#FOREI
LIBENT
育教僑華 頁三第張六第日八十月四年戌庚夏 WAH KIULYAT PO
報日僑華
2.2MAY 1970
CITY HALL
五期星日二廿月五年〇七九一曆公年九十五國民華中
英中會考附加數學(三)答案
in positive downward
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(讀)
堅道英文院撰答
Solution to H.K. Certificate of Education
Examination (English)
Additional Mathe, Paper III (Mechanics)
Section B
Si
中文中學售
試題預習專欄)
A
Component in the Ox direction
3500w A. + 29coad
*
52co
Component in the of 2.
13501aß
520227
35 • 2 - 52 •
2961 A
29.
--19 1b.ut. (negative Of direction)
Resultant moment anticlockwise direction
taken: to_ba positive)
-29 • 3eixα 52 • 2008/7 4:35
64 ft.lb.wt.
When the block is at equilibrium against elipping
P = Wainỡ + UWCOB
(1)
Take moment at C, the blook will overturn
P+2L > ¥• CE.
When
WCF LOOB A
HL(2tand + 1) c0s d ML(28ind
→→
COB X
(2)
From (1) the block will not slip but from (2) it will overturn.
11. a) Retardationg gino
0.4 Pt/
4000
b) Acceleration seen from the notion of the car
:20m
15 m.p.h.)2 - (30 m.p.h.)2
2 ( mile)
11.
2
Pulling force dus to the
4000 lb. (0.4 - 10)
800
Work done by this foron x y z 5280 KTIJDENT
物理科 (廿九)
鄧炳恩•
N. 直流電及歐姆定律問答計算題解
·肝()賞電镢s關閉時工=3安培,由歐姆定
A
R.與R,互联其「可 危阻
4x6
416-24in
(R2+R) × 3 = 12
R1 =42
R2 =4-24=1.6 (1)
R=6
(b) 當S 開啟時,設其電流為工,则
12
6716
=1.57 (安培),
Vor = Vig = B. C 点 之電位差
−1·57×6=942 ((**)
2解設電路中文
電流為工則
刻
4+8
0.576+8+8:5+9
0.5(安培)
4
樊a点之電位差,則
E=8!
Ve - Va = (8+05)x 0-5-8 = -3-75(伏特)
即a点之電位较e点高3.75 伏特
設VC為E電池两端之霓压则
V1c = E, - Ir1 = 4 − 0.5×0.5 = 3.75 (UK°*)
及 工 = E/ CR+Y)得
3解: V=E -IY
Loss of Kis, of the dar
41,250 T1.id.ut.
ERE
RTY
==
YEL
• (1)
营1几與3丸之導線至联,其電阻為
R
In order to produce a total positive moment.
The direction of the resultant of X and Y is ahown in the Magram.
From the diagram tan ✪ 19
86 59*
Resultant of X and Y is √(192 + 12) - 19.0261b.wt
The I intercept of the line of resultant fora. is negativa and ite magnitude is given by d:
19.026 x doos86
4000(442
,904,000 ft.pal.
90,750 ft.lb.wt.
Gain of P.B. - ngh
4000 x 32 x x 5200 x 100 ft.
105,600 ft.lb.wt
f) Work done, against resistance
10g x = 5280 ft.pdź - 26,400 ft.lb.ut.
By conservation or energy work, qané by the pulling force ie (105,600,+ 26,400 - 90,750) ft.Jb.
41,250 ft.lb.wt. the same as in (0),
19.026 I OPB
63.92 rt.
ft
M.K.S, unite
Forve
Angular speed v
258
0.2
they intercept 18 -63.92 ft.
2L
When P la just large enough to cause motion the bisok will begin to move up. Friotional Toros F is given by
A woond
2W
When F
it cannot
wtart the motion of the block upward. It is greater than the downward component of the block Wein or W. So the
9 block is in equilibrium. From the condition of equilibrium frictional force is downward
2W
and its magnitude is given by 2~ - ~ - ~
The minimum value or to pull the block. upward is
downward
when of Pㄨㄢˇ
35 radian/
When the spring is inside the ring angular speed is increased to 63 radian/s
Force towards 0 is mou
1x 0.2 x 63
81 kg.wt.
Normal reaction on the spring
81-25 - 56 kg.ut.
Moment of frictional force about
56 x 0.25 x 0.2 n.kg.uk.
2.8 a.kg.t..
Power to overcome friction.
56 x 0.25 x 0.2 1.63 176.4
velocity of truck after K, 18 projectea
If V2 is in the opposite direction change
in (1)
ty after projection wELL DIOSES
= 0.5 (5)
=15 (V)
0. 電流之各種效应填-
題
並联之電灯150 枝 每枝之電阻 120 歐姆美 两端之電压為100伏特,由一發電机供给之 使10%之功率耗於電浆内组主上馬力等於)
馬力 746 瓦特则萨發電机之功率為
240燭光之電灯20枝每燭光用電卜4 瓦特 其電压为220伏特設每度電之電費為0.28 无則所用之全電流為每時電費多 3.放電阻線於1 立升之水中歷時1分鐘蒸温 度增高10°C,電阻缘两端之電压为100伏特 則
阻缘之電阻為.
4在硝酸银溶液中通以1安培之電流已知镢 電光当量为0.00 11 83.则欲析]克之纯银 需時。
5.三侗串联三月聶耳電池與銅電量計相連結, 時内祈出铜 31.7克已知钢原子量為
634,锌原子量:65,則同一時間内電池所祈 出钢为 及溶解之锌為
6.其電流於1分40秒内析出银0-112克於两倍 時間内析出钾0.081克已知银之化学当量為
正切電流計與 - 電池及一银電量計相串 联,斜之偏向为5於1小時内前出银0:105元 无鼓银之電化当量为0.001118則此正切電流 計之常數
8.正切電流計之缘圆共2匝半径15厘米與铜 電量計相串联:設偏向为60,且其磁场强度H
18 奥斯特,则於15分鐘内 出之銅步 正切電流計由三匝線圈组成其直径为30 厘 米通以085安培之電流則美 中心所生之磁 增强度為
毅当地之地磁水平强度為 0.17 奥則磁針之偏向角度為
電池一電流計與正切電流計串联,正功電 洗計以直径40厘米之十匝線圈组成,其偏向 為42,電流計指針所示为0.5157 安培則 当地之地磁水平强度為。