WAEC SSCE - Physics - 2022
Ajụjụ 1 Ripọtì
The area under a force-time graph represents--------
Akọwa Nkọwa
It is a graph that represents the force that acts on a body in a certain amount of time.
Force on the object is represented by the y-axis and time taken by the object is represented by the y axis.
The area under the curve indicates the impulse, the change in momentum that the object undergoes.
Ajụjụ 2 Ripọtì
.An inductor of inductance 10 H is connected across an a.c circuit source of 50 V, 100 Hz. What is the current in the circuit? [π = 3.14]
Akọwa Nkọwa
Given Data: L = 10, V = 50, F = 100, I = ?
Inductive Reactance [XL ] = 2π FL → 2 * 3.14 * 100 * 10
XL
= 6,280
Current[I] = VXL → 506280
I = 0.0079
≈ 0.008A
Ajụjụ 3 Ripọtì
The distance between the fixed points of a mercury-in-glass thermometer is 30 cm. Determine the temperature when the mercury level is 10.5 cm above the lower fixed point
Akọwa Nkọwa
A mercury-in-glass thermometer measures temperature based on the expansion of the mercury in the bulb as the temperature changes. The length of the mercury column in the glass tube rises or falls as the temperature increases or decreases. In this question, the distance between the fixed points of the thermometer is given as 30 cm. The lower fixed point is the temperature at which the mercury bulb is completely immersed in melting ice and the upper fixed point is the temperature at which the mercury bulb is completely immersed in steam at standard atmospheric pressure. If the mercury level is 10.5 cm above the lower fixed point, then the length of the mercury column above the lower fixed point is 10.5 cm. To find the temperature corresponding to this length, we can use the following formula: (T - T_lfp)/(T_ufp - T_lfp) = l/l_fp where T is the temperature, T_lfp is the temperature of the lower fixed point (0 °C), T_ufp is the temperature of the upper fixed point (100 °C), l is the length of the mercury column above the lower fixed point (10.5 cm), and l_fp is the distance between the fixed points (30 cm). Solving for T, we get: (T - 0)/(100 - 0) = 10.5/30 T = (10.5/30) x 100 T = 35 °C Therefore, the temperature when the mercury level is 10.5 cm above the lower fixed point is 35 °C. The correct answer is: 35.0 ºC
Ajụjụ 4 Ripọtì
The half-life of a radioactive substance is 15 hours. If at some instance, the sample has a mass of 512 g, calculate the time it will take 78 of the sample to decay
Akọwa Nkọwa
Ajụjụ 5 Ripọtì
The engine of a car provides a forward force of 1240 N and the total resistive force on the car is 800N. If the mass of the car is 1220 kg, determine the distance the car has to travel from the rest before acquiring a speed of 4 m s−1 .
Akọwa Nkọwa
Acceleration[a] = [forward−resistive]forcemass
a = 1240−8001220 → 440−8001220
a = 0.36m/s2 , v = 4m/s, s = ?
Second equation of motion:
s is distance
u is initial velocity
v is final velocity
a is acceleration
v2 = u2 + 2as
42 = 02 + 2 * 0.36 * s
16 = 0.72s
22.2m = s
Ajụjụ 6 Ripọtì
In doping an intrinsic semiconductor to produce a p-type semiconductor,
Ajụjụ 7 Ripọtì
When the direction of vibration of the particles of a medium is perpendicular to the direction of propagation of a wave, the wave is said to be--------
Akọwa Nkọwa
When the particles of a medium vibrate perpendicular to the direction of wave propagation, the wave is called a transverse wave. In a transverse wave, the particles move up and down or side to side, but the wave itself moves forward. This type of wave is often seen in water waves, where the water moves up and down while the wave moves forward, or in light waves, where the electric and magnetic fields oscillate up and down perpendicular to the direction of the wave.
Ajụjụ 8 Ripọtì
Which of the following concepts is a method of heat transfer that does not require a material medium?
Akọwa Nkọwa
Radiation is a method of heat transfer that does not require a material medium. It is the process of emission of energy in the form of electromagnetic waves from a warm body to its surrounding environment. Radiation can travel through a vacuum and can also travel through transparent media like air, water, and glass. This means that heat can be transferred from one object to another through radiation even if they are not in contact with each other or if there is no physical medium between them.
Ajụjụ 9 Ripọtì
An example of a mechanical wave is---------
Akọwa Nkọwa
An example of a mechanical wave is "water waves". A mechanical wave is a type of wave that requires a medium (a substance or material) to travel through. When the medium is disturbed, it creates a disturbance that propagates through the medium, carrying energy with it. This disturbance can be in the form of oscillations or vibrations of particles in the medium. Water waves are a type of mechanical wave that propagates through water as a medium. When wind or a disturbance creates a ripple in the water, the ripple spreads out in a circular pattern, with the particles of water moving up and down as the wave passes through them. This motion of the water particles is the oscillation that carries the energy of the wave. On the other hand, radio waves, X-rays, and light rays are all examples of electromagnetic waves, which do not require a medium to travel through. Electromagnetic waves consist of oscillating electric and magnetic fields that propagate through space. They can travel through a vacuum and do not need a material medium to carry their energy. Therefore, the correct answer to the question is "water waves".
Ajụjụ 10 Ripọtì
The viscosity of a fluid depends on the following factors except the?
Akọwa Nkọwa
The viscosity of a fluid refers to its resistance to flow. It is affected by various factors such as the relative motion between the layers of the fluid, the nature of the material of the fluid, and the normal reaction in the fluid. However, the surface area of the fluid in contact is not a factor that affects viscosity. In simple terms, viscosity is affected by things like how thick or thin a fluid is and how easily its layers slide past each other. The larger the normal reaction in the fluid or the stickier the fluid, the higher the viscosity.
Ajụjụ 11 Ripọtì
The linear expansivity, α , and cubic expansivity, γ , of a material are related by the equation?
Akọwa Nkọwa
cubic and linear expansivity are related by this equation. γ=3α.
Ajụjụ 12 Ripọtì
Arrange the following radiations in order of increasing ionization of air.
I. Alpha
Il. Gamma
IIl. Beta
Akọwa Nkọwa
The ionization of air refers to the process of creating charged particles by adding or removing electrons from atoms or molecules in the air. The ability of a radiation to ionize air depends on its energy and type. Out of the given options, the correct order of increasing ionization of air is "II < III < I". Gamma radiation has the highest energy and is the most penetrating type of radiation, but it has the lowest ionizing power because it interacts weakly with matter. It can only ionize air when it directly hits an atom or molecule. Beta radiation has a medium energy and a higher ionizing power than gamma radiation. It can ionize air when it collides with an atom or molecule and transfers its energy to it, causing it to eject an electron. Alpha radiation has the lowest energy and the highest ionizing power among the three types of radiation. It can ionize air when it collides with an atom or molecule, transfers its energy to it, and ejects one or more electrons. Therefore, the correct order of increasing ionization of air is "II < III < I", where "II" represents gamma radiation, "III" represents beta radiation, and "I" represents alpha radiation.
Ajụjụ 13 Ripọtì
Akọwa Nkọwa
Two conductors with current flowing in the same direction will attract.
Ajụjụ 15 Ripọtì
Which of the following statements about a straight current-carrying wire placed in a uniform magnetic field is correct? The wire experiences ---------------
Akọwa Nkọwa
The greatest force is experienced when the conductor is at right angle to the field.
I.e at perpendicular or Ø = 90°
Ajụjụ 16 Ripọtì
Suppose three identical steel balls Q, R, and S, are placed on an undulating ground as illustrated in the diagram above. Which of the balls is/are in neutral equilibrium?
Akọwa Nkọwa
A system is in neutral equilibrium if its equilibrium is independent of displacements from its original position. A marble S on a flat horizontal surface is an example.
Ajụjụ 17 Ripọtì
Gamma rays are produced when?
