頁二第張六第日十初月九年酉辛醫夏 WAH KIU YAT PO 郭日僑華
R2-10a
1982
wall.
Form which
中學會考試題預習專欄
(Ans.)
物理
Take the
the resultant force
acting on the lower
end of the ladder,
ethe vertical
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PHYSICS (2)
MILL & DALE PRESS LTD
Suggested solutions to
Exercise one.
Take
10. s
upward.
direction
as positive.
Let the tension.
in the spring of the
spring balance be T, the acceleration of the elevator be a and the
mass of the block be
weight of the block Wang
component of R.
the horizontal
component of R frictional force
between the ladder
and the floor.
W weight of the ladder.
ii)Since the ladder is
in equilibrium, thei
fore,
-f=0
三期星
(b)Letv be the velocity
(1)
(2)
(8)when a=2.50s
Taking moment about
5.
of the block B and the
bullet after the
bullet lit the block
Since the bullet is
embedded in block B,
hence, by conservation
of momentum (0.01)(2000)-(0.01
日七月十年一八九一公年十七國民華中 育教備
1+10
11
25
The required quadratic
equation is
22
x-11x+25=0·
Solution
2sipA-coSĄ 3sinA+c OSÅ
+0.99)v
(ZainA=cosA)
COSA
v=20ms
(39inA+COSA
CosĄ
By conservation of
energy
2tanA-1 3LanA+1
2(-3)-1
-3(-2)+1
h=20m
Aus
$(0.01-0.99)(20)2 ~(0.01+0.00) għ
Solution:
51ogx-1=41og2-10g5
1
Solution:
Let the
G.P. be
2
2.2
of the
As shown in the force
diagram above take.
k about A.
moment
F(OB) 100g(AB)
where F is the
T-100N Substitute into (1)
m=8kg
Truc weight of the block 50N
(b) T=64N
horizontal force
required,
OA
**
1x100m 5m
OB 5-41m
2
VA OB
5-OB
√24 m
4,899 m
F{1)=100(10)(4.899)
F-4899N
(Ans.)
(ii)The minimum force.
T
should tangent min to the wheel as
shown in the diagram
below
Substitute into (1) 64-8(10)=8a
(Ans.)
Ans. The elevator
loves downwards.
with acceleration.
-2.
2.8 (or the
alevator has an
upvard deceleration
upward decelerat-
-ion Cas
(c)If the elevator moves
upwards with uniform
speed, then a-0
Fora (1)
Emin
T
mg
3W~8N; =0
From(3)
(3)
3x160(10)
=60ON (Ans.)
irection of
Ans: The
the force acting
on the upper
end of the
ladder is
perpendicular to
the vertical wal
with magnitude
600N.
The frictional force
fis horizonta
From (1)
N1 = GUON
)The resultast R
f
-600+16002
(Ans.)
(i)As shown in the force
diagram below
The bouyant force B will apply at the mid- point of the part of the rod submerged in
water.
ii)W-weighy of the rod
#1.2g
=12N
sweight at the end
of the rod to be
determined.
The volume of the rod
1.2 0,5xd
N
B
Solution PQ//BC
LAPN LABN
LAMP LANK-90
th
term ot the
ssion is 1.
8.
Solution:
LPAN=LBAN
• 4 APM ABN
AM AP AN AB
are similar.
Simarily AP P
PO
AB. BC
AB is the
ameter of
the
AM PO
AN BC
AM
AM=37cm
Area of APų=3⁄4×33x5sq.
cm
78{, Cm
AQR 90*
QAB
AO//OF
180°
¿AOP=¿QAB=62°
LAOP
AIAOP 51
1009
is tangent to ni tx the wheel at point
C which is one of the
extremes of diameter
AOC of the circular
wheel
monent about A
Taking00g(AB)
min
min
-489.9N (Ang.)
( the force is applied the 10kg block,
jokg 5 kg
lokg
where is the force
between the blocks
-the reaction of 5-kg block on 10kg block Let the acceleration of
the system be a ma
F-E1 =
R1 =5a
=108
0180, F-15N
- wg -80N
(Ans.)
Hence, the reading of the balance is 80N. (d)If the elevator moves
downwards with uniform
speed, hence, we have
но
1708 8N
Ans.
Also,
tam ex
T80N
(Ans.)
Therefore the reading
of the spring balance
is 80N.
)Let
(e)If the cable wire
breaks, then the
elevator will fall-
"down" vith acceleration
equal to the gravitat-
ional acceleration
Hence
T
Therefore ̈
the reading
of the spring balance
will then be ON
(1) (2)
(i)The force diagram
shown below i
solving R R, from (1) und (2) we have
R1 =5N
(Ans.).
(ii)Similarly, let the
force between the
blocks be and the
acceleration of the
L2
system be ama
15~R=5a'
where,
N. -Normal reaction.
acting on the upper
end from the smooth
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0.375
20°33' (Aus.).
The resultant at the end of the
ladder is 1708.8N making 20°33+
with the
vertical:
and
be the 1 velocities of the
block A after the
bullet emerges and the
bullet as it emerges
from block. A
respectively.
By conservation of
energy
{(2)v^-(2)(g)(0.2)
#20円
(Ans.
(ii)By conservation of
linear momentum
The initial
=(0.01)(24 mentum
24kgnís
The final momentum
A
►(2)(2)+0,01v,
*4 +0.01% 1
24-4+0,01v,
-1
2000ms (Ans.)
(where d is the density of water) Volume of rod
subuerged in water
The bouyant force B
Weight of water
displaced by the rod =(density of water)
x(volume of the rod submerged)xg
=20N
(Ans.)
(iii)Taking moment about
the hinge
(Bsing)(2.5)
=(Wsine)(5)+(wsino)
+(wsine)(6)
4.5x20=12x5+6w
v=2.533N (Ans.)
數學 (=)
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MATHEMATICS(Sy11.2) (2)
MILL & DALE PRESS. LTD.-
C. P. MAN
Solution to exercise
Solution:
A manufacturer's cate-
logued prices are 25% above cost, but buyers are allowed a discount of 85
off these prices. What is
the manufacturer'a net
gains per cent? Solution:
Let 6x be the cost
manufacturer, then the
catelogued price of the Banufacturersgx(1+25%)
$1.15x
cost of the buyer -81.25x(1-8%
=$1.15x
The gain per cent of the manufacturer
1.15x-x
-15%
100%
ед
1.
Solution:
Let p, i be the roots of
DE
2:
the equation x
CE
sin30 3/3 sin60
6.co
then p+q=1
pq=-5
p2+q2=(p+q)=-2pq
10 cm.
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HMT 5-491181
Section
Solution:
*) ABCDEP is regular
hexagon
OB=BC=0C=7cm
Height of the pyramid
3.873cm
(b)Area of ABÚC
**x7x7sin60”9q -21.22sq.cm.
Area of the hexagon AHCDEF
6x21,22sq.em
127.3sq.cn
Volume of the pyramid *=x127.3x3.873 cu.cm
1643cú, co
(c) ON= √72-3.52 cm
»6,062 cm
tan ONV-
3.873
6.062
LUNV=34°54
Angle between the
planes VBC and ABCDEF
Âa 32°54**
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