WAEC SSCE - Physics - 2015
Swali 1 Ripoti
All electromagnetic waves in a vacuum have the same
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All electromagnetic waves in a vacuum have the same speed. This is known as the speed of light, which is approximately 3.0 x 108 meters per second. Regardless of their frequency or wavelength, all electromagnetic waves travel at this same speed in a vacuum, such as in outer space. This is a fundamental property of electromagnetic waves and is a cornerstone of modern physics.
Swali 2 Ripoti
The slope of a straight-line velocity-time graph represents
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The slope of a straight-line velocity-time graph represents the uniform acceleration. In other words, if the graph is a straight line, the slope of the line represents the rate at which the velocity is changing, which is the acceleration. The steeper the slope, the greater the acceleration, while a zero slope represents zero acceleration, which means the velocity is constant. Therefore, is the correct answer: uniform acceleration.
Swali 3 Ripoti
An electric generator rated 12 kW, 2 kV distribute power through a cable of resistance 20\(\Omega\). Calculate the power loss in the cable.
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Swali 4 Ripoti
Two strings each inclined at an angle 30o to the horizontal are used to hold a bucket of water weighing 20N. Calculate the tension i n each string.
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Swali 6 Ripoti
Which of the following statements is correct? When capacitors of different values are connected in series
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When capacitors of different values are connected in series, the same charge flows through them. This is because the capacitors are connected in a series, so the same current flows through all of them. Since capacitors store charge, the same amount of charge must flow through each capacitor to maintain the same current. However, the voltage across each capacitor will be different, since the voltage drop across each capacitor will be proportional to its capacitance value. So, option A is correct.
Swali 7 Ripoti
When current flows through a moving-coil galvanometer, the force exerted on the pointer depends on the: i. strength of the current. ii. magnetic field strength. iii. length of the coil. Which of the statement(s) above is/are correct?
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Swali 8 Ripoti
A battery of emf 24V is connected to three resistor P, Q and R as illustrated in the diagram. Determine the voltage across the resistor Q
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Swali 9 Ripoti
What type of wave is emitted by a loud speaker?
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A loudspeaker emits a longitudinal wave. Sound waves are mechanical waves that travel through a medium by compressing and rarefying the particles of the medium. In a longitudinal wave, the displacement of the particles is parallel to the direction of wave propagation. When an electric current is applied to the speaker, it causes the speaker cone to vibrate back and forth, creating areas of compression and rarefaction in the surrounding air. These pressure waves then travel through the air to reach our ears, where they are detected as sound. Therefore, sound waves are examples of longitudinal waves.
Swali 10 Ripoti
A wire of cross-sectional area 0.60mm2 has a restivity of 7.2 x 10-7 \(\Omega\). Calculate the length of wire that will have a resistance of 3\(\Omega\).
Maelezo ya Majibu
The formula for calculating resistance is R = ρ(L/A), where R is the resistance of the wire, ρ is the resistivity of the wire material, L is the length of the wire, and A is the cross-sectional area of the wire. We are given the cross-sectional area and resistivity of the wire, as well as the desired resistance. We can rearrange the formula above to solve for L as follows: L = (RA)/ρ Substituting the given values, we get: L = (3 x 0.60 x 10^-6) / (7.2 x 10^-7) = 2.5 meters Therefore, the length of wire that will have a resistance of 3Ω is 2.5m. The correct option is (b) 2.50m.
Swali 11 Ripoti
What must be connected between the points P and Q to establish a steady current through the incomplete circuit diagram given?
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To establish a steady current through the incomplete circuit diagram, a cell (battery) must be connected between the points P and Q. A cell provides the potential difference (voltage) required for the current to flow in the circuit. Without a cell, the circuit is incomplete, and no current will flow. Therefore, the correct answer is option C, cell.
Swali 12 Ripoti
An alternating current supply is connected to an electric lamp which lights with the same brightness as it does with direct current source emf 6v. The peak potential difference of the a.c supply is
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Swali 13 Ripoti
Which of the following statements about a three-pin plug is/are correct. i. The brown wire is the live wire. ii. The fuse is connected to the live wire. iii. The earth wire is coloured blue
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Swali 14 Ripoti
An object is placed 20cm from a concave mirror of focal length 15cm. Calculate the linear magnification of the image formed
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The linear magnification of an image formed by a spherical mirror is given by the formula: magnification = height of image / height of object = -v/u where u is the object distance, v is the image distance and the negative sign indicates that the image is inverted. Given that the object is placed 20cm from a concave mirror of focal length 15cm, we can use the mirror formula: 1/f = 1/u + 1/v Rearranging this formula, we get: 1/v = 1/f - 1/u Substituting the given values, we get: 1/v = 1/15 - 1/20 1/v = (4 - 3)/60 1/v = 1/60 v = 60 cm Now, the magnification can be calculated as: magnification = -v/u = -60/20 = -3 Therefore, the linear magnification of the image formed is 3. Since the image is inverted, the sign of the magnification is negative. So, the correct answer is (d) 3.0.
Swali 15 Ripoti
Which of the following units is derived?
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The derived units are those that can be obtained from the combination of two or more base units. Among the options given, N (Newton) is a derived unit because it is obtained from the combination of kg (kilogram), m (meter), and s (second) through the formula F=ma, where F is the force in N, m is the mass in kg, and a is the acceleration in m/s^2. On the other hand, kg (kilogram), m (meter), and K (Kelvin) are base units because they are independent and cannot be derived from any other unit.
Swali 16 Ripoti
Which of the following units is the S.I. unit of heat capacity?
Maelezo ya Majibu
The S.I. unit of heat capacity is the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius. Heat capacity is measured in Joules per Kelvin (J/K) or Joules per degree Celsius (J/°C). Therefore, the correct option is JK-1, which stands for Joules per Kelvin.
Swali 17 Ripoti
In an experiment in which a molten substance is allowed to cool, the cooling curve above was obtained. The temperature 60oC is known as the
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The cooling curve shows the temperature changes of a molten substance as it cools. As the molten substance cools, its temperature gradually decreases until it solidifies. The temperature at which the substance changes from liquid to solid state is known as its melting point. In the given cooling curve, the temperature at which the curve plateaus and changes direction from a downward slope to a horizontal line is around 60°C. Therefore, 60°C is the melting point of the substance being tested. Therefore, the correct answer is "melting point".
Swali 18 Ripoti
A quantity of hot water at 100oC is added to 400g of water at 10oC until the temperature of the mixture is 60oC.
Calculate the mass of the hot water added. [Specific heat capacity of water =4200JKg-1 K-1]
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Swali 19 Ripoti
A uniform bar AB is balanced on a knife-edge which is 60cm from B by a mass of 22g which hangs at C 10cm from A. Calculate the mass of the bar.
