trial 2008 (klang)(section A)
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trial 2008 (klang)(section A)
trial
Section A [40 marks] Answer all questions in this section. 1. A stone is projected horizontally with a velocity 2.0 m s1 from a bridge 20 m above a river. Neglect air resistance. The stone hit the water at a point X. (a) Calculate the time taken by the object to hit the water. [2 marks] (b) Calculate the velocity of the stone just before it lands. [3 marks] (b) Calculate the distance of point X from the bridge. [2 marks] 2. A system undergoes forced oscillation and it experiences a light damping. (a) Sketch a graph of amplitude of forced oscillation against the frequency of the driving force. [2 marks] (b) Describe the variation in amplitude of forced oscillation with the frequency of the periodic driving force. [2 marks] 3. A source of sound which produces a sound of frequency 750 Hz is placed at the open end of a burette filled with water. The resonance occurs for the first time when the length of the air column is 10.0 cm and occurs for the second time when the length of the air column is 32.6 cm. Assume the end correction to be c. (a) Sketch a graph of displacement against distance to show how the air column in the burette vibrates to produce the first overtone. [2 marks] (a) Calculate the speed of sound in air. [3 marks] (b) Determine the value of the end correction, c. [2 marks] 4. A satellite of mass 300 kg orbits round the Earth with a period which is same with the period of rotation of the Earth. Assume the mass of the Earth to be 6.0 × 1024 kg. (a) Determine the radius of the orbit of the satellite. [2 marks] (b) Calculate the gravitational force on the satellite by Earth. [2 marks] 5. Three identical capacitors of capacitance 6 μF, initially uncharged are connected in series with a 12 V battery. (a) Calculate the charge on each capacitor. [2 marks] (b) The capacitors are disconnected from the battery and then connected to obtain a bigger capacitance, C. Determine the potential difference across the larger capacitor. [2 marks] (c) Determine the loss of the energy during the process. [2 marks] 6. (a) What is meant by mutual induction? [1 marks] (b) A solenoid P of radius 1.5 cm consists of 200 turns per cm. A current of 2.0 A passes through the wire. A solenoid Q which consists of 400 turns and of same radius is arranged coaxially with solenoid P. Calculate the mutual inductance between the two solenoids. [3 marks] 7. An operation amplifier has a high openloop gain of 105. It is used as a difference amplifier as it amplifies the difference between the voltages at the two inputs. (a) Draw a diagram to show an difference amplifier connected to a ± 9 V supply. In the diagram, label voltage V+ which is applied to the noninverting input and voltage V which is apply to the inverting input. [2 marks] (b) Determine the output voltage VO if the noninverting input, V+ and the inverting input V are 0.2 mV and 0.1 mV respectively. Explain your answer. [2 marks] 8. In a microwave oven of power 1000W, a magnetron tube is used to produce microwaves of frequency 2450 MHz. The microwave energy is directed into the cooking chamber where the food is placed to be heated. (a) What is the energy of a photon of microwave? [2 marks] (b) Determine the number of photons emitted per second. [2 marks] 
Giraffe 校长
 帖子数 : 542
1018
来自 : johor
注册日期 : 090722
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回复： trial 2008 (klang)(section A)
Answer:
Section A [40 marks]
1.a) vy = 0, Sy = 20 m, a = g = 9.8 ms1,
Sy = ut + ½ gt2
20 = 0 + ½ (9.8)t2
t2 = 20/ 4.9
t = 2.02 s
1.b) vy = u + gt
= 0 + (9.(2.02)
= 19.8 ms1
vx = ux = 2.0 ms1
v =
= 19.9 ms1
θ = tan1
= tan1
= 84º14’ below water su***ce
1.c) sx = ux t
= 2(2.02)
= 4.04 m
2.a)
2 b) At the beginning, the amplitude increases with the
frequency of the driving force that acts on the system Because
the system is light damped, the amplitude will increase to a finite
maximum value when the frequency approaches natural frequency
fO.
When the frequency of the forced oscillation increases beyond
the value of fO, the amplitude will decrease continuously.
3 a)
3 b) When first resonance occurs: L1 + c =  (1)
When second resonance occurs: L2 + c =  (2)
(2) – (1) L2 – L1 =
0.326 – 0.100 =
λ = 2(0.226)
= 0.452 m
Hence, speed of sound, v = f λ
= 750 (0.452)
= 339 ms1
3 c) From (1) L1 + c =
End correction, c = – L1
= (0.452) – 0.100
= 0.013 m
= 1.30 cm
4 Period, T
= 24 × 60 × 60 s
= 8.64 × 104 s
Mass of the Earth, M = 6.0 × 1024 kg.
4 a) To determine the radius of the orbit ,
the gravitational force provides centripetal force
= mR
R3 =
= =
= 7.5674 × 1022
R =
= 4.229 × 106 m
= 4.2 × 106 m
4 b) Gravitational force, F =
=
= 2237.7N
= 2.2 kN
5 a) The capacitors are connected in series
The effective capacitance Ceff can be determined by:
=
= =
Hence, Ceff = 2 μF.
