實四第五尊
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日八月三年七六九一曆公年六十五國民離率
中文中學會考試題預習專櫃
無限英文中學會考試題預習專欄
*
九六七年
物理科
L (十九)
○腐仲強。
( P h ) PHYSICS (19) B.K.Leung:
227609
第十八次預習試題解答
SOLUTion:
(1) Ai ka u Z zq ° ( h« tt Z i 9)
35.(a)
Frequency of the oog-yngol
48 x 150
c/s
zz x + y = y + z.
3+x
x+y+
証
30-C
x + y + z ax + b
60
3C-a
- 120
c/a
#1
答)一串相等之諸比,其各前項之和,與各後項之和之比,等於原
此和法定理
# 4 - ===
此定理可再擴-
為
pa+qc+ re+.
p b + q d + r
信和定理原東比等於...
बँ
a+c+e
b+ d + ft.
_ ( x + y) + (y + z ) + ( Z + x).__
+
'3a-b) + (3b-c)+(30-2)
2 (a+b+c)
2(x + y + z) = x + y + z
-
(j + x ) + ( x + y) − (y + z ) ___ ___ 22.x
(30 α)+(3a-b)-(3b-c)
2a-4b+46
2-24 +20
24
Z-20+ 20
From V *
ሰአ Where V is the velocity or sound," I 19 the frequency and is the wave-length_
V
et.
I'
rt.
-
#
1120 120
9-
Answer: The frequency of the note produced
is 120 and the wave-length ja
,b)
Fig. (1), snows the vibrating string by pluck-
ing at the middle. It forms a single loop
which is the fundamental.
Fig. (ii) shows the vibrating string by plucking at a quarter near one end of the string. It forms two loops which ja tha first overtone of (i).
From V-
Where V is the velocity of sound, f the "fre- quency, and the wave-length. For the first overtone, 4L where 1 is the length of
the closed tube
f-
1
१.उ It
(x + y) + (y + z) - ( 3 + x)
(à-b)+(36 0)-(30-a)
(y+3)+( 3+ X) = (X + Y)
{3} −c)+(3¢- a)-(3a- b)
ax+by+ cz
its 2c-4a+ub
*
2a+3h
ax + by + cz
a ( a − 2b + 2 c ) + b (b − 2 c + 2 a) + C ( c = 2a +28)
4+
-
x + y + z = ax + b y + c z
a+b+c
(2) 25=a+b+c
試証
(近)左方:
Q*+ & + C*
:
QE D
S2 + ( 5~a)*+ ($−b)+ (5-6)*= a*+ b2+ c*·
5*+ $*- Zas + a2 + s 2 z & s + b * + s* 2¢ $ + C*
45-25(a+b+ c} + a2 + L2+ p *
·4525x25 + q2+ b'+ c2
·a2+b2+c2 (Þ»)
Q ED
(3)(4)着等差級數之第項的欠倍等松其第8項的
gu št q p + q zQ Z Z
tu ka flu k ž Ť » J.
$p3 tp=a+(p-1) d.
依題意
綾瀬 分解
一只公差
d qi
·
$q = @ ty = a + if ild
p[a+ (p-1)d] =g[a+(q nd]
pa + (fp) d = ga+ lg gid
pa-ga+ (pg p + q) d
0
( p q ) a + ( p q ) ( p + q - 1) d = 0
a + ( p +
B p = q thun p-q
hot 13 b < p + q 24 torq
2
(f) bl- 28 40 16 $ pre 3 1983
=arpiq nd
144 SQ A
b x, z ž z y
(1) x+(8 11g=
(証)設此調和報數之對衣等差級數之首項
9] x+(-1) y=
(1)-(2). (4-8) g = ↓·↓ - G-P
Q-P
Q-P
AX X{1}= q x+( p. 1) * P Q ( P • q )
(p-1) (Q-P) --
PQ ( p q )
HO
Q(b−q)~p(Q-P}+(Q−P} _ pP_qQ • Q-P
~P Q ( p q )
¥ ¥ 1 w k < $ ( p + q) 2g = x + ( p + q -D g
Q-P
+ P - q Q - P + Q + ( p + q - 1) + P(-8)
z P − q Q - P + Q + ( p +q-1}{Q=P)
PQ (kg).
