REME LIBRARY
育教僑華 頁三第張四第日一廿月四年子壬腦夏
報日橋
WAH KIU YAT PO
CITY HALL
五期星 日二月六年二七九一屣公年一十六國民華中
英中會考物理科答案(二)
1922
中文中學會考 𧈢顒預習尃欄
PHYSICS (32)
SECTION A GENERAL PHYSICA
2. (a) än (experiment was performed to
assure the acceleration due to
gravity by means of a simple pariw dulum consising of a length of string and à spherical bob,
(1) Draw a diagram to show the effect-
hive length of the pendulums (A1) Bay why a sphersoal bob Luisabi (111) Show with the help of a diagram.
What is meant by one complete oHG-
(iv) How would the period of the pende
ulun. be affected if the gaan of the bob were doubled?.
(v) What would be the effect on the
period of the pendulum if its length were increased to four times the original length? (vi) a number of different lengthe of
pendulum were measured during the experiment. If one particular length of the pendulum gave an approximate period of 2 seconds, how many oscillations would you may should suitably have been timed?
(vii) Explain briefly why it is better to time a number of dacillations than just one oscillation. (viii) In this experiment, what is reger- ded as a small angle of amplitude?
(b) A pile-driver of mass 2 kg is raiømdi to a height of 5 and allowed to fall on to a pile of manu 8 Kg, to be driven into the ground. kebume that the acceleration due to gravity 2日 10m/20
(1) Calculate the velocity of the pile-
driver just before it strikes the pile.
(11) If the pile-driver and the pilu
move together after impact, calcu“ late the common velocity.
(111) If the system comen to rest aftez penetrating 0. Im into the ground, osiculate the average retarding force exerted by the ground,
(a) {}
(11) á spherical bob is used to minimiss
effecta of air resistance.
(111)
ہے
1 complete Oscillation
(iv) Period remains unchanged if mass
of bọb la doublec
{v} it becomes 4£
becomes 27
1.0. The period is doubled if the
length is increased to ¿ tiusm \ts original length
T20
- K
ረ
Ásauming least count
pr
atop watch
1
aec.
-
{ (vi) Approximate period is
% of accuracy deeired
Then time recorded
1 tloe recorded
1 asc.
This implies that the illations to be
Tired
(vii) It is better to
(viii)
5%
100
。
20 seca.
number of 080-
20
- 10
a araber 01
ogulilations than just one 08011- lation because this minimizes
experimental error due to reaction
time of experimenter and instrum mental error.
以<50 1 Considered
a11
2kg
100/2
((1) By principle of conservation of
GASTLY
2 x 10 x 10m/#
ANALAN DE MOD
(11) By principle of conservation of
I
MORRATURE
2 x 10 + 8 x 0 = (2 + 8) = conson
Common velocity m
20
-4
(111) Apply v2 - 42 - 240
C
甘 final velocity 。
13
tal velocity
/
velocity
8- distance travelled = 0.18
aw
• average socalaction
- 2 x & x 0.1
— Average retardation = 20 m/s2
Retarding foros -
-
10(20) → 10(10)
+
JOO HOVEOL∗
SMOTION B
REAT
Botes
Betarding foros
8g) = ma
kala) In an experiment to find the ace?--
/101ent of apparent expansion of
(1)
aliquad 4, the following results were obtained.
Mass of empty density bottle
- 55:5 €
Kass of density bottle when filled
IN LIQUE & et 3090 =14565 6
Mass of density bottle when filled. with liquid à at 100°C=143.0 Calculate the coefficient of apparens expansion or liquid di (11) Name two precautions which suzy
be taken in order to obtain godo resulte in carrying out this
b) Explain briefly the behaviour of ont
modecule of a liquid during the process of evaporation and subasqu- ent condensation of the liquid,
a) State the effect, if any, produoed
од
(1) the melting point of ice by an
increase in pressÜLTO,
(ii) the volume when ice changes to
water at 0°C,
(111) the pressure of a saturated
vapour in an anolosed space when the volums of the apude is dear- sased at constant temperature. (iv) the relative humidity of the air
in an enclosed room when the temperature of the air decreasing {v} the relative humidity of unsatu*=* ated air in an enclosed space when ita pressure is increased
et constant temperature.
(2)
a
empty bottle
at 30°
- 55.58
Rado - 145.5% *
物理科(卅一) 一陳新一
總複習!
