WAEC SSCE - Physics - 2010
Ajụjụ 2 Ripọtì
Which of the following statements best describes a rigid body? A body which
Ajụjụ 3 Ripọtì
Oxidation of the filament in a light bulb is prevented by the introduction of
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Oxidation is the process of losing electrons, and in a light bulb, the filament gets extremely hot and can react with the oxygen in the air, which can cause the filament to deteriorate and burn out. To prevent this oxidation of the filament, an inert gas is introduced into the bulb, which does not react with the filament even at high temperatures. Of the options given, argon is an inert gas and is commonly used to prevent oxidation of the filament in a light bulb. Therefore, the correct answer is option C - argon.
Ajụjụ 4 Ripọtì
In a thermos flask, heat loss by radiation is minimizes by the
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In a thermos flask, heat can be transferred by conduction, convection, and radiation. To minimize heat loss by radiation, the inner surface of the outer wall of the flask is usually silvered. This silvered surface is highly reflective and has low emissivity, meaning that it reflects radiation back into the flask and reduces the amount of heat lost to the surroundings. This is the reason why the silvered surface of a thermos flask appears mirror-like. Therefore, the correct answer is "silvered surfaces."
Ajụjụ 5 Ripọtì
an electron moves with a speed of 2.0 x 106ms-1 in a straight line. Calculate the wave length of the electron wave. [mass of an electron = 9.1 x 10-31 kg][Planck's constant = 6.6 x 10-34Js]
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Ajụjụ 6 Ripọtì
In a cathode ray tube, the function of the x-plates is to
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In a cathode ray tube, the x-plates are used to deflect the electron beam horizontally. The electron beam is generated by the cathode and accelerated by the anode towards the screen. The y-plates are used to deflect the beam vertically. By applying appropriate voltages to the x-plates and y-plates, the electron beam can be made to scan across the screen and form a visible image. Therefore, option A, which states that the x-plates deflect the electron beam horizontally, is the correct answer.
Ajụjụ 7 Ripọtì
The pressure P, volume V, and absolute temperature T of a given mass of an ideal gas change simultaneously. Which of the followng equations is correct about the gas?
Ajụjụ 8 Ripọtì
Which of the following statements about static friction is correct? It
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Dew point is not affected by
Ajụjụ 10 Ripọtì
The following devices use plane mirrors in their operations, except
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The device that does not use plane mirrors in its operation is binoculars. A periscope uses two plane mirrors to reflect light and allow a person to see over an obstacle or around a corner. A sextant uses a plane mirror to reflect the image of the sun or stars onto a scale, enabling sailors to navigate. A kaleidoscope uses multiple plane mirrors to reflect and multiply the image of small colored objects into a visually attractive pattern. However, binoculars use a combination of convex and concave lenses to magnify distant objects, rather than plane mirrors.
Ajụjụ 11 Ripọtì
A positively charged glass rod is placed near the cap of a positively charged electroscope. The divergence of the leaf is observed to
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When a positively charged glass rod is brought near the cap of a positively charged electroscope, the charge on the electroscope cap is repelled by the charged glass rod. As a result, the cap loses some of its charge and the electrons move down to the leaves, causing them to acquire a positive charge. The like charges on the leaves cause them to repel each other, causing them to diverge. Therefore, the divergence of the leaf will increase. Thus, the correct option is (b) increase.
Ajụjụ 12 Ripọtì
Neutrons are used to induce artificial radioactivity because they
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Neutrons are used to induce artificial radioactivity because they have no charge and can penetrate atomic nuclei, unlike positively charged alpha particles or negatively charged beta particles. When neutrons are absorbed by the nucleus of an atom, they can cause the nucleus to become unstable and undergo radioactive decay. This can result in the emission of various forms of radiation, such as alpha particles, beta particles, or gamma rays, depending on the type of decay that occurs. Therefore, neutrons can be used to induce artificial radioactivity in a target material, which can then be used for various purposes such as medical imaging, cancer treatment, and nuclear power generation.
Ajụjụ 13 Ripọtì
A galvanometer of internal resistance 50\(\Omega\) and full scale deflection of 20 mA is converted into a voltmeter by connecting a resistor of resistance 1950\(\Omega\) in series to it. Calculate the maximum voltage that can be read by the voltmeter
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The maximum voltage that can be read by the voltmeter can be calculated using the formula: V = I(G + S) Where V is the voltage, I is the current passing through the circuit, G is the galvanometer resistance, and S is the shunt resistance. In this case, the full scale deflection current of the galvanometer is 20 mA, which is 0.02 A. The galvanometer resistance is given as 50 Ω, and the shunt resistance is 1950 Ω. To calculate the current passing through the circuit, we can use Ohm's Law: V = IR Where R is the total resistance in the circuit. In this case, the total resistance is the sum of the galvanometer resistance and the shunt resistance: R = G + S = 50 Ω + 1950 Ω = 2000 Ω Substituting the values into Ohm's Law, we get: V = IR = 0.02 A × 2000 Ω = 40 V Therefore, the maximum voltage that can be read by the voltmeter is 40V. Option D is correct.
Ajụjụ 14 Ripọtì
The energy, E stored in an inductor of inductance L when current I passes through it is given by the equation
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Ajụjụ 15 Ripọtì
The ability of a wave to spread around corners is called
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The ability of a wave to spread around corners is called diffraction. When waves encounter an obstacle or an aperture, they bend around the corners or edges of the obstacle or aperture and spread into the region behind the obstacle or aperture. This phenomenon is known as diffraction. The degree of diffraction depends on the size of the obstacle or aperture and the wavelength of the wave. Diffraction is a common phenomenon observed in many types of waves, including sound waves, light waves, and water waves.
Ajụjụ 16 Ripọtì
A ray of light travels from air to water. The refeactive index of water is given by the expression
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Ajụjụ 18 Ripọtì
A solid of mass 0.5kg initially at 50oC melts at 60oC. Calculate the quantity of heat required to melt it completely. [specific heat capacity of the solid =2.0KJ kg-1K-1][Specific latent heat of fusion of solid =1.0 KJ kg-1]
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Ajụjụ 19 Ripọtì
Which of the following electromagnetic wave has the longest wavelength?
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Radio waves have the longest wavelength among the given options. Electromagnetic waves are waves that are created by oscillations in electric and magnetic fields. The wavelength of an electromagnetic wave is the distance between two consecutive peaks or troughs in the wave. The different types of electromagnetic waves have different wavelengths and frequencies, and they are arranged in a spectrum according to their wavelengths. Radio waves have the longest wavelength and the lowest frequency among the electromagnetic waves. They are used for communication, such as in radio and television broadcasting. Gamma rays, on the other hand, have the shortest wavelength and highest frequency among the electromagnetic waves, and are often produced by radioactive decay or nuclear reactions. Infrared and ultraviolet rays have intermediate wavelengths and frequencies between radio waves and gamma rays. Infrared waves are often associated with heat radiation, while ultraviolet rays are known for their ability to cause sunburn and skin damage.
