育教僑華 頁三第張六第日二初月九年巳丁屣复
WAH KIU YAT PO
報日僑
五期星
3(a)
育收僑谶
5. A force diagram is con- structed as following:
1978
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物理(二)
魯榮家
PHYSICS (2)
Answers to Exercise one.
1. From the given graph (a) In stage I, the truck takes the longest time, it takes 40 seconds, (b) Stage and stage V show
that the truck travel the greatest distance (400 m), (c) Stage 1 and Stage II.
show that the truck tra- vels in uniform speed. (d)(i) Stage I obtains the
greatest speed (20ms), (11) Since the slope of the graph represents the acceleration, therefore stage III shows the truck travels with greatest acceleration for its slope is the greatest.
(iii) Stage I shows that truck obtains the greatest kine ic energy for its spee1 is the greatest in this stage.
(e) Total distance travelled Total area under the graph
+
(10)(30) +(10)(20)
(10+20)(10)
+ 1(20)(40)
1300 (Ads)
(20)(20)
Correction: The masses
indicated in figure 2 are incorrect, they should be 0.5kg, 0,2kg and 0,1kg respectively.
Let
tension between the 0.5kg block and the 0.2kg block,
tension between the 0.2kg block and the 0.1kg block. acceleration of the system.
fr 2 and f, are the frictional fortes on the blocks 0.5kg, 0.2kg and 0.1kg respectively. Apply Newton's 2nd Law to each block separately
0.5 kg block a
alGM
-0-13
Let the mass per unit length
of the wire be ukgm
Mass of QA m 0,1бu
Mass of OB- (0,28-0,16)u ·
0.12
Total mass - 0.28u
Let x be the perpendicular distance of the centre of
gravity G from OA and that
y from (B.
Take moment about OA.
0.12u(0.06) -(0,28u)x
x= 0.0257m (Ans.) Take monent about OB
0,16u(0.08) -(0,28u)y
(5)
y= 0.0457m (Ans,)
As figure shown above, G is. the centre of gravity of the cylindrical box. The box will topple over when G is just outside the vertical line through C. Therefore, the position of G, which is vert- ically above C, is the limit- ing position.
GM = 0,6m
tane -
MC- 0.4m
33°41
(Ans.)
Let u. speed of water.
issued from the jet
20 ms
V speed
when: ANALY
ball,
cros sectional area of the jet. 1 cm
2
x 10 Apply the formula
2
20
2gh
2(10)(h)
(1)
here h height of the ball
above the jet,
Let d
density of water (- 1000 kg m22 )
Mass of water issued per sec,
dvá
Momentum destroyed per
Copper
Where T tension in the cord
weight of the
copper block -0.089g
weight of the iron block Bouyant force exerted on the copper block" Bouyant force exerted on the. iron block
Consider the copper block
T - W1 = B1
According to the Archi Principle
Bouyant force
des!
weight of equal volume of liquid displaced.
Volume of copper block mass of copper block
density of copper 0.089
日四十月十年七七九一曆公年六十六國民華中
with liquid A. However, lead
(II)nitrate cyrstals are
anhydrous crystals.
(g) Potassium nitrate or sodium
nitrate is an example. It gives oxygen only on heating. 211103
2K10 + On
2N2 NO 2
or, 2NaNO
02
The pale yellow liquid dini- trogen tetroxide vaporised on heating and a light brown gas of nitrogen dioxide is formed which becomes dark-brown on further heating to 100 °C. On cooling from 100 C to the room temperature, the changes are reversed.
20(1)
pale- yellow liquid
Solution to 4.2
210(e)
light brown and then dark-brown on heating
(a) Lolecular mess of CO
= 12 + 2x16 =44
ass of carton atom in 6.6 carbon dioxide is
6.6 Ex
12 41
1.86
But, this amount of carbon atom must have come from 2.1
of the caseous hydrocarbon. mass of hydro en in 2.1 € of A = (2.1 1.8) g
0.3.
--
Rolar ratio of CE
1.8
12
:
= 0.15 0.3
Hence, the empirical formula of A is CH.
mole of A is
22400 560
the molecular mass of: is 42.
conditions :· (1)
iron is at red-hot,
water is changed to
stean.
equation is
CH2=CH2+H2
conditions:
CH CH
CCH2
(1) a catalyst such as
platinum,
(ii) under pressure, (iii) heating is needed. (5) The equation is
02+2H20+2502
conditions: (i) cone 1,50, is used, (ii) heating it necessary.
solution turns from colourless to light brown. The products are bromine and sodium chloride.
2Br (aq) + 012(8)
Br2 (aq)+201 ̄(aq)
(2) The blue solution becomes.
colourless and copper is deposited on zinc rod which has partially dissolved." Zn(s) + Cu2+(aq).
