1965-04-07 — Page 15

百三第張四第

WAH KIU YAT PO #y

1965 中文中薈

試題頭商專欄

病乙年度正

考試題預習

欄

數學科

(十八) 。喬仲強。

物理科 (十八)

陸永熾·

1. Solution

(A)

模擬會攷試題(乙組數學試二)

(試題分甲、乙两部,甲部四題全做,乙部選做四題,共八題完寒! 作風題依設長度作園、窩出作法、本領証明、保留作銭)

[甲部(全做)

)(a)✯✯z‡o≤ A B C 14-$P. MI APAB¦APBC:APCA=1:2:3

泥)t△ABC之直角A之分角线,典由辩论中奌D时作斜边之垂线DE交

E. DA=DE

(A)在△ABC内作AD⊥BC,其垂趾為D,護由口至AB及AC各作垂线 DE及DF,其垂趾E及F,試証B.C.E,F共国.

(8)以直角三角形之一腰為直徑作,則由园與斜边交奌作此國 之切线,必平分他腰

I

3242X: (I+ sec'd tano) ( 1+ Tańß sin☺)

=(1+ and3ß tañə) (14 tamix sin()

在地上某兵測一塔,将仰角29,再向前行100呎再測,得迎角 萬480 求塔高

4. 14 y = 3am x + 4 ca x )] 1' (x*£ do° £ ̈°?%),£4!A÷

PA 143 sin x+4 cm x = 3.5

II部(選做問題)

5. 求作一個等边三角形,

,使其頂奌分别在三已知平行线上”

6. 两国内切於A,外国之宝 BC切内国控D.則<BADLDACT 9.樽形ABCD中,延長AD與BC交拾P,又对角线BDIAC 密拾 Q;

ASE PO & PÒ TH¶) 2 MÀ ABR.CD.

d. (W) IQR LABC II ‡ @‡‡‡; AŽ 611.

A=2Rain A sim B sin C

試試

(証明 (b+c) co+A+ (c+a) cm8+ {a+b) cos (= a+b+c.

1.3m 28 = 1 + 2 cor 20 (**«»* 360°è x }}}

10. 在塔之正西A桌,測得一之仰角為25,從A向南偏東560

全方向前行500呎露之B矣,則測將塔在日之比偏東

求塔高

a (cus A+ cos B) m 2 (22+ b) sin2 £

11. AABCT. E

第十七次試題解答

(1) AE fa

(10)

)+12-12-13

} } & = −7++ 6+ 7+13) (1+18−12−(5)

(1+2/7+6)=(2+2/T+3)

(1+16)=(2+{

(七)在口哩之路程中,苏司横驾驶平均速度為24哩小時器其點 初3小時之平均速度為25哩/時,試求在其餘路程中之半的速度, (解)共特時間=120÷4=(小時)

最初小時行路5哩X86哩

(2)(2)甲えぬえ善ええ牧入え60%申き支出為でき70%美甲 乙之支出之5:問甲之儲蓄版配之位等顯之分

(解)韻乙禾收入勇义,支出笃4; 则乙>作儲蓄顒弟(wy.

X 9 2 14 x 3 0.6x, £23 0.74 42 Kik 211 306x-075)

又因 0.6x=0.758

x=1284

饭两人储蓄之比為

答:甲之位儲蓄為乙儲蓄-20%

有一不準確之時鐘每日快30秒,若此鐘於今晨9時軻正

間次日上午此鐘好指時間為9時其正確之時間為何時?

解) 1小時 =

•=60160=7600 47,

1 14 4 th it s600+30=862044

今暑3時を翌是9時共歴 24小時(木準確之時間)故準

7% of På 3 xu 03 18

36:30: 3600 === 24. X

2 3 1 98 48 173 +91 98 -2404-80.44

答:正確時間為 8:048

(3)()()差因子分解能擴展至度数時,x+y

(u) to x+y=Q; A x, y +í § 2 ¤í ti

(A4) (

+ y2 = (x+yx) (x−yi)—o

(a) A transverat`vavéTMis ́one"thatˇeuuses the particels of a medium to vibrate at right angles to the direction in which

the wave is moving. Example: water wave,

A longitudinal wave is one that causes the particles of a medium to vibrate parallel to the direction in which the, wave is moving. Example: sound wave,

(b) When a sound wave falls upon a hard Burface TC 18 re- flected. Echoes are the results of relfection of these sound waves. Due to the persistence of sound in our ears, a distinct echo cannot be heard unless the reflecting sur-) face is more than 55 feet from the source of sound.