Ajụjụ 20 Ripọtì
In order to prevent bar magnets from losing their strength, they are stored wih keepers such that the two bars are placed with
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To prevent bar magnets from losing their strength, they are stored with keepers such that the two bars are placed with unlike poles facing each other and the keepers placed at their ends. This is because the magnetic field of a bar magnet flows from one pole to the other. When two magnets with the same pole are placed together, their magnetic fields interfere with each other and cancel out some of their magnetic properties. On the other hand, when two magnets with opposite poles are placed together, their magnetic fields combine and create a stronger magnetic field. By placing the keepers at the ends, the magnetic field of the bar magnet is confined to the magnet itself and doesn't interact with any external magnetic fields, which could cause it to lose its magnetism over time.
Ajụjụ 21 Ripọtì
Which of the following factors does not affect the electric resistance of wire?
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The factor that does not affect the electric resistance of a wire is mass. Electric resistance is the opposition to the flow of electric current through a conductor. The four factors that affect the resistance of a wire are length, temperature, cross-sectional area, and the type of material used to make the wire. Length: the longer the wire, the higher the resistance as the electric current has to travel further, experiencing more collisions with the wire's atoms and electrons, which leads to a higher resistance. Temperature: an increase in temperature leads to an increase in the wire's resistance. This is because as the temperature increases, the wire's atoms and electrons vibrate more, impeding the flow of current, thus increasing the resistance. Cross-sectional area: the larger the wire's cross-sectional area, the lower its resistance because there are more paths for the electric current to flow through, reducing the chances of collisions with atoms and electrons, thus reducing resistance. Type of material used: different materials have different numbers of free electrons, which affects their resistance. For example, copper has more free electrons than iron, and therefore has a lower resistance. Mass, on the other hand, does not affect the resistance of a wire. The mass of a wire is not related to the number of free electrons or the wire's ability to impede the flow of current.
Ajụjụ 22 Ripọtì
Which of the following properties of a long steel bar would alter if the bar were melted, recast into a cube and allowed to cool to its original temperature?
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The property that would alter if a long steel bar were melted, recast into a cube and allowed to cool to its original temperature is its volume. When the steel bar is melted and recast into a cube, the volume of the cube will be different from the volume of the original steel bar. This is because the shape of the cube is different from that of the bar. However, if the cube is allowed to cool to its original temperature, its volume will be the same as that of the original bar. The other properties such as density, specific heat capacity and electrical resistance will remain unchanged since they are intrinsic properties of the material and do not depend on the shape of the object.
Ajụjụ 23 Ripọtì
In the wiring of houses, the fuse is connected to the wire coloured
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The correct wire color for the fuse in the wiring of houses may vary depending on the country or region. However, in many countries such as the UK and Europe, the wire color for the fuse is brown. This is because brown is the color assigned for live wires in electrical wiring, and the fuse is connected to the live wire for protection against electrical overloading or short-circuiting. Blue is typically used for neutral wires, yellow/green for earth wires, and yellow for the live wire in some countries. It is important to always follow the wiring regulations and guidelines of your region and seek professional assistance if necessary when dealing with electrical wiring.
Ajụjụ 24 Ripọtì
An electric drill rated 400W is used to drill a hole in copper of mass 400g in 20s. Calculate the rise in temperature if all the heat produced is absorbed by the copper. [specific heat capacity of copper = 400Jkg-1K-1]
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To calculate the rise in temperature of the copper, we need to use the formula: heat energy = mass x specific heat capacity x temperature change We know the power of the drill (400W) and the time taken (20s), so we can calculate the total energy produced: total energy = power x time total energy = 400W x 20s total energy = 8000J We also know the mass of the copper (400g) and the specific heat capacity of copper (400Jkg-1K-1). To calculate the temperature change, we rearrange the formula: temperature change = heat energy / (mass x specific heat capacity) temperature change = 8000J / (0.4kg x 400Jkg-1K-1) temperature change = 50K Therefore, the rise in temperature of the copper is 50oC. So the correct option is: 50oC.
Ajụjụ 25 Ripọtì
Two cars X and Y travelling in opposite directions along the same high way at uniform velocities 110 kmh-1 and 90kmh-1 respectively pass each other at a certain point. The velocity of X relative to Y at the time they pass each other is
Ajụjụ 26 Ripọtì
A certain wave has a speed of 20ms-1. If the frequency of the wave is 0.25Hz, calculate the distance between successive crests of the wave
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The speed of a wave is given by the product of its frequency and wavelength, i.e., speed = frequency x wavelength Rearranging the above equation, we get: wavelength = speed / frequency Substituting the given values, we get: wavelength = 20 / 0.25 = 80m Therefore, the distance between successive crests of the wave is 80m. Answer: 4. 80.0m
Ajụjụ 27 Ripọtì
From the diagram, the correct equation for the refractive index n of the medium is
Ajụjụ 28 Ripọtì
The resonant frequency of an a.c circuit is 1000 kHz. If each of the capacitance and inductance in the circuit is reduced by 50%, and no other changes are made, the resonant frequency will become
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The resonant frequency of an a.c circuit is given by the formula: f = 1/(2π√(LC)) Where f is the resonant frequency, L is the inductance of the circuit, and C is the capacitance of the circuit. If each of the capacitance and inductance in the circuit is reduced by 50%, then the new values of L and C will be 0.5L and 0.5C respectively. Substituting these new values into the formula, we get: f' = 1/(2π√(0.5L * 0.5C)) f' = 1/(2π√(0.25LC)) f' = 1/(0.5π√(LC)) f' = 2f Therefore, the resonant frequency of the circuit will be doubled, which means it will become 2000kHz.
Ajụjụ 29 Ripọtì
a machine is said to be a third class lever when the
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In a third class lever, the effort is applied between the fulcrum and the load. This means that the effort arm is shorter than the load arm. Examples of third class levers include tweezers and forceps.
Ajụjụ 31 Ripọtì
A moving body of mass 25.0kg undergoes a uniform retardation of 20.0ms-2. Calculate the magnitude of the retardating force
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The force acting on an object is given by the formula F = ma, where F is the force, m is the mass and a is the acceleration. However, in this case, the object is undergoing retardation, which means its acceleration is negative. Therefore, we have a = -20.0ms-2. The mass of the object is given as 25.0kg. Using the formula F = ma, we can calculate the magnitude of the force as follows: F = m x a = 25.0kg x (-20.0ms-2) = -500N The negative sign indicates that the force is acting in the opposite direction to the motion of the object. However, since the question asks for the magnitude of the force, we take the absolute value, which is 500N. Therefore, the magnitude of the retardating force acting on the object is 500.00N. So, the answer is 500.00N.
Ajụjụ 32 Ripọtì
The amplitude of a sound wave determines its
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The amplitude of a sound wave determines its loudness. The loudness of a sound is the human perception of its intensity, which is directly related to the amplitude of the sound wave. The larger the amplitude of a sound wave, the louder the sound is perceived to be. However, it is important to note that the pitch, or the perceived highness or lowness of a sound, is determined by the frequency of the sound wave, not its amplitude.