(本復習依照會考試題形式) 考生編號
座位編號: 試
喝
(兩小時死巷)甲部總分_
本卷分甲乙两部:甲部十題全答,共 佔40分,答案須填入空格內,計算方法 應寫在每題下之空位内;乙部選答四题 共佔60分
試題紙須連同試卷繳交。 甲部
(答計算題而不寫計算方法者如
(一)在彈性限度内,在彈簧秤下端懸一
仟克之重物時,彈簧秤長度為50糖; 若改题15仟克之重物時,則其長度為 55糎.求在未懸物體時該彈簧秤 之長度應為
厘米。
[註](因篇幅問係,本復習將每題後之空
住略去
(3)
(二)一單擺,擺長為9呎,招幅為5: 錘重為言磅,則其週期為3.3秒。 (2)當擺長不变,振幅不变,而錘重增 秒 至兩倍時,週期將為 (b)當擺長不变,垂重不变,而振幅增 至兩倍時週期將為 秒。 (C)當擺長增至16呎,振幅号锥重皆
秒。 不要,則週期將為。
有一句截面之細玻璃 管,垂直放置,如圖一所 示,其B端封閉·M為水 12cm 銀柱,長12糖,5為乾. 土爆谷氨長8糎。若將 該管倒置時,如圖二 所示,若當時的大氣壓 力意76糖水銀柱高則 則空氣柱之長見應為
糖。
(四)兩接觸面互相作用之極限摩擦力
與接觸面之.
有間,但與接
觸面立
間立
數恒精
.無間;又與接觸面 功成正比,該比例常
設將一物體置於與水平面成30°角 土斜面上時,則適可滑動,該物體 與斜面間从之值為。
(3)在一部閉之實驗室內,在15°C時之相
41.700°C
mass = 143.95
(40
pooled to 30°C BASE = 143. Og
Consider the 3rd and 4th figures. If the temperature rises from 30°C to 100°C, the volums in figure 4 expanda to fill the bostie. (see figuro 3). The volume of space above liquid in (4)
- (145.5 143.0) x density of liquid
at 30°C
The volume of the liquid in (4)
(14) - 55.5) x density of liqulu
at 30°c
(*)
.. Coefficient of apparant expansion
-
Volume of space above liquid in (4). Volume of liquid in (4) x (100
2.5
(143.0
55.5) x 70
2.5
87.5 x 70
- 0.0004082/c
(11) Precantions.
{ 1. The stopper to the density bottl
must be really well fitting.
30)
2. after removing the bottle from the
heating unit (usually boiling water) the stoppar should te pr«B-
對濕度為60%。已知水蒸飽和 蒸氣壓力在15°C及20°C時分別為1271 毫米及174毫米;温度上升至20°C時 其相對濕度為.
(六)A為一軟铁棒在一線圈中,B為一電 池組。通電後,則铁棒將成為一 設電流在線圈中活 小小到同一向如圖中箭頭所示,
則铁棒之 甲端將為 三
乙端將為_ 極。 (b)-缐圈L三兩端接於微計之雨
接線柱FG如圖(b)所示。 當一磁棒IN極徐徐向左進入線
(b)
圈之中央時則電洗將 從接線柱. 通
過線圈流向接線柱
•
又當該磁棒 IN極從線之中央向 右退却時,則電 流將從接線柱 通過線圈流向
接線柱.
(七)有一冰塊,其重量為100克,溫度則
為°C,威於一容器中,(該容器3热) 容量可略去不計)。今以100°C3蒸氣 通入容器內將該冰塊熔解,並使 其温度升至40°C時,則需蒸氣重
克。(水汽化热為540卡/克冰熔 解热為80卡/克)
(八)有一房間,高10呎;房頂中央鉛直 垂懸-100燭光了燈。若該燈距 離地面8呎,則房頂距離該燈最近 之點之照度為
地面距離該 燈最近三點之照度則為.
(九)有一炮艇位於兩高山中鳴砲一响
在1秒鐘後則開砲响第一响回聲 自左方傳來,再過1秒鐘後則開另一 响回聲自右方傳來。設在空氣中音速 為330米/秒,則該兩高山間之距離 為
(+)
圖示一均匀之梯AB重20. 仟克,長3米,靠在一平滑而 鉛直之牆上。若梯與地面 所成主角為60%則地面對 梯反作用力為
R該反作用力與地面所 成主角為
甲部完,乙部在下期續
(祇算甲部3時間為壹小時).
sed to a good fit and then niper immediately. This would szaure that no exoena "liquid in drawn back on subsequent ogoling and hence affect the final weighing, (b) 'molecule in a liquid is performing Brownian motion all the time. When it possesses an amount of energy large enough for it to release itamìf from the attraction forces of the Dulk of the liquid, it leaves the Liquid surface and this is called evporation. These molecules usually gain epergy in the form of heat, after leaving the liquid. it loses energy when it is, cooled. Hence its velocity deoreases.
When it is suf- ficiently cooled it condenser on cold surfaces.
e)(1) an increase in Pressure causes
the melting point of toe to fall (11) When ice changes to water at 0°C
ita volume deoreadas.
(111) The pressure of a maturated vap- our in an enclosed space renaing constant when the volume of the apaus le decreased at constant temperature. (This applice to * pure vapour, not a gaa satur ated with a vapour)
(iv) Whan the temperature of the air decreases, the relative humadát of the air in the enclosed room increases,
(v) when the pressure of unsaturateu
air in so enclosed space is increased, the partial pressure of water-vapour increases algo and so you the relative humidity, as the temperature 18 conatan te
Page 15Page 16
No comments yet.
Private notes are available after approval.