報日僑華
Relation of velocity to wavelength and frequency In 1 second, a body vibrating n time sends
南教備華·真三第張六第日四廿月正年酉己曆里WAH KIU YAT PO
1969AFUD
試題預習
物理科
(十九)
·陸永熾·
PHYSICS (19)
The Illumination or Intensity of illumination. of a surface is the luminous flux rate of flow of light energy) received per unit area of the surface.
Units are (1) lumen per sq.ft.
(ii) lumen per sq.m.
The lumen/sq.ft. is the British unit or illumination. It is also known as foot-cande. or foot candle since it is equal to the illumination at a distance of 1 foot from a uniform point source of 1 candela.
The lumen/eq.metre is the metric unit of illumination. It is also known as lux and is. equal to the illumination at a distance of 1 metre from a uniform point source of 1 candela. This unit was formerly called the metre-candle The inverse-square law of illumination
The illumination of a place is directly proportional to the luminous intensity of the source
• I, and is inversely proportional to the square of the distance from the source.
To verify the Inverse-square law of Illumination
A drop of hot paraffin wax is put on the centre of a piece of stiff white absorbent paper. The wax spread to a size of a dollar coin and the paper is then clamped upright. This is a simple but effective Bunsen's grease-spot photometer. A bulb (1) illuminates one side of the photometer at a distance of dy. Two bulbs (I) of equal rate (12 = 211) are put together and illuminate. The other side of the phobometer at a distance of d such that there is a photometric balance on both side (equally. illuminated on both side), The experiment is
repeated with 3 bulbs (13), 4 bulbs (1), 5 bulbs (1) etc. If the inverse square law da true then the square of the distance d must be proportional to the number of lamps n in ISS
n,
is a constant]
The graph of d2 against n is 3 straight line.
Luminous intensity of the old lan
*
200 candela
Distance from the table
di 10 ft.
Luminous intensity of the combination of two lamps
21, 400 candela
Illumination on the table.
Ex
400
100 foot-
ft-cd.
Luminous intensity of new lamp.
12 160 candela
Distance of the new lamp from the tabl Since equal illumination is obtained
4-ft-ed.
out n waves or cycles. If A is the length of
each wave, the sound has gone a distance of wh Hence its velocity is V = nλ
Velocity - frequency x wavelength
Pressure
(b) Velocity of sound is proportional to Density
In accordance with Boyle's law, if pressure of air is halved its volume will be doubled. Hence the density is halved. Thus, at constant temperature the ratio of (Pressure/Density) will always remain constant, however much the pressure varies.
(i) Consequently, at the top of a cliff, if the
temperature remains the same, presaurs bás n effect on velocity. It is the same on the top and bottom.
(ii) If the temperature in a room werns and the pressure unaltered, the density varied. It becomes less dense, The ratio (Pressure/ Density) will therefors increase, and hence, the velocity of sound increases with temperature.
(c) (1) The volume control alters the intensity of Bound. The frequency is unaltered, Frequency determines the pitch of sound. The volume control has no effect on it.
(ii) Increase of temperature speeds up the
velocity of sound. The pitch is unaffected, (111) Tone control varies the frequency of sound
produced and thus the pitch of sound is altered by such a control.
(d) Since V = 17%
Frequency of sound one octave higher
V is kept constant
A is halved
·340 2x680
For a sound of one octave lower frequency becomes halved and wavelength is doubled
340
·680.
100 cm.
س
The first echo he heard travelled along the path AdC. The second one travelled along ABC.. They reached the observer 1 sec and 23 sec. after fire respectively. Both of these proceeded onward and were reflected to the observer by the cliffs simultaneously. They appeared as a united. echo which was heard 4 sec. after the rifle was fired.
Distance travelled by the first echo 14 x 1120]
#1680 ft.
Distance of the observer from this cliff
三期星
日二十月三年九六九一醫公年八十五國民華中
試題預習
玉鑛釗
1969KEUP
英文中學會考
化學科 (TA)
Aluminium, iron, magnesium and zinc are electropositive than hydrogen,
More
Copper is Lena electropositive than hytrogen.
Un heating, the other four substances sublime, and no new substance is formed
When potassium chlorate is heated, however it decomposas into two new substances potassium chloride and oxygen"
2X010
KOL
One atom of elemnt K will transfer the two electrons in ite outer shell to two atome of element. V. Each atom of element Y will accept one electron from an atom of X.
20X
21
The formula or the compound is therefore I,
salt dissociates into ions before the passage of the electric current.
