1972-11-18 — Page 25

REFERENCE LIBRARY

18 NOV 1972

CITY

日僑茶 WAH KITA HA☀

肖教儒者 頁一第張七第一日三十月十年子壬层复 WAH KID.YAT PO

橋

1913英文中學會考試題預習專欄:

(四)

1 lbf. is a force equal in size

to the pull of the earth on

mase of 1 pound.

1 kgf - g MES 1 gf = B

1 lbf -

newtons

Procedure in solving a problem

Draw a diagram to clarify the problem situation. Isolate the body and indi- cate all forces acting on. the body.

Disregard the forces exeru on other bodies.

1978

中文中學會考試題預習專欄]

數學科(四)

六期星 日八十月一十年二七九一服公年一十六國民華中

4(2x-5) (X−4)(x-1) — (2x-7)(2x-3)

張正邦。

第三次預習題解答

(21-5)

· (2x+1)(2X-3) (x−4) (X− 1)- - 5(2x-5)=

x=不使ら母島根

解下列各方程式

|

Calculate or tabulate the x- and y-components of all forces acting on the body Equate

| (2x−1)(3X−1)(4X+1)(6X+2)=0

(2x−1)(3x−1)(4X+1) {5X+2)=0

21-1

解

2=0

4X

本

ZF

X

may

52420

FRICTION

故原方程式之解為言当

2

物理科

KINE

CS

Fer

PHYSICS (4)

A forces definedTEN

terms of the effact

it produces when acted on a tedy,

A force changes; or tends to cr ge (1) the state of rest of a cocy (2) the sperd of a body, cr (3) the direction of motion of

a body.

Summary:

A force 19 that which changes or tends to change a body's stete of rest or of uniform motion. a straight line.

Newton's firs: Law of Motion Every body continues in its state of rest or of uniform motion in a straight line unless compelled by Some external forces to ast other- wise.

This is orten mow as the law of Inertia. Mass of a substance, therefore, may be defined as the quantity of ines tid of at.

If a body is 11 motion with a cor stant velocity,w there is no re- sultant force acting on

Momentum

A mʊving body possesses momentun. The momentum of a body is the pro- duct of the mass and the velocity. of a body.

Momentum my

Impulse

Impulse of a force is the change of momentum of a body.

Conservation of Homeż tum

The Principle of Conservation of, Momentun states that is two or more hodias collide together, tie sum of momenta befare collision equals to the sum of morerta afte- the collision.

If m, and 1, are the masses of tho

2

bodies,

u and us are the initial vel- ocities of them, and are the final velocit

V and 2 ties of then,

then:

Sum of momenta.

before collision. MqUq

Sum of momenta

after collision.

#22

Nearton's Secoud Law of Motion

This law states that the rate of change of momentum of a body is directly proportional to the re- sultant force acting on the body and is in tac same direction as t

Resultant Force =

The Absolute Units S.I

-my-mu

ma

unita Newton (N)

1 N is the force which causes

a mass OI 1 kg to accele- rate at 1 1/9/9

.g.s unit. dyne fdyn)

dyne is the force which

causes a mass of 1 gm to accelerate at 1 cm/s/s

.p.s. uniti - poundal (pd1).

pdl is the force which cau-

ses a mass of 15 to accelerate at 1ft/s/s

The Gravitational Units

is a force equal in size

to the pull of the earth on a mass of 1 kg.

1 g. is a force equal in size to

the pull of the earth on a mass of 1 gn.

Solve the equations

when two surfaces in contuct move, or tends to move, relative to one

another, a force acts in the plane}}

of the contacting surface in such

a direction as to oppose the mot-

tion. This force is known as the force of friction

Laws of Friction

1. Friction always opposes motion 2. When a a body lies on a surface, friction is equal and opposite to the applied force in the direction of motion. That is, friction can take any value * up to a certain maximum value called the limiting friction. 3. For a given pair of surfaces, the limiting friction is pro- portional to the normal force. N between the surfaces. The ration of f/N is called the coefficient of static friction.

Fimiting friction Normal force.

The limiting Triction is an dependent of the area of the contact surface.

Once the motion starts, the ratio of the friction to the normal reaction on the surface is known as the coefficient of siding friction.

Sliding friction is always

less than the static friction Sliding friction is independ-

ent of the speed within limits.

Exercise for this week

1. Two weights of 8 kg and 16 kg respectively hang on the ends of a cord over a small fric- tionless pulley. If the system released, what is the accel eration of the 16-kg weight?

2. (a) What is meant by the coel-

ficient of static and sliding friction.

(b) Describe an experiment to Jetermine the coefficient of sliding friction of a body on a plane.