Akọwa Nkọwa
Gamma rays are produced when energy changes occur within the nuclei of atoms. Gamma rays are high-energy electromagnetic radiation that are emitted by atomic nuclei during radioactive decay or nuclear reactions. These changes within the nuclei of atoms can result in the release of excess energy in the form of gamma rays. This process is known as gamma decay. Gamma rays have the highest energy and shortest wavelength in the electromagnetic spectrum, and they can penetrate most materials, making them useful in a wide range of applications, including medical imaging and cancer treatment.
Ajụjụ 18 Ripọtì
A freely suspended compass needle on the earth's surface settles in a plane called----------
Akọwa Nkọwa
A freely suspended compass needle on the earth's surface settles in a plane called magnetic meridian. This is because the needle of a compass always points towards the earth's magnetic north pole due to the magnetic field of the earth. The plane containing the magnetic north and south poles of the earth is called the magnetic meridian. This is different from the geographic meridian, which is the imaginary line on the earth's surface passing through the geographic poles. The magnetic declination is the angle between the magnetic meridian and the geographic meridian at a particular location, while isoganals are lines connecting points with the same magnetic declination.
Ajụjụ 19 Ripọtì
The speed of fast-moving neutrons in a nuclear reactor can be reduced by using?
Akọwa Nkọwa
In a nuclear reactor, neutrons are used to induce and sustain a chain reaction. Fast-moving neutrons, however, are not very efficient at inducing fission, so they need to be slowed down to increase the likelihood of fission. This process is called moderation. One way to moderate neutrons is by using materials with light nuclei that can scatter neutrons without absorbing them, such as graphite. Therefore, the correct answer is "graphite rods".
Ajụjụ 20 Ripọtì
Which of the folowing statements about light travelling from one material medium to another is not correct?
Akọwa Nkọwa
Refraction occurs when light travels from one medium to another which changes the speed at which the light travels.
This causes light to bend upon incidence with the interface of a new material.
Angle of refraction is smaller than angle of incidence when light ray travels from rarer medium to denser medium.
I.e light can either refract towards the normal(when slowing down while crossing the boundary) or away from the normal (when speeding up while crossing the boundary).
When light undergoes refraction, its frequency remains the same.
Ajụjụ 21 Ripọtì
A lamp is rated 240 V, 60 W. Determine the resistance of the lamp when lit?
Akọwa Nkọwa
P = 60w, V = 240v, R = ?
P = V2R
: 60 = 2402R
→ R * 60 = 57,600
R = 5
R = 960Ω
Ajụjụ 22 Ripọtì
From the principle of flotation, a body sinks in a fluid until it displace a quantity equal to its own?
Akọwa Nkọwa
Archimede's Principle states that a body immersed in a fluid experiences an upthrust equal to the weight of the fluid displaced, and this is fundamental to the equilibrium of a body floating in still water.
Ajụjụ 23 Ripọtì
The diameter of a brass ring at 30 °C is 50.0 cm. To what temperature must this ring be heated to increase its diameter to 50.29 cm? [ linear expansivity of brass = 1.9 x 10−5 K−1 ]
Akọwa Nkọwa
The problem involves the concept of linear expansivity. The diameter of the brass ring increases with an increase in temperature. We can use the formula for linear expansivity to solve the problem: ΔL = αLΔT where ΔL is the change in length, L is the original length, α is the linear expansivity and ΔT is the change in temperature. Since we are given the initial diameter, we can use it to find the original length (L) of the ring. L = πd/2 where d is the diameter. Substituting the values, we get: L = π × 50.0/2 = 78.54 cm The change in length (ΔL) is given by the difference in the final and initial lengths: ΔL = 50.29/2 - 50.0/2 = 0.145 cm Substituting the given values of α and L, we get: ΔL = αLΔT 0.145 = (1.9 × 10^-5) × 78.54 × ΔT Solving for ΔT, we get: ΔT = 0.145/(1.9 × 10^-5 × 78.54) = 116.3 °C Therefore, the temperature to which the ring must be heated is: 30 + 116.3 = 146.3 °C Rounding off to one decimal place, we get: 146.3 °C Hence, the correct option is 152.6 °C (Option A).
Ajụjụ 24 Ripọtì
Which of the following statement(s) is/are correct about a fixed mass of gas compressed in an inexpansible container;
I. The average speed of the molecules increases
II. The temperature of the gas increases
III. The molecules hit the walls of the container more often than before the compression
Akọwa Nkọwa
The kinetic-molecular theory of gases assumes that ideal gas molecules:
(1) are constantly moving;
(2) have negligible volume;
(3) have negligible intermolecular forces;
(4) undergo perfectly elastic collisions; and
(5) have an average kinetic energy proportional to the ideal gas's absolute temperature.
Ajụjụ 25 Ripọtì
Which of the following statements about electric potential energy is not correct?
Akọwa Nkọwa
The statement that is not correct is "The electric potential energy of a negatively charged particle increases when it moves to a point of lower potential." This statement contradicts the definition of electric potential energy. Electric potential energy is the energy that a charged particle possesses due to its position in an electric field. The potential energy of a charged particle is directly proportional to the potential difference between two points in the electric field, where the potential difference is defined as the change in potential energy per unit charge. Therefore, a negatively charged particle would move towards a point of higher potential energy since it moves towards a region of higher electric field. When a positively charged particle moves towards a point of higher potential energy, it gains electric potential energy, and when it moves towards a point of lower potential energy, it loses electric potential energy. The work done in taking a charged particle around a closed path in an electric field is zero because the potential difference between the starting point and ending point is zero.
Ajụjụ 26 Ripọtì
The correct relationship between the displacement, s, of a particle initially at rest in a linear motion and the time, t, is?
Akọwa Nkọwa
The correct relationship between the displacement, s, of a particle initially at rest in a linear motion and the time, t, is s α t2. This means that the displacement of a particle is directly proportional to the square of the time elapsed when the particle is moving in a straight line with a constant acceleration. In other words, if the time is doubled, the displacement will be four times greater; if the time is tripled, the displacement will be nine times greater, and so on. This relationship is also known as the second equation of motion, which is commonly used in physics to describe the motion of objects under constant acceleration.
Ajụjụ 27 Ripọtì
A galvanometer with a full-scale deflection of 20 mA is Converted to read 8 K by connecting a 395 Ω resistor in series with it. Determine the internal resistance of the galvanometer
Akọwa Nkọwa
r = resistance of the galvanometer
l = current through galvanometer = 20mA or 0.02A
V1 = p.d across the galvanometer = I X r = 0.02 X r
: V1 = 0.02r
V2 = p.d across the multiplier = 8 - 0.02r
R = resistance of the multiplier = 395Ω
where R = VI
--> 395 = 8−0.02r0.02
CROSS MULTIPLY
8−0.02r = 395 * 0.02
8−0.02r = 7.9
--> 0.02r = 8 - 7.9
∴ r = 0.10.02
= 5Ω
Ajụjụ 28 Ripọtì
A hunter fires a gun at a point 408 m away from a cliff. If he hears an echo 2.4 s later, determine the speed of the sound wave
Akọwa Nkọwa
When a sound is produced, it travels through the air as a sound wave. The speed at which sound waves travel through air is constant and depends on the air temperature. Assuming a temperature of 20°C, the speed of sound in air is approximately 340 m/s. Now, let's consider the situation in the problem. The hunter fires a gun and hears an echo 2.4 s later. The echo is the result of the sound waves reflecting off the cliff and traveling back to the hunter's ear. We know that the distance between the hunter and the cliff is 408 m. The sound wave has to travel this distance twice, once from the gun to the cliff and then from the cliff back to the hunter. So, the total distance traveled by the sound wave is 2 × 408 = 816 m. We also know that the time taken for the sound wave to travel this distance is 2.4 s. Using the formula speed = distance / time, we can calculate the speed of the sound wave as: speed = distance / time = 816 m / 2.4 s ≈ 340 m/s Therefore, the speed of the sound wave is approximately 340 m/s, which is the first option in the list.