The concentration of the solution 13 always increased the positive fons are re polled from the positive pole or ende. Hydrogen is frequent ly but not always liberated at the cathode during electrolysis
In the Contact Process, the sulphur trioxide la dissolved into concentrated sulphuric acid to form ole un
#230
The reason for this is that when sulphur trioxide is dissolved directly into water, fine! drops of sulphuric acid which form a mist will result and fill the factory.
Although lead is more electropositive than hydrogen, there is little apparent action because insoluble Lead sulphate forms a protective layer over the lead thereby prevent ing any further action from taking place
In the case of potassium ferricyanide, a deep: blue precipitate of Turnbull's blue is formed with a ferrous salt in solution. In the case. of ammonium hydroxide, a dirty green precipitate of ferrous hydroxide is formed The ferrous hydrazide rapidly absorbs oxygen, and as a result the parts in contact with si soon pass into brown ferrio bydroxide
A
The substance is probably a uznic nyve
Lement % by
ese figures are proportional to
#
No
of each
40 x
6.7
Hence the simplest COB (or HCHO)
༣།
40
“25སལུག་g
ecmpound 18
G Ammonia will burn in an atmosphere, or uxyge
to form nitroge, and water
8.94
200
100
The Illumination at the centre of the table is 4 ft-ed. The new lamp should be placed 8.94 ft. above the centre of the table.
ft above the ce
The lamp is 2 ft. below the mirror. There is an image of the lamp 4 ft. above the lamp,
Luminous intensity of lamp - Ij
- 100 c.p.
Luminous intensity of its image.
0.81,
80 c.p.
Distance of its image above the road - 24 ft. The illumination immediately below the lamp da due to both I, and 12.
Intensity of illumination)
100
20
20
-0.389 lwen/sq.ft.
Ans, The illumination of the road vertically.
below the lamp is 0.389 Iumen/sq.ft.
36. (a) A sound wave is a longitudinal wave; the
particles of the medium vibrate backward and forward in the same direction in which the wave is moving. The wave consists of alternate condensations (compressions) and rarefactions (region of reduced pressure).
The wavelength of a sound wave is the distance between two successive compressions or rarefactions. It is measured in length units. The total number of complete vibrations, (that is, the repetition of compression in the Bound wave per unit time is called the frequency of sound.
1680
840 ft.
Distance travelled by the second echo
2 x 1120 2800 ft.
Distance of the observer from this cliff
.2830.
1400
Hence, distance between the cliffs is (1400 • 840) - 2240 ft.
questions for next week
37
What le meant by a node and antinode of a stationary wave? braw a sketch in Ullustration What is the diatarice in terms of wave length, between
(1) two sug ce sa ive nodes.
(ii) two successive, antinodes, and
iii) an antinode and a neighbouring node?
X
∞
The above illustrates stationary waves of displacement, A, B, C, D, due to: (1) a closed pipe; (11) a violin string producing its Fundamental note; (111) the first overtone of the violin string; and (iv) an open pipe. Identify each of the waves, and give reasons your answer.
questions for next week
we will again review a number of multiple questions next wednesday.
A stationary wave is produced in air between s ama 11 loudspeaker emitting a note of 1000 cycles per sec and a hardboard reflector some distance away. (1) if the board at a node pr antinode? (ii) Starting from the board, four successive nodes are detected in moving throu a total distance of 51 cm, Calculate the velocity of sound in air.
(d): The following measurements were made in an
experiment to find, for å string under constant tension, the relation between the length of the vibrating portion and the frequency of the note which it emita:
Lon.
49.2 39.432.8 24.6
Length/sec 256 320 364512 ·
Freq.
Determine the relation between n and L, Nhất length of string would give a note of frequency 100 per sec? In what
ratio should the tension of a violin string be altered in order to increase the frequency of a note in the ratio of 5 to 21.
How much would the string have to be shortened to make the same change without altering the tension?
Uraw a sketch of the stationary wave in (1) a olosed pipe;
(ii) an open pipe of the same length,
when each is sounding its fundamental note khich produces a note of higher pitch? (b) Make labelled diagrams epresenting two sound
saves that:
(1) interfere completely with each other and
produce silence
(13) reforce each other in intervals and
produce. beata.
a) Lescribe, with suitable diagram, a method
measuring the velocity of sound in sir, and show how the result is calculated. d) An organ pipe emite a fundamental note or frequency 275 per sec. By blowing it more atrongly it vill emit a note of frequancy 550 per sec. Is it a closed or an open pipe and what is its length? Explain how your answer obtained. Velocity of sound 1100 ft./sec.