When the reflecting surface is less thân this distance the echo follova so closely upon the direct sound wave that they cannot be distinguished as separated sounds. It merely impresses as the original sound has been pro- longed, This affect is called reverberation

{C}

ན

Coz

3 sec. after

another

andher

1 min. ister

2. sec. after,

Let S be the distance of the ship from some cliffa ̄in ̄ft., and x be the speed of the ship in ft./sec.

Since sound cakes 3 sec. on the first occasion to travel from ship to cliff and back, the distance travelled;

(28 - 3x)

+ 1100 3

(1)

On the second occasion the sound takes 2 sec. for the double distance, therefore

{S

60x) - ($

T

62x)

- 1100 2 2

2S - 122x

2200

(ii) - (i)

(ii)

1100

- 1100/119 10. per sec,

9.25 ft./ec.

555 ft 1,

6.3 m.p.h.

* into (1)

28 · -

81

3300 + (3 × 9,25), 1663.9 ft.

·

Ana. Distance from the cliffs when the first giren is)

Bounded 1663.9 ft.

The speed of the ship is 6.3 n.p.4.

20120TOS

(*) Noise & Mu

Musical note

A noise Or

of its vibration:

have

combinat

irregular isduu

Every

(distingu

musical note depends on the regula rity at a regular rate, and

produce musical not e sounds. Noise in alconad <constantly varying pitch or

quality of a musical note

three characteristics that

(1) Pitch, (2) Intensity,Zand (6) quality.

設一為好求直线,使典曲线相交而通合方程式

(9x) a k

*1

4x-£x+($£−36)=0

E D

九二3話同時通会((2))

bk Ã3s y = 3 2 £#£11✨ #✨# x=-117* x=182, 此馬方程式之根, 答:x=-117 或1.92 (註)有時須設好加之直线总y=mx+龙方能適合

日七月四年

一腐公年四十五國民静中

(1) Pitch Pitch means the lowness or highness of a mum à ca

nate, The pitch of a musical note depends upon the fre

quency of vibration of its source; the greater the frequency of vibration, the higher will be the pitch of the sound produced

(2) Intensity The intensity of a sound or musical note is. defined as the rate of flow of energy per unit area per- It is directly pendicular to the direction of the wave, proportional to the square of the amplitude of the wave of the sound.

(3) quality. Generally a musical note consists of several different frequencies blended together. The strongest audible frequency present is called the fundamental and gives the note its characteristic pitch. The other fre-) quencies are called over-tones, and these determine the quality of sound.

A note whose frequency of vibration istvice ̄that_of_another acto is called the octave above the latter note,

150 x 102 (c) First frequency of siren

60

cycles/sec)

255 cycles/aec..

Let n, and a be the respective frequencies of_the_sounds] and N the frequency of the beats.

Then..

-

N.

[ 255

D= 256 cycles/stor

Second frequency of siren、

Ans.

• 150 103/60 cycles/secy

2575 cycles/set,

N

N 257 256*

14 beats/sec.

The frequency of the fork is 256 cycles per secu The raid of the beats is l1⁄2 beats per sec,

(questions (19)

1. (a) How does the ferquency of vibration of a stretched

string depend on: (i) ita length, and (i) che tension? (b) A wire of fixed length is correctly Luned to various

forks by altering ita tension, with these results: Frequency of fork, a, per seo. 250 320 382 512 Tension of wire T, ga. wt.

