WAEC SSCE - Physics - 2016
Pergunta 1 Relatório
An astronomical telescope has objectives and eye-piece lenses of focal length 3.5m and 5cm respectively. Determine the magnifying power of the telescope when in normal adjustment
Detalhes da Resposta
The magnifying power of a telescope is given by the formula: Magnifying power = (focal length of objective lens) / (focal length of eyepiece lens) Substituting the given values, we have: Magnifying power = (3.5 m) / (0.05 m) = 70 Therefore, the magnifying power of the telescope when in normal adjustment is 70.0. The correct option is (a) 70.0.
Pergunta 2 Relatório
In storing magnets, keepers are used to
Detalhes da Resposta
Keepers are used to reduce self-demagnetization of magnets. When a magnet is not in use, it can gradually lose its magnetic strength due to various factors, including temperature changes, vibrations, and exposure to other magnetic fields. This process is called self-demagnetization. A keeper is a piece of ferromagnetic material that is placed across the poles of a magnet to complete its magnetic circuit. This prevents the magnetic field from leaking out and helps maintain the magnet's strength over time. By providing a closed path for the magnetic flux, the keeper ensures that the magnetic energy of the magnet is conserved and not dissipated. Therefore, keepers are useful for storing magnets for prolonged periods to prevent self-demagnetization. They do not cancel the effect of the earth's magnetic field or protect the magnet from stray electric fields, nor do they increase the strength of the magnets.
Pergunta 3 Relatório
How many beta particle(s) are emitted in the radioactive decay of \(^{198}_{79}Au \leftarrow ^{198}_{80}Hg\)?
Detalhes da Resposta
In the given radioactive decay equation, beta decay occurs when a neutron in the nucleus of Hg-198 is converted into a proton, causing the atomic number to increase by one and the atomic mass to remain the same. In beta decay, a high-energy electron is emitted from the nucleus.
The equation for beta decay is:
n → p + e- + anti-νe
where n represents a neutron, p represents a proton, e- represents an electron, and anti-νe represents an antineutrino.
Therefore, in the given decay equation, one beta particle (electron) will be emitted as a neutron is converted to a proton.
Hence, the answer is 1.
Pergunta 4 Relatório
Two different materials, rubbed against each other acquired opposite charges when separated. This is an example of charging by
Detalhes da Resposta
The process of rubbing two different materials against each other and causing them to acquire opposite charges when separated is an example of charging by friction. When the two materials are rubbed together, electrons are transferred from one material to the other, causing one material to become positively charged and the other to become negatively charged. This is because electrons are negatively charged particles and when they are transferred from one material to the other, they create an imbalance of charges. This process is also known as triboelectric charging, and it is commonly observed in our daily lives, for example, when we rub a balloon on our hair and it becomes attracted to a wall.
Pergunta 5 Relatório
A student measures the volume of a liquid using a measuring cylinder. What else needs to be measured by the student in order to determine the density of the liquid?
Detalhes da Resposta
To determine the density of a liquid, the student needs to measure the mass of the liquid and the volume it occupies. The volume can be measured using the measuring cylinder, while the mass of the liquid can be measured using a balance. The mass of the cylinder itself is not necessary for determining the density of the liquid, as it does not contribute to the density calculation. The depth of the liquid in the cylinder is also not necessary, as the volume of the liquid can be measured regardless of its depth. However, the temperature of the liquid can affect its density, so it is important to measure the temperature and ensure that it is constant during the measurement.
Pergunta 6 Relatório
The image formed by a concave mirror is real, inverted and magnified, when object is placed
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Pergunta 7 Relatório
When a body is thrown vertically upwards, its velocity at the maximum height is
Detalhes da Resposta
When a body is thrown vertically upwards, its velocity at the maximum height is zero. This is because when the body is thrown upwards, it gains potential energy due to its height from the ground. As the body moves higher, its potential energy increases, but its kinetic energy decreases, which means its speed decreases. When the body reaches its maximum height, all of its kinetic energy is converted into potential energy, and the velocity becomes zero for a moment. Then the body starts to fall back down towards the ground, and as it falls, its potential energy decreases, but its kinetic energy increases, which means its speed increases. Therefore, the velocity at the maximum height is zero, and it is also the moment when the direction of the velocity changes from upward to downward.
Pergunta 8 Relatório
Total internal reflection. i. is a phenomenon of refraction of light. ii. occurs when light rays travel into a denser medium. iii. occurs when the critical angle is just exceeded. iv. is a phenomenon of reflection of light. Which of the statements above are correct?
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Pergunta 9 Relatório
At a birthday party, the celebrant pops a corked fruit wine. If the cork shoots out of the bottle at an angle of 40o to the horizontal and travels a horizontal distance of 4.50m in 1.25s. Calculate the initial speed of the cork.
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Pergunta 10 Relatório
Two different materials, rubbed against each other acquired opposite charges when separated. This is an example of charging by
Detalhes da Resposta
This is an example of charging by friction. When two different materials are rubbed against each other, their surfaces may exchange electrons. Electrons are negatively charged particles that can move from one material to another during friction. As a result, one material may lose electrons and become positively charged while the other material gains electrons and becomes negatively charged. This is why the two materials acquire opposite charges when separated. This process is known as triboelectric charging, which is the scientific term for charging by friction.
Pergunta 11 Relatório
The pair of musical instruments that works on the vibration of air in pipes is
Detalhes da Resposta
The pair of musical instruments that work on the vibration of air in pipes is the flute and the trumpet. The flute is a wind instrument that works by blowing air across a hole, which creates a vibrating column of air inside the instrument's pipe. As the player changes the position of their fingers on the flute's keys, different notes are produced by changing the length of the vibrating column of air. The trumpet, on the other hand, is a brass instrument that works by the player buzzing their lips into a small cup-shaped mouthpiece. This creates a vibrating column of air inside the instrument's long, curved pipe. Like the flute, different notes are produced by changing the length of the vibrating column of air by pressing the instrument's valves. Both the flute and the trumpet use the vibration of air in pipes to produce sound, but they achieve this in slightly different ways. The flute relies on blowing air across a hole, while the trumpet relies on buzzing lips into a mouthpiece.
Pergunta 12 Relatório
In the clinical thermometer, the function of the kink is to i. Prevent the back flow of mercury after use. ii. Ensure the back flow of mercury after use. iii. Maintain the thermometric property of mercury. iv. Allow the temperature attained to be read.
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Pergunta 13 Relatório
A uniform metre rule is balanced on a fulcrum placed at the 35cm mark by suspending a mass of 120g at the 10cm mark. Calculate the mass of the metre rule.
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Pergunta 14 Relatório
The diagram given, represents the graph of stress against strain for an elastic wire. The point Q on the graph is the
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Pergunta 15 Relatório
An aluminium rod of length 1.8m at 10°C is heated to proceed produce a difference in length of 0.007m. Calculate the temperature to which it is heated. (Linear expansivity of aluminium = 2.3 x 10\(^{-5}\)K\(^{-1}\)
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Pergunta 16 Relatório
The mouth piece of a telephone handset converts
Detalhes da Resposta
The mouthpiece of a telephone handset converts sound energy into electrical energy. When you speak into the mouthpiece, your voice creates sound waves. These sound waves cause a small diaphragm in the mouthpiece to vibrate. This vibration is converted into an electrical signal that can be transmitted through the telephone line to the receiver at the other end. In other words, the mouthpiece of a telephone handset works like a microphone, converting sound energy into an electrical signal that can be transmitted through the telephone system.