2+
2+
Cu(s) + Zn (aq)
The products are copper and zinc sulphate solution. (3) Yellow particles of sulph-
ur is observed. The prod- ucts are sulphur and water. 502(c)+2H25(aq) → 35+241 0
(4) The green powder of copper
(11)carbonate turns black
8.9d
where d is the density of water
(b) Lass of 1
0.105 EX
0,010
8.9d
42 g
0.01g
0.089g
Let the molecular formula of A be (CH2)n
Then, molecular mass of A
200
14
*:14 = 42
0.01g -0.079g ..... (1),
Similarly, consider the iron :block
B
Let me be the mass of the iron block.
volume of the iron block
7.88
mg
(1-7)ng
8718ng
1
...(2)
By equating (1) and (2)
0.079g 0.8718ng
lyn = 30 mg
Hence, the molecular formula
of A is (C) or Czile
(c) According to the equation for
the combustion of A
and water drops are seen on the upper part of test- tube. The limewater turris milky. The products are.
water vapour, copper(II)
oxide and carbon dioxide..
CuCO2 (s)->Cu)(s)+00, (6)
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1 #2
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B
902 600+ 61120
Duk
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vola 9 vols.
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the volume of Q, required
AS, T
Dif
E 7
3.
** AHM D-#
Ble
**
فيدة
2
am?
3 x 2 = 9 dm3
to burn 2 dm of A is
lence, the volume of air is
100
= 45 am 201 (d) from the informati in given,
0.079-
0.8718
0,0906kg (Ane.)
(e)
化學(二)
朱宏林
9 dm x
REAN
引得
A
A is most probably an alkene
DEA
MEJA SZT
called propene. Its structur- al formula is
APHE
B.
D-RIG
KATA
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A. 29
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FUM
AKM
Bu
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永乐西倒收件
樹水
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24 T.
0.2kg block;
0,1 gk block :
0.5kg
(•); (1) + (2)
- 0%
(1)
2a
(2)
0,la
(3)
|nal o
+
(3) B
8a
24 - (1 + fgt fz)
Since fung
(0,2)(0,5)(10)=IN (0,2)(0.2)(10)=0,4N (0.2)(0,1)(10) -0.2N
24 -(1+0.4+0,2) = 0,8a
-28ms
B
(b) From (3)
-2
(Ans.)
T2 0.2 0.1a:
-
(0.1)(28)
-0.2 +2.8 3N
-
(Ans.)
(0.8)(28)
Substitute into (1)
24
www
T
T
1.
- 9N (Ans.)
Ane : The tension between
the 0.5kg and the 0.2kg is
9N and that between 0.2kg
and 0.1kg is 3N,
周柯
The mementum destroyed per
second - Force to support
the weight of the ball.
dy An, ngoi (2)
where m mass of the
ball.
Substitute (1) into (2)
d(a-2gh)A ng:
2
1000(20-2x10xh)x 10°
3 x 10
h 5 m Substitute into (1)
2
■ 20-2(10)(5). 300
300 ms
17.32 ms
(1) The mass of
per second
dyA
ater issued
1000 x 17.32 x 1 x 1,732 kg. (Ans.)
10-4.
(ii) Momentum destroyed. per
second
dv2A
(1000)(17.32)2(1074)
-1
- 30 kg.ngTM (Ans.)
(or simply equal to the
weight of the ball -30N).
(iii) The height of the ball'
above the jet h
-5 (Ane.)
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Chemistry (2)
Solution to 2.1
(a) A is dinitrogen tetroxide” and
B is oxygen gas.
(b) The residue would be lead(II)
oxide.
4
(o) 2Pb(NO3)2 >2Fb0+2N20 (1)+0, (d) Formula mass of lead(11)nitr-
ate = 207 + 2(14+16x3)| = 331 From the equation in (c), 2. moles of lead (11)nitrate,will give 1 mole (or 22400 cm3) of oxygen gas at s.t.p..
volume, of oxygen evolved at s.t.p. by 3.31 g Fo(NO3)2
3 22400 cm X
3 112 cm
3.31 2.x
(e) The brown gas is nitrogen
dioxide which is a mixed oxide reacting with water to form a mixture of two acids: nitric acid and nitrous acid. 2NO2 + H20 - HO+ HNO2
(f) It is because copper(II)nit-
rate crystals are hydrated, containing water of crystal-
ล t
lisation. On heating, the
water of crystallisation will
come off too so as to react
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Solution to 9.3
(
(a) Formula mass of hydrated salt
=(24+32+16x4+18x). 120+18%
Formula pass of anhydrous salt =(24+32+16x4) = 120
(120+18x) g of hydrated salt sives 120 g anhydrous salt, we have the equation below:
120 2.46 g x
= 1.2 € 120+18% 120 + 18x = 246
18x126
x=7
Hence, the formula of the. hydrated salt is MgSO.7,0.
(b) (1) The equation is
250 + 02 -
conditions:
250
(i) a catalyst such as
platinum or vanad- ium(V)oxide,
(ii) heat to about 450°c.
(2) The equation is
C2H5O + CH, COOH
CH COOCH
conditions:
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cone H,50 is added which acts a catalyst
and dehydrating agent (ii)heating is necessary
(3) The equation is
356 + 41120 Fe204+ 4112.
HAKULAT
A 12, E43, CHE B 15. DI
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