1966-06-14 — Page 22

(1)

(9)

The required difference is 21 68

百六月四年年丙胶复

WAH KIU YAT PO

其中F英數先修專欄 ·金繼賢。

中中會考物理科答案 (續) 蔡淦明

期三星、日四十月六年六六九一四公年五十五网民靠中,

故液体之比重了為

EXERCISE 6 Answers)

W-W1 WW

(3)

I spent 20 minutes riding on the toe.

(4)

"The man's average speed was 240.p.a,

(5)

何謂速度白速率,其分别為何,該分別說明之 圖示一物体之速度為時

該液体之密度P為

W-W W. W

速度(买/秒)

式中為水密度

Pw= 1/8*=62.4/p

(2). The required sim 15 109. The requires product is 315000,

The required quotient la 50 eg. L.

(6) The cost is 816.80. (7) 21 hours 30 minutes.

(8)

The required difference la 1 con 364 lb. 12 oz.

(9) I need 36 penoles.

(10) The hall holde 960 students. There are 68, empty seate (11) He had left 80 eggs. He received (£1.

(12) He will take 2 hours 18 minutes 45 seconde

(13) He cycles 432 miles

(14) He kept 80 marbles.

(15) 6gbarte of milk are needëä.

LAN FACTORS

CHAFTER SEVEN

FACTORS AND

PRIME NUMBERS

Since 153, 3 and 5 are each called FACTOR of 15 and 16 called t80 MULTIPLE, of 3 and also the multiplə

OP..

(B) NUMBERS

(1) Number of which 2 is a factor,Tare called "EVEN

NUMBERS eigl 2, 4, 6, 8,

këtë

(11) Any number which is not even is called an ODD NUNDGU

e.c. 1 3. 6. 7.

如在時间為9秒時甚

·瞬時加速度為若干、

(b) 在時间為多秒時其

·瞬時加速度為若干,

(6)在時间為13秒時其瞬

時加速度為若干

時間(秒)

小在首9秒时间内此物体所移动之距离為若干;

A number 18 said

be primefiċlitis?divisible only by

解:物体递劝吁,其位移而所歷時间之比,称為速度, 若物件之運動三論其数值,而不同其方向,則其

·移动之路程而時间之比,称為速率,以算式表之

速度=

时间 路程

LUCY PRIMB NUMBERS..

itself and one:

(D) COMPOSITE NUMBERS

A number which is not prime is called a Composice Number3

It can be written as the product of two or more or ima

numbers, i.e. Prime Factors.

NOTE: eg。

etc.

| For example

that is

et cotara

(8) INDEX NOTATION..

:(1) 4 x 4.18 written as

(11)

the équaro of 4

called

4% 424 18 written as 4

he cube

quared

called. " 4 cubéd. 16 4x4x4x4x4x4 is written as 4o called

4 to the power 6 "or" the 6th power of In the symbol, 4o, 6 is called the index.

{{P} PRIME NUMBERS within 200

3

3.

23

29

5 31

17

19

47

53

69

61

83

97

201 103. 107:

109

127

137

139 149 151

157

179 181 -191 197

197

̇BXAMPLE-4

Express, 480 in prime factore

4802 x 240

在此西式中,时间、路程為意向量,位移為向量, 故知速度為向量,而速率則為世向量。 简言之,連率是光研究物体在单位时间内程方 之距离而速度則除研究此数值外,还研究其程 动之方向。

呎/秒 未速 (2村)= 44呎/秒

※時间 = 12-6

初速16秒瓜

時! 44-41

加速度一

2 120

2 x 60 2x2x2 x 2 x 30

2130

2 x 2 x 2×2× 2 2x2 x 2 x 2 x 2

15 x 3 x 6

Therefore

SXAMPLE

Find the sum of the prime numbers 3 between 50 and 70:

equired Bum

6167

ine roquzrea, sun or the prime numbers between 50: and 70 16 240-

来连(6秒)000/秒

初速10) 20呎/秒 未速(6吋)=44呎/秒

时间 = 6-0 = 6秒 加速度-44-20

=44 /秒

磅?