Pergunta 17 Relatório
The continuous stirring in the method of mixtures to determine the specific thermal capacity of a substance ensures
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Pergunta 18 Relatório
The electricity meters in houses measure energy unit consumed in
Detalhes da Resposta
Electricity meters in houses measure the amount of energy consumed by electrical appliances in a unit called "kilowatt-hour" (kWh). A kilowatt-hour is a measure of energy that is equivalent to using one kilowatt (1,000 watts) of power for one hour. This unit of measurement is used because it takes into account both the amount of power being used (in kilowatts) and the duration of its use (in hours). In contrast, volts measure the electrical potential difference between two points, amperes measure the current flowing through a circuit, and coulombs measure the quantity of electric charge. While these units are important in understanding how electricity works, they are not used to measure the amount of energy consumed by electrical appliances in households.
Pergunta 19 Relatório
A man and his friend have masses 70kg and 60kg respectively. If they are seated 1.0m apart, calculate the gravitational force of attraction between them. (G = 6.67 x 10-11 Nm2 kg-2)
Detalhes da Resposta
The gravitational force of attraction between two objects can be calculated using the formula F = Gm1m2/r2, where F is the force of attraction, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between the centers of the two objects. In this case, the masses of the man and his friend are 70kg and 60kg, respectively, and they are seated 1.0m apart. Therefore, using the above formula, we can calculate the gravitational force of attraction between them as: F = Gm1m2/r2 = (6.67 x 10-11 Nm2 kg-2) (70kg) (60kg) / (1.0m)2 ≈ 2.80 x 10-7 N Therefore, the gravitational force of attraction between the man and his friend is approximately 2.80 x 10-7 N. This force is very small, which is why we do not notice the gravitational force between two ordinary objects in our everyday lives. The reason why we feel the gravitational force of the Earth is because the Earth is a very large object with a very large mass, which produces a significant gravitational force.
Pergunta 20 Relatório
A ball is dropped from the top of a tower. Due to air resistance,it reaches terminal velocity. Which of the following statement(s) about its motion is/are correct? i. The acceleration of the ball is zero. ii. The net force on the ball is zero. iii. The velocity of the ball increases.
Detalhes da Resposta
When the ball is initially dropped from the top of the tower, it accelerates due to the force of gravity. However, as it falls, it encounters air resistance. Air resistance is a force that opposes the motion of an object through the air. Eventually, the force of air resistance becomes equal in magnitude to the force of gravity, and the ball stops accelerating. This is known as terminal velocity. At terminal velocity, the acceleration of the ball is zero. This is because the net force on the ball is zero. The force of gravity is balanced by the force of air resistance. Thus, statement (i) and (ii) are correct. However, statement (iii) is incorrect. At terminal velocity, the velocity of the ball does not increase. It remains constant. This is because the net force acting on the ball is zero, and Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on it. If the net force is zero, the acceleration is zero, and the velocity remains constant. Therefore, the correct option is (i and ii only).
Pergunta 21 Relatório
Given that v, f and \(\lambda\) are the velocity, frequency and wavelength of a wave respectively. Which of the following equations is correct?
Detalhes da Resposta
The correct equation is: f = \(\frac{v}{\lambda}\). This equation represents the relationship between the frequency, wavelength, and velocity of a wave. It states that the frequency of a wave is equal to the velocity of the wave divided by its wavelength. In other words, if you know the velocity and wavelength of a wave, you can calculate its frequency using this equation. Alternatively, if you know the frequency and velocity of a wave, you can calculate its wavelength using the same equation. The other equations given in the options are not correct. The first equation v = f2\(\lambda\) implies that the velocity of a wave is proportional to the square of its frequency and wavelength, which is not true. The second equation f = \(\frac{v}{\lambda^2}\) implies that the frequency of a wave is inversely proportional to the square of its wavelength, which is also not true. The last equation \(\lambda = \frac{f}{v^2}\) implies that the wavelength of a wave is inversely proportional to the square of its frequency and velocity, which again is not true.
Pergunta 22 Relatório
A pressurized perfume bottle is left on the window pane of a room. What happens to he gas molecules of the perfume on a sunny day? They
Detalhes da Resposta
On a sunny day, the gas molecules of the perfume inside a pressurized bottle left on a window pane will move more rapidly. This is because the increase in temperature from the sun's heat will cause the gas molecules to gain kinetic energy, which in turn increases their speed. As a result, they will collide more frequently with each other and with the walls of the bottle, exerting a greater pressure. However, the expansion of the gas molecules will be limited by the container's pressure and volume. Therefore, the perfume bottle will not expand or contract. The only observable change will be that the gas molecules inside the bottle will move more rapidly.
Pergunta 23 Relatório
The diagram given illustrates the waveform of a note produced by vibrations in an air column of length L. If the speed of sound in air is V, derive an expression for the frequency of the notes
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Pergunta 24 Relatório
Which of the following statements is not correct about steel and soft iron?
Detalhes da Resposta
The statement that is not correct is "steel is more magnetized than soft iron." In general, soft iron is more readily magnetized than steel because it has a higher magnetic permeability. This means that it can be magnetized more easily when exposed to a magnetic field. However, steel can be magnetized to a higher degree than soft iron, meaning that it can hold a stronger magnetic field. This is why permanent magnets are often made of steel, which can retain its magnetic properties over time. The statement that "soft iron more readily loses its magnetism than steel" is generally true. Soft iron has a lower coercivity than steel, which means that it is easier for its magnetic field to be disrupted or demagnetized. Steel, on the other hand, has a higher coercivity, which means that it is more resistant to losing its magnetism. Overall, both steel and soft iron have their unique magnetic properties, and each is better suited for different applications depending on their magnetic needs.
Pergunta 25 Relatório
Which of the following factors does not affect the e.m.f of a primary cell? Its
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Pergunta 26 Relatório
X-rays and infrared rays are both electromagnetic. Which common property do the waves have? Both waves
Detalhes da Resposta
The common property that X-rays and infrared rays share is that they are both electromagnetic waves and travel at the same speed in a vacuum. This means that they both propagate through space without the need for a medium, and they both travel at the speed of light, which is approximately 299,792,458 meters per second. However, X-rays have much higher frequencies and shorter wavelengths than infrared rays, which means they have higher energy and can penetrate through solid objects, while infrared rays have longer wavelengths and are primarily used for heating and sensing purposes.
Pergunta 27 Relatório
The two person of a body undergoing a uniformly accelerated motion are (10s, 10ms-1) and (30s, 50ms-1). On the velocity-time graph pf the body.
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Pergunta 28 Relatório
In a p-type semiconductor, the?
Detalhes da Resposta
In a p-type semiconductor, holes are the majority charge carriers. Semiconductors are materials that have conductivity between conductors (such as metals) and insulators (such as rubber). In a pure semiconductor, such as silicon or germanium, there are equal numbers of positive (holes) and negative (electrons) charge carriers, and the material is called intrinsic semiconductor. However, by adding impurities to the semiconductor material, we can change its electrical properties. When we add impurities such as boron, which has one less valence electron than the semiconductor material, the boron atom can bond with the semiconductor material leaving a "hole" in the crystal lattice where there should be an electron. These holes are positively charged and can move through the material like a particle. In a p-type semiconductor, the majority of charge carriers are these positively charged holes, while the minority carriers are negatively charged electrons. Therefore, when a voltage is applied, the holes are the ones that move and contribute to electrical conductivity, while the electrons do not move much. As a result, the electrical resistivity of the p-type semiconductor increases because the majority charge carriers (holes) have a lower mobility than electrons.