Ajụjụ 29 Ripọtì
Using venier calipers, which of the following readings gives the correct measurement for the length of a rod?
Akọwa Nkọwa
Vernier calipers commonly used in industry provide a precision to 0.01 mm (10 micrometres), or 0.01cm or one thousandth of an inch.
Usually, it is the total length of the main scale. Vernier calipers, generally, have a range of 300 mm
300 millimeters equal 30.0 centimeters (300mm = 30.0cm).
Ajụjụ 30 Ripọtì
The period of a 10 kHz radio wave traveling at 10
ms−1
is--
Akọwa Nkọwa
The formula for the period of a wave is given by T = 1/f, where T is the period and f is the frequency. In this case, the frequency of the radio wave is 10 kHz, which means that f = 10,000 Hz. We can use the formula v = λf, where v is the velocity of the wave, λ is the wavelength, and f is the frequency. Solving for λ, we get λ = v/f. Substituting the given values, we get λ = 10 m/s / 10,000 Hz = 0.001 m. Now, we can use the formula T = λ/v to find the period. Substituting the given values, we get T = 0.001 m / 10 m/s = 0.0001 s. Therefore, the period of the 10 kHz radio wave traveling at 10 m/s is 0.0001 seconds or 1.0 x 10^-4 s. The answer is the second option.
Ajụjụ 31 Ripọtì
The periodic rise and fall in the intensity of sound produced whn two notes of nearly equal frequencies are sounded together is called?
Akọwa Nkọwa
The periodic rise and fall in the intensity of sound produced when two notes of nearly equal frequencies are sounded together is called a "beat". When two sound waves of slightly different frequencies interfere with each other, they create a periodic variation in the sound intensity, resulting in a pulsating sound. This pulsation is the beat. The beat frequency is equal to the difference between the frequencies of the two notes. For example, if one note has a frequency of 440 Hz and the other note has a frequency of 442 Hz, the beat frequency would be 2 Hz. The other options, "doppler effect", "interference", and "resonance", are different phenomena in acoustics. The Doppler effect is the change in frequency of a sound wave due to the motion of the source or the observer. Interference is the combination of two or more waves that results in a new wave pattern. Resonance is the tendency of a system to vibrate with maximum amplitude at a specific frequency called its natural frequency. Therefore, the correct answer to the question is "beat".
Ajụjụ 33 Ripọtì
If the kinetic energy of an electron is 100 eV, what is the wavelength of the de-Broglie wave associated with it? [h = 6.6 x 10−34 J s, e = 1.6 x 1019 C, me , = 9.1 x 10−31 kg]
Akọwa Nkọwa
de broglie wavelength formula in terms of kinetic energy:
λ = h2mE√
λ = 6.6∗10−342∗9.1∗10−31∗100√
λ = 5.50 x 10−14 m
Ajụjụ 34 Ripọtì
Which of the illustrated graphs above represents a body moving with uniform retardation?
Akọwa Nkọwa
If a body moves with uniform retardation, the velocity-time graph is a straight line.
Ajụjụ 36 Ripọtì
The basic principle of operation of a beam balance is------------------
Akọwa Nkọwa
The basic principle of operation of a beam balance is the principle of moments. This principle states that when an object is placed on one side of a beam balance, it creates a moment (or a turning force) around the pivot point, which is the center of the beam. This moment is counteracted by an equal moment created by a weight placed on the other side of the beam. The weight on the other side is moved along a calibrated scale until the beam is in a horizontal position, indicating that the moments on both sides are equal. This means that the weight of the object can be determined by reading the position of the weight on the calibrated scale.
Ajụjụ 37 Ripọtì
Three cells each of emf, 1.0 V, and internal resistance, 2 Ω , are connected in parallel across a 3Ω resistor. Determine the current in the resistor.
Akọwa Nkọwa
Given Data: Emf = 1v, r = 2 ohms, R = 3 ohms, I = ?
3 resistance in parallel = 1r1 + 1r2 + 1r3
1rT= 12 + 12 + 12
1rT = 1+1+12
1rT = 32
cross multiply
rT = 23 or 0.67
E = I(R+r)
1 = I(3+0.67)
1 = I(3.67)
1 ÷ 3.67 = I
0.27 = I
: I ≈ 0.30A
Ajụjụ 38 Ripọtì
Which of the following statements-about electromagnetic waves is not correct?
Akọwa Nkọwa
The statement that is not correct is: "They are longitudinal". Electromagnetic waves are actually transverse in nature, which means that the electric and magnetic fields oscillate perpendicular to the direction of wave propagation. Longitudinal waves, on the other hand, have oscillations parallel to the direction of wave propagation. The other three statements are correct: electromagnetic waves carry energy as they travel through space, they travel with the speed of light, and the electric and magnetic fields are at right angles to each other.
Ajụjụ 39 Ripọtì
The efficiency of a wheel and axle is 100% and the ratio of their radii is 5:1. calculate the effort required to lift a load of mass 20kg using this machine, [g = 10 m s−2 ]
Akọwa Nkọwa
To calculate the effort required to lift a load of mass 20kg using a wheel and axle machine with 100% efficiency and radii ratio of 5:1, we need to use the principle of moments. The principle of moments states that the sum of the moments acting on a system in equilibrium is equal to zero. In this case, we can assume that the system is in equilibrium since the wheel and axle are not rotating, and the load is not moving. Let R1 and R2 be the radii of the wheel and axle, respectively. Since the ratio of their radii is 5:1, we have R1 = 5R2. Let E be the effort required to lift the load and G be the weight of the load. We have G = 20kg × 10 m/s² = 200N. The moment of the effort E about the axle is ER1, and the moment of the weight G about the wheel is GR2. Since the machine is in equilibrium, we have: ER1 = GR2 Substituting R1 = 5R2 and G = 200N, we get: E(5R2) = 200N(R2) Simplifying the equation, we get: E = 40N Therefore, the effort required to lift the load of mass 20kg using this machine is 40N.
Ajụjụ 40 Ripọtì
The joule is equivalent to?
Akọwa Nkọwa
Joule as the unit of work done or energy used.
Force = mass × acceleration
Work = Force × Distance
Work = mass x acceleration x distance
Joule = kg × ms2 × m
→ kg m2 s−2
Ajụjụ 41 Ripọtì
Molecules move in random motion within a liquid. The total internal energy of the liquid depends on all of the following except its?
Akọwa Nkọwa
The answer is "melting point" because the total internal energy of a liquid depends on various factors, such as its temperature, mass, and specific heat capacity. However, the melting point of a substance is not related to the total internal energy of the liquid, but rather to the temperature at which the solid form of the substance changes into a liquid state. Therefore, "melting point" is the odd one out, and does not affect the total internal energy of the liquid.
Ajụjụ 42 Ripọtì
A small object of mass 50 g is released from a point A. Determine the velocity of the object when it reaches a point B, a vertical distance of 30m below A. [g = 10 ms−2
Akọwa Nkọwa
We can solve this problem using the principle of conservation of energy. When the object is at point A, it has some gravitational potential energy given by mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above some reference level. When the object falls to point B, it loses some potential energy, which is converted into kinetic energy. We can write this as: mgh = (1/2)mv^2 where v is the velocity of the object at point B. Rearranging the equation, we get: v = √(2gh) Substituting the given values, we get: v = √(2 × 10 × 30) = √600 = 24.5 m/s Therefore, the velocity of the object when it reaches point B is 24.5 m/s. So, the correct answer is "24.5 m/s".
Ajụjụ 43 Ripọtì
Akọwa Nkọwa
A plane mirror forms a virtual image of an object. The image is always erect (upright), of the same size as the object, and laterally inverted (left and right are switched). Therefore, the correct option is: "erect and bigger than the object". This statement is not correct as the image formed by a plane mirror is of the same size as the object and not bigger.
