1969-01-15 — Page 15

15 JAN DOY

1969Tum

宜教佑華、莫三第張四第日七廿月一十年申戊魔复 YAM KID YAT PC

FALL

三期星

日五十月一年九六九一番公年七十五酵民泰中

CARUAR=#1

Force tending to prevent sliding (5001n30′′)Ibwt.,

育華僑華

196925FUL?

and the normal force (5000830°) 1b.wt.

Therefore, the total normal foros is

物理科

陸永熾

PHYSICS (11)

(4)

my cause

The component of ng causing the block to slidel

19 ng sin 459. The normal force acting on the plane 19 mgcos450, Therefore, the friction offered to the sliding is 0.5 mgcos45°. The resultant force,

Since F

ggin45°

ZAN

agsin450 0.5 m2c0845 angsin45

The acceleration, a, produced is

842 ft/sec ( 32 ft/sec2) 346 cm/sec (8.980 cm/sec2)

4.(11) The mess of the moving system,

M = (30 • 20) gm. ^

The force that produces motion, F (30- 20) gm.wt.

- 10 g dynas

Since ma

= acceleration of

the systen

10 B/50

0.2 g

6.4 ft/sec2

196 cm/sec2

D.(1) The Barth attracts the 10-lb

mass towards it by a forde

or 10 lb.vt. or 10g pdl. This force is transmitted equally to the other parts of an inextensible string, If the cord has at negligible weight, the tension in the cord is 10 lb.wt.

The body experiences an

upthrust when it is

immersed in water, wVOR magnitude is measured by:

the eight of water.

displaced,

W

[(1008309)+(50cos30°)] 1b.wt.

Friction opposing motion,

• (100% in30* + 5000#30~)

Since it slides up with uniform spesa,

there is no acceleration on the body The

problem can be treated as a statics problem, 50sin30° - M (100sin30°+50cda30o)

• 25 • 14 (50 • 25 No3)

503-25

50-25,3

2√3-1

2+3 0.659

The coefficient of sliding friction between the body and the plane is 0.659;

Let x cm of the plate"

be in water and

(2.5x) cm. in

mercury.

By the Law of Flotation:

WWater

mercury)

The upthrust of water on the plate the upthrust of mercury on the plate the weight of the plate

thrust of water on it Wt. of water displaced

Upthrust of mercury on it

1234x.gm.st.

1234(2.5 - */

x 13.6 gm.D.

1234x 1234 x 13,6(2.5 -x) • 1234 x 2.5 x

13.6 x 2.5 - 13,6× - 2.5 x 2.8

12.6x-2.5(13.6

-27

Ans 2:14 cm. of it is in water and 0.36 cm.

in mercury-

化學科(十一)

ferric

ion Khrown!

ferrous

מט1

(green)

EDU

Sulphur dioxide may be prepared in the laboratory by the action of medium concentrated hydrochloric acid on sodium sulphite. The apparatus used 15 similar to that used in 4.1. when hydrochloric acid; 19 poured down the thistle furmel, effervescence Occurs and sulphur dioxide is evolved.

Na2503 sodium

sulphide

2001 • H2O + So, ↑

The gas is then passed through a little water to remove the hydrochloric acid vapour, and the gas

is then dried by passing it through concentrated sulphuric acid,

The gas is finally colected by Qunuwaso delivery.

Two physical properties.

(1) It is easily liquefied under pressure. (11) It is very soluble in water.

TWO chemical properties:

(1) It is an acid anhydride.

2KMnO

When the gas dissolves in water, it forme

a weak, unstable acid, sulphurous acid, which turns blue litmus red, and forms salts with alkalis and basic oxides.

50

·H2O

12503 2NaOH

H2503 Sulphurous acid

Na2303 + 2820

* NaHSO, + H2O

It is a reducing agent

It will reduce permanganates to manganous salts, the purple solution being decolourised in the process,

potassium permanganate (purple

solutionj

2820

Two industrial usesi

2:30

manganou

sulphate

(colourless)

solution)

(1) It is used in vary Large quantities in the

anufacture of sulphuric acid,

1) It is used in the manufacture of calda um

bisulphite for the treatment of wood-pulp in the manufacture of paper.

