1970-05-21 — Page 23

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.