Ajụjụ 44 Ripọtì
The power of the lens is +2.5D. What is the radius of the curvature?
Akọwa Nkọwa
The power of a lens, P is defined as the reciprocal of the focal length of the lens, f in meters, expressed in diopters (D). Therefore, we can use the formula P = 1/f to find the focal length of the lens. For a thin lens, the radius of curvature, R of each surface of the lens is related to the focal length, f, by the formula: 1/f = (n - 1) × (1/R1 - 1/R2) Where n is the refractive index of the lens medium, R1 is the radius of curvature of one surface and R2 is the radius of curvature of the other surface. For a single curved surface of a thin lens, the formula reduces to: 1/f = (n - 1) × (1/R) where R is the radius of curvature of the curved surface. Rearranging the formula, we get: R = 1 / ((n-1) x f) Substituting P = 2.5D = 2.5m⁻¹ in place of f, and assuming the refractive index of the lens medium is 1, we get: R = 1 / ((1-1) x 2.5) = undefined Therefore, it is not possible to determine the radius of curvature of the lens from the given information. The answer is undefined.
Ajụjụ 45 Ripọtì
A quantity of water at 0 °C is heated to 30 °C. For each degree rise in temperature, its density will---------
Akọwa Nkọwa
As the temperature of water rises, its density changes due to thermal expansion or contraction. Water is most dense at 4°C, and its density decreases both below and above this temperature. As a result, when water is heated from 0°C to 30°C, its density will decrease with each degree of temperature rise. Therefore, the correct option is "fall steadily."
Ajụjụ 46 Ripọtì
An object is placed at different distances, u, from a converging lens of focal length, 15.0cm. For what value of u does the lens act as a microscope?
Akọwa Nkọwa
Objects at a distance greater than two times of the focal length produces a real image, inverted and diminished in size.
→ 2 * 15 = 30cm
: At >30cm this is applicable
Ajụjụ 47 Ripọtì
A body of mass, M, moving with velocity, V, has a wavelength, X, associated with it. This phenomenon is called----------
Akọwa Nkọwa
The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material.
Ajụjụ 48 Ripọtì
In which of the following situations is friction not useful?
Akọwa Nkọwa
friction, force that resists the sliding or rolling of one solid object over another.
Examples of friction in our daily life
- Driving of a a vehicle on a surface.
- Applying brakes to stop a moving vehicle.
- Skating.
- Walking on the road.
Ajụjụ 49 Ripọtì
State three properties of a semiconductor.
Semiconductors are materials that have an electrical conductivity between that of a conductor and an insulator. Three properties of a semiconductor are:
- Conductivity: Semiconductors have intermediate electrical conductivity, which means they are not good conductors like metals but not as poor as insulators. They can conduct electricity under certain conditions, such as when exposed to light or heat, or when impurities are added to them (a process called doping).
- Energy gap: Semiconductors have a small energy gap between their valence band (the highest energy level occupied by electrons) and their conduction band (the lowest energy level that electrons can occupy and still be free to move). This energy gap determines the conductivity of the material and can be manipulated through doping or other processes.
- Temperature sensitivity: Semiconductors are sensitive to changes in temperature, which can affect their electrical conductivity. As the temperature of a semiconductor increases, the energy of its electrons increases and they become more likely to move into the conduction band, increasing the material's conductivity. However, if the temperature gets too high, the semiconductor may begin to behave like a conductor and lose its unique properties.
Overall, these properties make semiconductors useful in a wide range of electronic devices, including transistors, diodes, and solar cells.
Akọwa Nkọwa
Semiconductors are materials that have an electrical conductivity between that of a conductor and an insulator. Three properties of a semiconductor are:
- Conductivity: Semiconductors have intermediate electrical conductivity, which means they are not good conductors like metals but not as poor as insulators. They can conduct electricity under certain conditions, such as when exposed to light or heat, or when impurities are added to them (a process called doping).
- Energy gap: Semiconductors have a small energy gap between their valence band (the highest energy level occupied by electrons) and their conduction band (the lowest energy level that electrons can occupy and still be free to move). This energy gap determines the conductivity of the material and can be manipulated through doping or other processes.
- Temperature sensitivity: Semiconductors are sensitive to changes in temperature, which can affect their electrical conductivity. As the temperature of a semiconductor increases, the energy of its electrons increases and they become more likely to move into the conduction band, increasing the material's conductivity. However, if the temperature gets too high, the semiconductor may begin to behave like a conductor and lose its unique properties.
Overall, these properties make semiconductors useful in a wide range of electronic devices, including transistors, diodes, and solar cells.
Ajụjụ 50 Ripọtì
(a)(i) State Coulomb's law of electrostatics.
(ii) The electron and proton of a hydrogen atom are separated by a mean distance of 5.2 x 10?11 m.
Calculate the magnitude of the electrostatic force between the particles.
[e = 1.6 x 10?19 C, (4? E0 )?1 = 9.0 x 109 mF1 ]
(b) The diagram below shows a potential divider circuit.
i. Show that Vout = Vin (R1R1+R2)
ii. If VinVout = 2.5 and R2 = 30? , calculate R1
iii. Define the volt.
(c) Explain why wood is not suitable for use as the core of transformers.
(d) State one application for the cathode ray tube.
(a)(i) Coulomb's law of electrostatics states that the force of attraction or repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Mathematically, F = kQ1Q2/r^2, where F is the electrostatic force, Q1 and Q2 are the charges on the objects, r is the distance between them, and k is Coulomb's constant.
(ii) Using Coulomb's law, the magnitude of the electrostatic force between the electron and proton of a hydrogen atom can be calculated as follows:
F = k(Q1Q2/r^2) = (9.0 x 10^9 Nm^2/C^2)(1.6 x 10^-19 C)(1.6 x 10^-19 C)/(5.2 x 10^-11 m)^2 = 8.2 x 10^-8 N.
(b)(I) Proof of Vout
Let the current through the circuit be I
Vin
= I(R1 + R2)
OR
I = VinR1+R2
Vout = IR2
II. Calculation of R1
;
Vout
= Vin(R1R1+R2)
OR
VoutVin=R1R1+R2
12.5=R1R1+R30R1 = 20Ω
(ii) The Volt The unit of potential difference between two points when one joule of work is done in taking one coulomb of charge between the points
(c) Wood is not suitable for use as the core of transformers because it is not a good conductor of electricity. Transformers require a material with a high magnetic permeability, which allows for efficient transfer of energy between coils. Wood does not have the necessary magnetic properties and can also be damaged by the heat generated during operation.
(d) One application for the cathode ray tube (CRT) is in television and computer displays. A CRT works by producing a beam of electrons that is directed towards a phosphorescent screen, causing it to emit light and create an image. By controlling the path of the electron beam, the image can be manipulated and displayed on the screen. Although CRT technology has largely been replaced by LCD and LED displays, it is still used in some specialized applications such as medical imaging and oscilloscopes.
Akọwa Nkọwa
(a)(i) Coulomb's law of electrostatics states that the force of attraction or repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Mathematically, F = kQ1Q2/r^2, where F is the electrostatic force, Q1 and Q2 are the charges on the objects, r is the distance between them, and k is Coulomb's constant.
(ii) Using Coulomb's law, the magnitude of the electrostatic force between the electron and proton of a hydrogen atom can be calculated as follows:
F = k(Q1Q2/r^2) = (9.0 x 10^9 Nm^2/C^2)(1.6 x 10^-19 C)(1.6 x 10^-19 C)/(5.2 x 10^-11 m)^2 = 8.2 x 10^-8 N.
