1970-01-01 — Page 43

REFERENCE LI LANK

→2JAMO

„CITY HAUL 頁11第刊特年新 日四廿月一十年酉己麽夏

WAH KIU YAT PO

報日僑華

四期星

日一月一年〇九一届公年九五國民華中

tension of cord -V pal.

(接第三張第二頁)

The motion of m2

is downwards

!$%#*

$#

僑

$$

#26

- T-ma

32

BAL

2

8 x 32 T - 8 x

170英文中學會考試題預習專欄

堅道英文書院主編

*****

T - 8 x 32 (1)

6

化學科

(JL)

* #%$# », -4# THE# $%$# $%$# $%$# $%$#$%$# $%$# $*£# $*£## $ %$#&%$#$%$#*%

- 8 x 32 x

pd1

7

8 x 6

-

1b.wt.

7

6.85 lb.wt.

170英文中學會考試題預習專欄

堅道英

物理科

PHYSICS (9)

Solutions for Exercise

(九)

20,000 cm/sec.

Ans. The 6-1b masa accelerates upward at 4.57

ft/Bac

The 8-1b mass accelerates downward at 4.57 ft/sec2. Tension of the cord is 6.85 lb.wt,

4. C,

30mph

4ỵ≈ 15 mph

The masses of the truck and the car are mand m2 respectively, of which

4000 lb

- "2

= 2000 lb

sulphide is heated with fuming nitric acid, 1.268 Lead sulphate is obtained. Indicate how these data may give the percentage composition of the anhydride of sulphuric acid.

Are there any assumption made in your calculation? Explain how you may justify it experimentally.

2. What is meant by the terms "unsaturated solution" "saturated solution" and "supersaturated solution"?

Describe exactly how you would proceed to determine the solubility of potassium nitrate *** water at 15 0..

3. Starting from sodium hydroxide, how would you prepare specimens of: (a) metallic sodium, (b) sodium nitrate, (c) aedium carbonate, (d) sodium bicarbonate

(e) sodium bisulphate?

4. Under what conditions would eklerine react with

ammonium hydroxide, (b) tin, (c) aluminium,

) sulphur dioxide, and (c) carben menexide?

** √2 × 980 x 13.4.

162 cm/s00

velocity of the bob at its lowest, sition is 162 cm/sec

W= m

20 gm. Initial velocity of it, u 20000 cm/Bac Isolating the projectile and indicating all forces acting on it, it is found that the only force is its weight W-mg applying Newton's second Law,

Mass of the projectile m-

F ma

-ng-ma

(Downward force is negative)

Therefore a

=

-8

請

-980 cm/Bec2

Ane, Its acceleration is downward and is equal to

the acceleration of gravity.

At the top of the trajectory (flight), its

velocity becomes zero. (v

By equation of motion,

v=u+at

- 20000 + (-980)+

980t

- 20000

لی

0)

Their initial velocities are

30 mph

u2 - 15 mph

MANOL BUG COLLL8Jon, the combination moves_on_ with the same velocity, v.

Applying the Principle of Conservation of momentum;

1

{1,

+

مرج

Ang. The

2gh

4000 x 30 + 2000 x 15

4000 + 2000

0.04 kg.

N = 20 kg

120 + 30

- 25 mph

Ans. The combination moves with 25 mph.

Solution for Exercise 8.

Initial velocity of the car, u'm u After 30 sec., it attains a speed of

V = 40 mph

t = 20.4 acc..

After 30 sec., the projectile 18

Sutat

- 20000 x 30 x 980 x 30 x 30

- 600,000 441,000

159,000 cm. above the starting point.

and the velocity then in

yu+at

* 20000 980 x 30 -9,400 cm/seo

(negative means the projectile is dropping down)

2. Mata of the two weights are m, and my

2

* 88 ft/sec

3.

ts acceleration is

T . 11

t

x 88

3

.30

88

.45

25 ft/sea2

The force, F, which produces such an acceleration

Let the speed of the bullet just before hitting the wooden blook be um/sec and the velocity of the combinater after the impact be v m/seo.

(i) By the Conservation of momentum

***PLE

112

{m + M)v

m + K

·

0.04.u

8 1b.

my

B2

* 16 lb.

When the system is 20188992,

m2 moves downward while m

upward. Both move with the

same accelerayion, a

The force that causes motion is F

F (16 8) 1b.wt.

8 x 32 pål.

Applying Newton's 2nd law of motion

F-

1:1

8 x 32

(16+8)

32

- 10.7 ft/sad

m

Ans. The 16 1b. yaight accelerates downward with

10.7 ft/seo

3. B, ana m2 are the weighs an

the ends of a cord, m, and'

are 6 and 8 lb. respectively.

The mass of the moving system is

(mq + m2) » 14 ib.

The force that produces motion 10

M

m.

F = (82 – 14) £

2 x 32 pdl.

Therefore, its acceleration is

巍

*? £3

214

-4.578ft/seo2

Isolate M2 Forces acting are: äta ona weight- m2g pal.

m29

is

Fma

=

1120 x

88

1120 x 88

pal

1b.wt.

0.04 20

4u 2004

น

501

m/seo

32 x 45

is force is constant as the acceleration so far remains constant.

Sance, Power rate of working

The block rises h metres after impact

- 2 - 2 cos 30°

2(1

СОВ 30°) Du

Potential Energy

E

(m + M) gh

P

F's

--

Force x velocity

(1) When its speed is 20 mph

- 20.04 x 9.8 x 2(1

- aos 30°) m➡icules

Kinetic Energy just after impaot

Power =

1120 x 88 32 x 45 1120 x 88 x 88 32 x45 x 3 x 550

3.65 h.p.

88.

x ft.lb./seo

K

horse-power

ii) When its speed is 30 mph

Power =

1120 x 88 x 88

32 x 45 x 2 x 550 5.48 h.p.

2. By Gous avation

of Energy, the kinetic energy possesses by the bob at its lowest point B must

equal to the

potential energy

at A with respect to

a

B

Eat B

=

1

2

tx 20.04 x

501

By Conservation of Energy

Fik

200

x 20.04 x

501

≈ 20.04 x 9.8 x (1-06830°)2

5012

Ep

mgh

Eat A

where m is the mass of the bob

v is the velocity of the bob

h is the height raised

h - 100(1 - cos30°) cm.

11

x 22 x 9.8 (1 - 009 30°)

501 x 2√√9.8 x 0.134

➡ 1650 m/sec

(ii) The lost energy of the bullet is calculated."

by the difference in biwakin anamau hu Đ

and after impact

a mu • }{m + M)

- * 0.04 x 16502 - x 20.04 x

x 16502 (0.04 - 20.04)

☐ 43000 joules

501

'<ii) The resistance of wood is

0.02F

43000

2,150,000 newtonia

1650 5012