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Swali 21 Ripoti
When an air column is set into vibrations, the distance between two consecutive nodes of the resultant wave is
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When an air column is set into vibrations, it produces sound waves which propagate through the air. In a stationary wave, nodes are the points that appear to be standing still and do not undergo any vibration. They are formed due to the interference between the incident wave and the reflected wave. The distance between two consecutive nodes of a stationary wave is half of the wavelength. This is because a node is formed at the point where the incident wave and reflected wave interfere destructively, resulting in a minimum amplitude or zero displacement. Thus, the distance between two consecutive nodes is equal to half the distance between a node and an antinode (the points of maximum amplitude or displacement), which is half of the wavelength. Therefore, the correct answer is "half of the wavelength".
Swali 22 Ripoti
in a nuclear fission process, the mass defect was 1.2 x 10-5kg. Calculate the energy released. [speed of the light in vacuum = c = 3.0 x 108ms-1]
Maelezo ya Majibu
The energy released in a nuclear fission process can be calculated using Einstein's famous equation E = mc^2, where E is the energy released, m is the mass defect, and c is the speed of light. Substituting the given values, we have: E = (1.2 x 10^-5 kg) x (3.0 x 10^8 m/s)^2 E = (1.2 x 10^-5 kg) x (9.0 x 10^16 m^2/s^2) E = 1.08 x 10^12 J Therefore, the energy released in the nuclear fission process is 1.08 x 10^12 J. So the answer is (c) 1.08 x 10^12 J.
Swali 23 Ripoti
The diagram given illustrates a wave form. Determine the speed of the wave
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Swali 24 Ripoti
Which of the following statements describes a body in a stable equilibrium correctly?
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Swali 25 Ripoti
A battery of emf 24V is connected to three resistor P, Q and R as illustrated in the given diagram. Determine the voltage across the resistor Q.
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Swali 27 Ripoti
A bullet fired vertically upwards reaches a height of 500m. Neglecting air resistance, calculate the magnitude of the initial velocity of the bullet. [g = 10ms-2]
Maelezo ya Majibu
When a bullet is fired vertically upwards, it will continue to move upward until the force of gravity acting downwards slows it down to a stop and then pulls it back down towards the ground. The maximum height the bullet reaches before it falls back down is called the maximum height or the vertex. At the maximum height, the final velocity of the bullet is zero, and the initial velocity is what we are required to find. We can use the equation of motion that relates the initial velocity, final velocity, acceleration and displacement to solve for the initial velocity. This equation is: vf2 = vi2 + 2as where vf is the final velocity (zero), vi is the initial velocity (what we want to find), a is the acceleration due to gravity (-10ms-2 since it is acting in the opposite direction to the motion of the bullet), and s is the displacement (500m). Substituting the known values into the equation, we get: 0 = vi2 + 2(-10ms-2)(500m) Simplifying, we get: vi2 = 10000 Taking the square root of both sides, we get: vi = 100ms-1 Therefore, the magnitude of the initial velocity of the bullet is 100.0ms-1.
Swali 28 Ripoti
Which of the following concepts is a method of heat transfer that does not require a material medium?
Maelezo ya Majibu
The correct answer is radiation. Radiation is a method of heat transfer that does not require a material medium. It involves the transfer of heat through electromagnetic waves, such as infrared radiation. Radiation can occur through a vacuum, which means that it can take place even in the absence of a medium or matter. Examples of radiation include the warmth you feel from the sun, the heat emitted by a fire, and the energy released by a light bulb.
Swali 29 Ripoti
The pressure applied to an enclosed fluid to one end id transmitted equally throughout the fluid and to the container walls. This statement is
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The statement mentioned in the question is related to Pascal's principle. Pascal's principle states that the pressure applied to an enclosed fluid in a container is transmitted equally throughout the fluid and to the container walls. This principle is based on the fact that fluids are incompressible, which means that they cannot be compressed or squeezed into a smaller volume. Therefore, any pressure applied to one end of the fluid is transmitted equally to all parts of the fluid and to the walls of the container. This principle is used in many applications, including hydraulic systems, where a small force applied to one end of a piston is transmitted to a larger force at the other end, allowing heavy loads to be lifted with ease.
Swali 30 Ripoti
A girl standing some distance from the cliff claps her hands and hears the echo 0.8s later. calculate her distance from the cliff. [speed of sound in air = 320 ms-1]
Maelezo ya Majibu
When sound waves travel, they can bounce off surfaces and return to the listener's ear, creating an echo. The distance between the listener and the reflecting surface can be calculated from the time taken for the sound to travel to the surface and back, and the speed of sound. In this problem, the time taken for the sound to travel to the cliff and back (i.e., the time for the echo to be heard) is given as 0.8s. The speed of sound in air is given as 320 ms-1. Let's call the distance between the girl and the cliff "d". When she claps her hands, a sound wave travels the distance "d" to the cliff and then back again to her ear, covering a total distance of 2d. The time taken for this round trip is equal to the time for the echo to be heard, which is 0.8s. So, we can use the formula: distance = speed x time to calculate the distance "d" between the girl and the cliff. Rearranging the formula, we get: distance = (speed x time) / 2 Substituting the values given in the problem, we get: distance = (320 ms-1 x 0.8s) / 2 distance = 128m Therefore, the girl is 128 meters away from the cliff. So, the correct option is (b) 128m.
Swali 31 Ripoti
The diagram given illustrates two charged particles W and X. The charges on W and x are respectively
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Swali 32 Ripoti
A body is projected with an initial velocity U at an angle O to the horizontal. The time taken by it to reach its maximum height is given by the expression
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When a projectile is projected with an initial velocity U at an angle O to the horizontal, it follows a parabolic path. The projectile reaches the maximum height at a certain point, then falls back to the ground. At the highest point, the vertical component of the velocity becomes zero. Let's assume that the time taken by the body to reach the maximum height is t. At the maximum height, the vertical component of the velocity becomes zero. Therefore, using the first equation of motion: v = u + gt where v = 0 and u = Usinθ 0 = Usinθ - gt gt = Usinθ t = Usinθ/g Therefore, the time taken by the body to reach its maximum height is given by the expression: t = Usinθ/g Thus, the correct option is: \(\frac{u \text{sin}\theta}{g}\)
Swali 33 Ripoti
Which of the following statements about mass and weight are correct? i. Weight is the force of gravity on a body. ii. The mass of a body is the quantity of matter in the body. iii. The weight of a body is greatest at the equator. iv. The mass of a body is greatest at the poles
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Swali 34 Ripoti
The mass of the earth is 6.0 x 1024kg and that of the moon is 7.0 x 1022kg. If the distance between them is 4.0 x 108m, calculate the force of attraction between them. [G = 6.7 x 10-11Nm2kg-2]
Maelezo ya Majibu
The force of attraction between two masses is given by the formula: F = G(m1m2)/d^2 where: F = force of attraction G = gravitational constant (6.7 x 10^-11 Nm^2/kg^2) m1 = mass of the first object (in kg) m2 = mass of the second object (in kg) d = distance between the centers of the two objects (in meters) Given: Mass of the earth (m1) = 6.0 x 10^24 kg Mass of the moon (m2) = 7.0 x 10^22 kg Distance between them (d) = 4.0 x 10^8 m Gravitational constant (G) = 6.7 x 10^-11 Nm^2/kg^2 Substituting the values into the formula: F = G(m1m2)/d^2 F = 6.7 x 10^-11 x (6.0 x 10^24) x (7.0 x 10^22) / (4.0 x 10^8)^2 F = 1.8 x 10^20 N Therefore, the force of attraction between the earth and the moon is 1.8 x 10^20 N. Answer option (D) is correct.
