育教僑華頁一第張八第 日八廿月十年午戊展篾
WAH KIU YAT PO
報日僑華
二期星
1979
-aSino
日八廿月一十年八七九一层公年七十六國民華中
57.6 (Ans)
· COSLFOT
- \( w r3) {P} {x} ({r}
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Py=APSinoD
a$in 0
PQ=PQSino"
zasin
2x17
(c) To - 2T1
數學
Mathematics 8
Solution to exercise
(A)
Section A.
1. Solution:
and. u are the roots
3x2+x-5=0。
APOT 430100
4 PUB=1400+43019-570%
· ́¿POB-57°21' -1.001°
Area of "sector
P0B={(4)*x1.
x1.001sq.cm)101
h2 = 3r2
- 5x
(b){}}
115.2N (Anae)
N the reaction from
the smooth wall
of the cylinder.
weight of the ball.
reaction between the
bells,
and are the roots of
(PS)
22
pq
(p+q)-2pg:
(-4)2-2(-3)
Solution:
-x(x-y)-(x-y)
~(xy)(x2-1).
7. Solution:
CB=OP
TP is the tangent.to the circle
£ OPT=90°
20TP+20PB+/0BP-90°
| 2(4OTQ+ LOBP)=90°
LOTQ+2OBP÷45°
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物理
Physics (8):
Suggested solutions to
exercise four
1(z)Take the direction to
tie right am positi
Initial momentum of
the target (100)(-3)
-300 kg Da
Initial momentum of
8. Solution:.
Let SX be the cost of A.
_X(1+2%)(1+15%)-X=190:
1.58X-X=190
a bullet
(0.01)(600)
6 kg as
Assume n bullets are required to stop the moti of the target,
as shown in the figure,
the centre of mass of the combined. body is Gend is x cm from Co
Hence,
moments of the weights of the "cone and the hemiu sphere about. Garn equal in magnitude. (nr2h) (f) (g) (3 + fr -x)
(g)(ăh
= (= πr3){f} (g) (x)
hh + r
(a) The direction of the "react-
ion from the smooth wall of the cylinder on the upper ball is normal to the wall
(b)
and towards the centre of the
ball。
As shown in the diagram, BA » 2x (radius of the bally
0.2m
2xx
BP
(diameter of the byhender) 2 x(radius of the ball) 2 x 0,15 2x0.1
AP BA2B72
-1013
sing- 字
AB
Since the balls are equilibrium,
(Anse)
=(x+y)(x+1)(x-1)
Solution:
Sun of interion angles. of the
pentagon
=(2x5–4)x900-540
Let X he the size of the
greatest angle
then 540-2x(50+X)
X=166.
The greatest interion
angle is 166°
Solution:
Length of the arc of the
240
sector =2↑(6) 560
*8cm.
radius of the base of: the
circulas code
XSi n47-
Sin25
¿K, Sin47°
AC
Sin1094
。
XSin108 Sings
n10g
2(a)
● Sin250 Sin25
the momentum of n bullets
-X-500
Since the target atops
finally, therefore, the final momentus of-the
(ii) The system is instable: equilibrium because the Cage is below. 0,
3.f
Profit of A-$500x20%
-$100.
zero.
300
kg
Section. B.
9. Solution(a)
Let AU-Xm
ACB=180
8 = 50 (Ans.)
(b) The energy lost during
the impact
}(100)(3)2
18 k
RainO:
1(0,01)(600) (50)
$50 +90000: 904503 (ans.)
As shown in the figure,
and: are the tensions:
in the strings.
Let a be the acceleration
of the 18kg-block, then the acceleration of the 8-kg.
block is 2a.
Oil
(1.5)(10)
R 17.321N (Ans.)
Height of tlie cons
4.47cm
Solution:
Sin(30-10")= Sin(30-10°) --
30o
39-10-30
Solution:
750
$510
-870
',160°,400°,520°,
10-33153,133, 173号:25350
2934
* /_ ACB=ZRQP=ZYAN=0°
AP=ABSin00
25X-5
X=1,004
AB÷1,004m
1xsin57 Sin25o
737mm
(b) Area of & ABC -
x1.004x1.737Sin10o: squ
829sqi.
10. Solution (a)
A÷0B-4ca
TansTOB
ZTCB-14 2
OT14-41cm=/17cm
工友頂力合辦
裝修拾生意 水晶聯
私家偵察局
英國偵操機會
Let f = desity of copper Weight of the copper cone -(= x r2h)(P)(g)
Tr')(S)(g)
3cm
Weight of the copper hemi-
sphere
1cm
Since the centre of wass of
combined body is at
therefore, moments of the
weights of the cone and the
hemi-sphere about. O are equal in magnitude,
(1⁄2 ïx2h) (†)(g)(;h)
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18g
= 8(2a)
16a
(1)
(2)
*** 18a
(3)
Substitute (2) into (3)
18g 2T, 186
9g
9a
Substitute (1) into (4)
9%
16 a
25a
-(10) 3.6 us
(Anse)
of the 18kg-
acceleration of i
2
block is 3.6 me and that of
the 8-kg-hlock is 7.2 ms
(b) T1 × 16(3,6)
|漏天補防 水
水台漏水
Wiber
Let de density of water
relative density of
the wooden block.
Upthrust cn the block due
to nil
(0.5)(d)(g)(84)
where A is the cross-sect-
ional area of cuhe,
Upthrust on the block due
to water
= (d)(g)(2A)
Weight of the wooden block
(P)(d)(g)(10A)
(F)(d)(z)(10A) (0.5)(d){g)(84)
(d)(B)(2A)
108- 4..+2
車磨打職
(AB)
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