1600 2500 3600 6400 Using graph paper, plot a against square root of T, and from the graph find the tension when the frequency is 480/aac,

(c) When a sozometer"wira ̋is adjusted to a length of 60 cm.

and plucked it produces the same note as a tuning fork of frequency 256 vibrations/sec.. What is the frequency gof a tuning fork which would be in tune with the same

wire adjusted to a length of 40 cm...and_at_the_aame tension?

What is meant. by reagpance?' (b) eserine and explain now you would measure the velcoity

of bound in air using a tuning-fork of known frequency. (c) A vertical glass tube is filled with water which ia

allowed to run away alowly at the lower end. At the came time a tuning fork is maintained in vibration above, the upper end which is open,As the tube as emptied, two positions of resonance of the air column are observed, the distances between the corresponding water surfaces being 2,75 cm. Find the frequency of the fork if,⠀⠀⠀ during the experiment, the velocity of sound in air in the tube above la 330 metres per sec State, with an explanation, how the above dis "after_if_the temperature of the sir rises.

(8)解聯立方程式:

(AB) A✯ TO

(3) 74 PA(4).

< y=5+^{3), 10(x++)-

જ

(9) at

(正)依,反比

ance will

-(20)

Mp ay(x++)=

骨(+2)=

各值不茑使合母為零 故善真根

或之

(4)

4-2, y=-5.

比值茑龙

則

y = Pex

a+b== ka

(temp)

左右式

QED.

(b) h log. (X+4)+ loga (y + 1) = loga 8 + loge &+ loga Y.

(解)由対較定律得

loga (x2+4) (y^+ 1) = loga #xy.

*******

(x2y—«xy+4)+(x2-4xy +44

x, y 1k B 1 B £1

=+2, y=±1 (121 K 7A)

(4)绘

肉域(取値由 33棋軸ー 表示等位) 玉砕棒如何利用薗待以解村式なズールアー

·0,

2

40-12

(5)-泳池長60呎闊4呎 -端深度3呎他端9呎,其底 為一傾斜、平面、問當地水浸至池底一半時,其亚容之水有若干加 6? (134% = 448 ^_-)

解)設如右图EF為底 ABCD之对邊

中央联线则EP=$X6呎=3呎 而EFGH之面積为古x60x24方呎

A 1080x67-6720 1A (7)

108023

(6) 某人以鱼服 8銍之價格售出票面5鎊股息7%

止股票40服後将丐得之欸購入票面10先令、股息

9%之股票,市價為每股24克先令問購之股票若干?又其股 息增诚若干?

解) 售得之款亏8X420=3675铸

購得新股票數為3675.

比对股息減少

原有股息5×77×4

新股票股息支x92x3000

1201

(7)不許用

(解) 分子=2】

$7000

= log (int x 22")

3

= log(27) = log = log = log(2)=3log

2/

原式一般 (2)

a2 (k=k+?)

(10) 一大車於144哩→途程中,若速度每時增加60哩則可早到

20分鐘求其原来速度

解)設原来速度骂x哩小時,依題意得式,

去分母,

432(x+6)=4]*% ++(x+6)

6x-2592=0

(x+44) (x−48)=0.

x=48-64 (7-639)

原來速度為48哩/小時

(11)AMBA: (a+d} (b−c)+(8=d)" (c-a)+(c-d)^(a−b)

{A})/Rk = (a^-zad+d'})(B-6)+(8-2bd+d'} {c=Q}}+{c=2¢d+d')} {ab}

=d2 [(b−c) + (c-a)+(a−b)] − 2d { alb-e) + (c-a) + C(a-b)};

+ a(b>c}+b^(c=Q)+c2(a−b} [it d' d=#{\*X]

=a^(b−c) + b^(c-a)+c2 (a, b) =a (f-c) + b c-ab + ac-bc

a^(b−c) + l'a(b−c)-a(btc)

• (b−c) [ a2+ be-ab-ac]

- (b-c) [a(a+b)-cle-f)]

· (b−c) (a− b ) (a−c)

-- (bnc) (c-a) (a_b) (%) (註)本題亦可用輪接对精式分解之

REFERENCE. LIBRARY ART

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