of ABCD-
Area of ABYXD
between the gutae // s.)
16. Solutio
·60
gec
60
60
(5)
kg
12-12
time taken
hour
2,24 sq ch
* 5600s
Area of ADAC ×1,35x
<= x(x+4)
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As shown in the figure,
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Total work
Work by A
Work by
Work by C
Normal week
100 x
12 x
40 x
IPT
Certain week
200 x
42 x • (1+150%) x = = 70x
200x-18x-70x = 112x
18 x
had to inc
ek's work hy 112x-40x
40x
全僑稱許
his
100% or 1804
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