Ajụjụ 33 Ripọtì
A nuclide represented as \(^{70}_{32}X\) has a neutron-proton ratio of
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The neutron-proton ratio of a nuclide is the ratio of the number of neutrons (N) to the number of protons (Z) in its nucleus. For the nuclide \(^{70}_{32}X\), the number 70 represents the total number of nucleons (protons + neutrons) in its nucleus, while 32 represents the number of protons. Therefore, the number of neutrons can be calculated as: N = 70 - 32 = 38 The neutron-proton ratio of this nuclide can then be calculated as: N/Z = 38/32 = 1.19 Therefore, the neutron-proton ratio of this nuclide is approximately 1.2, which corresponds to.
Ajụjụ 36 Ripọtì
A vertical string, suspended from a fixed point and having a small mass attached to the free end is set into oscillations. Which of the following statements about the system are correct? i. The potential energy of the mass is a minimum at the middle of the swing ii. Its kinetic energy is a maximum at the middle of the swing iii. The sum of the potential and kinetic energies is constant throughout the swing
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Ajụjụ 37 Ripọtì
The advantages of the thermoelectric thermometer include the following characteristics except that it
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The thermoelectric thermometer is a type of temperature measuring device that utilizes the Seebeck effect to measure temperature. It consists of two different metals or semiconductors connected together, which generate a small voltage when there is a temperature difference between the two junctions. This voltage is directly proportional to the temperature difference between the two junctions, making it an effective and reliable way of measuring temperature. The advantages of the thermoelectric thermometer include its ability to measure rapidly changing temperatures, measure high and low temperatures, and measure temperature almost at a point. Additionally, it is fairly sensitive, which allows it to detect even small changes in temperature. Therefore, the statement that is not true is that the thermoelectric thermometer is not fairly sensitive.
Ajụjụ 38 Ripọtì
a thermometer records 680 mmHg at steam point and 440 mmHg at ice point. The temperature it records at 380 mmHg is
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To find the temperature at 380 mmHg, we need to interpolate between the steam and ice points. We can use the formula: t = ((P - Pice) / (Psteam - Pice)) x (Tsteam - Tice) + Tice where: - t is the temperature we want to find - P is the pressure we want to use for interpolation (380 mmHg in this case) - Pice is the pressure at the ice point (440 mmHg) - Psteam is the pressure at the steam point (680 mmHg) - Tice is the temperature at the ice point (0oC) - Tsteam is the temperature at the steam point (100oC) Plugging in the values, we get: t = ((380 - 440) / (680 - 440)) x (100 - 0) + 0 t = (-60 / 240) x 100 t = -25oC Therefore, the temperature recorded at 380 mmHg is -25oC.
Ajụjụ 39 Ripọtì
The average speed of an object is determined by dividing the sum of its initial and final speeds by two only when the acceleration is
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The statement is true only when the acceleration is uniform. Uniform acceleration means that the object is changing its velocity by an equal amount in the same direction during equal intervals of time. In this case, the average speed can be calculated by dividing the sum of initial and final velocity by 2. However, if the acceleration is not uniform, the average speed cannot be calculated using this formula. In that case, we need to use other formulas that take into account the non-uniform acceleration of the object.
Ajụjụ 41 Ripọtì
Which of the following factors will increase the sensitivity of moving coil meter?
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A moving coil meter works by measuring the magnetic field produced by an electric current passing through a coil of wire. The sensitivity of the meter refers to how easily it can detect small changes in the current. A strong spring, which opposes the motion of the meter's pointer, will actually decrease the sensitivity of the meter, making it less responsive to small changes in current. A low number of turns on the coil or a small area of the coil will also reduce sensitivity, since there will be less magnetic field produced for a given current. Therefore, the factor that will increase the sensitivity of a moving coil meter is a soft iron core. This is because a soft iron core will amplify the magnetic field produced by the current, making it easier to detect small changes. The iron core will also help to concentrate the magnetic field, further increasing the sensitivity of the meter.
Ajụjụ 42 Ripọtì
When an object is placed at the principal focus of a concave mirror, the location of the image formed is
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When an object is placed at the principal focus of a concave mirror, the location of the image formed is at infinity. This is because light rays that are reflected from the mirror are parallel to the principal axis of the mirror, and do not converge or diverge after reflection. Therefore, they appear to be coming from a point at infinity, and the image formed by the mirror appears to be located at this point.
Ajụjụ 43 Ripọtì
What is the density of a fuel of relative density 0.72? [Density of water = 1000kg m-3]
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Relative density, also known as specific gravity, is defined as the ratio of the density of a substance to the density of a reference substance. In this case, the reference substance is water, which has a density of 1000 kg/m3. Therefore, the density of the fuel can be calculated by multiplying the relative density by the density of water: Density of fuel = Relative density x Density of water Density of fuel = 0.72 x 1000 kg/m3 Density of fuel = 720 kg/m3 Hence, the density of the fuel is 720 kg/m3. Therefore, the correct option is: 720 kgm-3
Ajụjụ 44 Ripọtì
a body is projected horizonatlly from the top of a cliff 45m above the ground. if the body lands at a distance 30m from the foot of the cliff, calculate the speed of projection. [g = 10ms-2]
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Ajụjụ 46 Ripọtì
The types of motion which a cylindrical drum rolling down an inclined plane undergoes are
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A cylindrical drum rolling down an inclined plane undergoes rotational and translational motion. Rotational motion refers to the spinning motion around its own axis, which the drum exhibits while rolling down the inclined plane. Translational motion refers to the linear motion of the center of mass of the drum, which moves along the inclined plane. Therefore, the correct answer is "rotational and translational."
Ajụjụ 47 Ripọtì
Specific heat capacity 36.4 Jkg-1 K-1 of a body means that when 36.4J of heat energy is supplied to a body of
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The specific heat capacity of a substance is defined as the amount of heat energy required to raise the temperature of a unit mass of the substance by 1 Kelvin (or 1 degree Celsius). Therefore, if the specific heat capacity of a body is given as 36.4 Jkg\(^{-1}\)K\(^{-1}\), it means that when 36.4 Joules of heat energy is supplied to a body of mass 1kg, the temperature of the body will rise by 1 Kelvin. Therefore, the correct answer is: "mass 1kg, the temperature rises by 1K".
Ajụjụ 48 Ripọtì
Which of the following sources of energy is/are exhausible? i. Solar ii. Fossil fuels iii. Tidal power
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The term "exhaustible" means that the source of energy is finite and will eventually run out. Solar energy is considered inexhaustible because it comes from the sun and will be available as long as the sun exists. On the other hand, fossil fuels, such as coal, oil, and natural gas, are considered exhaustible because they are finite resources and will eventually be depleted. Tidal power is also considered inexhaustible because it relies on the gravitational pull of the moon and the sun on the Earth's oceans, which will continue as long as the Earth continues to orbit the sun. Therefore, the answer is: ii and iii only.