Swali 35 Ripoti
Which of the following radiations constitutes a stream of electrons?
Maelezo ya Majibu
Cathode rays constitute a stream of electrons. Cathode rays are a stream of negatively charged particles (electrons) emitted from the negative electrode or cathode in a vacuum tube, also called a cathode ray tube. The electrons are accelerated by an electric field and focused into a beam that can be used for various applications such as in televisions and computer monitors. Therefore, is the correct answer.
Swali 36 Ripoti
Two angles of a pentagon are in the ratio 2:3. The others are 60o each. Calculate the smaller of the two angles
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Swali 37 Ripoti
The crackling noise produced by aluminum roofing sheets on a house during a hot sunny day is a result of
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The crackling noise produced by aluminum roofing sheets on a house during a hot sunny day is a result of the expansion of the sheets. The heat from the sun causes the metal to expand, and as a result, the sheets push against each other, making the crackling noise. This is because metal expands when heated and contracts when cooled. So during a hot day, the metal expands, and as it cools down in the evening, it contracts, which may also produce a similar noise.
Swali 38 Ripoti
In an experiment in which a molten substance is allowed to cool, the cooling curve above was obtained. The temperature 60oC is known as the
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The cooling curve shows how the temperature of a substance changes over time as it cools down. The temperature at which a substance changes from a liquid to a solid is known as its melting point. In the given cooling curve, the temperature at which the slope of the curve changes is known as the melting point. Therefore, the correct answer is the melting point. So, the temperature 60oC is the melting point of the substance.
Swali 39 Ripoti
Red light travels faster than blue light when both of them pass through a glass prism because the
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Swali 40 Ripoti
In a school compound a girl walks 40km due east from the laboratory to the staff room to submit her report. Then she returns and walks 30m due north to her classroom. Determine the magnitude of her displacement from the10m laboratory
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Swali 41 Ripoti
An object floats in a fluid when it displaces it's
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An object floats in a fluid when it displaces its weight of the fluid in which it floats. This is known as Archimedes' principle. The principle states that any object, wholly or partially immersed in a fluid, experiences an upward buoyant force equal in magnitude to the weight of the fluid displaced by the object. If the buoyant force is greater than or equal to the weight of the object, the object will float in the fluid.
Swali 42 Ripoti
\(^{226}_{88}Ra\) emits two alpha particles to produce polonium(po) nuclide. The correct formula for the nuclide?
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Swali 43 Ripoti
Materials that can be stretched and still return to their original forms when the stresses are removed are said be
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Materials that can be stretched and still return to their original forms when the stresses are removed are said to be elastic. When a force is applied to an elastic material, it deforms in response to the force, but once the force is removed, the material returns to its original shape. This is due to the elastic material's ability to store the energy of deformation and then release it when the force is removed. Examples of elastic materials include rubber bands, springs, and certain types of metals.
Swali 44 Ripoti
An object is placed 15cm from a diverging lens of focal length 12cm. The image of the object formed by the lens in
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Swali 45 Ripoti
The resultant force in a couple is
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In a couple, two equal and opposite forces act on a body, but they are not in the same line. The resultant force of a couple is defined as the algebraic sum of the forces, which is zero because the forces are equal in magnitude and opposite in direction. However, the couple produces a turning effect or torque on the body, which is not zero, and depends on the magnitude of the forces and the distance between them. So the answer is (a) zero.
Swali 46 Ripoti
The diagram above illustrates a block ans tackle pulley system in which an effort E supports a load of 100.0N. If the efficiency of the machine is 75%, calculate the value of E
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Swali 47 Ripoti
A body of mass 1000kg is released from a height of 10m above the ground. Determine its kinetic energy just before it strikes the ground .[g = 10ms-2]
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Swali 48 Ripoti
A swimming pool is 2.1m deep. If it contains water of refractive index 1.3, calculate the apparent depth of pool
Maelezo ya Majibu
When light enters from one medium to another, its speed changes and it bends. This is called refraction. The refractive index (n) of a medium is the ratio of the speed of light in vacuum to the speed of light in the medium.
The formula for calculating the apparent depth (d') of an object in a medium is given as:
d' = d/n
where d is the actual depth of the object and n is the refractive index of the medium.
In this case, the actual depth of the pool is given as 2.1m and the refractive index of water is 1.3.
Therefore, the apparent depth of the pool can be calculated as follows:
d' = 2.1m / 1.3 = 1.6m
Thus, the apparent depth of the pool is 1.6m. Therefore, the correct option is C.
Swali 49 Ripoti
The angle which the direction of the earth magnetic field makes with the horizontal at a palace is know as.
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The angle which the direction of the earth magnetic field makes with the horizontal at a place is known as the "angle of dip". This angle represents the inclination of the Earth's magnetic field lines with respect to the horizontal plane at a particular location. It is also known as the magnetic inclination or magnetic dip angle. The angle of dip varies at different locations on the Earth's surface, and it can be used to determine the position of the magnetic poles.
Swali 50 Ripoti
The barometric reading at a place is 73.5m Hg. Calculate the pressure at a point 30m below the surface of water contained in a reservoir at the place. [g = ms-2, density = 1.2 x 104kg m-3, density of water = 1.0 x 103m-3]
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Swali 51 Ripoti
A body is projected at an angle of 30° to the horizontal with a velocity of 150 ms\(^{-1.}\). Calculate the time it takes to reach the greatest height [Take g = 10 ms\(^2\) and neglect air resistance]
At the greatest height the vertical component of velocity is zero. Time to reach the greatest height:
\[ t = \frac{u\sin\theta}{g}. \]With \(u = 150\ \text{m s}^{-1}\), \(\theta = 30^\circ\), \(g = 10\ \text{m s}^{-2}\):
\[ t = \frac{150 \times \sin 30^\circ}{10} = \frac{150 \times 0.5}{10} = \frac{75}{10} = 7.5\ \text{s}. \]Time to reach greatest height \(= 7.5\ \text{s}\).
