1967-04-19 — Page 17

鴨教僑華

真一第張五第

日十初月三年宋丁唇宴

WAH KIU YAT PO

#

英文中學會考試題預習專欄

#

物理科

(廿五)

梁海明-

九院压

WADHWA LA A

Solution.

47.(a)

PHYSICS (25) B.M. Leung

A high resistande võltmeter is connected to the cell ter- minale and the reading noted. This will give the e.m.f. of the cell. A l-ohm resistor is now connected to the cell ter- minals and the voltmeter read- ing taken again. (Fig) Further readings are taken using 2,3

R

6. ato, ohms, and we use the following forme

ula to calculate the internal resistance of the cell.

R

R(EV)

ohme

The above formula is derived as follower

Lat R.

E

A

V

=

internal resistance of the call in ohme e.m.f. of the cell in volts.

p.d. across cell terminals when cell

is producing current in a resistance Rohme.

Since this is a simple series circuit, the

current is the same through both the cell antd

the external resistor. The current through the

high-resistance voltmeter is negligible.

P.d. driving current through the external

registor R = V volts.

F.d. driving current through the cell

- "Lost voltage" - (EV) volts. hence the current is given by I.

V

and also by

I

AMD.

V

EV R

£

R(E-V)

R

nh

(E-V)

41

JAMINA 2 WA

配位則成統嶼

*RY 23

千敬游

-

18 APR 1967

CITY HALL.

自然科學研究

經常舉辦學藝性活動

最近參觀金錢水瓶廠

- BRA*CXEX

·標本 - 放映科學影片,召州本

,行跟外活坳,如搜集,孟氏教育基金委设

科學之見開,經常學 木月十2日下午: 時 緻搬談秕祀设财自然入。茲敬就正江下i - BERLFa

川味

|+QWW - EKIS EXIRONTE 在阿時間內,與該記著麼戲會,其示佛 分始被京然項活動。調心向教育阴影,

【恆及招赉碓老師帶領 頂及,致爾外經營 「人,由徐鎛弱,郭志 @日,本人也就管晃 一,於是日下午三時二 首先需

「單位:計有玻璃

WHY-BAHAN 20 SE 10

• EARN-MEISTER -TK-

-

! 求水準。

各生列各款所熱水瓶 人才HR需變當中,

SARHES() #4

+

#

中文中學會考試題預習專欄

一九农七年

數學科 (廿五 )

第廿四次類習試題解答

(1) 341A : ca, 2(a+b-2x)=5{(a−x)(k-x)

*

日九十月四年七六九一座公年六十五嵗民赛甲

堅道書院發展

直要工作。 十三小翠,供漁民子弟此,此驾照 GERPRINZDAZĀK-OR }

|民予弟花費佻學校攻。 此外,魚統處並設置 學號,協做

分入,則由魚經流些爆助,而中帆讲一 其他小學攻者,出有些四一去,ㄘ, 上月底,留有魚類統營處將聞心而在

各學校,現有學生的三千六百人。 (26) BERGFEMALEYN 另有九百餘人座獎學金

推行漁民教育

三千子弟就讀

#

49x+126 = 63x

#x=1 # *-- -子(不合理) HAB# 4=7 & AB=AC+CB=9+4x7=96

答:两地距離96里,甲乙二人速度分別為每時中里及黑

(5)在同一坐標軸上,描绘下列两图(x取值由~3至2

lü) y=12x-34

A

喬悅强。

(i) y = x2-5x

(解)先製成下表: 4

問此两图线之交奌横坐標,是何方程式之根?又其根為何?

! === x2 - 5 x * === x*(x-5)

x-3-2-1

4-72-28-6

o 0 -4-12-18 | -16

} ટ્ 3 4 5

6 7 3698

(1= hely 2(0-x)-6{{4−x)(b−x)+2(t−x)= 0 分解

: 2[a-x--x) { $a-x-257-x)=0

Wafa-x-16-x-0

zla-x-

4(a-x)=b+x.

* x=(4a-b)

+

(i) ra-x-25Fx=0

Jaz

=

Q-x=4(b-x)

x=(4b-a)

因方程式两边根式均為正值,故不須驗箕知萬算根

答:x=(40-b)或(48-a).

