1973-04-17 — Page 24

其四第張六第日五十月三年丑癸屬

WAH KIU YAT PO

報日橋

二期星

日七十月四年三七九一届公年二十六國民華中育教僑華

2500

(接第六張第一頁)

500+

It can be seen that the p.d.

500 + 1000

The circuit for (b) can similarly be drawn.

across the lamp is 12 V. The pow er of the Tamn is

Vi = 12

6 W.

The effective resistance is

2. КЛ

The current in the

cuit is

∙100 2 K

50 mA

The current thro the voltmeter is

x 50 mA

25 mA

The voltmeter reading is

25 mA x

2 x n = 50 V.

The circuit for the charging of the battery is given in fig 14.

KA

2 A

Hence the lamp is a 6W, 1.2V lamp.

Solution to ex..

1. The intemal resistance of the cell is

1)8. 1.2A

The circuit is given in fig 15

toy

5062

500

ww

10062

The effective resistance is

Current in the circuit is

x

12500-230- 12 mA.

Current in the voltmeter is

4 mA

5:00 500 * 12 mA

Voltmeter reading

= 4 mA x 1000 JL

3. For 1 gm of hydrogen to be liberated, quantity of electrici ty passing the voltmeter is

1.04 x 10

coul.

= 96500 Coul. 1 Faraday.

147 KJ of energy is required to decompose water to produce 1 gm of hydrogen.

The work done by the battery =96500 x 6 J = 579 KJ

The chemical energy released when gm. of hydrogen burns is 147 KJ:

The fraction of electrical energy converted into chemical energy is

476 x 100% = 74%

4. The copper voltmeter in parall- el with the silver voltmeter must. have less current thro' it than the series copper voltmeter. If X is the mass of copper deposited in the parallel copper voltmeter,

then 3X is the mass of moter.

deposited in the series

If t is the time taken in electro- lysis in hours, then for the silver voltmeter,

108

965.00 x 11t

for the parallel copper voltmeter

96500 12t

& for the series voltmeter.

3x =

31:51

5 x (1+1)t

(2)

(3)

(1) —(2)

X

$31.5

54

108

gua. in 1 hour.

(2)

(3)

the total mass of copper deposited

31.5:

is

4X

J

·1973英文中學會考試題預習專欄!

MBALI NA

數學科(廿五)

MATHEMATICS (25)

Solution to exercise 24

APQC

/ KXH =

XAB +

Text. Lor

A)

= 1 rt. / + = rt. L.

Zrt. L.

From A ABF

AB + BP

or AB + AC

AR 2AD

Similarly, BC + BA 2BE

CA + CB > 20F

Adding 2(AB+BC÷CA) (2 (AD+BE+OF

AB+BC+CA AD+BE+CF.

BXC =

BPA

= 90°

900

90

(90. – a)

ABP

•/• (prove (provec

BC AB

(side of a

square

(AAS)

BOX

In As

/ BXC = / APB

A ABCX = AABP

BX AP

PX = BX BP.

Similarly, we can prove that

AP AP

A ABP A BOX

AP

PB

PX

CDQ

(AAS)

QX QC XC.

APQX is isosceles and right

angled at X.

/ QFX ZAPQ

450

= 90°

QPX900

(2) Draw KH/BC cutting AY at H

In A AYC, AK = KC (diag., sq

HK

AH = HY

KH CY (const.

(intercept thm.)

(mid-pt. thm.)

__XAB = = _ _CAB

I rt. L

LXBArt.

取同情僑函聲宣將粉波箱本百,且便國行雅辦

收箱日件明佈地,廣,報中實收而,於 條腿報告將者址凡告以設之有效價四

(diag. of

/ AKB

/ AKH = / ACB = 1⁄2 It./.

XXH / AKB - / AKH

rt./.

▲ KHX, / KHX + / HXK + / XKH

2 t..

/ KHX + 3 rt./. + — I. L.

- 2 It./.

/ KHX = & rt./

KHX / KXH

KH = XX

HK = LOY

KX = √CY

CY = 2KX

Inequalities

(1) Ext. of

opp. /.

1

any int.

(2) Any two sides together > the

third side.

(3) The greater side has the

greater opp. L.

Example (1);

If AD, BE and CF are the three medians

ABC prove that 2(AD + BE + CF) > AB + BC + CA

>AD + BE + OF. Proof: From A EBC, BE + BAC BC. From A FCA +BA CA From A DAB, AD + BC AB.

CF

(AB+BC÷CA)

2(AD+BE+CF) > (AB+BO+CA )

Adding AD+BE+CF >

Produce AD to P such that DP join BP

和大素較,登等。

AD

centres of a triangle The perpendicular bisectors of the 3 sides of a A are con- current (circumcentre theorem) The circumcentre is equidist- ant from the three vertices. (2) The bisectors of the internal

angles of a triangle are concurrent (In centre theorem). The in- centre is equidistant from the 3 sides. (3) The bisector of one intemal angle and two other external angles bisectors are con- current. (Ex-centre theorem). The ex-centre is equidistant from the three sides.

(4) The +

three medians of a tri- angle are concurrent (centroid theorem). The centroid is

2.

of the way along each median measured away from the vertex. (5) The three altitudes of a tri- angle are concurrent. (ortho- centre theorem)

Example (2)

Y, Z are the mid-points of the sides of AABC prove that the. orthocentre of A XYZ is the cir cumcentre of A ABC.

Proof Let O be the orthocentre

of A XY2.

Join XO and produce it to out YZ at P.

Zare mid points of AC and

XP 1 YZ

BC

YZ A AB. (mid pt. XPYZ and YZ AB

PX LAB

Since X is the mid pt. of AB

(given)

FX bisects AB at rt./.

i.e. OX bisects AB at rt.·

Similarly Oy bisects AC at rt./.

Hence O is the circumcentre of

ABC.

本速免康在營養聲幼代藏訪訪物介訪聘房各集

*效告比票院明查郵求尋買件翻轉

Areas

Parallelograms on the same base. (or equal bases) and between the same parallels are equal in area.

In a right angled triangle, the area of the square on the hypoten- use equals the sum of the areas of the squares on two adjacent sides.

Example (3)

K is the mid-pt. of the dia- gonal BD of the quadrilateral ABCD, prove that the difference between the areas of ABC ADC is equal to twice the

area of AKO Proof: A KEA, KDA are on equal

bases (*.* BK = KD) and of he same altitude.

AKBA AKDA

AKBA

▲ AND AKDA AAKN

AAND

AAND = 2▲ AKN

▲ AKN + AKBA ▲ ANB -AAND - 2 ΔΑΚΝ

- 2 ACKN Similarly, ANEC – ▲CND Adding ▲ ABC AADC = 2 A AKC

B

Exercise 25

CAM.

(1) În ⠀⠀ ABC, AB AC, AM is a

median, prove that BAM (2) If AD, BE, CF are medians of

ABC, prove that

AD+BE+CF (AB+BC+CA )

(3) As shown in the figure, Q is

any point on AC, prove that area of XQRD area of PQYB.

R

鞋大效收廉用費話

要。姓不老德有效百宏康其廠,查上弍十個每廣價非➟襴└招親通伴失物出職職職出如刊別襄見各 司名欲遇利信·發大,利各盛期元個,段告目常設刊等生事发虱侶物件讓業務員售有出類告特界