Maelezo ya Majibu
At the greatest height the vertical component of velocity is zero. Time to reach the greatest height:
\[ t = \frac{u\sin\theta}{g}. \]With \(u = 150\ \text{m s}^{-1}\), \(\theta = 30^\circ\), \(g = 10\ \text{m s}^{-2}\):
\[ t = \frac{150 \times \sin 30^\circ}{10} = \frac{150 \times 0.5}{10} = \frac{75}{10} = 7.5\ \text{s}. \]Time to reach greatest height \(= 7.5\ \text{s}\).
Swali 52 Ripoti
(a) What is a projectile?
(b) Give the reason why the horizontal component of the velocity of a projectile remains the same at every point of its flight.
(a) A projectile is a body that is given an initial velocity and then moves freely under the action of gravity alone (its own weight), following a curved (parabolic) path. Examples are a thrown ball or a fired bullet.
(b) The horizontal component of the velocity remains constant throughout the flight because, neglecting air resistance, there is no horizontal force acting on the projectile. Gravity acts vertically downward only, so it changes the vertical component of velocity but has no component along the horizontal direction; hence the horizontal velocity is unchanged (zero horizontal acceleration).
Maelezo ya Majibu
(a) A projectile is a body that is given an initial velocity and then moves freely under the action of gravity alone (its own weight), following a curved (parabolic) path. Examples are a thrown ball or a fired bullet.
(b) The horizontal component of the velocity remains constant throughout the flight because, neglecting air resistance, there is no horizontal force acting on the projectile. Gravity acts vertically downward only, so it changes the vertical component of velocity but has no component along the horizontal direction; hence the horizontal velocity is unchanged (zero horizontal acceleration).
Swali 53 Ripoti
(a)(i) what is resonance?
(ii) State two eXwilples of resonance.
(iii) Differentiate between loudness and intensity of sound.
(b) When a ray is refracted through a rectangular glass prism, which of the following properties of the ray will change? Wavelength, frequency and speed.
(c)
(i) Copy the diagram above in your answer booklet.
(iii) On the copied diagram, sketch the pattern of the waves immediately after passing through the opening.
(d) A diverging lens of focal length 18.0 in is used to view a shark that is 90.0 in away from the lens. If the image formed is 1.0 m long, calculate the:
(i) image distance;
(ii) length of the shark.
(a)(i) Resonance. Resonance occurs when a body is set into vibration by a periodic force (or by impulses from a nearby vibrating body) whose frequency equals the natural frequency of the body, so that the body vibrates with maximum (very large) amplitude.
(a)(ii) Two examples of resonance.
- The loud sound produced when the vibrating prongs of a tuning fork are held over an air column (resonance tube) whose length is adjusted until the sound is loudest.
- A bridge vibrating strongly when soldiers march across it in step at a frequency equal to the natural frequency of the bridge (sympathetic vibration).
(a)(iii) Loudness versus intensity of sound. Intensity is a measurable physical quantity: the sound energy flowing per second through unit area held perpendicular to the direction of travel (unit \(\text{W m}^{-2}\)); it depends only on physical quantities (energy, time and area). Loudness is the physiological sensation produced in the ear by the sound; it depends on the intensity but also on the sensitivity and response of the individual ear, so it is subjective and not directly measurable.
(b) Refraction through a rectangular glass prism. When light passes from air into glass, the speed changes (it decreases) and the wavelength changes (it decreases in the same ratio). The frequency does not change. So: speed changes, wavelength changes, frequency unchanged.
(c) Diffraction through the wide opening. The given diagram shows straight (plane) wavefronts, drawn as parallel vertical lines, travelling towards a barrier that has a wide opening. Because the width of the opening is much larger than the wavelength, the wavefronts emerging on the far side remain almost straight and parallel across the middle of the gap, and only curve (bend) slightly round the two edges of the opening. The required sketch of the pattern immediately after passing through the opening is shown below.
(If the opening were instead made very narrow, comparable with the wavelength, the emerging waves would spread out strongly as almost circular arcs.)
(d) Diverging lens forming the image of a shark. Data: focal length \(f = -18.0\ \text{cm}\) (negative for a diverging lens), object distance \(u = 90.0\ \text{cm}\), image length \(= 1.0\ \text{m}\).
(d)(i) Image distance. Using the lens formula with \(f\) negative for a diverging lens:
\[ \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \]\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{90} = \frac{-5 - 1}{90} = \frac{-6}{90} = -\frac{1}{15} \]\[ v = -15\ \text{cm} \]The image distance is \(15\ \text{cm}\); the negative sign shows the image is virtual, upright and on the same side as the object, as expected for a diverging lens.
(d)(ii) Length of the shark. The magnification links the image and object sizes to their distances:
\[ m = \left|\frac{v}{u}\right| = \frac{\text{length of image}}{\text{length of shark}} \]\[ \text{length of shark} = \frac{\text{length of image} \times u}{|v|} = \frac{1.0 \times 90}{15} = 6.0\ \text{m} \]The shark is \(6.0\ \text{m}\) long.
Maelezo ya Majibu
(a)(i) Resonance. Resonance occurs when a body is set into vibration by a periodic force (or by impulses from a nearby vibrating body) whose frequency equals the natural frequency of the body, so that the body vibrates with maximum (very large) amplitude.
(a)(ii) Two examples of resonance.
- The loud sound produced when the vibrating prongs of a tuning fork are held over an air column (resonance tube) whose length is adjusted until the sound is loudest.
- A bridge vibrating strongly when soldiers march across it in step at a frequency equal to the natural frequency of the bridge (sympathetic vibration).
(a)(iii) Loudness versus intensity of sound. Intensity is a measurable physical quantity: the sound energy flowing per second through unit area held perpendicular to the direction of travel (unit \(\text{W m}^{-2}\)); it depends only on physical quantities (energy, time and area). Loudness is the physiological sensation produced in the ear by the sound; it depends on the intensity but also on the sensitivity and response of the individual ear, so it is subjective and not directly measurable.
(b) Refraction through a rectangular glass prism. When light passes from air into glass, the speed changes (it decreases) and the wavelength changes (it decreases in the same ratio). The frequency does not change. So: speed changes, wavelength changes, frequency unchanged.
(c) Diffraction through the wide opening. The given diagram shows straight (plane) wavefronts, drawn as parallel vertical lines, travelling towards a barrier that has a wide opening. Because the width of the opening is much larger than the wavelength, the wavefronts emerging on the far side remain almost straight and parallel across the middle of the gap, and only curve (bend) slightly round the two edges of the opening. The required sketch of the pattern immediately after passing through the opening is shown below.
(If the opening were instead made very narrow, comparable with the wavelength, the emerging waves would spread out strongly as almost circular arcs.)
(d) Diverging lens forming the image of a shark. Data: focal length \(f = -18.0\ \text{cm}\) (negative for a diverging lens), object distance \(u = 90.0\ \text{cm}\), image length \(= 1.0\ \text{m}\).