(o) Block A weighs 4 Kg und Llock B 2 kg. The coefficient. of sliding friction between all surfaces 18 0.25. Find the force F necessary to drag A to the left at constant speed,

(1) if B rests on A and moves

with it,

(ii) if B is held at rest, (iii) if A and B are connectT-

ed by a flexible cord passing around a friction less pulley.

ballson of total mass 2000 Ib. floaty motionless over the earth's surface. If 200 1b of sand ballast is throwa overboard, with what acceler ation does the balloon start to rise and what will be its velocity after 9 seconds?

SUGGESTED ANSWERS TO ISSUES 3 and 4 WILL BE GIVEN ON THE NEXT WEEK.

X+1

a+1

v) a (a+1) ✯ ✯.

a(x+1)+x(a+1)=2a(a+1)

^x+a tax +x=2a2+20

20X+X−20 + Q

x(2a + 1) = a(za+1) x = a

{3. ((a+b)x-c] = [(a−b)x+c]2

解 ((a+b)x-c] = [(a−b>x+c]2

解

(Q+b)X−C = ±((a−b)X+c] (a+b)x-c = [(a - b)x+<}

ax+bx=c = ax-bx +c

2bx=2c^

(a+b)X-C — -[(a-b)x+c]

ax+bx_c = −aX+bx

20x=0 X=0 其根為告及口

3x+2x-5=

(X+1)(X−1)_X(X−1) \). X(X+1)(X−1) k£8.

(3X+2)(x−1) − × (x-5)=(x-3)(x+1)

3X −X-2-X2 + $X= x2−2x-3

x2+6x+1=0. X; = −3 ± 2'√5

X=-3±25 不使分母萬0枚 其根

5-x+29-3x

18+3x+†

-x+2-9-3x=- 18+3x-3x

X+དཱ+3w-q= $ * auxu

X+2

3(x-3)= ¶ + 3(x-3)(x+2)

1%, 9(x+2)(x-3) RAIL

9(X-3)+3(x+2)=2(x+2)(x−3) + 5×3

9x-27+3x+6=2x-2x-12 +15

X=7x+12=0, (X−3)(x−4)*

X-3-0

» X-4=0

X=3使ら母0奴不取

-4x-3-6x-23-8X-10

2x-7

4x+(X+2)+2(x=4)=0

-X-2= (x+1)(X-2) =∞

X=2 使分母為。故不取

淇根為-1

8% 2x-1 2x+5

4x+3_~2(X-1) 2(x-1)(4x+3) 以2(オーリ)(41+3)乘み端

· · 16X2(X-1) — (2×−1}{4X43) +2x+5=0

16x3-16x3-8x2-2x+3 + 2 x + 5 = 0

16X? —24X - 8 •

12 x3-3x2+1 = 0

(X-1){X−1)(2x+1)=0)

13 -x-1=0

2X+1=0

・メニ使分母0.放不敗)

प

其根為一位

x+a

+

b(x+b)___a(x+6)

a+b

ab

94 1% ab(X+a)(x+b) R &*%

ax2+2xa+a3+bx2+2b2x+b3

a(x+a)2+b(x+b)=(a+b){x+a)(x+b)

ax2+a2x+abx‍+a2b+by+abx+b'xtab

a2x +2 abx + bx= a3+a*b

x(a-b) = (a−b)2(a+b) Y--(a+b),

X26

10 x=ax+2bx-2ab+ B-x2 + 3<2

A? x(x-a)+26(x-6) + b2= y2, 3c2

x+2b+b=x__ 3< X-2 X-26 (X+26)(X-26)(メーユ()+(b-x)(X-20)

+3C2(x-26)=0

X3 - > < x2 - 4b2 x+46°*c+b2x

−2b2°C −X3+2CX2+3C2X−6 bc=0

(b2=C2)X=26°C -26c2 (b+c)(b−c)x=2bc(b−c)

b+c

第四次預習題

求下列各式的道

1.36240° 2 Sm 150°

2. Dan (-120°)sec(-330° )Gt (-210)

Sin 120 Cs330 + cos 120° Sun 150

Sim2 (-300") + Couc (-210");

22-10 5 125 (213) (st(16-20, (645) Inbet at(if +x) at (1h-x)

14 37-13-4x-3-(x-2-[1-x 6.42 3X-|2-1 __ 4X-4+1-6X-21 8X-00 7. 2x

2x-12x-3

X-4

Sin (A) & (A-1), G(RH)

S(A-180°) G(A-40) @[16+]

CAC (180°-A) sec( 90°↑ R)

* Cot (90°-A) = -LeCA CACA

P. Gs 420 Sin 398 +cos(-29070°

Sm (-330) = C

X-4+x-1-2-2-2-2-2-2-3 4.36 18 Sun 240° Jam 330" + (0) (-15)

(x-4)(x-1)=(2x-1)(-2x-3) 10* Sun 780° Sun 126 +6 120 & Sm#D30

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