報日僑華
四期星
7) The resultant R
+ N
MATHEMATICS (2)
言敏僑齡 其三第張六第 日八廿月八年未己
WAH KIU YAT PO
m = 8kg
1980
True weight of the block
= BUN (Ans.)
中學會考試題預習專欄
(b)T G4N
Substitute into (1)
物理
(=)
-8(10) BA
E
--2ms-2 (Ans.)
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PHYSICS (2)
Ans.
日八十月十年九七九一年八十六國民蘸中
1et !=X}={%=h
Afsk coser28°
AK=b-rosec14 al=h•cosc¢5"
COR/PAN=A
2
600
- 1708.8N
+ 1600*
(Ane.)
2
Solution to Exercise
SECTION A
1. Let p and q he the rants of
22
Also,
tane
*
0.375
Ng
0
x -x-5- 0, then
pegs and pq=-5
p2+q2=(p+q)2=2?q =1-2(-5)=11
AZER
ཡས
Suested solutions to
Exercise one
100s-2)
(Take g
(a)(1)
The elevator moves downwards with acceleration 2me-2,
(Or the elevator has an upward decelerat- ion 2ms-*)
(c) If the elevator moves
upwards with uniform
speed, then a
From (1)
0
20°33' (Ane.)
Ans: The resultant at the
end of the ladder
is 1708.8N making 20 33 with the vertical.
4.(a)(i) Let v
EXTRA X
1009
I
As shown in the force diagram above, take moment
about A.
F(OR)
*
100g(AB)
where F is the horizontal
force required.
OA
x10m 5m
OB - 5 - 4 - 12
2
2
AB-
OA
OB
52
OB
/240
F(1)
L
4.899m 100(10)(4,899) F - 4899N (Ans.) (ii) The minimum force F
min
should tangent to the wheel as shown in the diagram "below
4m
(AC)
→Fmin
100g(AB)
min
☐ 489.9N
min
T
-
mg = 0
•
T = mg
- 80N (Ans.)
Hence, the reading of the balance is 80N.
(d) If the elevator moves
downwards with uniform speed, hence, we have also
á 0
.. T = BON (Ans.)
Therefore the
reading of the
spring balance is 8ON. (e) If the cable wire breaks,
then the elevator will fall down with
acceleration equal to the gravitational
acceleration gi
Hence
T-mg"
..TO
-ng
Therefore, the reading of the spring balance will then be ON.
3.(i) The force diagram is
shown below:
where
Ni
(Ans.)
R
min
is tangent to the
wheel at point C which is -one of the extremes of
diameter AOC of the
circular wheel.
Taking moment about A
min
P (10) 1000(4.899)
(b)(i) If the force is applied to the 10kg block,
E roky skg
where Ri
is the force
between the blocks
the reaction of 5-kg
block on 10kg block
Let the acceleration of the
accelergi
system be a s
- 10a B
F
R1 5a
also, F- 15N
N2
10 kg
f
.(1)
.(2)
solving R, from (1) and (2)
We have
R1 5N (Ans.)
(ii) Similarly, let the force between the blocks be
Ro and the acceleration of
"the system be a ma
15 -
BZ R2
From which
B
* 581
- 10a'
10N (AB)
2. Take the
upward
direction as
positive
Let the tension
in the spring of
-2
the spring balance be T, the acceleration of the elevator be a and the mass of the block be m.
weight of the block
W
&m
Normal reaction ebe acting on the upper and 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\\s** weight of the ladder.
(ii) Since the ladder is
in equilibrium, therefore,
N1
-
f = 0.
(1)
Ng
W - 0
:(2).
W x 3
3W
-
8N1
·
0.... (3)
Taking moment about B
N x 8 = 0
From (3)
N1
3 x 160(10)
8
- 600N (Ane.)
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)
f - N, - 600N (Ans.)
(iv) From (2)
Ng
-W-1600N
The coefficient of sliding
friction between the
ladder and the horizontal floor
be the and
1 "A velocities of the block A after the bullet energes and the bullet as it emerges from block A respectively.
By conservation of energy
4(2)▼2 - (2)(k)(0.2)
V
-1
- 2ms (Ans.)
(ii) By conservation of linear momentum
The initial momentum
(0.01)(2400) 24kgmg-1
The final momentum
(2) A
+
(2)(2) + 0.01v、
- 4 + 0,011
.*. 24 - 4 + 0.011
24
?
The required equation
2 ig x*-11x+25 = }
2.. 51ugx-1 = 41og2-1og5
51ogx-log10
5
19470
5
*
416g2-10g5
4
= 1o¢ ད་
x = 2
3. 2ṣin(28-10°)=1
sin(20-10°)=}
20-10°-30°, 150",
A =
3000
or 510° 20°, 80", 200°
or 260°
10.
hecusre28"
b'cosec14
[Pax=58°50'
AF
Al
hreagee28",
h*cosee's"
[PAR=79°18"
•*. {XA!=79°18' - 98°59'
=20°19'
.". The angle between the tracks on the hill
is 20° 19'
130*
3cm
B
(a) Area of // gram AUCU =3x4xsin150
6 $4.00.
sq.cm.
(b) ac2=32+4°-2x5x4xcoa150"
= 45.78
AC - 6.766 cm.
