育教僑華 頁三第張六第 日七十月四年戌庚夏 WAH KIU YAT PO
郭日僑華
四聖星
日一日月五年〇七九一圈公九十五國民華中
(担第六張第二頁)
有数僑華 育
2
are the forces on the two strings shown in the diagram. Let a ba the acceleration. indicated by the arrows
會考試題預習專欄
In M.K.S. units.
2g
2. a
(1)
英中會考附加數(三)答案
數學科
(廿九)
喬仲强•
-4a
(2)
堅道英文書院撰答
3a
(3)
Solve for
Solution to H.K. Certificate of Education
Examination (English)
48
Section A
Additional Mathematics
Paper III (Nechanice)
EFFORT
·LOAD
When the system rotatea for one revolution
A will go down 24.7 in.
B will go up 5. inv
C will go down 47 in
D will go up
-57-47
241
Velpoity ratio
- F-8 (net force
towards the right)
19.8 seuton 13.1 newton (1 place of
Equation of motion in the vertical and horizontal directions are
-vein30
voo 30
When
t(vain30 - get)
27130
(1)
(2)
decimal)
(a~b) (a–c) * (b+c) (b_a) * (c_a) (c-b
a2(b−c)+f2(c-a)+c2(a−b).
(a-b)(b-c)(c-a).
✩3=a2(b-c)+bc-ab2+ac2- be
-a (b-c) + (bc-bc2) (ab2 ac2)
-a (b_c)+bc(b−c)−a(b+c)(b−c) -(b−c) [ a2+bc_a(b+c)]
=(b-c) [a(a-b)-c(a−b)]
-(b-c)(a−b) (a-c) — ~(a−b) (b−c)(c-a)
~(ab)(b−c) (c~a) ___ 1.
(amb)ch=c) (Ca):
Q.E.D.
(1) (a) & 21 of 4] Aà* ({a}={a2+2£ał, x
<a2(b-c)=-1(a−b) 17 A
(7) £8 x2+(a+1) x2+ (b−3)x−2 ft M (x+2)*4*
Th, & a, b ził.
x++之形式(中為文學常數)
2) 81X x2+4x+4=XX £F-24 18 $21,
X+(a+1)x+(bz
比較两边數,得
8-3-4-14p
-(x2+4x+4)(x+p)
x3 + 4 | 2x2 + 4 x + 4p
+p +4pl
a+1=4+p
4.
由(シ)得ヤ=代入)
a+1==4-2=32
When
Mechanical advantage - 48 x 70%
Effort to load 1 ton - 2240
200
66 — 10.
2. Let the final velocity of combined wase be
By the Principle of Conservation of Nowen
* 20 + 2m
Loss of energy
| (2x)2 + }(2m)u2 - __(3m) ( 4 ) )
Time for sound to travel tha
Therefore
Vantor diagram for velocity is
2m
y is the resultent velocity of u and the velocity of water.
Now in order to reaca o 10 2 EKARZA 21-60
1200
1200
x 60
10
代入(元)
3=4+4(-1⁄2)=
2:8
(就)本題不能用「餘數定理解= 第廿九次預習題
(1) A ABCY, ADLBC. DŽEM €18=240 221
AB+BD=IC.
(2) 14 QI) ABCD AD, BC 2 BL 3A 2' 7 E; X AB,]
• DC π à ? F % LE, LF270 Wire du LEGF== (<A+zC)
(3) AABC = = | AD, BE CF H3
BHxB= +CH×Cr=BC,
(4) ► ABC«, <A✨✨|  ¥917 a »©<FE#TD §•
AD2+ABX AC—BDX CD,
(5)/4 == A + ABC +, # BC & FR D, E ij § 12 BD= DE EC, E AB+AC=AD+AE+4DE,
(6)以銳角三角形底边BC為直径作园,從A作切线
AT, ABL ADAT, KADA DE LAB Ả An
<Œ3⁄4KE ✯ SE A ADESAABC.
7) 0, P笃定直线龙同侧两已知园,求作一直线 與七平行而在該两园内都相等之法.
8) P满见直线外一定桌,求在(线取A,B两奌,使
LAPB=定角且AB=ka.
Let R be the reaction at the ato For the rollar et equilibri
W - Room + Poos (90 -
RocEP + Pain(Ø +
(1)
Rsing - Pein(90 – £ - 0)
∙Poos (P + B)
(2)
Substitute R from (2) into (1)
Poes(" + 9)cond
Poste cos(
Pain(
otprin{p + 0)
- cots
und. § are smaller than 90
Top the vector diagram and using comine
ов60
100
84 2√21/
Using eine rule
sin60
(3)
2sin60
390
least value of Pia obtain from (3):
putting
COM
Googm
2009
завда
Take momen
500 * 10 300 x 12 - (200
300)r
8:6
Take moment along OA 500 x 10sin60 - (200 50
300)reinu
Bine
0.504
2 Kg
Section B
8. Solve by graphical method.
39 weight.
Reaction
Reaction 11.75 kg.
FORCE VECTOR DIAG
Tension 7:45 Ka
A diagram of the set-up of the bar and the string is drawn. The three foroes must pass the point of intersection of the line for the weight of the bar and BC. Joining it to A will give the direction of reaction at A. It is 390 with ACTA force vector diagram for the three forces are drawn, each being parallel to the corresponding lines on the first diagram. Reaction? at A is measured to be 11.75 kg and tension of atring 7.45 kg (Both are to the nearest 0.05 kg,
Maximum load for tension 200 kg is
200 7.45
10kf
· 268.4 kg.