Ajụjụ 49 Ripọtì
simple harmonic oscillator has a period of 0.02s and amplitude of 0.25m. Calculate the speed in ms-1 at the centre of the oscillation
Ajụjụ 50 Ripọtì
Which of the following instruments is best for measuring the diameter of a thin constantan wire?
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The micrometer screw gauge is the best instrument for measuring the diameter of a thin constantan wire. Micrometers are highly sensitive instruments that can measure small distances accurately. They are commonly used in precision engineering to measure the thickness of wires and the diameter of cylinders. The micrometer screw gauge has a spindle that moves as a thimble is turned, and the distance moved is indicated on a graduated scale. This allows for very precise measurements of small distances, such as the diameter of a thin wire. Callipers, metre rule, and vernier callipers are useful for measuring larger distances but are not as accurate as a micrometer screw gauge for measuring small distances.
Ajụjụ 51 Ripọtì
(a) State two factors which affect the angle of deviation of a ray of light through a triangular glass prism.
(b) Seven virtual images of an object are formed when two plane mirrors are inclined at an angle 0 to each other. Calculate the value of 0.
(c) By means of a ripple tank, a student was able to generate series of transverse waves by varying the frequency of the dipper and all the waves so generated covered a distance of 0.80 m in 0.2s.
(i) Determine the speed, v, of the waves.
Copy and complete the table given in your answer booklet.
(iii) Plot a graph with f on the vertical axis and \(\lambda ^{-1}\) on the horizontal axis.
(iv) What does the slope of the graph represent?
(a) Two factors which affect the angle of deviation of a ray through a triangular glass prism:
- The angle of incidence of the ray on the first refracting face.
- The refracting (apex) angle of the prism together with the refractive index of the glass (which itself depends on the colour/wavelength of the light).
(b) Angle between the two plane mirrors.
The number of images formed by two plane mirrors inclined at an angle \(\theta\) is
\[N=\frac{360^\circ}{\theta}-1.\]With \(N=7\):
\[7=\frac{360^\circ}{\theta}-1 \;\Rightarrow\; \frac{360^\circ}{\theta}=8 \;\Rightarrow\; \theta=\frac{360^\circ}{8}=45^\circ.\](c)(i) Speed of the waves.
\[v=\frac{\text{distance}}{\text{time}}=\frac{0.80\ \text{m}}{0.2\ \text{s}}=4.0\ \text{m s}^{-1}.\](ii) Completed table. Using \(\lambda=\dfrac{v}{f}=\dfrac{4.0}{f}\) and \(\lambda^{-1}=\dfrac{1}{\lambda}\):
| \(f\) /Hz | \(\lambda\) /m | \(\lambda^{-1}\) /m\(^{-1}\) |
|---|---|---|
| 2.0 | 2.00 | 0.50 |
| 4.0 | 1.00 | 1.00 |
| 6.0 | 0.67 | 1.50 |
| 8.0 | 0.50 | 2.00 |
| 10.0 | 0.40 | 2.50 |
(iii) Graph of \(f\) (vertical axis) against \(\lambda^{-1}\) (horizontal axis).
The points \((0.50,2.0),(1.00,4.0),(1.50,6.0),(2.00,8.0),(2.50,10.0)\) lie on a straight line passing through the origin, since \(f=v\,(\lambda^{-1})\).
(iv) Meaning of the slope.
Reading two points on the line of best fit, e.g. \((2.50,10.0)\) and \((0.50,2.0)\):
\[\text{slope}=\frac{f}{\lambda^{-1}}=\frac{10.0-2.0}{2.50-0.50}=\frac{8.0}{2.0}=4.0.\]Since \(f=v\times\lambda^{-1}\), the slope \(=f\lambda=v=4.0\ \text{m s}^{-1}\). The slope of the graph represents the speed (velocity) of the waves.
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(a) Two factors which affect the angle of deviation of a ray through a triangular glass prism:
- The angle of incidence of the ray on the first refracting face.
- The refracting (apex) angle of the prism together with the refractive index of the glass (which itself depends on the colour/wavelength of the light).
(b) Angle between the two plane mirrors.
The number of images formed by two plane mirrors inclined at an angle \(\theta\) is
\[N=\frac{360^\circ}{\theta}-1.\]With \(N=7\):
\[7=\frac{360^\circ}{\theta}-1 \;\Rightarrow\; \frac{360^\circ}{\theta}=8 \;\Rightarrow\; \theta=\frac{360^\circ}{8}=45^\circ.\](c)(i) Speed of the waves.
\[v=\frac{\text{distance}}{\text{time}}=\frac{0.80\ \text{m}}{0.2\ \text{s}}=4.0\ \text{m s}^{-1}.\](ii) Completed table. Using \(\lambda=\dfrac{v}{f}=\dfrac{4.0}{f}\) and \(\lambda^{-1}=\dfrac{1}{\lambda}\):
| \(f\) /Hz | \(\lambda\) /m | \(\lambda^{-1}\) /m\(^{-1}\) |
|---|---|---|
| 2.0 | 2.00 | 0.50 |
| 4.0 | 1.00 | 1.00 |
| 6.0 | 0.67 | 1.50 |
| 8.0 | 0.50 | 2.00 |
| 10.0 | 0.40 | 2.50 |
(iii) Graph of \(f\) (vertical axis) against \(\lambda^{-1}\) (horizontal axis).
The points \((0.50,2.0),(1.00,4.0),(1.50,6.0),(2.00,8.0),(2.50,10.0)\) lie on a straight line passing through the origin, since \(f=v\,(\lambda^{-1})\).
(iv) Meaning of the slope.
Reading two points on the line of best fit, e.g. \((2.50,10.0)\) and \((0.50,2.0)\):
\[\text{slope}=\frac{f}{\lambda^{-1}}=\frac{10.0-2.0}{2.50-0.50}=\frac{8.0}{2.0}=4.0.\]Since \(f=v\times\lambda^{-1}\), the slope \(=f\lambda=v=4.0\ \text{m s}^{-1}\). The slope of the graph represents the speed (velocity) of the waves.
Ajụjụ 52 Ripọtì
State one reason each why cathode rays:
(a) are not electromagnetic waves;
(b) cast sharp shadows of objects in their path;
(c) can rotate a light paddle wheel inside a discharge tube.
(a) Cathode rays are not electromagnetic waves because they are deflected by both electric and magnetic fields, showing that they are streams of negatively charged particles (electrons); electromagnetic waves carry no charge and are undeflected by such fields.
(b) They cast sharp shadows of objects in their path because they travel in straight lines from the cathode.
(c) They can rotate a light paddle wheel inside a discharge tube because they are material particles possessing mass and momentum (kinetic energy), which they transfer to the vanes on striking them.
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(a) Cathode rays are not electromagnetic waves because they are deflected by both electric and magnetic fields, showing that they are streams of negatively charged particles (electrons); electromagnetic waves carry no charge and are undeflected by such fields.
(b) They cast sharp shadows of objects in their path because they travel in straight lines from the cathode.
