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日五十月四年〇八九一愿公年九十六國民奉中
磅,距由肝癌员白產士利安• 四十五分開跑,各郁馬蹄香 我爲九塲,下千一時。新力 本遞宋一十九日】 小人物
週末賽馬配磅評定
跑道移欄四班長途改式式五 公尺
打比賽爆冷列正選
一四〇一好好嘢
「啊,昨已進行移搽, 深比賽將在移探的跑百利: 二七
· 草地跑還受天爾影 好事成雙 一二九:羅|天之恩子
|定進行,因此-第四踮
四街
脚
一二七一切射突雄
DR
上配售周資格列爲正醒:
来所隔目的香港打
|天化時已修正爲二二孖 年比 -颇有一四媟冷以将爲第四班搿越級林
月,在节四四正落思
·翁嫩評磅,至於兩場港发 新码及香港打比赛,江湖好薰 中褘配獨佔五匹之多。狂惑
艾特醌磅列下·
-波斯砷
(1).
心
Ho
七四八八九〇六二入我
嫩嫩品較上過三疏落,但
歡我
快狀
做
歡呼快試餘勢強
試態:
勇兆
昨晨沙田煨後試跑,
沙田
1:1 2003
之下勇况初起,石果山 一無爹泡揚,勇將軍勱说 一跡象,小缸關攔少, ▲ 已其正估“甜惠有好轉
操
機會搶先出試,三段一分 - 告東尼登早抵變,眾多
事賽場六晚明 報綜操狀各 導合練態駒
蔡卓有取
咩味
多,配絡不足造人
水跟
。將態 脢盘 优势金慢
時空 有三操復
操近瑙,力的職 屉燕欠,不
郡说强 初多未進立
,足人不及缺稍,鑑,足操 操無
蛤蛋愛
「操繞耳質,祝動及水
進工
無進展,飛來了有起色,勇品未足得平常。
勇輔
:(第六)好運,
郭朱邦
,上味不過不失,砌成,稍有神殿,在高操,
護士女足狀態已术,更树動案之下。
步機亲近况平平,强烈一在導拣糖冷落,狀 亦無大作爲,安經最務繕米足,愛我動升班後, 拭测易醌走勢仍由“勇其勞正毀,肥見習生史·书 - 高超大傑
育教輕馬,頁三第張六第日一初月三年申庚夏 WAH KIU YAT PO
门
老敏
勇新
來詨騎齢误榮陀氧
(七):森獠
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好好强(由第五珽升上)
·盡此寶穿小成功一由第六班升上
(+) BEN
DER
勇俠、擔心,斋雷利,好巓意(由撼四班降
CTY ARE
勇馳升第二班
》 耍多好,
上選日夜燈之後,財會昨許定六四嶲升班(CF),接雙武
四匹馬降班,名單如下
·勇由三联升上}
培養
(七)先在
(大) 好開心・
(H) BE
(四) 有希望
·快鬟登,牌得嘅(由第四斑升上)
勝利將
*保海威欄或霞飛 歡燜約勝之
嶺英 大麻 蘇糖 財 拍者信㞊
報日僑華
心孩
二期星
鼠神龙絁銀前衛藝矼英勝火磻
門半快福 預兼茶駒
1111#
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蘇米雞賊能沈
康米泡在布磁齦 沈摩
頭館 古聯歡 伯發航鲜星心
「超山百小精清洗猛 石永先得特验证
郵發 紅。
郭新婚摩與沈陳蜇蹾
雞雞活
心旗臉火 二第
號寶
眠
减我
第二場
【香草美人鍋標近期揲速-
足,初
好巡二八步多,仍可再提,
二八, 來逃足,狀低氣及水
永老佔 紅線 天高 慳 友金天利大康添味好 旋職壓勝友先發漢 超冷恰
撟丁林军巡瑞或 翼于必翩
·燕山·發費城美
〇三 三九九八八八八八八八八八八八
繫強段 勁光袋=
·狀態仍是紙。
.長懮動,扭脚整齊而有勁,
由
尙末 我領苦行
李勇 七
主再,
共下末
持快李
新香 試森东錦
試出氣
第 三對均提足星勇,號 「搭威武、歡呼、真希 第十分經鑿,再乘再起
。
,
三八 鹿隼健仍足,大致保森_末段坎凼沙四,由脚頗勁,身型收好,脚恩大致 ,上週末棄惠不出留學段出现打,甚獷,但氣勢則稍欠,狀態者, 如意建由李廂主材,怏走半吧五十.一秒人
·我勝由國文就主試,半唯一分等二秒七、宋】
,好開心楜穗從未鬈
西班爭
氣蟹:
過一
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·南億-
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四維
快的
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中學會考試題預習專欄
Since the jazi
rfter collision is therefore
3(1)(v,) =(1)(1))(0.5)
物理
(Ans)
(八) 明書出版社魯榮家提供資料
STES(28)
Take the direction to- wards the
he posit
tive.