Pergunta 29 Relatório
A 10\(\mu\)C charge moves with a velocity of 1.0 x 105 ms-1 at right angle to a uniform magnetic field of flux density 5.0 x 10-4T. Calculate the force on the charge
Detalhes da Resposta
The force experienced by a charged particle moving in a magnetic field is given by the formula F = Bqv, where F is the force, B is the magnetic field flux density, q is the charge on the particle, and v is the velocity of the particle. Using the given values, we can calculate the force as: F = Bqv = (5.0 x 10^-4 T)(10^-5 C)(1.0 x 10^5 m/s) = 5.0 x 10^-4 N Therefore, the correct answer is: 5.0 x 10^-4 N.
Pergunta 30 Relatório
The dimensions of momentum are
Detalhes da Resposta
The dimensions of momentum are MLT-1. Momentum is a physical quantity that measures the amount of motion an object has. It is defined as the product of an object's mass and velocity. Mass is a measure of the amount of matter in an object, while velocity is a measure of the speed and direction of motion of the object. The dimension of mass is M (in kilograms), the dimension of velocity is LT-1 (in meters per second), and the dimension of momentum is the product of mass and velocity, which is MLT-1. Therefore, the dimension of momentum is MLT-1.
Pergunta 31 Relatório
Which of the following statements correctly defines the boiling point of a liquid? The temperature at which the
Detalhes da Resposta
The boiling point of a liquid is the temperature at which its saturated vapor pressure becomes equal to the atmospheric pressure. At this temperature, the molecules of the liquid gain enough energy to overcome the intermolecular forces holding them together and escape into the atmosphere as vapor. As the vapor pressure increases, it eventually reaches the same level as the atmospheric pressure, causing bubbles to form throughout the liquid and resulting in the characteristic boiling behavior. correctly defines the boiling point of a liquid.
Pergunta 32 Relatório
A mercury-in-glass thermometer read 4cm at ice point and 29cm at steam point. Calculate the temperature when the mercury level is at 9cm
Detalhes da Resposta
To find the temperature when the mercury level is at 9cm, we need to first calculate the length of the mercury column from the ice point to the 9cm mark. The total length of the mercury column from the ice point to the steam point is 29cm - 4cm = 25cm. To find the length of the mercury column from the ice point to the 9cm mark, we can use the proportion: length of mercury column from ice point to 9cm mark / total length of mercury column = temperature at 9cm / temperature range where the temperature range is the difference between the temperatures of the steam point and ice point. Plugging in the values we know, we get: length of mercury column from ice point to 9cm mark / 25cm = (temperature at 9cm - 0oC) / (100oC - 0oC) Simplifying this equation, we get: length of mercury column from ice point to 9cm mark = (temperature at 9cm / 100) x 25cm Now we can solve for the temperature at the 9cm mark by plugging in the given value for the length of the mercury column at the 9cm mark: 9cm - 4cm = (temperature at 9cm / 100) x 25cm Simplifying this equation, we get: temperature at 9cm = 100 x (9cm - 4cm) / 25cm temperature at 9cm = 20oC Therefore, the temperature when the mercury level is at 9cm is 20oC.
Pergunta 33 Relatório
The diagram given illustrates a force-distance graph for the motion of a wooden block. Determine the work done on the block when moved through a distance of 5m.
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Pergunta 35 Relatório
Two magnets are used to magnetize a soft iron bar PQ as illustrated in the diagram given. The correct polarity at P and Q respectively
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Pergunta 36 Relatório
What determines the polarity at the ends of an electromagnet? The
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Pergunta 37 Relatório
An aluminium cable of diameter 4 x 10-3m and resistivity 3.0 x 10-8\(\omega\)m has a resistance of 21\(\Omega\). Calculate the length of the cable. (\(\pi = 3.14\))
Detalhes da Resposta
The resistance of a conductor is given by the formula:
R = ρL/A
where R is resistance, ρ (rho) is resistivity, L is the length of the conductor, and A is the cross-sectional area of the conductor.
We are given the resistance of the cable (R = 21 Ω), its diameter (d = 4 x 10-3 m), and its resistivity (ρ = 3.0 x 10-8 Ωm). We can use the diameter to find the cross-sectional area of the cable, using the formula:
A = πd2/4
Substituting the given values:
A = π x (4 x 10-3)2/4 = 1.26 x 10-5 m2
Now we can rearrange the formula for resistance to solve for the length of the cable (L):
L = RA/ρ
Substituting the given values:
L = 21 x 1.26 x 10-5 / 3.0 x 10-8
Simplifying the expression, we get:
L = 8.82 x 103 m
Therefore, the length of the cable is 8.82 x 103 m.
So the correct answer is) 8.8 x 103 m.
Pergunta 39 Relatório
An object of volume 400cm3 and 2.5gcm-3 is suspended from a spring balance with half its volume immersed in water. Determine the reading on the spring balance. (Density of water = 1gcm-3)
Detalhes da Resposta
The reading on the spring balance is 800g. To understand why, we need to consider the principle of buoyancy, which states that an object partially or completely submerged in a fluid experiences an upward force equal to the weight of the fluid displaced by the object. In this case, the object has a volume of 400cm3 and a density of 2.5gcm-3, which means its weight is: 400cm3 x 2.5gcm-3 = 1000g When the object is partially submerged in water, it displaces a volume of water equal to half its own volume, or 200cm3. Since the density of water is 1gcm-3, the weight of this displaced water is: 200cm3 x 1gcm-3 = 200g According to the principle of buoyancy, the object experiences an upward force equal to the weight of the displaced water, which is 200g. Therefore, the reading on the spring balance will be the weight of the object minus the upward buoyant force: 1000g - 200g = 800g So the reading on the spring balance is 800g.
Pergunta 40 Relatório
The distance between two successive trough of a wave is 0.4m. If the frequency of the source is 825Hz, Calculate the speed of the wave
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Pergunta 41 Relatório
A plank is placed along the path on which a wheel barrow containing sand is pushed at a construction site to prevent it from sticking as illustrated in the diagram given. The large area of the plank prevents the wheel barrow from getting stuck by
Detalhes da Resposta
The plank helps to prevent the wheelbarrow from getting stuck by decreasing the pressure on the ground. When the wheelbarrow is pushed over sand or soft ground, its weight can cause it to sink into the ground, making it difficult to move. The plank provides a wider surface area for the weight of the wheelbarrow to be distributed over, which reduces the pressure exerted on the ground. This makes it easier for the wheelbarrow to move over the soft ground without sinking in. So, the correct option is "decreasing the pressure on the ground."
Pergunta 42 Relatório
Which of the following statements is correct about the effect of dissolved salt on the freezing point of water?
Detalhes da Resposta
When salt is dissolved in water, it lowers the freezing point of water. This is because the salt ions attract some of the water molecules, which makes it harder for the water molecules to come together and form ice crystals. As a result, a lower temperature is needed for water to freeze in the presence of salt than in pure water. This phenomenon is used to de-ice roads and sidewalks during the winter. By spreading salt on icy surfaces, the salt dissolves in the thin layer of water on top of the ice, which lowers the freezing point and causes the ice to melt.