(c) They can rotate a light paddle wheel inside a discharge tube because they are material particles possessing mass and momentum (kinetic energy), which they transfer to the vanes on striking them.
Ajụjụ 53 Ripọtì
(s) State the Heisenberg Uncertainty Principle.
(b) The product of the uncertainties in Heisenberg Uncertainty Principle is equal to or greater than a constant. State the mathematical expression for this constant.
Ajụjụ 54 Ripọtì
(a) State two factors which affect the mass of elements deposited during electrolysis.
(b) List two non-electrolysis.
(a) Two factors that affect the mass of an element deposited during electrolysis (Faraday's laws):
- The quantity of electricity passed, i.e. the product of current and time (\(Q = It\)); the mass is directly proportional to it.
- The electrochemical equivalent of the element (which depends on its relative atomic mass and valency).
Other acceptable factor: the duration of current flow at a fixed current.
(b) Two non-electrolytes (substances that do not conduct electricity by ionic dissociation):
- Distilled (pure) water.
- Sugar solution (also acceptable: kerosene, ethanol, pure alcohol).
Akọwa Nkọwa
(a) Two factors that affect the mass of an element deposited during electrolysis (Faraday's laws):
- The quantity of electricity passed, i.e. the product of current and time (\(Q = It\)); the mass is directly proportional to it.
- The electrochemical equivalent of the element (which depends on its relative atomic mass and valency).
Other acceptable factor: the duration of current flow at a fixed current.
(b) Two non-electrolytes (substances that do not conduct electricity by ionic dissociation):
- Distilled (pure) water.
- Sugar solution (also acceptable: kerosene, ethanol, pure alcohol).
Ajụjụ 55 Ripọtì
When a lead-acid accumulator is fully charged, evolution of gases occurs at the electrodes. Name these gases and the respective electrodes at which thcy are given off.
Gases evolved are:
(i) Hydrogen at the cathode (negative lead plate).
(ii) Oxygen at the anode (positive lead peroxide plate).
Akọwa Nkọwa
Gases evolved are:
(i) Hydrogen at the cathode (negative lead plate).
(ii) Oxygen at the anode (positive lead peroxide plate).
Ajụjụ 56 Ripọtì
(a) Differentiate between plane polarization and interference as applied to waves.
(b) List two uses of polaroids.
(a) Plane polarisation compared with interference
- Plane polarisation is the restriction of the vibrations of a transverse wave to a single plane (one direction) perpendicular to the direction of travel. It shows that light is a transverse wave and occurs only with transverse waves.
- Interference is the superposition of two waves of the same frequency travelling in the same region, producing points of reinforcement (constructive) and cancellation (destructive). It occurs with both transverse and longitudinal waves and demonstrates the wave nature of light.
In short, polarisation concerns the plane of vibration of a single wave, while interference concerns the combined effect of two coherent waves.
(b) Two uses of polaroids
- In sunglasses (polaroid spectacles) to reduce glare from reflected sunlight.
- In camera filters and in the control of light intensity, e.g. in liquid-crystal displays and photographic work.
Akọwa Nkọwa
(a) Plane polarisation compared with interference
- Plane polarisation is the restriction of the vibrations of a transverse wave to a single plane (one direction) perpendicular to the direction of travel. It shows that light is a transverse wave and occurs only with transverse waves.
- Interference is the superposition of two waves of the same frequency travelling in the same region, producing points of reinforcement (constructive) and cancellation (destructive). It occurs with both transverse and longitudinal waves and demonstrates the wave nature of light.
In short, polarisation concerns the plane of vibration of a single wave, while interference concerns the combined effect of two coherent waves.
(b) Two uses of polaroids
- In sunglasses (polaroid spectacles) to reduce glare from reflected sunlight.
- In camera filters and in the control of light intensity, e.g. in liquid-crystal displays and photographic work.
Ajụjụ 57 Ripọtì
(a) Write down the names of two particles used in explaining the wave nature of matter.
(b) State the wave characteristics which are exhibited by the particles named in (a) above.
(a) Two particles used to explain the wave nature of matter:
- The electron.
- The neutron (the proton is also acceptable).
(b) Wave characteristics exhibited by these particles:
- Diffraction (they bend around obstacles/through gaps, as in electron diffraction).
- Interference (they can superpose to give reinforcement and cancellation).
These effects confirm that moving particles have an associated de Broglie wavelength \(\lambda = \dfrac{h}{mv}\).
Akọwa Nkọwa
(a) Two particles used to explain the wave nature of matter:
- The electron.
- The neutron (the proton is also acceptable).
(b) Wave characteristics exhibited by these particles:
- Diffraction (they bend around obstacles/through gaps, as in electron diffraction).
- Interference (they can superpose to give reinforcement and cancellation).
These effects confirm that moving particles have an associated de Broglie wavelength \(\lambda = \dfrac{h}{mv}\).
Ajụjụ 58 Ripọtì
(a) State three conclusions that can be drawn from Rutherford's experiment on the scattering of alpha particles by a thin metal foil in relation to the structure of the atom
The diagram above illustrates th3 energy levels of an electron in an atom. If an excited electron moves from \(n_2\) to \(n_\theta\), calculate the:
(i) frequency;
(ii) wavelength of the emitted radiation. [ \(h = 6.6 \times 10^{-34}\) Js; le V = .6 \(\times 10^{-19}\) J; C = 3.0 \(\times 10^8\) ms\(^{-1}\)]
(c) The following nuclear equations represent two types of radioactivity.
\({}^{226}_{88}R_a \to {}^{222}_{86}R_n + {}^a_2a\) (Equation A)
\({}^{14}_7N + {}^4_2a \to {}^{17}_8O + {}^1_1p\) (Equation B)
Identify each type and explain briefly the difference between them
(a) Three conclusions from Rutherford's alpha-particle scattering experiment
- Most of the atom is empty space, because the vast majority of alpha particles passed straight through the thin metal foil undeflected.
- Each atom has a very small, dense core (the nucleus) which carries the whole positive charge, because a few alpha particles were deflected through large angles and a very small number were bounced almost straight back.
- Nearly all the mass of the atom is concentrated in this tiny nucleus, while the electrons occupy the large space surrounding it.
(b) Transition from \(n_2\) to \(n_0\)
From the energy-level diagram: \(n_2 = -2.0\ \text{eV}\) and \(n_0 = -12.0\ \text{eV}\). The energy of the emitted photon is the difference between these levels:
\[ E = E_{n_2} - E_{n_0} = (-2.0) - (-12.0) = 10.0\ \text{eV} \]
Convert to joules using \(1\ \text{eV} = 1.6\times10^{-19}\ \text{J}\):
\[ E = 10.0 \times 1.6\times10^{-19} = 1.6\times10^{-18}\ \text{J} \]
(i) Frequency from \(E = hf\):
\[ f = \frac{E}{h} = \frac{1.6\times10^{-18}}{6.6\times10^{-34}} = 2.42\times10^{15}\ \text{Hz} \]
(ii) Wavelength from \(c = f\lambda\):
\[ \lambda = \frac{c}{f} = \frac{3.0\times10^{8}}{2.42\times10^{15}} = 1.24\times10^{-7}\ \text{m} \]
(about 124 nm, in the ultraviolet region.)