(b)(I) Proof of Vout
Let the current through the circuit be I
Vin
= I(R1 + R2)
OR
I = VinR1+R2
Vout = IR2
II. Calculation of R1
;
Vout
= Vin(R1R1+R2)
OR
VoutVin=R1R1+R2
12.5=R1R1+R30R1 = 20Ω
(ii) The Volt The unit of potential difference between two points when one joule of work is done in taking one coulomb of charge between the points
(c) Wood is not suitable for use as the core of transformers because it is not a good conductor of electricity. Transformers require a material with a high magnetic permeability, which allows for efficient transfer of energy between coils. Wood does not have the necessary magnetic properties and can also be damaged by the heat generated during operation.
(d) One application for the cathode ray tube (CRT) is in television and computer displays. A CRT works by producing a beam of electrons that is directed towards a phosphorescent screen, causing it to emit light and create an image. By controlling the path of the electron beam, the image can be manipulated and displayed on the screen. Although CRT technology has largely been replaced by LCD and LED displays, it is still used in some specialized applications such as medical imaging and oscilloscopes.
Ajụjụ 51 Ripọtì
(a) State two factors that affect the rate of evaporation of a liquid.
(b) Explain the term latent heal.
(c) Explain each of the following phenomena:
(i) On a dry day, water in a clay pot is cooler than water in a rubber container:
(ii) Cooking of food is faster in a pressure cooker than in an ordinary pot.
(d) A 40 V electric heater is used to supply a current of 12 A for 1400 seconds to a body mass of 1.5 kg at its melting point. The body melts and its temperature rises by 60°C in an extra 1.2 minutes, Calculate the:
(i) Latent heat of fusion of the body
(ii) Specific heat capacity of the body.
(e) State two differences between evaporation and boiling.
(a) Two factors that affect the rate of evaporation of a liquid are: 1. Temperature: Higher temperatures generally result in faster rates of evaporation, as the heat energy increases the kinetic energy of the liquid molecules, causing them to move faster and escape into the air more readily. 2. Surface area: A larger surface area of the liquid exposed to the air can increase the rate of evaporation, as there are more liquid molecules available to escape into the air. (b) Latent heat refers to the amount of heat energy that is required to change the phase of a substance without changing its temperature. For example, when ice is heated, it melts and changes from a solid to a liquid, but its temperature remains constant at 0°C until all the ice has melted. The energy absorbed during this phase change is known as the latent heat of fusion. (c) (i) On a dry day, water in a clay pot is cooler than water in a rubber container because the water in the clay pot evaporates faster due to the porous nature of the clay, which allows air to circulate through it. As the water evaporates, it removes heat from the remaining water, causing it to cool down. (ii) Cooking of food is faster in a pressure cooker than in an ordinary pot because the pressure inside the cooker increases the boiling point of water. This means that the food can be cooked at a higher temperature than in an ordinary pot, resulting in faster cooking times. (d) (i) Latent heat of fusion can be calculated using the formula Q = mL, where Q is the heat energy absorbed, m is the mass of the substance, and L is the latent heat of fusion. In this case, Q = (40 V x 12 A x 1400 s) = 672,000 J, and m = 1.5 kg. Assuming that the temperature of the substance was constant during melting, L can be calculated as L = Q/m = 672,000 J / 1.5 kg = 448,000 J/kg. (ii) Specific heat capacity can be calculated using the formula c = Q/(mΔT), where c is the specific heat capacity, ΔT is the change in temperature, and all other variables are the same as in part (i). In this case, Q = (40 V x 12 A x 72 s) = 34,560 J and ΔT = 60°C. Substituting these values into the formula, we get c = 34,560 J / (1.5 kg x 60°C) = 384 J/(kg·°C). (e) Two differences between evaporation and boiling are: 1. Evaporation occurs at the surface of a liquid, while boiling occurs throughout the liquid. 2. Evaporation occurs at a range of temperatures below the boiling point of the liquid, while boiling occurs only at the boiling point. Evaporation is a slower process than boiling, as it only occurs at the surface of the liquid, while boiling involves the rapid formation of bubbles throughout the liquid.
Akọwa Nkọwa
(a) Two factors that affect the rate of evaporation of a liquid are: 1. Temperature: Higher temperatures generally result in faster rates of evaporation, as the heat energy increases the kinetic energy of the liquid molecules, causing them to move faster and escape into the air more readily. 2. Surface area: A larger surface area of the liquid exposed to the air can increase the rate of evaporation, as there are more liquid molecules available to escape into the air. (b) Latent heat refers to the amount of heat energy that is required to change the phase of a substance without changing its temperature. For example, when ice is heated, it melts and changes from a solid to a liquid, but its temperature remains constant at 0°C until all the ice has melted. The energy absorbed during this phase change is known as the latent heat of fusion. (c) (i) On a dry day, water in a clay pot is cooler than water in a rubber container because the water in the clay pot evaporates faster due to the porous nature of the clay, which allows air to circulate through it. As the water evaporates, it removes heat from the remaining water, causing it to cool down. (ii) Cooking of food is faster in a pressure cooker than in an ordinary pot because the pressure inside the cooker increases the boiling point of water. This means that the food can be cooked at a higher temperature than in an ordinary pot, resulting in faster cooking times. (d) (i) Latent heat of fusion can be calculated using the formula Q = mL, where Q is the heat energy absorbed, m is the mass of the substance, and L is the latent heat of fusion. In this case, Q = (40 V x 12 A x 1400 s) = 672,000 J, and m = 1.5 kg. Assuming that the temperature of the substance was constant during melting, L can be calculated as L = Q/m = 672,000 J / 1.5 kg = 448,000 J/kg. (ii) Specific heat capacity can be calculated using the formula c = Q/(mΔT), where c is the specific heat capacity, ΔT is the change in temperature, and all other variables are the same as in part (i). In this case, Q = (40 V x 12 A x 72 s) = 34,560 J and ΔT = 60°C. Substituting these values into the formula, we get c = 34,560 J / (1.5 kg x 60°C) = 384 J/(kg·°C). (e) Two differences between evaporation and boiling are: 1. Evaporation occurs at the surface of a liquid, while boiling occurs throughout the liquid. 2. Evaporation occurs at a range of temperatures below the boiling point of the liquid, while boiling occurs only at the boiling point. Evaporation is a slower process than boiling, as it only occurs at the surface of the liquid, while boiling involves the rapid formation of bubbles throughout the liquid.
Ajụjụ 52 Ripọtì
You are provided with a battery of e.m.f, E, a standard resistor, R, of resistance 2 ? , a key, K, an ammeter, A, a jockey, J, a potentiometer, UV, and some connecting wires.
(i) Measure and record the emf, E, of the battery.
(ii) Set up the circuit as shown in the diagram above with the key open.
(iii) Place the jockey at the point, U, of the potentiometer wire. Close the key and record the reading, i, of the ammeter.
(iv) Place the jockey at a point T on the potentiometer wire UV such that d = UT = 30.0 cm.
(v) Close the circuit, read and record the current, I, on the ammeter,
(vi) Evaluate I1 .
(vi) Repeat the experiment for four other values of d = 40.0 cm, 50.0 cm, 60.0 cm and 70.0 cm. In each case, record I and evaluate I1 .
(vii) Tabulate the results
(ix) Plot a graph with d on the vertical axis and I on the horizontal axis stalling both axes from the origin (0,0).
(x) Determine the slope, s, of the graph.
(xi) From the graph determine the value I1 , of I when d = 0. (ci) Given that=s, calculate 8.
(xii) State two precautions taken to ensure accurate results.
(xii) Given that E?
= s, calculate ??
.
(b)(i) Write down the equation that connects the resistance, R, of a wire and the factors on which it depends. State the meaning of each of the symbols.
(ii) An electric fan draws a current of0.75 A in a 240 V circuit. Calculate the cost of using, the fan for 10 hours if the utility rate is $ 0.50 per kWh.
(a) OBSERVATIONS
(i) Value of E correctly read and recorded to at least 1 d.p in volts.
(ii) Value of i correctly read and recorded to at least 1 d.pin amperes.
(iii) Five values of d correctly determined and recorded to at least 1 d.p in cm.