CaCO3 • H20 • 2002 • Ca(H503)2

The calcium bisulphite dissolves out the lignin from the wood-pulp, leaving the sell

Wuestions for next week (12)

Next week we will examine a number of multiple choice questions on sulphur and sulphur compounds.

Let the speed of the bullet just before hitting the block bey m/sec and the velocity of the combination after impact be V m/sec, [(1) by the conservation of momentum.

(1) The speed of the bullet before

hitting is 165 m/sec.

(11) The lost of energy of the bullet 18.

536 Joules.

(iii) The resistance free of wood 18

26600 nawtonë.

uestions for next week

21. (a) Describe and explain briefly one useful

application with increasing temperature of (1) a solid (ii) ■ liquid. Also describe briefly one case of expansion with temperature being nuisance, and explain how the effects of this expansion may be reduced or eliminated.

_ber ̧ =" (m^+ (M) V.

0.04V 0.04

4V

Vol, of iron-volume of water displaced

Upthrust

7.8 x 62.

10 x 62,4

10

Ib.wt

7.8 x 62.4

The tension in the cord is

10. 7.8 8.72 lb

10

20. (a)

W = 10/10.ND

- 32 ft/sec2 (upwards)

The resultant force must

be (TW)g pdl (upwards

-101g 10 a

(T

T- 20 lb.wt.

The horizontal push 100 15,wt.

resolved into two components:-

The effective push (100c0830) b.wt., and the normal force on the plane (1008 in30)1b.wt.

The weight 50 lb.wt. is resolved into;=

The block rises n m after

h - 2 - 2coa30o

201 – cos30.

The potential energy then

P.E.(mM)gh

• 2,04 x 2(1-cos30) x 9.8 joules The kinetic energy of the combination Just after impact

M)V2

·04 x

512

conservation of energy

x 2.04 x

512

• 512 x 22 x 9.8(1 – cos30°)

- 51 - 2 No9.8 x 0.134

165 m/sec.

x9.8

(11) The lost or energy or the bullet in

calculated by the difference of the K.E. before and after Impact

mv2 = 3{m + M) V2

1 x 0.04x1652-}x2.04x4x9.8x0.134

536 joules.

(111) If the resistance of wood 18 r. KONUVIL

0.025 536

536

0.02

26600 newtons

(b) A flask 19 fitted with a tube and is inverted

so that the end of the tube is under water. The amount of air in the flask 19 adjusted that the water level comes to some convenient point in the tube. How does the arrangement work ás a thermomenter?

Explain why this type of thermounter 15 useless practically,

A Centigrade thermometer reads 1,5°C, in melting ice and 99.4°C. in steam from water Boiling at 76 cm. pressure, Assuming the bore to be uniform what is the correct temperature when the thermometer reads: (1) 20°C. (11) -10°c9-***

Two rods of iron and brass respectively mis -made of equal length at a certain temparature. At 1590. the brass rod is found to be shorter than the iron by 0.015 cm,, the latter being 50,2 cm. long. "What was the temperature at which the lengths were equal? The coefficients of linear expansion of iron and brass are 12 and 19 x 10 per degree respectively.

22. (a) Explain what is meant by saying that the

cdefficient of linear expansion of steel 25 0.000012 per deg.C.

(b) Describe, with the aid of a diagram, cim

method by which you could determine the coefficient of linear expansion of steel. Explain how to calculate the result,

The distance between two points, masurSU WILA a steel tape-measure at 20°C, appears to be exactly 500 yds. The tape-measure, however, measures correctly at 100% What is the actual distance between the points?

(d) A barometric height, as read at 12°C by z Draam scale correct at 0°C., is 76.52 cm. What Ir the actual height of mercury colum at 12907 (Coef. of linear expansion of brass is 0.000019 per deg...)

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