Pergunta 43 Relatório
The p.d across a parallel-plate capacitor is 103V. If the distance between the two plates is 10cm. Calculate the magnitude of the electric field strength between the plates.
Detalhes da Resposta
The electric field strength between the plates of a parallel-plate capacitor is given by: E = V/d Where V is the potential difference across the plates, and d is the distance between the plates. Substituting the given values, we get: E = 10^3/0.1 E = 10^4 V/m Therefore, the magnitude of the electric field strength between the plates is 1.0 x 10^4 V/m. is the correct answer.
Pergunta 44 Relatório
Which of the following substances lowers the surface tension of water?
Detalhes da Resposta
Detergent lowers the surface tension of water. Surface tension is the property of water that causes it to form droplets and resist external forces that can break those droplets apart. It's due to the cohesive forces between water molecules. However, when a substance is added to water, it can disrupt these cohesive forces, which lowers the surface tension of water. Detergent contains molecules that have both hydrophilic (water-attracting) and hydrophobic (water-repelling) regions. When added to water, the hydrophobic regions of detergent molecules will orient themselves towards the air-water interface, while the hydrophilic regions will interact with the water molecules. This reduces the cohesive forces between water molecules, leading to a decrease in surface tension. Metal, sand, and paper do not have properties that lower surface tension. Metal and sand are insoluble in water and do not interact with water molecules in a way that lowers surface tension. Paper is made of cellulose, a hydrophilic material that does not disrupt cohesive forces between water molecules.
Pergunta 45 Relatório
Resonance occurs when one vibrating object causes a second object to vibrate at its own natural
Detalhes da Resposta
Resonance is a phenomenon that occurs when one object vibrates and causes another object to vibrate at the same frequency. Every object has a natural frequency at which it likes to vibrate. When an object is forced to vibrate at a different frequency, it does not vibrate as strongly. However, if the frequency of the vibration matches the natural frequency of the object, the object will vibrate with a larger amplitude, or strength, which is known as resonance. To illustrate this, think of a swing. If you push a swing at a random time, it will move back and forth, but it won't go very high. However, if you push the swing at the right time, it will go higher and higher with each swing. This is because you are pushing the swing at the same frequency as its natural frequency, causing it to resonate and swing with greater amplitude. In summary, resonance occurs when an object is forced to vibrate at its natural frequency, causing it to vibrate with greater strength or amplitude. The key factor in resonance is the frequency of the vibration, rather than the speed, intensity, or amplitude.
Pergunta 46 Relatório
The phenomenon by which two light atomic nuclear combine to form a heavy nuclide with the release of energy is known as
Detalhes da Resposta
The phenomenon by which two light atomic nuclei combine to form a heavier nucleus with the release of energy is called nuclear fusion. In nuclear fusion, the nuclei of two atoms come together to form a single, heavier nucleus. This process releases a large amount of energy in the form of light and heat. Nuclear fusion is the process that powers the sun and other stars. Scientists are trying to harness this process as a source of energy on Earth, but it is very difficult to achieve because it requires very high temperatures and pressures to overcome the repulsion between the positively charged nuclei.
Pergunta 47 Relatório
Let \(\Delta\)x be the uncertainty in the measurements of position and \(\Delta\)p the uncertainty in measurement of momentum. The uncertainty principle relation is given as
Detalhes da Resposta
The uncertainty principle is a fundamental principle in quantum mechanics that states that it is impossible to precisely determine both the position and momentum of a particle at the same time. The principle is expressed mathematically as: \(\Delta\)x . \(\Delta\)p \(\geq\) h where \(\Delta\)x is the uncertainty in the position measurement and \(\Delta\)p is the uncertainty in the momentum measurement, and h is Planck's constant. This principle implies that the more precisely you know the position of a particle, the less precisely you can know its momentum, and vice versa. In other words, there is a trade-off between the precision of these measurements. The principle also implies that there is a fundamental limit to how precisely we can know both the position and momentum of a particle at the same time. This limit is determined by Planck's constant, which is a fundamental constant of nature. Therefore, it is impossible to make measurements with arbitrary precision in quantum mechanics, and there will always be some uncertainty associated with our measurements. It is worth noting that the uncertainty principle is not a limitation of the experimental apparatus or the measurement technique, but rather a fundamental property of nature. It is a consequence of the wave-particle duality of matter in quantum mechanics.
Pergunta 48 Relatório
(a) Explain the following terms:
(i) mass defect;
(ii) binding energy of a nucleus.
(b)(i) Assuming the wave nature of an electron, what is the effect of decreasing the speed of a photoelectron on its; (\(\alpha\)) wavelength? (\(\beta\)) energy?
(ii) A particle of friasS 4.4 x 10\(^{-23}\) kg moves with a velocity of 10\(^5\)ms\(^{-1}\). Calculate its wavelength. (h = 6.6 x 10\(^{-34}\) Js)
The diagram above shows part of a radioactive decay series. Use it to answer the following questions.
(i) Name a pair of isotopes.
(ii) Name the isotopes with which the series starts.
(iii) Write down a nuclear equation for two ekgmples of each of: (\(\alpha\)) alpha decay; (\(\beta\)) beta decay.
(a)(i) Mass defect. The mass defect is the difference between the sum of the separate masses of all the protons and neutrons (nucleons) in a nucleus and the actual (smaller) measured mass of the nucleus:
\[ \Delta m = \big(Z m_p + (A-Z) m_n\big) - M_{\text{nucleus}} \](a)(ii) Binding energy of a nucleus. It is the energy equivalent of the mass defect (\(E = \Delta m\, c^2\)); that is, the energy released when the free nucleons come together to form the nucleus, or equivalently the energy that must be supplied to separate the nucleus completely into its individual nucleons.
(b)(i) Effect of decreasing the speed of a photoelectron (de Broglie relation \(\lambda = h/mv\), kinetic energy \(E = \tfrac{1}{2}mv^2\)):
- (\(\alpha\)) Wavelength: since \(\lambda = \dfrac{h}{mv}\), decreasing \(v\) increases the wavelength.
- (\(\beta\)) Energy: since \(E = \tfrac{1}{2}mv^2\), decreasing \(v\) decreases the kinetic energy.
(b)(ii) de Broglie wavelength. Data: \(m = 4.4 \times 10^{-23}\ \text{kg}\), \(v = 10^{5}\ \text{m s}^{-1}\), \(h = 6.6 \times 10^{-34}\ \text{J s}\).
\[ \lambda = \frac{h}{mv} = \frac{6.6 \times 10^{-34}}{(4.4 \times 10^{-23})(10^{5})} \]\[ \lambda = \frac{6.6 \times 10^{-34}}{4.4 \times 10^{-18}} = 1.5 \times 10^{-16}\ \text{m} \](c) The decay-series graph. The graph plots nucleon number \(A\) (vertical, 208 to 232) against proton number \(Z\) (horizontal, Pb 82, Bi 83, Po 84, At 85, Rn 86, Fr 87, Ra 88, Ac 89, Th 90). Each down-left diagonal step (\(Z\) falls by 2, \(A\) falls by 4) is an alpha decay; each horizontal step to the right (\(Z\) rises by 1, \(A\) unchanged) is a beta decay. This is the thorium (4n) series running from Th-232 down to Pb-208.