Charge on each capacitor are the same, Q
Q = Ceff V
= 2 ×106 × 12
= 2.4 ×105 C
5 a) To obtain larger capacitance, the capacitors are connected in parallel
The new effective capacitance C can be determined by:
C = C1 + C2 + C3
= 3(6× 106)
= 1.8 × 105 F
V =
=
= 1.333 V
5 b) Energy stored in a capacitor, U = CV2
In the first connection, U1 = (2 × 106) (12)2 = 1.44 × 104 J
In the second connection, U2 = (1.8 × 105)(1.333)2 = 1.60 × 105 J
Loss in energy = U2 – U1
= (1.44 – 0.1599) × 104
= 1.28 × 104 J
6 a) Mutual induction is the phenomenon where an electromotive
force (e.m.f.) is induced in a conductor when the current in a
neighbouring conductor is changing.
6 b) Rp = 1.5 cm = 1.5 ×102 m
= = 2.0 × 104 turns m1
I = 2.0 A
NQ = 400 turns
Mutual inductance, M =
=
= 0.15 H
7 a)
7 b) Ao = 105
V+ = 0.2 mV
V = 0.1 mV
Vo = Ao (V+ – V)
= 105 (0.2 – 0.1) mV
= 10 V
Saturation occurs when Vo ≥ 9.0 V, Hence the output voltage = 9.0 V
8 a) Energy of a photon, E = hf
= 6.63 × 1034 × 2450 × 106
= 1.62 × 1024 J
8 b) Power = = J s1
Total energy radiated per second = 1000 J
Hence,
number of photons emitted per second =
=
= 6.173 × 1026 photons
= 6.2 × 1026 photons
Section A [40 marks]
1.a) vy = 0, Sy = 20 m, a = g = 9.8 ms1,
Sy = ut + ½ gt2
20 = 0 + ½ (9.8)t2
t2 = 20/ 4.9
t = 2.02 s
1.b) vy = u + gt
= 0 + (9.(2.02)
= 19.8 ms1
vx = ux = 2.0 ms1
v =
= 19.9 ms1
θ = tan1
= tan1
= 84º14’ below water su***ce
1.c) sx = ux t
= 2(2.02)
= 4.04 m
2.a)
2 b) At the beginning, the amplitude increases with the
frequency of the driving force that acts on the system Because
the system is light damped, the amplitude will increase to a finite
maximum value when the frequency approaches natural frequency
fO.
When the frequency of the forced oscillation increases beyond
the value of fO, the amplitude will decrease continuously.
3 a)
3 b) When first resonance occurs: L1 + c =  (1)
When second resonance occurs: L2 + c =  (2)
(2) – (1) L2 – L1 =
0.326 – 0.100 =
λ = 2(0.226)
= 0.452 m
Hence, speed of sound, v = f λ
= 750 (0.452)
= 339 ms1
3 c) From (1) L1 + c =
End correction, c = – L1
= (0.452) – 0.100
= 0.013 m
= 1.30 cm
4 Period, T
= 24 × 60 × 60 s
= 8.64 × 104 s
Mass of the Earth, M = 6.0 × 1024 kg.
4 a) To determine the radius of the orbit ,
the gravitational force provides centripetal force
= mR
R3 =
= =
= 7.5674 × 1022
R =
= 4.229 × 106 m
= 4.2 × 106 m
4 b) Gravitational force, F =
=
= 2237.7N
= 2.2 kN
5 a) The capacitors are connected in series
The effective capacitance Ceff can be determined by:
=
= =
Hence, Ceff = 2 μF.
Charge on each capacitor are the same, Q
Q = Ceff V
= 2 ×106 × 12
= 2.4 ×105 C
5 a) To obtain larger capacitance, the capacitors are connected in parallel
The new effective capacitance C can be determined by:
C = C1 + C2 + C3
= 3(6× 106)
= 1.8 × 105 F
V =
=
= 1.333 V
5 b) Energy stored in a capacitor, U = CV2
In the first connection, U1 = (2 × 106) (12)2 = 1.44 × 104 J
In the second connection, U2 = (1.8 × 105)(1.333)2 = 1.60 × 105 J
Loss in energy = U2 – U1
= (1.44 – 0.1599) × 104
= 1.28 × 104 J
6 a) Mutual induction is the phenomenon where an electromotive
force (e.m.f.) is induced in a conductor when the current in a
neighbouring conductor is changing.
6 b) Rp = 1.5 cm = 1.5 ×102 m
= = 2.0 × 104 turns m1
I = 2.0 A
NQ = 400 turns
Mutual inductance, M =
=
= 0.15 H
7 a)
7 b) Ao = 105
V+ = 0.2 mV
V = 0.1 mV
Vo = Ao (V+ – V)
= 105 (0.2 – 0.1) mV
= 10 V
Saturation occurs when Vo ≥ 9.0 V, Hence the output voltage = 9.0 V
8 a) Energy of a photon, E = hf
= 6.63 × 1034 × 2450 × 106
= 1.62 × 1024 J
8 b) Power = = J s1
Total energy radiated per second = 1000 J
Hence,
number of photons emitted per second =
=
= 6.173 × 1026 photons
= 6.2 × 1026 photons
Giraffe 校长
 帖子数 : 542
1018
来自 : johor
注册日期 : 090722
年龄 : 27
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