(d)(i) Image distance. Using the lens formula with \(f\) negative for a diverging lens:
\[ \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \]\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{90} = \frac{-5 - 1}{90} = \frac{-6}{90} = -\frac{1}{15} \]\[ v = -15\ \text{cm} \]The image distance is \(15\ \text{cm}\); the negative sign shows the image is virtual, upright and on the same side as the object, as expected for a diverging lens.
(d)(ii) Length of the shark. The magnification links the image and object sizes to their distances:
\[ m = \left|\frac{v}{u}\right| = \frac{\text{length of image}}{\text{length of shark}} \]\[ \text{length of shark} = \frac{\text{length of image} \times u}{|v|} = \frac{1.0 \times 90}{15} = 6.0\ \text{m} \]The shark is \(6.0\ \text{m}\) long.
Swali 54 Ripoti
State three ways of increasing the rate of cooling of a cup of hot tea.
Three ways of increasing the rate of cooling of a cup of hot tea:
- Increase the exposed surface area, for example by pouring the tea into a wider, shallower cup or into a saucer, so that more evaporation and heat loss can take place.
- Blow air across the surface (or place it in a draught/fan), which speeds up evaporation and convection by removing the warm layer of air/vapour above the tea.
- Lower the temperature of the surroundings, for example by placing the cup in a cool place or standing it in cold water, so that the temperature difference driving heat loss is larger.
(Stirring the tea and using an uncovered, poorly lagged cup also increase the cooling rate.)
Maelezo ya Majibu
Three ways of increasing the rate of cooling of a cup of hot tea:
- Increase the exposed surface area, for example by pouring the tea into a wider, shallower cup or into a saucer, so that more evaporation and heat loss can take place.
- Blow air across the surface (or place it in a draught/fan), which speeds up evaporation and convection by removing the warm layer of air/vapour above the tea.
- Lower the temperature of the surroundings, for example by placing the cup in a cool place or standing it in cold water, so that the temperature difference driving heat loss is larger.
(Stirring the tea and using an uncovered, poorly lagged cup also increase the cooling rate.)
Swali 55 Ripoti
(a) Define specific latent heat of vaporization.
(b) (i) What are renewable energy sources?
(ii) List four renewable energy sources.
(c) Explain why tomatoes keep longer when kept in a moist jute bag in a clay pot.
(d) A box has a volume of 0.28 m\(^3\) and is 70% filled with iron fillings at 25°C. Calculate the:
(i) total mass of the iron fillings;
(ii) energy required to melt 10% of the iron fillings. [Density of iron = 8.00 x 10\(^3\) kgm\(^{-3}\); specific latent heat of fusion of iron = 1.38 x 10\(^5\) Jkg\(^{-1}\), specific heat capacity of iron = 460 Jkg\(^{-1}\) k\(^{-1}\), melting point of iron = 1500°C] 10.
(a) Specific latent heat of vaporisation is the quantity of heat required to change unit mass of a liquid into vapour at constant temperature (its boiling point).
(b)(i) Renewable energy sources are sources of energy that are naturally replenished and are not used up (do not run out) when used.
(b)(ii) Four renewable sources: solar energy, wind energy, hydro (water/tidal) energy, and biomass (biogas). Geothermal energy is also acceptable.
(c) The moist jute bag round a clay (earthenware) pot stays wet; water continually evaporates from its surface. Evaporation takes latent heat from the pot and its contents, so the tomatoes inside are cooled and kept below the surrounding temperature, slowing ripening and decay, so they keep longer.
(d) Volume of iron fillings \(= 70\% \times 0.28 = 0.196\ \text{m}^3\).
(i) Total mass \(= \rho V = 8.00\times10^{3} \times 0.196 = 1568\ \text{kg} \approx 1.57\times10^{3}\ \text{kg}.\)
(ii) Mass to be melted \(= 10\% \times 1568 = 156.8\ \text{kg}.\) It must first be raised from 25°C to the melting point 1500°C, then melted:
\[ Q = mc\,\Delta\theta + mL_f. \] \[ mc\,\Delta\theta = 156.8 \times 460 \times (1500-25) = 156.8 \times 460 \times 1475 = 1.06\times10^{8}\ \text{J}. \] \[ mL_f = 156.8 \times 1.38\times10^{5} = 2.16\times10^{7}\ \text{J}. \] \[ Q = 1.06\times10^{8} + 2.16\times10^{7} = 1.28\times10^{8}\ \text{J}. \]Energy required \(\approx 1.28\times10^{8}\ \text{J}\).
Maelezo ya Majibu
(a) Specific latent heat of vaporisation is the quantity of heat required to change unit mass of a liquid into vapour at constant temperature (its boiling point).
(b)(i) Renewable energy sources are sources of energy that are naturally replenished and are not used up (do not run out) when used.
(b)(ii) Four renewable sources: solar energy, wind energy, hydro (water/tidal) energy, and biomass (biogas). Geothermal energy is also acceptable.
(c) The moist jute bag round a clay (earthenware) pot stays wet; water continually evaporates from its surface. Evaporation takes latent heat from the pot and its contents, so the tomatoes inside are cooled and kept below the surrounding temperature, slowing ripening and decay, so they keep longer.
(d) Volume of iron fillings \(= 70\% \times 0.28 = 0.196\ \text{m}^3\).
(i) Total mass \(= \rho V = 8.00\times10^{3} \times 0.196 = 1568\ \text{kg} \approx 1.57\times10^{3}\ \text{kg}.\)
(ii) Mass to be melted \(= 10\% \times 1568 = 156.8\ \text{kg}.\) It must first be raised from 25°C to the melting point 1500°C, then melted:
\[ Q = mc\,\Delta\theta + mL_f. \] \[ mc\,\Delta\theta = 156.8 \times 460 \times (1500-25) = 156.8 \times 460 \times 1475 = 1.06\times10^{8}\ \text{J}. \] \[ mL_f = 156.8 \times 1.38\times10^{5} = 2.16\times10^{7}\ \text{J}. \] \[ Q = 1.06\times10^{8} + 2.16\times10^{7} = 1.28\times10^{8}\ \text{J}. \]Energy required \(\approx 1.28\times10^{8}\ \text{J}\).
Swali 56 Ripoti
(a) Define nucleon number
(b) A radioactive isotope of Americium (Am —241) decays into a nucleus of Neptunium (Np — 237) and an alpha (\(\alpha\)) particle as shown in the nuclear equation below. \(^{241}_{95}Am\) \(\to\) \(^{237}_{c} + ^b_a \alpha\)
(i) State the number of neutrons in the nucleus of Americium — 241.
(ii) Determine the values of a,b and c.
(c)(i) Why are y-rays not deflected by electromagnetic field?
(ii) State two properties of gamma rays that make them suitable for sterilizing medical equipment.
(d) A sample of radioactive substance was found to be left with h of its initial count rate after 110 years. Calculate its decay constant.
(a) Nucleon number (mass number) is the total number of protons and neutrons contained in the nucleus of an atom.