故調和級数々草が8項
Patt
pQ-qP
PQ (P-q)
PQ (+-)
D.E.D.
pQ-gp
(4) BIMA 2x 4x-5 | X-3x-1-5=0-
(23) 76541 à, 946$ 2(x-3x-1)-5[X-3x-)-3=6,
([x-3x-1-3)( 2}X=3x-1 + 1) =0
1/2(本田根城取正号】
2(17 3x-1)2-5) x2-3x-1-3=0
Jx-3
x-3x-1=9
3x-10=0
分解
(x-5)(x+2)=0
67 @ * * ALE}
x=5✯-2 (71
£ x=5-2
船以正常速度航行36哩,順流較逆流少1小時,香速度減 為言則順流較逆流少3小時,求船速及水流速度
(公民)設船速北の時水流速度yo/則順流速度 mary/時
呼哩時依題意得式
x y z xy
18004
( 3 x 5 4 ) ( 3 x + 5}}
Tby' = 48.
x-y=724
1
I
1 T 2LN m
J
*
3x330 410.6 412.5
(b)(1) The frequency can be raised to 200 ̄by:
shortening the length of the string.
From the formula:
Where r 18 the frequency; L is the length.. Tis the tension and m is the masa per unit length of the string.
Suppose we maintain the tension constant?
I then
V = 330 m/sec.) L =.0.6
Answer: The frequency of this note is 412.5'
(o) Let Li2 be the lengths of the two wires
respectively. ff be the frequencies and
TT be the tensions of the two wires for
1
the shorter wire to give a note an octave. higher than the longer,
b: 2:3
if 2:1
From f,-
20
150-200 2-100 cm
100 x 150
.. b
200
75
(11)
cm
T
L
#
2
a change of tension on
an incss contre tension
to raise the frequency to 200
From
·T
200 x 20
theatring
420
wit.
Answer: The new length in case (1) la
75 cm. The new tension in case (11) is 5 kr.wt. 35-2
(0) Assume the velocity of sound in air to be
1120 ft.per sec. In resonance WA have
V =45(L+Q,6x)
ft/sec
where V-1120
36. (a)
f=512
I-2
1/6
∙11204
L- 5.36
ft.
512(L + 0.1)
in.
We can use the formula V=2f(L2-Li) to
second position of resonance. Vr120
PUBLIC LA
512
-1120
2x512 1120x12
2x512 13.13
Answer: 5.36 in. of the tube has been drawn
Fight)
Fighti)
out for the first resonance, and 13.13
again.urthercer for resonanca
occur.
SIT
鼈
2
2L
꿈(는)
-4
16
Answer: The m
2
m
3
2
TE
2N
ratio of their tenatons for the shorter wire to give a note an onteve higher than the longer là 16:9.
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9x-254=booy
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第十九次預試題
(1) ABCD3141⁄2 10B) ZD4 18 29 § 2 * × TLC ¿D 28 AU ŽKY @ DA CBR X Y Z LAY-
(2)P為梯形 ABCD之对角线或美遇P作PQ無梯梢两应平行而变
BC* Q * AADQ= 2ARBC
(3)梯形两底各為20吋及12时 其高18时 引是两腰使相遇 求册成小三角形之面積
(4) ABC ABD A B COBI both CARE 14 2 8D
於F DA延线交 BC妁E.则AE BF共园 (5)
lan 75 B sin 75 ± 41 (£Ã££*)
jol li of an 35 + cot 75 m új
$20
seco+ Tan &
serb) - "tan &
4 sec = x + + + 2
(B)若三角形ABC中 200 B sin A cscC
試証 △ABC為等腰三角形
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