在英制中

WW

設此銅球空心部份之体槓為7.

48F] oz $//fe = 43 14 7 = (445-3251/0.8

=150厘米3

8.911150-7)=445

100 R$

备該銅球空心部份之体積為100厘米

ک

三动有一定方形之物体重1千元静置於水平面上 物体用平面间之静摩擦係數為04其滑动 摩擦係數為219)作用於物体2摩擦嗎 若干个(小)砍便物体用始運动,至少高力若干。 5,設有200克重之水平力作用於物体上則所產 生之摩擦力為若于,(0)物体用始作加速運动 後,作用於物体之摩擦力為若干,

(D)有一均匀之棒AB;長320厘米,重7千克 C端

以一軟绳懸之,此棒浮於水中如右面所示棒 之四份一见没於水中試求绳之力及棒之 绣体積(水,密度,克/厘米

北厘米

(1) HIGHEST COMMON FACTOR "H.C.P.1"

(A) & number which is a factor of twe or more given aunders

in called a COMMON PACTOR, of these numbers,

(B) The greatest or highest number which is a common factor

of two or more given numbers te called their HIGHEST COMMON FACTOR or H.C.P.

EXAMPLB

Find the common factora of 36 40 48

Answer The requirse common factors are 3.

BXAMPLE 2

Find the 8,C;

72 and 488;

36

288

The highest power of a whịch le a factor of eman number the higher power of 3 which is a factor "of each numbe

100 and there 3000

common factor

36.

Answer: The requi 60 H CP of 36 72 and 388 1= 36

EXAMPLU: 3

0437. And,667 by the Dav10100 Method

432

230

·667 4373

·307.

$230

207

207

"Answer

23

EXAMPLE

Pind the largest number which numbers 504 292 and 2016.

504

126

required aumber,

CA)某散假如其爲两個成身

(B)两個或多個已知數。

DIEAR36 072 ► 288147

例文用輾棺相除洗

each of the

2396

(因數及質因數)

張芳24

最大公

吕知數的公因數成公約數。

5729

10] a 16-12 = 4 加速度=

02 6 7 27 10 3 ) = 35 (20044)

32.0呎/秒

0至6札移动之距离=3216

6至9秒移动之距离= 44x3

= 13呎

049秒移动距离=192+135

合:四冬(6)4 -1105/27 10.304呎

(三)就述阿基米得原理及如何應用此原理测定一

疸体密度

一支酒之銅球在空氣中重445克在- 比重為 0.8之液体中曾325克,銅之比重為8.9試求 此球内空心部份之体穫。

各物体在流体中所減輕之重量(或称浮力)

手於该物体排用流体之重量此称阿基米 得原理。

在應用此原理测定液体之密度時可依下 列步驟!

(1)取任一固体,此固必須,一在水及該液体

中不溶解昔古化学变化。能沉入水 五該液体中

(2)測得固体在空式中電設為內

(13) 測得固体在液体中之現曾設為Wri

(4) 谢得固体在水中之視電設為2, 根据阿基米得原理

而固体同体精之液体重=W-W

(a) 若量任何力加於此物体上,且此物体

是靜置於水平面上,則根据牛頓浮三運 动定律,其摩擦力為零、

1000×0.4 = 400 K. (C)若加力前此物件為静置者

則 200克-静摩擦力

故此物体不会移动,且其時;摩擦力再作

用力相等,即200元

若加力前此物体经已在運動中

則

作用力(200克)一动摩擦力

故此物体亦不因之星士加速度,而其時之摩

擦力仍為200元

7000$0.2=200 £

(11) 設經之張為T. 污力為F.棒重為W 以B為轉軸

FICO - WOB=0

ΣFy=F+T-W=0

CB=-320-40

= 280厘米

08=1320=160 kt

W-7仟元

280 F - 160×7 =0

解得

F+T-1

4克

1 = 7-4

3. 仟元

在水中部份体横==水密度

棒之飨体積=400ov.4

4000厘米

=16000 厘米

后海冬 (bì40克(6) 200克 (0)200克