(b) The decay is \({}^{241}_{95}\text{Am} \to {}^{237}_{c}\text{Np} + {}^{b}_{a}\alpha\).
(i) Number of neutrons in \(\text{Am}\text{-}241 = \text{mass number} - \text{proton number} = 241 - 95 = 146\).
(ii) An alpha particle is a helium nucleus, \({}^{4}_{2}\text{He}\), so \(a = 2\) and \(b = 4\).
Conserving proton (atomic) number: \(95 = c + 2 \Rightarrow c = 93\).
Check with mass number: \(241 = 237 + 4\), which balances. Hence \(a = 2,\ b = 4,\ c = 93\).
(c)(i) Gamma rays are not deflected by an electric or magnetic field because they are electromagnetic waves and carry no electric charge; only charged particles experience a deflecting force in such fields.
(c)(ii) Two properties that make gamma rays suitable for sterilising medical equipment:
- They have very high penetrating power, so they can reach and destroy micro-organisms even inside sealed or packaged equipment.
- They are strongly lethal to bacteria and other micro-organisms, and they do so without leaving any residue or making the equipment radioactive.
(d) The sample is left with \(\dfrac{1}{32}\) of its initial count rate after \(110\) years. Writing the fraction as a power of one half:
\[\frac{N}{N_0} = \frac{1}{32} = \left(\frac{1}{2}\right)^{5},\]so \(n = 5\) half-lives have elapsed. The half-life is
\[t_{1/2} = \frac{\text{total time}}{n} = \frac{110}{5} = 22\ \text{years}.\]The decay constant is
\[\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{22} = 3.15\times10^{-2}\ \text{year}^{-1}\ (\approx 3.2\times10^{-2}\ \text{year}^{-1}).\]Examination note: the neutron count is always mass number minus proton number, so use \(241 - 95 = 146\), not \(241 - 90\). For the half-life step, express the remaining fraction as \(\left(\tfrac{1}{2}\right)^{n}\) to read off the number of half-lives before using \(\lambda = \dfrac{\ln 2}{t_{1/2}}\).
Maelezo ya Majibu
(a) Nucleon number (mass number) is the total number of protons and neutrons contained in the nucleus of an atom.
(b) The decay is \({}^{241}_{95}\text{Am} \to {}^{237}_{c}\text{Np} + {}^{b}_{a}\alpha\).
(i) Number of neutrons in \(\text{Am}\text{-}241 = \text{mass number} - \text{proton number} = 241 - 95 = 146\).
(ii) An alpha particle is a helium nucleus, \({}^{4}_{2}\text{He}\), so \(a = 2\) and \(b = 4\).
Conserving proton (atomic) number: \(95 = c + 2 \Rightarrow c = 93\).
Check with mass number: \(241 = 237 + 4\), which balances. Hence \(a = 2,\ b = 4,\ c = 93\).
(c)(i) Gamma rays are not deflected by an electric or magnetic field because they are electromagnetic waves and carry no electric charge; only charged particles experience a deflecting force in such fields.
(c)(ii) Two properties that make gamma rays suitable for sterilising medical equipment:
- They have very high penetrating power, so they can reach and destroy micro-organisms even inside sealed or packaged equipment.
- They are strongly lethal to bacteria and other micro-organisms, and they do so without leaving any residue or making the equipment radioactive.
(d) The sample is left with \(\dfrac{1}{32}\) of its initial count rate after \(110\) years. Writing the fraction as a power of one half:
\[\frac{N}{N_0} = \frac{1}{32} = \left(\frac{1}{2}\right)^{5},\]so \(n = 5\) half-lives have elapsed. The half-life is
\[t_{1/2} = \frac{\text{total time}}{n} = \frac{110}{5} = 22\ \text{years}.\]The decay constant is
\[\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{22} = 3.15\times10^{-2}\ \text{year}^{-1}\ (\approx 3.2\times10^{-2}\ \text{year}^{-1}).\]Examination note: the neutron count is always mass number minus proton number, so use \(241 - 95 = 146\), not \(241 - 90\). For the half-life step, express the remaining fraction as \(\left(\tfrac{1}{2}\right)^{n}\) to read off the number of half-lives before using \(\lambda = \dfrac{\ln 2}{t_{1/2}}\).
Swali 57 Ripoti
(a) (i) Define uniform acceleration.
(ii) Write an equation that relates linear velocity, angular velocity and radius
of path in circular motion.
(b) Two forces 30 N and 40 N act at right angles to each other. Determine by scale drawing, the magnitude and direction of the resultant force,.using a scale of 1 cm to 5 N.
(c) Explain why ships are usually refilled with sand and water after they have been emptied of their cargo.
(d) A crate of drinks of mass 20 kg is placed on a plane inclined at 30° to the horizontal. If the crate slides down with a constant speed, calculate the:
(i) co-efficient of kinetic friction;
(ii) magnitude of the frictional force acting on the crate. [ g = 10 ms\(^{-2}\)]
(a)(i) Uniform acceleration is a constant rate of change of velocity with time; equal changes of velocity occur in equal intervals of time.
(a)(ii) In circular motion the linear (tangential) velocity, angular velocity and radius are related by \( v = \omega r \).
(b) The two forces are perpendicular, so the resultant is the diagonal of the rectangle (right-angled triangle):
\[ R = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \text{N}. \]Direction (angle \(\theta\) the resultant makes with the 40 N force):
\[ \tan\theta = \frac{30}{40} = 0.75 \;\Rightarrow\; \theta = 36.9^\circ. \]By scale drawing (1 cm to 5 N), the 40 N force is 8 cm and the 30 N force is 6 cm at right angles; the diagonal measures 10 cm, i.e. 50 N at about 37° to the 40 N force.
(c) An empty ship floats high, raising its centre of gravity and making it unstable and easily rolled over by waves. Refilling with sand and water (ballast) lowers the centre of gravity and increases stability, and gives the ship enough weight and draught to sit properly in the water so it is not tossed about.
(d) Sliding down at constant speed means the friction force balances the component of weight along the plane.
(i) Coefficient of kinetic friction: for motion at constant speed on an incline, \(\mu = \tan\theta = \tan 30^\circ = 0.577\).
(ii) Frictional force \(= mg\sin\theta = 20 \times 10 \times \sin 30^\circ = 20 \times 10 \times 0.5 = 100\ \text{N}\).
Maelezo ya Majibu
(a)(i) Uniform acceleration is a constant rate of change of velocity with time; equal changes of velocity occur in equal intervals of time.
(a)(ii) In circular motion the linear (tangential) velocity, angular velocity and radius are related by \( v = \omega r \).