(c) Types of radioactivity
Equation A: \(^{226}_{88}\text{Ra} \to\ ^{222}_{86}\text{Rn} + ^{4}_{2}\alpha\). This is natural (spontaneous) radioactivity, specifically alpha decay: an unstable nucleus disintegrates on its own, emitting an alpha particle and forming a new element.
Equation B: \(^{14}_{7}\text{N} + ^{4}_{2}\alpha \to\ ^{17}_{8}\text{O} + ^{1}_{1}p\). This is artificial (induced) transmutation: a stable nucleus is deliberately bombarded by an incoming particle (an alpha particle), changing it into a different nucleus.
Difference: In A the disintegration is spontaneous, happening by itself with no external cause; in B the nuclear change is artificially induced by bombarding the target nucleus with a fast-moving particle.
Akọwa Nkọwa
(a) Three conclusions from Rutherford's alpha-particle scattering experiment
- Most of the atom is empty space, because the vast majority of alpha particles passed straight through the thin metal foil undeflected.
- Each atom has a very small, dense core (the nucleus) which carries the whole positive charge, because a few alpha particles were deflected through large angles and a very small number were bounced almost straight back.
- Nearly all the mass of the atom is concentrated in this tiny nucleus, while the electrons occupy the large space surrounding it.
(b) Transition from \(n_2\) to \(n_0\)
From the energy-level diagram: \(n_2 = -2.0\ \text{eV}\) and \(n_0 = -12.0\ \text{eV}\). The energy of the emitted photon is the difference between these levels:
\[ E = E_{n_2} - E_{n_0} = (-2.0) - (-12.0) = 10.0\ \text{eV} \]
Convert to joules using \(1\ \text{eV} = 1.6\times10^{-19}\ \text{J}\):
\[ E = 10.0 \times 1.6\times10^{-19} = 1.6\times10^{-18}\ \text{J} \]
(i) Frequency from \(E = hf\):
\[ f = \frac{E}{h} = \frac{1.6\times10^{-18}}{6.6\times10^{-34}} = 2.42\times10^{15}\ \text{Hz} \]
(ii) Wavelength from \(c = f\lambda\):
\[ \lambda = \frac{c}{f} = \frac{3.0\times10^{8}}{2.42\times10^{15}} = 1.24\times10^{-7}\ \text{m} \]
(about 124 nm, in the ultraviolet region.)
(c) Types of radioactivity
Equation A: \(^{226}_{88}\text{Ra} \to\ ^{222}_{86}\text{Rn} + ^{4}_{2}\alpha\). This is natural (spontaneous) radioactivity, specifically alpha decay: an unstable nucleus disintegrates on its own, emitting an alpha particle and forming a new element.
Equation B: \(^{14}_{7}\text{N} + ^{4}_{2}\alpha \to\ ^{17}_{8}\text{O} + ^{1}_{1}p\). This is artificial (induced) transmutation: a stable nucleus is deliberately bombarded by an incoming particle (an alpha particle), changing it into a different nucleus.
Difference: In A the disintegration is spontaneous, happening by itself with no external cause; in B the nuclear change is artificially induced by bombarding the target nucleus with a fast-moving particle.
Ajụjụ 59 Ripọtì
(a) List two examples each of substances with:
(i) low viscosity;
(ii) high viscosity.
(b) When is a liquid said to be viscostatic?
(a) Examples
- (i) Low viscosity: water and kerosene (also petrol, ethanol).
- (ii) High viscosity: glycerine and honey (also engine oil, palm oil, tar).
(b) Viscostatic liquid
A liquid is said to be viscostatic when its viscosity remains almost constant (does not change appreciably) with change in temperature. Such liquids, e.g. certain multigrade lubricating oils, keep a steady flow behaviour whether hot or cold.
Akọwa Nkọwa
(a) Examples
- (i) Low viscosity: water and kerosene (also petrol, ethanol).
- (ii) High viscosity: glycerine and honey (also engine oil, palm oil, tar).
(b) Viscostatic liquid
A liquid is said to be viscostatic when its viscosity remains almost constant (does not change appreciably) with change in temperature. Such liquids, e.g. certain multigrade lubricating oils, keep a steady flow behaviour whether hot or cold.
Ajụjụ 60 Ripọtì
(a) Explain the terms: (i) inertia; (ii) inertial mass.
(b) List three factors which affect the rate of evaporation of water in a pond.
(c) Two ice cubes pressed together for some time were found to stick together when the pressure was removed. Explain this observation.
(d) Two vertical capillary tubes of the same diameter are lowered into beakers situated at the same level, containing liquids A and B of densities 9.2 x 10\(^2\) kgm\(^{-3}\) and 1.30 x 10\(^3\) kgm\(^{-3}\) respectively. A suction pump is used to withdraw air from the top of the liquid columns in the tubes by means of a T-piece arrangement until the liquid in A rises to a height of 26.0 cm. Calculate the height of the liquid in tube B.
(a)(i) Inertia is the property of a body by which it resists any change in its state of rest or of uniform motion in a straight line.
(a)(ii) Inertial mass is the measure of the inertia of a body; it is the ratio of the resultant force acting on the body to the acceleration it produces, \(m = \dfrac{F}{a}\).
(b) Three factors affecting the rate of evaporation of water in a pond:
- Temperature of the water (and of the surroundings).
- Surface area exposed to the air.
- Movement of air (wind) and the humidity of the surrounding air.
(c) Ice cubes sticking together (regelation): the pressure between the two cubes lowers the melting point of the ice at the contact surfaces, so a thin film of water forms there. When the pressure is removed the melting point rises again and this water re-freezes, joining the two cubes into one.
(d) Capillary tubes calculation. The two tubes have the same diameter and are joined to the same reduced pressure through the T-piece, so the pressure supporting each column is the same. Hence \(\rho_A g h_A = \rho_B g h_B\), giving
\[ h_B = \frac{\rho_A h_A}{\rho_B} = \frac{(9.2\times10^{2})(26.0)}{1.30\times10^{3}} \] \[ h_B = \frac{23920}{1300} = 18.4\ \text{cm} \]The liquid in tube B rises to 18.4 cm.
Akọwa Nkọwa
(a)(i) Inertia is the property of a body by which it resists any change in its state of rest or of uniform motion in a straight line.
(a)(ii) Inertial mass is the measure of the inertia of a body; it is the ratio of the resultant force acting on the body to the acceleration it produces, \(m = \dfrac{F}{a}\).
(b) Three factors affecting the rate of evaporation of water in a pond:
- Temperature of the water (and of the surroundings).
- Surface area exposed to the air.
- Movement of air (wind) and the humidity of the surrounding air.
(c) Ice cubes sticking together (regelation): the pressure between the two cubes lowers the melting point of the ice at the contact surfaces, so a thin film of water forms there. When the pressure is removed the melting point rises again and this water re-freezes, joining the two cubes into one.