4(a+b-2x)=25(a−x) (b-x) 然後依二次方程式解法能之,不過所得之根必須驗算

à)✯✯✯

витой амп.

(b) 12x=17x-6x*

(解)以空乘全式,並穆嘎 12×3- 17x3+

+6=0

V+IRI IRTIR

FTV HV = ITRAR

分解

( 3x3 - 2) { 4x3

()

x=

然後選擇適當之比例(橫軸每吋之單位,鍛軸每吋表40單位) **** 1* kit v z. 17 –“âr the #" (Cubical Parabola).

再因 y=12x-34為一直线,此直线可由(0-34)(7.50)泱 定之由两方程式谨去y得 x3-5x=12x-34

X-5x-12x+34=0為阿求之方程式 從图中得此方程式之根為x-2.95 1.90 36.05

By

4=12x-34

(b)(i) From the first diagram,

1.11.1 - (2r + 8 )x0.16 2r+ R - 55/4

From the second diagram,

2.2

0.2R

R 11

put R 11 into (1)

55

20 *

11

ohms.

. r 1.375

27

456

.(1)

± 2/3

Rojba

(a) 3(1+1)

(15-12)

212 3/5-6

3-2

原式

3(1+E)

2/

7- 2570

3(15-2)

Answer: The internal resistance of thế

Daniell cells is 1.375 ohms.

(11) From (1) we have found that

-1.3750 R 112

Therefore, if one cell is used, the current of the original circuit ie

___ \ { }+2[70+ } [2+2{20}

9 3

انادا

3(1+52) (7+2/10)

2/35+2)

(7-2/70}{7+2570) #{$5-2}{45+2)

2/10+4/2

= √ { 3 +352 + 4/5),

"(Vxhq) @} (TA+"{B } { '[A' -'[A® + '] 8* }={'TA}'+ (†8}' — A+B

故有+巨之有理化因式為F-YAB+YE

19-972+476

(F+7)(F-13-4+{4}

(3)(a) * 9x-12x2 + 10 = 11⁄2

(23)

+(19-Tz+ 2]Z) [4]

答:態方程式 ズーラズー12ス+34=根3-3.95.1.40 6.0 6)化 アスター×2+10:毎分項式

(x+1)*, (X-3)-

(c)

1.1.

1

апр.

1-3754

1.1

I ·

amo.

0.089

amp.

HONG

Answer: The current is 0.089 ampera

if one cell is used original eiranit.

in the

Let 8 1.05 microhms per inch cube.

Length or the aluminium wire.

A area of cross-section of aluminium.

0.2°π

으로가

#2- 0.7

1:00

=

cu.in.

microns per inon cube.

yds.

A area of cross-section of copper

→

au.in.

4

GL

From R.

་

A

"2 0.7 x 100 x

0.22

yds.

1.05 x 0.1

266.7

yas.

Answer: The length or copper wire 18

266.7 yds so that it has the same resistance as copper.

148.(a) Let R be the resistance of the

galvanometer.

80.505 1.

Since the full scale of the ammeter 18 0.015 ampere, and the new full saale

ie 1.5 amperes. the scale is magnified by

100%

R

From R

8

I

RR

R(N= 1 )

+

92-1252+10-

(解)因原分式為假分式(分子次數高於分母),故須先化带式

-7x+4

-X-2+ (x+}}2 (x-3)

(35)=9%

652-2

-7x+4

A

B

+

+

Roxy (x+1)2 (x−3)

−12/%+10

x+1 (20+1) X-3

~12[X+ 4

答:平方根薦

6X-4+

6

x

±(352~2+)

3.

-21+4=0 × 4*

-7x+4= A (x-3)(x+1)+ B(x−3)+ C (x+ 1}*

7+4=8*(-4)

7. C = -

---

两边龙之俦數

0= A+C

17

源分式

→ X-2+

16(x+1)-

AA=4

(答)

16(x-3)

(*: (£)=3(Tas)

x-3x2-2x-3x+1=0.