(c)(i) A pair of isotopes. Isotopes have the same proton number but different nucleon numbers. From the graph, examples are the two lead points \(^{208}_{82}\text{Pb}\) and \(^{212}_{82}\text{Pb}\) (both at Z = 82); equally acceptable: \(^{216}_{84}\text{Po}\) and \(^{212}_{84}\text{Po}\) (Z = 84), or \(^{228}_{88}\text{Ra}\) and \(^{224}_{88}\text{Ra}\) (Z = 88), or \(^{232}_{90}\text{Th}\) and \(^{228}_{90}\text{Th}\) (Z = 90).
(c)(ii) Isotope with which the series starts. The top-right point, at Z = 90 and A = 232, is thorium-232, \(^{232}_{90}\text{Th}\).
(c)(iii) Nuclear equations (two examples of each).
(\(\alpha\)) Alpha decay (emits \(^{4}_{2}\text{He}\); Z falls by 2, A by 4):
\[ ^{232}_{90}\text{Th} \rightarrow\ ^{228}_{88}\text{Ra} + ^{4}_{2}\text{He} \]\[ ^{224}_{88}\text{Ra} \rightarrow\ ^{220}_{86}\text{Rn} + ^{4}_{2}\text{He} \](\(\beta\)) Beta decay (emits \(^{0}_{-1}e\); Z rises by 1, A unchanged):
\[ ^{228}_{88}\text{Ra} \rightarrow\ ^{228}_{89}\text{Ac} + ^{0}_{-1}e \]\[ ^{212}_{82}\text{Pb} \rightarrow\ ^{212}_{83}\text{Bi} + ^{0}_{-1}e \]Detalhes da Resposta
(a)(i) Mass defect. The mass defect is the difference between the sum of the separate masses of all the protons and neutrons (nucleons) in a nucleus and the actual (smaller) measured mass of the nucleus:
\[ \Delta m = \big(Z m_p + (A-Z) m_n\big) - M_{\text{nucleus}} \](a)(ii) Binding energy of a nucleus. It is the energy equivalent of the mass defect (\(E = \Delta m\, c^2\)); that is, the energy released when the free nucleons come together to form the nucleus, or equivalently the energy that must be supplied to separate the nucleus completely into its individual nucleons.
(b)(i) Effect of decreasing the speed of a photoelectron (de Broglie relation \(\lambda = h/mv\), kinetic energy \(E = \tfrac{1}{2}mv^2\)):
- (\(\alpha\)) Wavelength: since \(\lambda = \dfrac{h}{mv}\), decreasing \(v\) increases the wavelength.
- (\(\beta\)) Energy: since \(E = \tfrac{1}{2}mv^2\), decreasing \(v\) decreases the kinetic energy.
(b)(ii) de Broglie wavelength. Data: \(m = 4.4 \times 10^{-23}\ \text{kg}\), \(v = 10^{5}\ \text{m s}^{-1}\), \(h = 6.6 \times 10^{-34}\ \text{J s}\).
\[ \lambda = \frac{h}{mv} = \frac{6.6 \times 10^{-34}}{(4.4 \times 10^{-23})(10^{5})} \]\[ \lambda = \frac{6.6 \times 10^{-34}}{4.4 \times 10^{-18}} = 1.5 \times 10^{-16}\ \text{m} \](c) The decay-series graph. The graph plots nucleon number \(A\) (vertical, 208 to 232) against proton number \(Z\) (horizontal, Pb 82, Bi 83, Po 84, At 85, Rn 86, Fr 87, Ra 88, Ac 89, Th 90). Each down-left diagonal step (\(Z\) falls by 2, \(A\) falls by 4) is an alpha decay; each horizontal step to the right (\(Z\) rises by 1, \(A\) unchanged) is a beta decay. This is the thorium (4n) series running from Th-232 down to Pb-208.
(c)(i) A pair of isotopes. Isotopes have the same proton number but different nucleon numbers. From the graph, examples are the two lead points \(^{208}_{82}\text{Pb}\) and \(^{212}_{82}\text{Pb}\) (both at Z = 82); equally acceptable: \(^{216}_{84}\text{Po}\) and \(^{212}_{84}\text{Po}\) (Z = 84), or \(^{228}_{88}\text{Ra}\) and \(^{224}_{88}\text{Ra}\) (Z = 88), or \(^{232}_{90}\text{Th}\) and \(^{228}_{90}\text{Th}\) (Z = 90).
(c)(ii) Isotope with which the series starts. The top-right point, at Z = 90 and A = 232, is thorium-232, \(^{232}_{90}\text{Th}\).
(c)(iii) Nuclear equations (two examples of each).
(\(\alpha\)) Alpha decay (emits \(^{4}_{2}\text{He}\); Z falls by 2, A by 4):
\[ ^{232}_{90}\text{Th} \rightarrow\ ^{228}_{88}\text{Ra} + ^{4}_{2}\text{He} \]\[ ^{224}_{88}\text{Ra} \rightarrow\ ^{220}_{86}\text{Rn} + ^{4}_{2}\text{He} \](\(\beta\)) Beta decay (emits \(^{0}_{-1}e\); Z rises by 1, A unchanged):
\[ ^{228}_{88}\text{Ra} \rightarrow\ ^{228}_{89}\text{Ac} + ^{0}_{-1}e \]\[ ^{212}_{82}\text{Pb} \rightarrow\ ^{212}_{83}\text{Bi} + ^{0}_{-1}e \]Pergunta 49 Relatório
(a) Explain the term net force.
(b) Define the principle of conservation of linear momentum and state one example of it.
(c) A ball of mass 200 g released from a height of 2.0 m hits a horizontal floor and rebounds to a height of 1.8 in. Calculate the impulse received by the floor. (g = 10 ms\(^{-2}\)).
(d) A body of mass 20 g performs a simple harmonic motion at a frequency of 5 Hz. At a distance of 10 cm from the mean position, its velocity is 200 cms\(^{-1}\). Calculate its:
(i) maximum displacement from the mean position;
(ii) maximum velocity;
(iii) maximum potential energy. (g = 10 ms\(^{-2}\) \(\pi\) = 3.14)
(a) Net force is the single resultant force obtained when all the forces acting on a body are combined (added as vectors). It is the force that determines the body's acceleration, \(F_{net}=ma\).
(b) Principle of conservation of linear momentum: in a closed system on which no external force acts, the total linear momentum before an interaction (collision or explosion) equals the total linear momentum after it. Example: the recoil of a gun when a bullet is fired - the forward momentum of the bullet equals the backward momentum of the gun.
(c) Impulse received by the floor.
Mass \(m = 200\,\text{g}=0.2\,\text{kg}\).
Speed on hitting the floor: \(v_1=\sqrt{2gh_1}=\sqrt{2\times10\times2.0}=\sqrt{40}=6.32\,\text{ms}^{-1}\) (downward).
Speed on rebound: \(v_2=\sqrt{2gh_2}=\sqrt{2\times10\times1.8}=\sqrt{36}=6.0\,\text{ms}^{-1}\) (upward).
Taking upward as positive, impulse on the ball
By Newton's third law the impulse received by the floor is equal and opposite: 2.46 Ns directed downward.