(d) Capillary tubes calculation. The two tubes have the same diameter and are joined to the same reduced pressure through the T-piece, so the pressure supporting each column is the same. Hence \(\rho_A g h_A = \rho_B g h_B\), giving
\[ h_B = \frac{\rho_A h_A}{\rho_B} = \frac{(9.2\times10^{2})(26.0)}{1.30\times10^{3}} \] \[ h_B = \frac{23920}{1300} = 18.4\ \text{cm} \]The liquid in tube B rises to 18.4 cm.
Ajụjụ 61 Ripọtì
A force of 40 N is applied at the free end of a wire fixed at one end to produce an extension of 0.24 mm. If the original length and diameter of the wire art., 3 m and 2.0 mm respectively, calculate the: (a) stress on the wire; (b) strain in the wire.
Given: \(F = 40\ \text{N}\), extension \(e = 0.24\ \text{mm} = 0.24\times10^{-3}\ \text{m}\), original length \(L = 3\ \text{m}\), diameter \(d = 2.0\ \text{mm} = 2.0\times10^{-3}\ \text{m}\).
(a) Stress \(= \dfrac{F}{A}\), where the cross-sectional area \(A = \dfrac{\pi d^{2}}{4}\).
\[ A = \frac{\pi (2.0\times10^{-3})^{2}}{4} = 3.14\times10^{-6}\ \text{m}^{2} \] \[ \text{Stress} = \frac{40}{3.14\times10^{-6}} = 1.27\times10^{7}\ \text{N m}^{-2} \](b) Strain \(= \dfrac{\text{extension}}{\text{original length}}\):
\[ \text{Strain} = \frac{0.24\times10^{-3}}{3} = 8.0\times10^{-5} \]Strain has no unit.
Akọwa Nkọwa
Given: \(F = 40\ \text{N}\), extension \(e = 0.24\ \text{mm} = 0.24\times10^{-3}\ \text{m}\), original length \(L = 3\ \text{m}\), diameter \(d = 2.0\ \text{mm} = 2.0\times10^{-3}\ \text{m}\).
(a) Stress \(= \dfrac{F}{A}\), where the cross-sectional area \(A = \dfrac{\pi d^{2}}{4}\).
\[ A = \frac{\pi (2.0\times10^{-3})^{2}}{4} = 3.14\times10^{-6}\ \text{m}^{2} \] \[ \text{Stress} = \frac{40}{3.14\times10^{-6}} = 1.27\times10^{7}\ \text{N m}^{-2} \](b) Strain \(= \dfrac{\text{extension}}{\text{original length}}\):
\[ \text{Strain} = \frac{0.24\times10^{-3}}{3} = 8.0\times10^{-5} \]Strain has no unit.
Ajụjụ 62 Ripọtì
(a) Distinguish between perfectly elastic collision and perfectly inelastic collision.
(b) Sketch a distance — time graph for a particle moving in a straight line with:
(i) uniform speed;
(ii) variable speed.
(c) A body starts from rest and travels distances of 120, 300 and 180m in successive equal time intervals of 12 s. During each interval the body is uniformly accelerated. (i) Calculate the velocity of the body at the end of each successive time interval.
(ii) Sketch a velocity-time graph for the motion.
(a) Perfectly elastic vs perfectly inelastic collision
| Perfectly elastic collision | Perfectly inelastic collision |
|---|---|
| Both momentum and kinetic energy are conserved. | Momentum is conserved but kinetic energy is not; some is converted to heat, sound and deformation. |
| The colliding bodies separate (do not stick) after impact. | The colliding bodies stick together and move as one body after impact. |
| The relative velocity of separation equals the relative velocity of approach. | The bodies move with a single common velocity after impact (relative velocity of separation = 0). |
(b) Distance–time graphs
For uniform speed the distance increases by equal amounts in equal times, so the graph is a straight line of constant gradient through the origin. For variable speed the speed keeps changing, so the gradient keeps changing and the graph is a curve (here the gradient increases, showing increasing speed).
(c) Body starting from rest
The body covers 120 m, 300 m and 180 m in three successive equal intervals of \(t = 12\ \text{s}\), being uniformly accelerated within each interval. For uniform acceleration the average velocity over an interval equals the mean of the velocities at its two ends, so the end-of-interval velocities follow directly.
(i) Velocity at the end of each interval
Interval 1 (\(u = 0\)): \(s = ut + \tfrac{1}{2}at^2\)
\[120 = 0\times 12 + \tfrac{1}{2}a(12)^2 = 72a \;\Rightarrow\; a = \tfrac{5}{3}\ \text{m s}^{-2}\] \[v_1 = u + at = 0 + \tfrac{5}{3}\times 12 = 20\ \text{m s}^{-1}\]Interval 2 (\(u = v_1 = 20\ \text{m s}^{-1}\)):
\[300 = 20\times 12 + \tfrac{1}{2}a(12)^2 = 240 + 72a \;\Rightarrow\; a = \tfrac{5}{6}\ \text{m s}^{-2}\] \[v_2 = 20 + \tfrac{5}{6}\times 12 = 30\ \text{m s}^{-1}\]Interval 3 (\(u = v_2 = 30\ \text{m s}^{-1}\)):
\[180 = 30\times 12 + \tfrac{1}{2}a(12)^2 = 360 + 72a \;\Rightarrow\; a = -2.5\ \text{m s}^{-2}\] \[v_3 = 30 + (-2.5)\times 12 = 0\ \text{m s}^{-1}\]So the velocities at the end of the successive intervals are 20 m s\(^{-1}\), 30 m s\(^{-1}\) and 0 m s\(^{-1}\) (at \(t = 12\ \text{s}, 24\ \text{s}\) and \(36\ \text{s}\) respectively). Each result checks against the average speed: \(120/12 = 10 = \tfrac{0+20}{2}\), \(300/12 = 25 = \tfrac{20+30}{2}\), \(180/12 = 15 = \tfrac{30+0}{2}\).
(ii) Velocity–time graph
The graph is made of three straight segments (constant acceleration in each interval): rising from 0 to 20 m s\(^{-1}\) in the first 12 s, rising more gently to 30 m s\(^{-1}\) at 24 s, then falling to rest at 36 s.
Akọwa Nkọwa
(a) Perfectly elastic vs perfectly inelastic collision
| Perfectly elastic collision | Perfectly inelastic collision |
|---|---|
| Both momentum and kinetic energy are conserved. | Momentum is conserved but kinetic energy is not; some is converted to heat, sound and deformation. |
| The colliding bodies separate (do not stick) after impact. | The colliding bodies stick together and move as one body after impact. |
| The relative velocity of separation equals the relative velocity of approach. | The bodies move with a single common velocity after impact (relative velocity of separation = 0). |
(b) Distance–time graphs
For uniform speed the distance increases by equal amounts in equal times, so the graph is a straight line of constant gradient through the origin. For variable speed the speed keeps changing, so the gradient keeps changing and the graph is a curve (here the gradient increases, showing increasing speed).