(iv) Five values of I correctly read and recorded to at least 1 d.p in amperes and in trend.
Trend; As d increases, I decreases
(v) Five values of I1 correctly evaluated to at least 3 s.f.
(vi) Composite table showing at least d, I, and I−1 .
(ii) P = IV
Power P = 240×0.75100
0.18kw
Energy consumed in 10hr = 0.18 x 10
= 1.8kWh
Cost of energy = 1.8 x 0.5 = $0.9
OR
Cost of energy = P x cost x time100
0.7×240×101000
= $0.9
Akọwa Nkọwa
(a) OBSERVATIONS
(i) Value of E correctly read and recorded to at least 1 d.p in volts.
(ii) Value of i correctly read and recorded to at least 1 d.pin amperes.
(iii) Five values of d correctly determined and recorded to at least 1 d.p in cm.
(iv) Five values of I correctly read and recorded to at least 1 d.p in amperes and in trend.
Trend; As d increases, I decreases
(v) Five values of I1 correctly evaluated to at least 3 s.f.
(vi) Composite table showing at least d, I, and I−1 .
(ii) P = IV
Power P = 240×0.75100
0.18kw
Energy consumed in 10hr = 0.18 x 10
= 1.8kWh
Cost of energy = 1.8 x 0.5 = $0.9
OR
Cost of energy = P x cost x time100
0.7×240×101000
= $0.9
Ajụjụ 53 Ripọtì
The diagram above illustrates the trajectory of a fired missile from point P at 250 ms-1
If the missile point Q after 40 s, calculate the distance |PQ|
|PQ| = Range = Ux.T
= UT Cosθ
= 250 x 40 Cos 42º
= 7431 m (7.4 km)
Akọwa Nkọwa
|PQ| = Range = Ux.T
= UT Cosθ
= 250 x 40 Cos 42º
= 7431 m (7.4 km)
Ajụjụ 54 Ripọtì
A 50 N force is applied to the free end of a spiral spring of force constant, 100 N m−1 . Calculate the work done by the force to stretch the spring.
Calculation of workdone by the force to scratch the string
W = ½ ke\(^{2}\) = \(\frac{f^{2}}{2k}\)
W = \(\frac{50^{2}}{2 \times 100}\)
W = 12.5 J
OR
e = \(\frac{F}{K} = \frac{50}{100}\) = 0.5m
w = ½ = ke\(^{2}\)
w = ½ x 100 x 0.5\(^{2}\)
W = 12.5 J
Akọwa Nkọwa
Calculation of workdone by the force to scratch the string
W = ½ ke\(^{2}\) = \(\frac{f^{2}}{2k}\)
W = \(\frac{50^{2}}{2 \times 100}\)
W = 12.5 J
OR
e = \(\frac{F}{K} = \frac{50}{100}\) = 0.5m
w = ½ = ke\(^{2}\)
w = ½ x 100 x 0.5\(^{2}\)
W = 12.5 J
Ajụjụ 55 Ripọtì
You are provided with a battery of e.m.f, E, a standard resistor, R, of resistance 2 Ω , a key, K, an ammeter, A, a jockey, J, a potentiometer, UV, and some connecting wires.
(i) Measure and record the emf, E, of the battery.
(ii) Set up the circuit as shown in the diagram above with the key open.
(iii) Place the jockey at the point, U, of the potentiometer wire. Close the key and record the reading, i, of the ammeter.
(iv) Place the jockey at a point T on the potentiometer wire UV such that d = UT = 30.0 cm.
(v) Close the circuit, read and record the current, I, on the ammeter,
(vi) Evaluate I1 .
(vi) Repeat the experiment for four other values of d = 40.0 cm, 50.0 cm, 60.0 cm and 70.0 cm. In each case, record I and evaluate I1 .
(vii) Tabulate the results
(ix) Plot a graph with d on the vertical axis and I on the horizontal axis stalling both axes from the origin (0,0).
(x) Determine the slope, s, of the graph.
(xi) From the graph determine the value I1 , of I when d = 0. (ci) Given that=s, calculate 8.
(xii) State two precautions taken to ensure accurate results.
(xii) Given that Eδ = s, calculate δ .
(b)(i) Write down the equation that connects the resistance, R, of a wire and the factors on which it depends. State the meaning of each of the symbols.
(ii) An electric fan draws a current of0.75 A in a 240 V circuit. Calculate the cost of using, the fan for 10 hours if the utility rate is $ 0.50 per kWh.
(a) OBSERVATIONS
(i) Value of E correctly read and recorded to at least 1 d.p in volts.
(ii) Value of i correctly read and recorded to at least 1 d.pin amperes.
(iii) Five values of d correctly determined and recorded to at least 1 d.p in cm.
(iv) Five values of I correctly read and recorded to at least 1 d.p in amperes and in trend.
Trend; As d increases, I decreases
(v) Five values of I1 correctly evaluated to at least 3 s.f.
(vi) Composite table showing at least d, I, and I−1 .
(ii) P = IV
Power P = 240×0.75100
0.18kw
Energy consumed in 10hr = 0.18 x 10
= 1.8kWh
Cost of energy = 1.8 x 0.5 = $0.9
OR
Cost of energy = P x cost x time100
0.7×240×101000
= $0.9
Akọwa Nkọwa
(a) OBSERVATIONS
(i) Value of E correctly read and recorded to at least 1 d.p in volts.
(ii) Value of i correctly read and recorded to at least 1 d.pin amperes.
(iii) Five values of d correctly determined and recorded to at least 1 d.p in cm.
(iv) Five values of I correctly read and recorded to at least 1 d.p in amperes and in trend.
Trend; As d increases, I decreases
(v) Five values of I1 correctly evaluated to at least 3 s.f.
(vi) Composite table showing at least d, I, and I−1 .
(ii) P = IV
Power P = 240×0.75100
0.18kw
Energy consumed in 10hr = 0.18 x 10
= 1.8kWh
Cost of energy = 1.8 x 0.5 = $0.9
OR
Cost of energy = P x cost x time100
0.7×240×101000
= $0.9
Ajụjụ 56 Ripọtì
(a)(i) What is meant by the term artificial radioactivity?
(ii) Complete the table below
| Emission | Nature | Charge | Ionizing |
| High speed electron | Moderately ionizing | ||
| Neutral | Negligible ionizing ability | ||
| Alpha particles | Positive |
(b) In an x-ray tube, an electron is accelerated from rest towards a metal target by a 30 kV source. Calculate the kinetic energy of the electron. [e=1.6 x 10?19 C]
(c) The table below shows the frequencies of radiations incident on a certain metal and the corresponding kinetic energies of the photoelectrons.
| Frequency x 1014 (Hz) | 6.8 | 8.0 | 9.2 | 10.0 | 11.0 |
| Kinetic energy x 10?19 (j) | 0.8 | 1.6 | 2.4 | 2.9 | 3.8 |
(i) Plot a graph of kinetic energy, K.E, on the vertical axis and frequency, f, on the horizontal axis starting both axes from the origin (0,0).
(ii) From the graph, determine the:
i. Planck's constant;
ii. Threshold frequency of radiations;
iii. Work function of the metal.