(d) Simple harmonic motion.
\(m=20\,\text{g}=0.02\,\text{kg}\), \(f=5\,\text{Hz}\Rightarrow \omega=2\pi f=2\times3.14\times5=31.4\,\text{rads}^{-1}\).
At \(x=10\,\text{cm}=0.1\,\text{m}\), \(v=200\,\text{cms}^{-1}=2.0\,\text{ms}^{-1}\).
(i) Maximum displacement (amplitude) A. Using \(v=\omega\sqrt{A^2-x^2}\):
\[ v^2=\omega^2(A^2-x^2)\Rightarrow A^2 = x^2 + \frac{v^2}{\omega^2}=0.01+\frac{4}{(31.4)^2}=0.01+0.00406=0.01406 \]\[ A = 0.119\,\text{m}\;(\approx 11.9\,\text{cm}) \](ii) Maximum velocity.
\[ v_{max}=\omega A = 31.4\times0.119 = 3.73\,\text{ms}^{-1} \](iii) Maximum potential energy (equals the total energy of the oscillation):
\[ P.E._{max}=\tfrac{1}{2}m\omega^2A^2=\tfrac{1}{2}(0.02)(31.4)^2(0.01406)=0.139\,\text{J} \]So \(P.E._{max}\approx 0.14\,\text{J}\).
Detalhes da Resposta
(a) Net force is the single resultant force obtained when all the forces acting on a body are combined (added as vectors). It is the force that determines the body's acceleration, \(F_{net}=ma\).
(b) Principle of conservation of linear momentum: in a closed system on which no external force acts, the total linear momentum before an interaction (collision or explosion) equals the total linear momentum after it. Example: the recoil of a gun when a bullet is fired - the forward momentum of the bullet equals the backward momentum of the gun.
(c) Impulse received by the floor.
Mass \(m = 200\,\text{g}=0.2\,\text{kg}\).
Speed on hitting the floor: \(v_1=\sqrt{2gh_1}=\sqrt{2\times10\times2.0}=\sqrt{40}=6.32\,\text{ms}^{-1}\) (downward).
Speed on rebound: \(v_2=\sqrt{2gh_2}=\sqrt{2\times10\times1.8}=\sqrt{36}=6.0\,\text{ms}^{-1}\) (upward).
Taking upward as positive, impulse on the ball
By Newton's third law the impulse received by the floor is equal and opposite: 2.46 Ns directed downward.
(d) Simple harmonic motion.
\(m=20\,\text{g}=0.02\,\text{kg}\), \(f=5\,\text{Hz}\Rightarrow \omega=2\pi f=2\times3.14\times5=31.4\,\text{rads}^{-1}\).
At \(x=10\,\text{cm}=0.1\,\text{m}\), \(v=200\,\text{cms}^{-1}=2.0\,\text{ms}^{-1}\).
(i) Maximum displacement (amplitude) A. Using \(v=\omega\sqrt{A^2-x^2}\):
\[ v^2=\omega^2(A^2-x^2)\Rightarrow A^2 = x^2 + \frac{v^2}{\omega^2}=0.01+\frac{4}{(31.4)^2}=0.01+0.00406=0.01406 \]\[ A = 0.119\,\text{m}\;(\approx 11.9\,\text{cm}) \](ii) Maximum velocity.
\[ v_{max}=\omega A = 31.4\times0.119 = 3.73\,\text{ms}^{-1} \](iii) Maximum potential energy (equals the total energy of the oscillation):
\[ P.E._{max}=\tfrac{1}{2}m\omega^2A^2=\tfrac{1}{2}(0.02)(31.4)^2(0.01406)=0.139\,\text{J} \]So \(P.E._{max}\approx 0.14\,\text{J}\).
Pergunta 50 Relatório
(a) Explain the terms:
(i) thermal equilibrium;
(ii) fundamental interval.
(b) List two uses of the hydraulic press.
(c) Name the material used to reset the steel index in the Six's maximum and minimum thermometer.
(d)(i) A nursing mother prepared her baby's milk mixture at 85°C, in a feeding bottle. In order to cool it to 40°C, she immersed the bottle in an aluminium bowl of heat capacity 90 JK\(^{-1}\) containing 500 g of water at 26°C. If the mass of the mixture is 300g, calculate the specific heat capacity of the mixture. [Neglect heat losses and heat capacity of the bottle; specific heat capacity of water = 4200 J kg\(^{-1}\) K\(^{-1}\)]
(ii) (\(\alpha\)) Name two ways through which the bottle losses heat.
(\(\beta\)) Name two industrial processes in which heat exchanger is used.
(a)(i) Thermal equilibrium: the state reached when two (or more) bodies in thermal contact are at the same temperature so that there is no net flow of heat between them.
(a)(ii) Fundamental interval: the temperature range between the lower fixed point (ice point) and the upper fixed point (steam point) of a thermometer scale.
(b) Two uses of the hydraulic press:
- Lifting heavy loads (e.g. hydraulic jack / car lift).
- Compressing or baling materials such as cotton, scrap metal and refuse (also forging and stamping metals).
(c) Material used to reset the steel index in the Six's maximum and minimum thermometer: a magnet (the steel indices are drawn back with a small magnet).
(d)(i) Specific heat capacity of the mixture.
Heat lost by the hot milk mixture = heat gained by the water + heat gained by the aluminium bowl.
Mixture: \(m=300\,\text{g}=0.3\,\text{kg}\), cools \(85^{\circ}\text{C}\to40^{\circ}\text{C}\), \(\Delta\theta=45\,\text{K}\).
Water: \(0.5\,\text{kg}\), warms \(26^{\circ}\text{C}\to40^{\circ}\text{C}\), \(\Delta\theta=14\,\text{K}\).
Bowl heat capacity \(=90\,\text{JK}^{-1}\), warms by \(14\,\text{K}\).
So \(c \approx 2.27\times10^{3}\,\text{Jkg}^{-1}\text{K}^{-1}\).
(d)(ii)(\(\alpha\)) Two ways the bottle loses heat: conduction and convection (radiation and evaporation are also acceptable).
(d)(ii)(\(\beta\)) Two industrial processes using a heat exchanger: electricity generation in power stations (steam condensers/boilers) and refrigeration / air-conditioning (also petroleum refining and food pasteurization).
Detalhes da Resposta
(a)(i) Thermal equilibrium: the state reached when two (or more) bodies in thermal contact are at the same temperature so that there is no net flow of heat between them.
(a)(ii) Fundamental interval: the temperature range between the lower fixed point (ice point) and the upper fixed point (steam point) of a thermometer scale.
(b) Two uses of the hydraulic press:
- Lifting heavy loads (e.g. hydraulic jack / car lift).
- Compressing or baling materials such as cotton, scrap metal and refuse (also forging and stamping metals).
(c) Material used to reset the steel index in the Six's maximum and minimum thermometer: a magnet (the steel indices are drawn back with a small magnet).
(d)(i) Specific heat capacity of the mixture.
Heat lost by the hot milk mixture = heat gained by the water + heat gained by the aluminium bowl.
Mixture: \(m=300\,\text{g}=0.3\,\text{kg}\), cools \(85^{\circ}\text{C}\to40^{\circ}\text{C}\), \(\Delta\theta=45\,\text{K}\).