(b) The two forces are perpendicular, so the resultant is the diagonal of the rectangle (right-angled triangle):
\[ R = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \text{N}. \]Direction (angle \(\theta\) the resultant makes with the 40 N force):
\[ \tan\theta = \frac{30}{40} = 0.75 \;\Rightarrow\; \theta = 36.9^\circ. \]By scale drawing (1 cm to 5 N), the 40 N force is 8 cm and the 30 N force is 6 cm at right angles; the diagonal measures 10 cm, i.e. 50 N at about 37° to the 40 N force.
(c) An empty ship floats high, raising its centre of gravity and making it unstable and easily rolled over by waves. Refilling with sand and water (ballast) lowers the centre of gravity and increases stability, and gives the ship enough weight and draught to sit properly in the water so it is not tossed about.
(d) Sliding down at constant speed means the friction force balances the component of weight along the plane.
(i) Coefficient of kinetic friction: for motion at constant speed on an incline, \(\mu = \tan\theta = \tan 30^\circ = 0.577\).
(ii) Frictional force \(= mg\sin\theta = 20 \times 10 \times \sin 30^\circ = 20 \times 10 \times 0.5 = 100\ \text{N}\).
Swali 58 Ripoti
(a) What is doping in a semi conductor?
(b) Draw the symbol for OR gate.
(a) Doping in a semiconductor
Doping is the deliberate addition of a small, controlled amount of impurity (a dopant) to a pure (intrinsic) semiconductor such as silicon or germanium in order to increase its electrical conductivity. The impurity increases the number of mobile charge carriers (free electrons or holes) available for conduction.
- Adding a pentavalent impurity (e.g. phosphorus or arsenic) donates extra free electrons, producing an n-type semiconductor.
- Adding a trivalent impurity (e.g. boron or aluminium) creates extra holes, producing a p-type semiconductor.
(b) Symbol for the OR gate
The OR gate has two (or more) inputs and one output. Its output is HIGH (logic 1) when at least one of its inputs is HIGH, and LOW (logic 0) only when all inputs are LOW. It is drawn with the standard curved-back distinctive shape:
The logic operation is written as \( Q = A + B \), giving the truth table:
| A | B | Q = A + B |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
Maelezo ya Majibu
(a) Doping in a semiconductor
Doping is the deliberate addition of a small, controlled amount of impurity (a dopant) to a pure (intrinsic) semiconductor such as silicon or germanium in order to increase its electrical conductivity. The impurity increases the number of mobile charge carriers (free electrons or holes) available for conduction.
- Adding a pentavalent impurity (e.g. phosphorus or arsenic) donates extra free electrons, producing an n-type semiconductor.
- Adding a trivalent impurity (e.g. boron or aluminium) creates extra holes, producing a p-type semiconductor.
(b) Symbol for the OR gate
The OR gate has two (or more) inputs and one output. Its output is HIGH (logic 1) when at least one of its inputs is HIGH, and LOW (logic 0) only when all inputs are LOW. It is drawn with the standard curved-back distinctive shape:
The logic operation is written as \( Q = A + B \), giving the truth table:
| A | B | Q = A + B |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
Swali 59 Ripoti
(a) Define:(i) the coulomb; (ii) resistance.
(b) State the three effects of an electric current.
(c) State the standard international colour convention for the insulating material covering the following electrical wires in a three-pin plug: (i) live; (ii) neutral; (iii) earth.
(d)
A water melon of mass 5.0 kg is suspended on a uniform rod of mass 4.0 kg and 4.0 in long as illustrated in the diagram above. If the rod is in equilibrium by the action of the force between the charges +q and -q, calculate the:
(i) anti-clockwise moment;
(ii) value of q. [ g = 10 ms\(^{-2}\)]
(a)(i) The coulomb is the quantity of electric charge that passes a point in a circuit when a steady current of one ampere flows for one second: \( 1\ \text{C} = 1\ \text{A}\times1\ \text{s} \).
(a)(ii) Resistance is the opposition offered by a conductor to the flow of electric current through it. It is the ratio of the potential difference across the conductor to the current through it, \( R = \dfrac{V}{I} \), measured in ohms \( (\Omega) \).
(b) Three effects of an electric current:
- Heating (thermal) effect.
- Magnetic effect.
- Chemical effect (electrolysis).
(c) Standard international colour convention:
- (i) Live: brown.
- (ii) Neutral: blue.
- (iii) Earth: green-and-yellow (striped).
(d) From the diagram, the fulcrum divides the uniform rod so that the \( 5.0\ \text{kg} \) watermelon hangs at the left end, a distance \( L_1 = 1.0\ \text{m} \) to the left of the pivot. The rod is \( 4.0\ \text{m} \) long and uniform, so its weight (\( 4.0\ \text{kg} \)) acts at its centre, which is \( L_2 = 1.0\ \text{m} \) to the right of the pivot. The right end carries \( +q \), a distance \( 3.0\ \text{m} \) to the right of the pivot, with \( -q \) fixed \( 2.5\ \text{m} \) directly below it. Take \( g = 10\ \text{m s}^{-2} \).
(i) Anti-clockwise moment (produced by the watermelon at the left end):
\[ W_{\text{melon}} = 5.0\times10 = 50\ \text{N} \]\[ \tau_{\text{acw}} = 50\times1.0 = 50\ \text{N m} \](ii) Value of \( q \): the clockwise moments are produced by the rod's weight and by the attractive electrostatic force \( F \) which pulls the right end downward.
\[ W_{\text{rod}} = 4.0\times10 = 40\ \text{N},\qquad \tau_{\text{rod}} = 40\times1.0 = 40\ \text{N m} \]Taking moments about the pivot (anti-clockwise = clockwise):
\[ 50 = 40 + F\times3.0 \]\[ F = \frac{50-40}{3.0} = \frac{10}{3.0} = 3.33\ \text{N} \]This is the Coulomb force between the two charges \( 2.5\ \text{m} \) apart:
\[ F = \frac{1}{4\pi\varepsilon_0}\frac{q^{2}}{r^{2}} = \frac{9\times10^{9}\,q^{2}}{(2.5)^{2}} \]\[ 3.33 = \frac{9\times10^{9}\,q^{2}}{6.25} \]\[ q^{2} = \frac{3.33\times6.25}{9\times10^{9}} = \frac{20.8}{9\times10^{9}} = 2.31\times10^{-9} \]\[ q = \sqrt{2.31\times10^{-9}} = 4.8\times10^{-5}\ \text{C} \]Therefore \( q \approx 4.8\times10^{-5}\ \text{C}\ (48\ \mu\text{C}) \).
Maelezo ya Majibu
(a)(i) The coulomb is the quantity of electric charge that passes a point in a circuit when a steady current of one ampere flows for one second: \( 1\ \text{C} = 1\ \text{A}\times1\ \text{s} \).
(a)(ii) Resistance is the opposition offered by a conductor to the flow of electric current through it. It is the ratio of the potential difference across the conductor to the current through it, \( R = \dfrac{V}{I} \), measured in ohms \( (\Omega) \).
(b) Three effects of an electric current:
- Heating (thermal) effect.