(a)() Meaning of artificial radioactivity;
The process by which a stable nucleus is bombarded with a neutron to make it unstable and so disintegrates/decays with the emission of particles/radiation and. energy.
| Emission | Nature | Charge | Ionizing |
| Beta (\(\beta\)) | High speed electron | Negative | Moderately ionizing |
| Gamma (\(\gamma\)) | Electro-magnetic radiation | Neutral | Negligible ionizing ability |
| Alpha particles | Helium nucleus | Positive | Highly ionizing |
(b) The kinetic energy of the electron can be calculated using the formula: KE = qV, where q is the charge of the electron and V is the potential difference. Substituting the given values, we get:
K.E = eV
KE = (1.6 x 10^-19 C)(30,000 V)
KE = 4.8 x 10^-15 J (c)
(i)
To calculate the slope of the graph, you need to determine the change in the dependent variable (kinetic energy) divided by the change in the independent variable (frequency). In this case, you can choose any two points on the graph and calculate the slope using the following formula:
slope = (kinetic_energy2 - kinetic_energy1) / (frequency2 - frequency1)
Let's take two points from the given data, for example:
Point 1: (frequency1, kinetic_energy1) = (6.8 x 10^14 Hz, 0.8 x 10^-19 J)
Point 2: (frequency2, kinetic_energy2) = (8.0 x 10^14 Hz, 1.6 x 10^-19 J)
Now, we can calculate the slope:
slope = (1.6 x 10^-19 J - 0.8 x 10^-19 J) / (8.0 x 10^14 Hz - 6.8 x 10^14 Hz)
slope = 1 x 10^-5 J Hz^(-1).
To determine Planck's constant from the given graph and slope, we can use the equation:
slope = h / e
where h is Planck's constant and e is the elementary charge (1.602176634 x 10^-19 C).
From the previous calculation, the slope of the graph is 1 x 10^-5 J Hz^(-1).
Let's substitute the values into the equation to solve for Planck's constant:
1 x 10^-5 J Hz^(-1) = h / (1.602176634 x 10^-19 C)
To isolate h, we can rearrange the equation:
h = slope * e
Substituting the values:
h = (1 x 10^-5 J Hz^(-1)) * (1.602176634 x 10^-19 C)
Evaluating the expression:
h ≈ 1.602176634 x 10^-24 J·s
Therefore, from the given graph and slope, the approximate value of Planck's constant is 1.602176634 x 10^-24 J·s.
(ii) To determine the threshold frequency of radiation from the given information, we need to use the concept of the photoelectric effect and the relationship between the kinetic energy of photoelectrons and the frequency of incident radiation.
According to the photoelectric effect, electrons are ejected from a metal surface when illuminated by electromagnetic radiation of sufficient energy. The minimum frequency of radiation required to eject electrons is known as the threshold frequency.
The relationship between the kinetic energy of photoelectrons and the frequency of incident radiation is given by the equation:
K.E. = h * (frequency - threshold_frequency)
where K.E. is the kinetic energy of the photoelectrons, h is Planck's constant, frequency is the frequency of incident radiation, and threshold_frequency is the threshold frequency.
From the graph, we have the slope, which is equal to h, and the kinetic energy corresponding to each frequency. We can select any point on the graph where the kinetic energy is non-zero and solve for the threshold frequency.
Let's choose the point (frequency, kinetic energy) = (6.8 x 10^14 Hz, 0.8 x 10^-19 J) from the given data.
0.8 x 10^-19 J = slope * (6.8 x 10^14 Hz - threshold_frequency)
Substituting the slope value:
0.8 x 10^-19 J = 1.602176634 x 10^-24 J·s * (6.8 x 10^14 Hz - threshold_frequency)
To solve for the threshold frequency, we can rearrange the equation:
threshold_frequency = 6.8 x 10^14 Hz - (0.8 x 10^-19 J / (1.602176634 x 10^-24 J·s))
Calculating the threshold frequency:
threshold_frequency = 6.8 x 10^14 Hz - 4.992706701 x 10^4 Hz
threshold_frequency ≈ 6.799500729 x 10^14 Hz
Therefore, the threshold frequency of radiation is approximately 6.799500729 x 10^14 Hz.
(iii)
To determine the work function of the metal, we can use the equation:
Work function = h * threshold_frequency
where h is Planck's constant and threshold_frequency is the threshold frequency of radiation.
From the previous calculations, the approximate value of Planck's constant is 1.602176634 x 10^-24 J·s and the threshold frequency is approximately 6.799500729 x 10^14 Hz.
Substituting these values into the equation, we can calculate the work function:
Work function = (1.602176634 x 10^-24 J·s) * (6.799500729 x 10^14 Hz)
Work function ≈ 1.090589631 x 10^-9 J
Therefore, based on the given information, the approximate value of the work function of the metal is 1.090589631 x 10^-9 J.
Akọwa Nkọwa
(a)() Meaning of artificial radioactivity;
The process by which a stable nucleus is bombarded with a neutron to make it unstable and so disintegrates/decays with the emission of particles/radiation and. energy.
| Emission | Nature | Charge | Ionizing |
| Beta (\(\beta\)) | High speed electron | Negative | Moderately ionizing |
| Gamma (\(\gamma\)) | Electro-magnetic radiation | Neutral | Negligible ionizing ability |
| Alpha particles | Helium nucleus | Positive | Highly ionizing |
(b) The kinetic energy of the electron can be calculated using the formula: KE = qV, where q is the charge of the electron and V is the potential difference. Substituting the given values, we get:
K.E = eV
KE = (1.6 x 10^-19 C)(30,000 V)
KE = 4.8 x 10^-15 J (c)
(i)
To calculate the slope of the graph, you need to determine the change in the dependent variable (kinetic energy) divided by the change in the independent variable (frequency). In this case, you can choose any two points on the graph and calculate the slope using the following formula:
slope = (kinetic_energy2 - kinetic_energy1) / (frequency2 - frequency1)
Let's take two points from the given data, for example:
Point 1: (frequency1, kinetic_energy1) = (6.8 x 10^14 Hz, 0.8 x 10^-19 J)
Point 2: (frequency2, kinetic_energy2) = (8.0 x 10^14 Hz, 1.6 x 10^-19 J)
Now, we can calculate the slope:
slope = (1.6 x 10^-19 J - 0.8 x 10^-19 J) / (8.0 x 10^14 Hz - 6.8 x 10^14 Hz)
slope = 1 x 10^-5 J Hz^(-1).
To determine Planck's constant from the given graph and slope, we can use the equation:
slope = h / e
where h is Planck's constant and e is the elementary charge (1.602176634 x 10^-19 C).
From the previous calculation, the slope of the graph is 1 x 10^-5 J Hz^(-1).
Let's substitute the values into the equation to solve for Planck's constant:
1 x 10^-5 J Hz^(-1) = h / (1.602176634 x 10^-19 C)
To isolate h, we can rearrange the equation:
h = slope * e
Substituting the values:
h = (1 x 10^-5 J Hz^(-1)) * (1.602176634 x 10^-19 C)
Evaluating the expression:
h ≈ 1.602176634 x 10^-24 J·s
Therefore, from the given graph and slope, the approximate value of Planck's constant is 1.602176634 x 10^-24 J·s.
(ii) To determine the threshold frequency of radiation from the given information, we need to use the concept of the photoelectric effect and the relationship between the kinetic energy of photoelectrons and the frequency of incident radiation.
According to the photoelectric effect, electrons are ejected from a metal surface when illuminated by electromagnetic radiation of sufficient energy. The minimum frequency of radiation required to eject electrons is known as the threshold frequency.
The relationship between the kinetic energy of photoelectrons and the frequency of incident radiation is given by the equation:
K.E. = h * (frequency - threshold_frequency)
where K.E. is the kinetic energy of the photoelectrons, h is Planck's constant, frequency is the frequency of incident radiation, and threshold_frequency is the threshold frequency.
From the graph, we have the slope, which is equal to h, and the kinetic energy corresponding to each frequency. We can select any point on the graph where the kinetic energy is non-zero and solve for the threshold frequency.
Let's choose the point (frequency, kinetic energy) = (6.8 x 10^14 Hz, 0.8 x 10^-19 J) from the given data.
0.8 x 10^-19 J = slope * (6.8 x 10^14 Hz - threshold_frequency)
Substituting the slope value:
0.8 x 10^-19 J = 1.602176634 x 10^-24 J·s * (6.8 x 10^14 Hz - threshold_frequency)
To solve for the threshold frequency, we can rearrange the equation:
threshold_frequency = 6.8 x 10^14 Hz - (0.8 x 10^-19 J / (1.602176634 x 10^-24 J·s))
Calculating the threshold frequency:
threshold_frequency = 6.8 x 10^14 Hz - 4.992706701 x 10^4 Hz
threshold_frequency ≈ 6.799500729 x 10^14 Hz
Therefore, the threshold frequency of radiation is approximately 6.799500729 x 10^14 Hz.