Water: \(0.5\,\text{kg}\), warms \(26^{\circ}\text{C}\to40^{\circ}\text{C}\), \(\Delta\theta=14\,\text{K}\).
Bowl heat capacity \(=90\,\text{JK}^{-1}\), warms by \(14\,\text{K}\).
So \(c \approx 2.27\times10^{3}\,\text{Jkg}^{-1}\text{K}^{-1}\).
(d)(ii)(\(\alpha\)) Two ways the bottle loses heat: conduction and convection (radiation and evaporation are also acceptable).
(d)(ii)(\(\beta\)) Two industrial processes using a heat exchanger: electricity generation in power stations (steam condensers/boilers) and refrigeration / air-conditioning (also petroleum refining and food pasteurization).
Pergunta 51 Relatório
(a) Explain briefly dielectric strength.
(b) An electromagnetic wave has its wavelength shorter than those of radiowave and microwave but longer than that of visible light.
(i) Identify the wave.
(ii) Name one suitable detector for the wave.
(iii) Name one source of the wave.
(c) An oil drop carrying a charge of 1.0 x 10\(^{-19 }\)C is found to remain at rest in a uniform electric field of intensity 1200 NC\(^{-1}\). Calculate the weight of the oil drop.
(d) An RLC series circuit consists of a 100\(\Omega\) resistor, 0.05 H inductor and a 25 \(\mu\) capacitor. A 220 V, 50 Hz mains voltage is applied across the circuit. Calculate the:
(i) impedance;
(ii) current. (\(\pi\) = 3.14)
(a) Dielectric strength: the maximum electric field (or maximum p.d. per unit thickness) that an insulator (dielectric) can withstand without breaking down and allowing charge to conduct through it.
(b) The wave is infrared radiation.
- (i) Infrared (heat) radiation.
- (ii) Detector: a thermopile / bolometer / blackened thermometer (a special infrared-sensitive photodiode is also acceptable).
- (iii) Source: any hot body, e.g. the Sun, an electric heating element or a warm object.
(c) Weight of the oil drop. Since the drop is at rest, the electric force balances the weight:
\[ W = qE = (1.0\times10^{-19})(1200) = 1.2\times10^{-16}\,\text{N} \](d) RLC series circuit. \(R=100\,\Omega\), \(L=0.05\,\text{H}\), \(C=25\,\mu\text{F}=25\times10^{-6}\,\text{F}\), \(f=50\,\text{Hz}\).
\[ X_L = 2\pi f L = 2(3.14)(50)(0.05)=15.7\,\Omega \]\[ X_C = \frac{1}{2\pi f C}=\frac{1}{2(3.14)(50)(25\times10^{-6})}=\frac{1}{7.85\times10^{-3}}=127.4\,\Omega \](i) Impedance:
\[ Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{100^2+(127.4-15.7)^2}=\sqrt{10000+12477}=\sqrt{22477}=150\,\Omega \](ii) Current:
\[ I=\frac{V}{Z}=\frac{220}{150}=1.47\,\text{A} \]Detalhes da Resposta
(a) Dielectric strength: the maximum electric field (or maximum p.d. per unit thickness) that an insulator (dielectric) can withstand without breaking down and allowing charge to conduct through it.
(b) The wave is infrared radiation.
- (i) Infrared (heat) radiation.
- (ii) Detector: a thermopile / bolometer / blackened thermometer (a special infrared-sensitive photodiode is also acceptable).
- (iii) Source: any hot body, e.g. the Sun, an electric heating element or a warm object.
(c) Weight of the oil drop. Since the drop is at rest, the electric force balances the weight:
\[ W = qE = (1.0\times10^{-19})(1200) = 1.2\times10^{-16}\,\text{N} \](d) RLC series circuit. \(R=100\,\Omega\), \(L=0.05\,\text{H}\), \(C=25\,\mu\text{F}=25\times10^{-6}\,\text{F}\), \(f=50\,\text{Hz}\).
\[ X_L = 2\pi f L = 2(3.14)(50)(0.05)=15.7\,\Omega \]\[ X_C = \frac{1}{2\pi f C}=\frac{1}{2(3.14)(50)(25\times10^{-6})}=\frac{1}{7.85\times10^{-3}}=127.4\,\Omega \](i) Impedance:
\[ Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{100^2+(127.4-15.7)^2}=\sqrt{10000+12477}=\sqrt{22477}=150\,\Omega \](ii) Current:
\[ I=\frac{V}{Z}=\frac{220}{150}=1.47\,\text{A} \]Pergunta 52 Relatório
The diagram above illustrates a cathode ray tube. Identify the components X, Y, and Z.
Identifying the components of the cathode ray tube. Following the electron beam from left (the electron gun) to right (the screen end):
- X - the filament/cathode (heater of the electron gun). X is connected to the coil at the far left. When heated by the low-voltage supply it emits electrons by thermionic emission, providing the source of electrons for the beam.
- Y - the anode (accelerating and focusing anodes). Y points to the cylindrical electrodes held at a high positive potential. They accelerate the electrons and focus them into a fine beam directed along the axis of the tube.
- Z - the deflecting plates. Z points to the pair of parallel plates. A potential difference applied across them sets up an electric field that deflects the electron beam, steering the spot on the fluorescent screen.
In summary: X = cathode/filament (electron source), Y = accelerating anode, Z = deflecting plates.
Detalhes da Resposta
Identifying the components of the cathode ray tube. Following the electron beam from left (the electron gun) to right (the screen end):
- X - the filament/cathode (heater of the electron gun). X is connected to the coil at the far left. When heated by the low-voltage supply it emits electrons by thermionic emission, providing the source of electrons for the beam.
- Y - the anode (accelerating and focusing anodes). Y points to the cylindrical electrodes held at a high positive potential. They accelerate the electrons and focus them into a fine beam directed along the axis of the tube.
- Z - the deflecting plates. Z points to the pair of parallel plates. A potential difference applied across them sets up an electric field that deflects the electron beam, steering the spot on the fluorescent screen.
In summary: X = cathode/filament (electron source), Y = accelerating anode, Z = deflecting plates.
Pergunta 53 Relatório
A projectile is fired with a velocity of 20 ms\(^{-1}\) at an angle of 40° to the horizontal. Determine the components of the velocity of the projectile at its maximum height.
The initial velocity components are \(u_x = u\cos\theta\) (horizontal) and \(u_y = u\sin\theta\) (vertical), with \(u = 20\ \text{m s}^{-1}\), \(\theta = 40^\circ\).
At maximum height the vertical component of velocity is momentarily zero (the projectile stops rising). Only the horizontal component remains, and it is unchanged throughout the flight:
\[ v_x = u\cos\theta = 20 \times \cos 40^\circ = 20 \times 0.766 = 15.3\ \text{m s}^{-1}. \] \[ v_y = 0. \]At maximum height: horizontal component \(\approx 15.3\ \text{m s}^{-1}\), vertical component \(= 0\).
Detalhes da Resposta
The initial velocity components are \(u_x = u\cos\theta\) (horizontal) and \(u_y = u\sin\theta\) (vertical), with \(u = 20\ \text{m s}^{-1}\), \(\theta = 40^\circ\).
At maximum height the vertical component of velocity is momentarily zero (the projectile stops rising). Only the horizontal component remains, and it is unchanged throughout the flight:
\[ v_x = u\cos\theta = 20 \times \cos 40^\circ = 20 \times 0.766 = 15.3\ \text{m s}^{-1}. \] \[ v_y = 0. \]At maximum height: horizontal component \(\approx 15.3\ \text{m s}^{-1}\), vertical component \(= 0\).