DAŽE PRESS LIÐ.)
solutions to
ECHANICS AND PROPERTIES OF
ATTER
Since the strings are long, inexten- sible and of negli- gible mass, there-
fore there are two
simple pendulums.
The pendulums have
same. Length, refore, they have
the same period. As the spheres are released at the same time. the time taken for the spheres to
each their loweste
position are the
same and are each
equal to one quarter of the period. fore the spheres will
have a head-on colli- sion at their lowest positions.
where-
(1) Let the velocit- ies of A and B before
collision be v Cand
respectively.
By conservation of
energy
迎
(3 m (Ans).
(Initial acmentum
(2)(3) + (1)(-5)
1. kgas:
FIhel momentum:
(2)(v,) +(1)(4)
conservation of linesr
(Ans):
(111) Let the maximur height, reached by A after collision be a
onservation of energy -= (2)(10)(h)
The average force on
A during the callici the rate of change
momentum of Aw
Change in, momentum
-(2)(-372)-(
the force on
-9/0.2
-45 N
(Ans):
(The negative sign indi- icates the force on
to the left.):
(v) mergy lost
3(2)(3)-
(孟(2) (3)+
10.25.J
4(1)(5)2
4(1)(4)2
(Ans)
The energy lost in this impact will dissipiated
in the form of heat and sound.
B(1)Kug)== (1) (10) (1.25) 2(a) Let the reactions at
sion-be
-10
Cine)
velociles
olli-
respoc-
Pand Qbe Ry and R2 respectively.
As ahowm in the force diagram below:
·Mg(3) mg(QB x) = 0 (50)(10)(3)-(75)(10)(3-x)
(Ans.): When the man is
1 meter from B. the
planck will tilt. memass of the man.
75 46
"= mass of the planck
50 kg
Since the planck is in equilitrium the result- ant of forces acting on the planck is zero.
rg - 1g=0 750 -500
1250 -- (1)
ne sum of moment
about any point must be
make moment about:P; R2(6) me(5) - Mg(3) R-(6) ~ (75)(10)(5)
-(50) (10)(3) = 0 = 875 N (Ans.) From (1),
1250 - R2 1250 875 375 $ (Ans.) Suppose: the planck
will tilt when the män is x meters from B. Hence' the reaction at
Take moment about Q
=口
3(a) The factors. that
affect the efficiency of a single string. pulley
are: the effic-
1. Frio
按鹨哩魚
我仍五立
Lency lowers when
friction increases Weight of pulley and string: the efficiency lowers when the weight of pulleys and string.
increases.
The efficiency lowers if the portions of the string are not vertical. (b)(i) The required veldc-
(Ans.) ity ratio is 5. (ii) The upthrust of
water on the cube B
the weignt of water displaced
(iii)
(1000)(10)(10) 10 N
(Ans.)
wekter
Let the pulling: force
exerted by the string on the cube be 1.
Since V.R.
5 x 10
80%
Therefore
10
80%
T = 2 x 105 N Let the acceleration of
the cube be a,
apply hewton's 2nd Law
(1 + 5) −ng = ma where m = mass of the
cube
(2.5x1000)(10)
水麵
= 2.5 x 104 kg
..(2x10+105)-(2.5x104)(10)
(2.5 x 104)a
-2
2 ms (Ans.) (iv) The cube will finally
stay at the water surface with part of its volume immersed.
Let the volume of the cube jumersed in
water be ve
a. hence
mg = 0
(T + B') Where B is the upthrust exerted on
the cube.
(2-5x10) (10)
- 20x105
0.5x105 N
But B1 = (1000)(g) (V') .*. (1000)(g) (V')
0.5x105
Therefore, the cube finally stay at the water surface with half
of it's volume immersed.
(v)Inorder to push the
whole cube out of water, the pulling force exerted on the block musta least equal to the wei- ght of the block in air.
ence,
the pulling force
2.5 x 10 N.
Let the effort required be 1
Agains
2.
2.5x
80%
.25 x 104N
the additional force: required
6.25x104-5x104
1.25 x 104N
(Ans.)