(0.01) (v1)
4.
5sinA-3ainA 5sina-3sinÃ
1
COBA
-(58ina-3sint)
-(5sinA+3cosa)
COSA
(e)
6.766
Stund-3
sin150"
Stanar
5(~)-3
5(-)+3
-1
- 2000ms' (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)▼
- 20s 1
By conservation of energy †(0.01 + 0.99)(20)2
(0.01 + 0.99)gh
(Ans.)
.*. h
= 20m
5. (i) As shown in the force
diagram below
W
ouyant force B will apply at the mid-point of tle part of the rod submerged in water. (11)
weight of the rod .1.2g 12N
weight at the end of the rod to be determined.
The volume of the rod
2.4 3
NO સ (where d is the density of water)
Volume of rod submerged in water
2.4
I
2,3
K
ď
The bouyant force B
weight of water displaced by the rod
(density of water)
* (volume of the rod submerged) *
- d x 7 x 10
- 20N (Ans.}
(iii) Taking moment about
the hinge
(Beine)(2,5) ►*(Weine)(3)
+(waine)(6)
2.5 x 20.
12 x 3 + 6w
W 2.3BN (Ans.)
5.
AD is the diameter of the circle
.*. [aq8=90°
[QAB=180"-40"-28"
=62"
AQ // op LAUP=LUAB=62°
.*. [AQP=d[AOP = 31" .
6. Cost price of the article
=8180+(1+445}
7.
-8125
The required selling price
=$125x(1+40%)
=6175
BE 1 EC 5
*. A ABE=A ABC
AD 1
DB >
'. ABDE~? A ABE
3
"IGA ABC
1
Similarly,
A ADF- ACEP = AANC
16
•*. ADEF¬(1~3×37) ▲ ABE
16
4
7
Алис
16
i..
ADEF AABC
7
16*
**
A. Simple interest on, 82500
at 4 p.a. for 3 years
=62500×45×3*
=8300
Compound interest on $2500 at 4o p.a. for 3 years #$2500x(1+4%)-' _82500
-$312.15
Compound interest
exceeds simple
interest hv
$312.15-8300 $12.15.
1980
中學會考試題預習專欄
SECTION #
sin/CAB
4
sin/CAB
Asin150"
6.766
[CAB = 17° 12'
數字推理(三)
數字推理練習三
數值的推理與應用
選出下列每题的正確答案,並在它的下圖畫
1.下列去做數中,那一個最大?
A
*12 15
BC1
D
21
2.若
是由1至10的任何一個歡,曲+4這個 歡一定是
B等於-14
C4
A**5
DX14
E大於4
3.器以是奇數,K是偶數,下到那一個數是奇
R7
A (2H+K)
C (2H+2K )
8 (H+2K } D(2H+4K 1
E (4H+2K }
4. 椰子糖一堆,五粒一數,五粒一數,最後餘 F3 - MTAM-MRÆMF@#9#B 7
A25 B27 C36 D38 E42 5、光门是由10 至20 的任何一個鰍,100卡口的
#KTËÆTLAR - 100 7
D10
B 6
C8
E12
A 5 6. SM#NŁMMA2-AKAN SÆ8 RS
的值是多少?
P 24
Q30 R 40
S 42
T50
7.茶数除以5、除點是3·某黢的4倍悠以感,
餘款是多少?
P 2
Q 3
RA
$7
T12
B. AXB = · PAF BAR-IN Æ991 ?
PA✩BESK
RA必是整数
QAÆBEDE S日必是分數
T&A是整数则B是分欺
9.有橙子、蘋果、梨子、香蕉四種生果·若華
*****A-ORAL 不同的組合?
MAST★ALÆ
P2# Q3 RA# 96T16
10.弟弟高Xcm,哥哥比他高10cm,姊姊比哥哥
#4 cm MTAA—****?
***{X+12) cm Q###{X+ & Jen R汛洛琉臭
S****
T姊姊比弟痱滿冬们
11. X 5. Ú018 ME #69 H · FFB-BRŁD
AX+2
-
BX-2
cx+2
E
4
12. ***EXER · EN ZAIDI BËJË TU
* · MS*ARASIMA?
A2X
DX+2
BX
E****
-X C
13. YRÅLENA 10 - SGAE✨Æ 120 - M
THR-RA TEAGMH***?
A120~+ 4
C120+4 — } }
E120X(4-1)
B120+ (41) D120+(4+1}X4
14. *2+IMLA EM #:* - ***97UM. E
#949 - BATAM-IRANAN V
A戤色的紙股眾
Bach MAEL
C不可能杆不能種蛾的弧戲
D肚色的比黃色的少2張
ELEME✯#CM #12 #R
15. 14) MA 40 • †₺160 · MARES
少了
A(160—40) + 2
B160+40 2
C160-40+ 2 E(160+40) + 2
D1602 40
JA
2) T
(3) B
(4) D
(S) B
(B) Q
:7) P
帶
D
12 A*
(8) T OS JO T 03 0
C 13 E
W
*
黧
T- mg - 10 40 .(1)
when a
- 100N
- 2.5m8-2
Substitute into (1)
"N2
.600
1600
-0.375 (Ans.)
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