Pergunta 54 Relatório
State the dimension of; (a) impulse; (ii) acceleration; (iii) work
(a) Impulse. Impulse \(=\) force \(\times\) time \(=\) change in momentum. Its dimension is that of momentum:
\[ [\text{impulse}] = MLT^{-1}. \](ii) Acceleration. Acceleration \(=\) velocity/time \(= \dfrac{LT^{-1}}{T}\):
\[ [\text{acceleration}] = LT^{-2}. \](iii) Work. Work \(=\) force \(\times\) distance \(= (MLT^{-2}) \times L\):
\[ [\text{work}] = ML^{2}T^{-2}. \]Detalhes da Resposta
(a) Impulse. Impulse \(=\) force \(\times\) time \(=\) change in momentum. Its dimension is that of momentum:
\[ [\text{impulse}] = MLT^{-1}. \](ii) Acceleration. Acceleration \(=\) velocity/time \(= \dfrac{LT^{-1}}{T}\):
\[ [\text{acceleration}] = LT^{-2}. \](iii) Work. Work \(=\) force \(\times\) distance \(= (MLT^{-2}) \times L\):
\[ [\text{work}] = ML^{2}T^{-2}. \]Pergunta 55 Relatório
(a) What is doping?
(b) Explain how doping improves the conductivity of a semiconductor
(a) Doping is the deliberate addition of a small, controlled amount of a suitable impurity (a pentavalent or trivalent element) to a pure (intrinsic) semiconductor in order to increase the number of free charge carriers and hence its electrical conductivity.
(b) How doping improves conductivity:
A pure semiconductor such as silicon has very few free charge carriers, so it conducts poorly.
- Adding a pentavalent impurity (e.g. phosphorus, arsenic) donates extra free electrons to the crystal, producing an n-type semiconductor in which electrons are the majority carriers.
- Adding a trivalent impurity (e.g. boron, aluminium) creates vacancies called holes that act as positive carriers, producing a p-type semiconductor in which holes are the majority carriers.
In both cases the impurity greatly increases the concentration of mobile charge carriers, so more charge can flow for a given voltage and the conductivity rises sharply.
Detalhes da Resposta
(a) Doping is the deliberate addition of a small, controlled amount of a suitable impurity (a pentavalent or trivalent element) to a pure (intrinsic) semiconductor in order to increase the number of free charge carriers and hence its electrical conductivity.
(b) How doping improves conductivity:
A pure semiconductor such as silicon has very few free charge carriers, so it conducts poorly.
- Adding a pentavalent impurity (e.g. phosphorus, arsenic) donates extra free electrons to the crystal, producing an n-type semiconductor in which electrons are the majority carriers.
- Adding a trivalent impurity (e.g. boron, aluminium) creates vacancies called holes that act as positive carriers, producing a p-type semiconductor in which holes are the majority carriers.
In both cases the impurity greatly increases the concentration of mobile charge carriers, so more charge can flow for a given voltage and the conductivity rises sharply.
Pergunta 56 Relatório
(a) Define critical angle.
(b) How are anti-nodes created in a stationary wave?
(c) The angle of minimum deviation of an equilateral triangular glass prism is 46.2°. Calculate the refractive index of the glass.
(d) An illuminated object is placed in front of a concave mirror and the position of a screen is adjusted in front of the mirror but no image is obtained on the screen. Give two possible reasons for this observation.
(e) An illuminated object is placed at a distance of 75 cm from a converging lens of focal length 30 cm.
(i) Determine the image distance.
(ii) If the lens is replaced by another converging lens, the object has to be moved 25 cm further away to have its sharp image on the screen. Determine the focal length of the second lens.
Detalhes da Resposta
None
Pergunta 57 Relatório
State three different materials that can be used to demonstrate Brownian motion.
Materials that can demonstrate Brownian motion:
- Smoke particles suspended in air (in a smoke cell), viewed under a microscope.
- Pollen grains suspended in water.
- Fine dust or graphite particles suspended in water or air.
Note: a suspension of gamboge or sulphur particles in water is also acceptable. In each case the tiny visible particles are seen to move in continuous, random, zig-zag paths because they are bombarded unequally by the much smaller, fast-moving molecules of the surrounding fluid.
Detalhes da Resposta
Materials that can demonstrate Brownian motion:
- Smoke particles suspended in air (in a smoke cell), viewed under a microscope.
- Pollen grains suspended in water.
- Fine dust or graphite particles suspended in water or air.
Note: a suspension of gamboge or sulphur particles in water is also acceptable. In each case the tiny visible particles are seen to move in continuous, random, zig-zag paths because they are bombarded unequally by the much smaller, fast-moving molecules of the surrounding fluid.
Pergunta 58 Relatório
An electron enters perpendicularly into a uniform magnetic field which has a flux density of 0.12 T This results in a magnetic force of 9.6 x 10\(^{-2}\) N on the electron. Calculate the speed of the electron as it enters the magnetic field. (e =1.6 x \(10^{19}\) C)
An electron moving perpendicular to a uniform magnetic field experiences a force
\[F=Bev\sin\theta,\qquad \theta=90^\circ,\ \sin 90^\circ=1.\]Making the speed the subject:
\[v=\frac{F}{Be}.\]Substituting \(F=9.6\times10^{-12}\text{ N}\), \(B=0.12\text{ T}\) and \(e=1.6\times10^{-19}\text{ C}\):
\[v=\frac{9.6\times10^{-12}}{0.12\times1.6\times10^{-19}}=\frac{9.6\times10^{-12}}{1.92\times10^{-20}}=5.0\times10^{8}\text{ m s}^{-1}.\]The speed of the electron as it enters the field is \(5.0\times10^{8}\text{ m s}^{-1}\).
Detalhes da Resposta
An electron moving perpendicular to a uniform magnetic field experiences a force
\[F=Bev\sin\theta,\qquad \theta=90^\circ,\ \sin 90^\circ=1.\]Making the speed the subject:
\[v=\frac{F}{Be}.\]Substituting \(F=9.6\times10^{-12}\text{ N}\), \(B=0.12\text{ T}\) and \(e=1.6\times10^{-19}\text{ C}\):
\[v=\frac{9.6\times10^{-12}}{0.12\times1.6\times10^{-19}}=\frac{9.6\times10^{-12}}{1.92\times10^{-20}}=5.0\times10^{8}\text{ m s}^{-1}.\]The speed of the electron as it enters the field is \(5.0\times10^{8}\text{ m s}^{-1}\).
Pergunta 59 Relatório
List three uses of rockets.
Uses of rockets:
- Launching artificial satellites and spacecraft into orbit (space exploration).
- As military weapons, i.e. carrying missiles/warheads.
- Scientific research, e.g. carrying instruments to study the upper atmosphere and outer space.
Other acceptable uses: signalling and distress flares, and fireworks/display.
Detalhes da Resposta
Uses of rockets:
- Launching artificial satellites and spacecraft into orbit (space exploration).
- As military weapons, i.e. carrying missiles/warheads.
- Scientific research, e.g. carrying instruments to study the upper atmosphere and outer space.
Other acceptable uses: signalling and distress flares, and fireworks/display.