- Magnetic effect.
- Chemical effect (electrolysis).
(c) Standard international colour convention:
- (i) Live: brown.
- (ii) Neutral: blue.
- (iii) Earth: green-and-yellow (striped).
(d) From the diagram, the fulcrum divides the uniform rod so that the \( 5.0\ \text{kg} \) watermelon hangs at the left end, a distance \( L_1 = 1.0\ \text{m} \) to the left of the pivot. The rod is \( 4.0\ \text{m} \) long and uniform, so its weight (\( 4.0\ \text{kg} \)) acts at its centre, which is \( L_2 = 1.0\ \text{m} \) to the right of the pivot. The right end carries \( +q \), a distance \( 3.0\ \text{m} \) to the right of the pivot, with \( -q \) fixed \( 2.5\ \text{m} \) directly below it. Take \( g = 10\ \text{m s}^{-2} \).
(i) Anti-clockwise moment (produced by the watermelon at the left end):
\[ W_{\text{melon}} = 5.0\times10 = 50\ \text{N} \]\[ \tau_{\text{acw}} = 50\times1.0 = 50\ \text{N m} \](ii) Value of \( q \): the clockwise moments are produced by the rod's weight and by the attractive electrostatic force \( F \) which pulls the right end downward.
\[ W_{\text{rod}} = 4.0\times10 = 40\ \text{N},\qquad \tau_{\text{rod}} = 40\times1.0 = 40\ \text{N m} \]Taking moments about the pivot (anti-clockwise = clockwise):
\[ 50 = 40 + F\times3.0 \]\[ F = \frac{50-40}{3.0} = \frac{10}{3.0} = 3.33\ \text{N} \]This is the Coulomb force between the two charges \( 2.5\ \text{m} \) apart:
\[ F = \frac{1}{4\pi\varepsilon_0}\frac{q^{2}}{r^{2}} = \frac{9\times10^{9}\,q^{2}}{(2.5)^{2}} \]\[ 3.33 = \frac{9\times10^{9}\,q^{2}}{6.25} \]\[ q^{2} = \frac{3.33\times6.25}{9\times10^{9}} = \frac{20.8}{9\times10^{9}} = 2.31\times10^{-9} \]\[ q = \sqrt{2.31\times10^{-9}} = 4.8\times10^{-5}\ \text{C} \]Therefore \( q \approx 4.8\times10^{-5}\ \text{C}\ (48\ \mu\text{C}) \).
Swali 60 Ripoti
A wire of length 2.00 m and radius 1.0 mm is stretched by 25.0 mm on application of a force of 103 N. Calculate the Young's modulus for the wire.
Swali 61 Ripoti
(a) Write the de Broglie equation.
(b) Explain the significance of the equation.
(a) de Broglie equation:
\[ \lambda = \frac{h}{p} = \frac{h}{mv}, \]where \(\lambda\) is the wavelength associated with a particle, \(h\) is Planck's constant, \(p = mv\) is the momentum, \(m\) the mass and \(v\) the velocity of the particle.
(b) Significance: the equation expresses the wave-particle duality of matter. It shows that every moving particle (such as an electron) has a wavelength and can therefore behave like a wave, exhibiting effects such as diffraction and interference. The wavelength is inversely proportional to momentum, so only very small (light, slow-enough) particles have wavelengths large enough for wave behaviour to be observable; ordinary large bodies have wavelengths far too small to detect. This idea underlies electron diffraction and the electron microscope.
Maelezo ya Majibu
(a) de Broglie equation:
\[ \lambda = \frac{h}{p} = \frac{h}{mv}, \]where \(\lambda\) is the wavelength associated with a particle, \(h\) is Planck's constant, \(p = mv\) is the momentum, \(m\) the mass and \(v\) the velocity of the particle.
(b) Significance: the equation expresses the wave-particle duality of matter. It shows that every moving particle (such as an electron) has a wavelength and can therefore behave like a wave, exhibiting effects such as diffraction and interference. The wavelength is inversely proportional to momentum, so only very small (light, slow-enough) particles have wavelengths large enough for wave behaviour to be observable; ordinary large bodies have wavelengths far too small to detect. This idea underlies electron diffraction and the electron microscope.
Swali 62 Ripoti
The circuit above consists of an a. c. voltage input, a diode, a resistor and a voltmeter.
(a) Identify the circuit.
(b) Draw the waveform for the output voltage.
(a) Identification of the circuit
An a.c. supply feeding a single diode in series with a load resistor, with a voltmeter connected across the resistor, is a half-wave rectifier. The diode conducts only when it is forward-biased. During the positive half-cycle of the a.c. input the diode conducts and the current passes through the resistor, so the input voltage appears across it. During the negative half-cycle the diode is reverse-biased and blocks the current, so the output falls to zero. Only one half of every complete a.c. cycle therefore reaches the load, which is why the arrangement is called a half-wave rectifier.
(b) Waveform for the output voltage
The output consists of the positive half-cycles only. During each positive half-cycle the voltage follows the sine curve of the input, rising to a peak value \(V_{p}\) and returning to zero; during each negative half-cycle the diode does not conduct, so the output stays at zero, leaving a flat gap before the next positive hump appears. The graph below shows the sinusoidal a.c. input (dashed) together with the resulting half-wave rectified output (solid) over two complete cycles.
Key features of the output waveform:
- It is a series of separated positive half-sine humps (the negative half-cycles are removed).
- The peak of each hump equals the peak input voltage \(V_{p}\).
- Between successive humps the output is zero for a full half-cycle.
- The output is a pulsating d.c. (unidirectional) voltage with the same period as the a.c. input.
Maelezo ya Majibu
(a) Identification of the circuit
An a.c. supply feeding a single diode in series with a load resistor, with a voltmeter connected across the resistor, is a half-wave rectifier. The diode conducts only when it is forward-biased. During the positive half-cycle of the a.c. input the diode conducts and the current passes through the resistor, so the input voltage appears across it. During the negative half-cycle the diode is reverse-biased and blocks the current, so the output falls to zero. Only one half of every complete a.c. cycle therefore reaches the load, which is why the arrangement is called a half-wave rectifier.
(b) Waveform for the output voltage
The output consists of the positive half-cycles only. During each positive half-cycle the voltage follows the sine curve of the input, rising to a peak value \(V_{p}\) and returning to zero; during each negative half-cycle the diode does not conduct, so the output stays at zero, leaving a flat gap before the next positive hump appears. The graph below shows the sinusoidal a.c. input (dashed) together with the resulting half-wave rectified output (solid) over two complete cycles.
Key features of the output waveform:
- It is a series of separated positive half-sine humps (the negative half-cycles are removed).
- The peak of each hump equals the peak input voltage \(V_{p}\).
- Between successive humps the output is zero for a full half-cycle.
- The output is a pulsating d.c. (unidirectional) voltage with the same period as the a.c. input.