(c) Body starting from rest
The body covers 120 m, 300 m and 180 m in three successive equal intervals of \(t = 12\ \text{s}\), being uniformly accelerated within each interval. For uniform acceleration the average velocity over an interval equals the mean of the velocities at its two ends, so the end-of-interval velocities follow directly.
(i) Velocity at the end of each interval
Interval 1 (\(u = 0\)): \(s = ut + \tfrac{1}{2}at^2\)
\[120 = 0\times 12 + \tfrac{1}{2}a(12)^2 = 72a \;\Rightarrow\; a = \tfrac{5}{3}\ \text{m s}^{-2}\] \[v_1 = u + at = 0 + \tfrac{5}{3}\times 12 = 20\ \text{m s}^{-1}\]Interval 2 (\(u = v_1 = 20\ \text{m s}^{-1}\)):
\[300 = 20\times 12 + \tfrac{1}{2}a(12)^2 = 240 + 72a \;\Rightarrow\; a = \tfrac{5}{6}\ \text{m s}^{-2}\] \[v_2 = 20 + \tfrac{5}{6}\times 12 = 30\ \text{m s}^{-1}\]Interval 3 (\(u = v_2 = 30\ \text{m s}^{-1}\)):
\[180 = 30\times 12 + \tfrac{1}{2}a(12)^2 = 360 + 72a \;\Rightarrow\; a = -2.5\ \text{m s}^{-2}\] \[v_3 = 30 + (-2.5)\times 12 = 0\ \text{m s}^{-1}\]So the velocities at the end of the successive intervals are 20 m s\(^{-1}\), 30 m s\(^{-1}\) and 0 m s\(^{-1}\) (at \(t = 12\ \text{s}, 24\ \text{s}\) and \(36\ \text{s}\) respectively). Each result checks against the average speed: \(120/12 = 10 = \tfrac{0+20}{2}\), \(300/12 = 25 = \tfrac{20+30}{2}\), \(180/12 = 15 = \tfrac{30+0}{2}\).
(ii) Velocity–time graph
The graph is made of three straight segments (constant acceleration in each interval): rising from 0 to 20 m s\(^{-1}\) in the first 12 s, rising more gently to 30 m s\(^{-1}\) at 24 s, then falling to rest at 36 s.
Ajụjụ 63 Ripọtì
An X-ray tube operates at a potential of 2500 V. If the power of the tube is 750 W, calculate the speed of the electron striking the target. [e = 1.6 x 10\(^{-19}\) C; mass of electron = 9.1 x 10\(^{-3}\) kg]
Given: accelerating p.d. \(V = 2500\ \text{V}\), \(e = 1.6\times10^{-19}\ \text{C}\), electron mass \(m = 9.1\times10^{-31}\ \text{kg}\). (The mass is taken as \(9.1\times10^{-31}\ \text{kg}\), the standard electron mass; the exponent printed as \(10^{-3}\) is a typographical slip.)
The work done by the accelerating field is converted to kinetic energy of the electron:
\[ eV = \tfrac{1}{2}mv^{2} \] \[ v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2(1.6\times10^{-19})(2500)}{9.1\times10^{-31}}} \] \[ v = \sqrt{\frac{8.0\times10^{-16}}{9.1\times10^{-31}}} = \sqrt{8.79\times10^{14}} \] \[ v \approx 2.97\times10^{7}\ \text{m s}^{-1} \]The electron strikes the target at about \(3.0\times10^{7}\ \text{m s}^{-1}\). (The power rating of 750 W is not needed for the speed; it would fix the tube current.)
Akọwa Nkọwa
Given: accelerating p.d. \(V = 2500\ \text{V}\), \(e = 1.6\times10^{-19}\ \text{C}\), electron mass \(m = 9.1\times10^{-31}\ \text{kg}\). (The mass is taken as \(9.1\times10^{-31}\ \text{kg}\), the standard electron mass; the exponent printed as \(10^{-3}\) is a typographical slip.)
The work done by the accelerating field is converted to kinetic energy of the electron:
\[ eV = \tfrac{1}{2}mv^{2} \] \[ v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2(1.6\times10^{-19})(2500)}{9.1\times10^{-31}}} \] \[ v = \sqrt{\frac{8.0\times10^{-16}}{9.1\times10^{-31}}} = \sqrt{8.79\times10^{14}} \] \[ v \approx 2.97\times10^{7}\ \text{m s}^{-1} \]The electron strikes the target at about \(3.0\times10^{7}\ \text{m s}^{-1}\). (The power rating of 750 W is not needed for the speed; it would fix the tube current.)
Ajụjụ 64 Ripọtì
A stone is projected vertically upward with a speed of 30ms\(^{1}\) from the top of a tower of height 50 m. Neglecting air resistance, determine the maximum height it reached from the ground. [g = 10 ms\(^-2\)]
Given: \(u = 30\ \text{m s}^{-1}\) (upward), tower height \(= 50\ \text{m}\), \(g = 10\ \text{m s}^{-2}\).
Step 1 - rise above the top of the tower. At the highest point the velocity is zero, so using \(v^{2} = u^{2} - 2gh\):
\[ 0 = 30^{2} - 2(10)h \] \[ h = \frac{900}{20} = 45\ \text{m} \]Step 2 - height above the ground. Add the height of the tower:
\[ H = 50 + 45 = 95\ \text{m} \]The stone reaches a maximum height of 95 m above the ground.
Akọwa Nkọwa
Given: \(u = 30\ \text{m s}^{-1}\) (upward), tower height \(= 50\ \text{m}\), \(g = 10\ \text{m s}^{-2}\).
Step 1 - rise above the top of the tower. At the highest point the velocity is zero, so using \(v^{2} = u^{2} - 2gh\):
\[ 0 = 30^{2} - 2(10)h \] \[ h = \frac{900}{20} = 45\ \text{m} \]Step 2 - height above the ground. Add the height of the tower:
\[ H = 50 + 45 = 95\ \text{m} \]The stone reaches a maximum height of 95 m above the ground.
Ajụjụ 65 Ripọtì
(a)
When a positively charged conductor is placed near a candle flame, the flame spreads out as shown in the diagram above. Explain this observation.
(b) A proton moving with a speed of 5.0 x 10\(^{5}\) ms\(^{-1}\) enters a magnetic field of flux density 0.2 T at an angle of 30° to the field. Calculate the magnitude of the magnetic fcrce exerted on the proton. [Proton charge = 1.6 x 10\(^{-19}\) C]
(c)
The diagram above illustrates a 9.0 V battery of internal resistance 0.5 \(\Omega\) connected to two resistors of values 2.0 \(\Omega\) and R \(\Omega\). A\(_1\) A\(_2\) and A\(_3\) are ammeters of negligible internal resistances. If Al reads 4.0 A, calculate the:
(i) equivalent resistance of the combined resistors 2.0 \(\Omega\) and R \(\Omega\);
(ii) currents through A\(_1\) and A\(_3\) ; (iii) value of R.
Akọwa Nkọwa
None