(香倒數方程式,除以8集項

分蛋

(x2++)-3(x+4) -

· 44 x2++=41-2, sa

(y+2)-3y-2=0,

(注)着$=4,則 化簡

x2-4x+1=0

1 x 4 t₤16 ~ 4 × 1 × 1

= zuti 13

.

x=2±3

x=-1±√1-4*1*1

2

±(-1±[3i)

LIBR

(4)甲乙二人,分由A,B两地同時相向而行,相遇時甲較已買行风

里、相遇後,申再嶝小時到B.乙再多小時到A. 求两地之距

·離及二人之速度

(解)設如右图,C篤二人相遇之處

A

→

C

則 AC-CB=12里又設甲乙二人之速度分別為每時不禁坚

c8-4x=44% Ac-7y=544

浓题量得或 544 1426 12

549

*

814-49x=126.

$!Y*=49x2

(1)

$(2)

A

以3) 代入(4) 去分母化簡

1 10

0.5050

--(4)

81%

(49x+126)*

=49X*

..

(2)

49x+126

B

第廿五次預習試題

̇17ABCD為正方形,延長AB以及P, A

ABX AB, BPBD DP BC, XC 於R及Q,試証 RQ=QP

2)設0萬△ABC内任意奌.D, E,F分

別為AB. BCCA之中奌;又PQ.R分

co. Ao, B80 ≥ 4 § à DP, EQ, FR=#

3) 在△ABC中試証

B

(a) (b+c-a) tan A = (c+a+b) tan = (a+b-c) tan &

b) a (cos A+ cos 8) = 2(a+b) sin2 G.

(4)(Q) ABC PLAK * L B C D F D LC.

BENDCAC延織を試右圏証明

(a+b) sin C === ccot 1⁄2 (4-8)

AABCH LAK LBS CARDAL

CD=CB.試從右图証明

(a-b) cos c

<=c sin ±(4-B)

(5) 7-3, E9A 1+ sec 20° cot 30° cot 40, Bl

(須用額角函数)

(6)(0) AB為半図え器AP平分

試証 2x + sin 2x —

vb t y = sin 2x 14 (x 1

口至1.6)年利用き以解方程式

2x + sin 2******

#

D

- OL505 = ( 100

49.995.

with full scale.

If the ammeter is converted into a voltmeter cale of 15 volts, then a resistor must be connected in series. Let R be the resistance connected,

then

15 0.015 (49.995+ R)

R 950.005

8.0159

Answer: The realstance or the ammeter is.

29-995 ohms The resistance connected in series ie 950.005 ohms.

-(-b)---In order to convert an ammeter into a ÑO IT—- meter, a resistance is connected into series.

Get Rohms be the

resistance connected..

8 1000 be the re- sistance of the am meter. The current of the circuit in

[R=100

130MA 2:48.

(49x+126) — 81×49==(63)

rul. 8o8l8 18 120 M. or u.00012 amp.

By Ohm's Law,

2.4 0.00012 ( 100+ R)

19900

Answer: The resistance connected should be

19900 ohme.

Topics for Revision thie week)_ Comparison of resistanoes by the meter bridge i comparison of e.m.f.s by the potentiometer.

QUEBT Lona...my

49.(a) Prove the relation between the resistannes in the arme of a balanced Wheatstone (meter) bridges would

to find the resistance of a length of wire as ac curately as possible

་་་མི་མིན་ས་སྐོ

(b) Explain how you would use a cell, voltmeter

and ammeter in order to find the valus of an unknown resistor. Indicate the points on which the accuracy of the method depends. 50. (a) When a certain cell is connected to a coll of 10 ohms resistance à current of 0,1 amp. flows through it, A 5 ohms coil is then con-‚' „nected_in_parallel-with-that-of-the-10-ohms

and the current through the coil becomes 0.18 amp. Determine the resistance and e.m.r. of the coil.

(b). A potentiometer wire.18 100.com long and has 1 resistance of 2.5 ohms. A Leclanche dell (e.m.fl 1.4 volts) being tested gives a ba lance point 70 cm. along the wire. Calculate. the current flowing in the potentiometer.