(iii)
To determine the work function of the metal, we can use the equation:
Work function = h * threshold_frequency
where h is Planck's constant and threshold_frequency is the threshold frequency of radiation.
From the previous calculations, the approximate value of Planck's constant is 1.602176634 x 10^-24 J·s and the threshold frequency is approximately 6.799500729 x 10^14 Hz.
Substituting these values into the equation, we can calculate the work function:
Work function = (1.602176634 x 10^-24 J·s) * (6.799500729 x 10^14 Hz)
Work function ≈ 1.090589631 x 10^-9 J
Therefore, based on the given information, the approximate value of the work function of the metal is 1.090589631 x 10^-9 J.
Ajụjụ 57 Ripọtì
(a)(i) State the law of inertia.
(ii) Use the law stated in (a)(I) to explain how wearing a safety belt in a moving vehicle could reduce the possibilities of severe injuries when the vehicle is involved in a collision.
(b)(I) Define the term moment of a force.
(ii) A uniform plank measures 2 m long from its ends point A to point B. If the weight of the plank is 54 N and it rests on a knife edge 0.50 m from end B and point A is supported by a vertical string so that AB balances horizontally:
i. Draw a force diagram for the arrangement;
ii. Determine the tension, T, in the string.
iii. Determine the force, F, acting on the knife edge
(c) State three differences between solid friction and viscosity
(d) State one method of increasing the velocity ration of a pulley system.
(a)(i) The law of inertia states that an object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity unless acted upon by an external force.
(a)(ii) Wearing a safety belt in a moving vehicle could reduce the possibilities of severe injuries when the vehicle is involved in a collision because of the law of inertia. When a car suddenly stops in a collision, the passengers in the car tend to continue moving forward at the same speed the car was moving before the collision. If they are not restrained by a seat belt, they will keep moving forward and could collide with the dashboard, steering wheel, or windshield. However, if they are wearing a seat belt, the belt applies a force to the passenger in the opposite direction, keeping them from moving forward, and reducing the risk of injury.
(b)(i) The moment of a force is the turning effect of the force about a pivot and is given by the product of the force and the perpendicular distance from the pivot to the line of action of the force.
(b)(ii)
i. The force diagram for the arrangement is a diagram that shows all the forces acting on an object in a particular situation. For this arrangement, the force diagram will show the weight of the plank acting downwards at the center, the tension T acting upwards at point A, and the reaction force F acting upwards at the knife edge.
ii. To determine the tension, T, in the string, we can use the principle of moments. The sum of the clockwise moments about the knife edge must be equal to the sum of the anticlockwise moments about the knife edge. Thus,
54 N x 0.50 m = T x 1.50 m.
Therefore, T = 18 N.
iii. To determine the force, F, acting on the knife edge, we can use the principle of moments. The sum of the clockwise moments about the knife edge must be equal to the sum of the anticlockwise moments about the knife edge and at equilibrium, the sum of upward forces = the sum of forces acting downward
T + F = 54
T = 18 N
18 + F = 54
F = 36 N
(c) Three differences between solid friction and viscosity are:
1. Solid friction is the force that opposes the motion of an object along a solid surface, whereas viscosity is the force that opposes the motion of an object through a fluid.
2. Solid friction depends on the nature of the surfaces in contact, whereas viscosity depends on the properties of the fluid, such as its density and viscosity.
3. Solid friction is independent of velocity, whereas viscosity is directly proportional to velocity.
(d) One method of increasing the velocity ratio of a pulley system is to increase the number of pulleys in the system. The velocity ratio of a pulley system is the ratio of the distance moved by the effort to the distance moved by the load. The more pulleys there are in the system, the greater the distance the effort can move for a given distance moved by the load, thus increasing the velocity ratio.
Akọwa Nkọwa
(a)(i) The law of inertia states that an object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity unless acted upon by an external force.
(a)(ii) Wearing a safety belt in a moving vehicle could reduce the possibilities of severe injuries when the vehicle is involved in a collision because of the law of inertia. When a car suddenly stops in a collision, the passengers in the car tend to continue moving forward at the same speed the car was moving before the collision. If they are not restrained by a seat belt, they will keep moving forward and could collide with the dashboard, steering wheel, or windshield. However, if they are wearing a seat belt, the belt applies a force to the passenger in the opposite direction, keeping them from moving forward, and reducing the risk of injury.
(b)(i) The moment of a force is the turning effect of the force about a pivot and is given by the product of the force and the perpendicular distance from the pivot to the line of action of the force.
(b)(ii)
i. The force diagram for the arrangement is a diagram that shows all the forces acting on an object in a particular situation. For this arrangement, the force diagram will show the weight of the plank acting downwards at the center, the tension T acting upwards at point A, and the reaction force F acting upwards at the knife edge.
ii. To determine the tension, T, in the string, we can use the principle of moments. The sum of the clockwise moments about the knife edge must be equal to the sum of the anticlockwise moments about the knife edge. Thus,
54 N x 0.50 m = T x 1.50 m.
Therefore, T = 18 N.
iii. To determine the force, F, acting on the knife edge, we can use the principle of moments. The sum of the clockwise moments about the knife edge must be equal to the sum of the anticlockwise moments about the knife edge and at equilibrium, the sum of upward forces = the sum of forces acting downward
T + F = 54
T = 18 N
18 + F = 54
F = 36 N
(c) Three differences between solid friction and viscosity are:
1. Solid friction is the force that opposes the motion of an object along a solid surface, whereas viscosity is the force that opposes the motion of an object through a fluid.
2. Solid friction depends on the nature of the surfaces in contact, whereas viscosity depends on the properties of the fluid, such as its density and viscosity.
3. Solid friction is independent of velocity, whereas viscosity is directly proportional to velocity.
(d) One method of increasing the velocity ratio of a pulley system is to increase the number of pulleys in the system. The velocity ratio of a pulley system is the ratio of the distance moved by the effort to the distance moved by the load. The more pulleys there are in the system, the greater the distance the effort can move for a given distance moved by the load, thus increasing the velocity ratio.
Ajụjụ 58 Ripọtì
You are provided with a metre rule, a weight hanger, slotted masses, M, a piece (if string, a weighing balance and a knife edge. Use the diagram above as a guide to perform the experiment.
(i) Using the weighing balance, determine and record the mass, Mo, of the unloaded metre rule.
(ii) Determine and record the mass, m, of the weight hanger.
(ii) Suspend the metre rule horizontally on the knife edge. Adjust the knife edge to a point G on the metre rule where it balances horizontally.
(iv) Record the distance, d = AG.
(v) Suspend the weight hanger securely at a point, P, on the metre rule such that AP= 5 cm. Keep the hanger at this point throughout the experiment
(vi) Add a mass, M = 20 g to the hanger, adjust the knife edge to a point K on the metre rule such that it balances horizontally as shown in the diagram above.
(vii) Determine and record the distance z = AK.
(vii) Record M and evaluate y - (z - 5), x - (d - z] and v = xy
(ix) Repeat the experiment for M = 40 g, 60 g, 80 g and 100 g. In each case, evaluate y, x and v.
(x) Tabulate the results.
(xi) Plot a graph with M on the vertical axis and v on the horizontal axis, sinning both axes from the origin (0,0).
(xii) Determine the slope, s, of the graph.
(xii) Determine the intercept, c, on the vertical axis.
(xiv) State two precautions taken to ensure accurate results.
(b) (i) Under what condition is an object said to be in a stable equilibrium
(ii) Auniform beam of weight 50 N has a body of weight 100 N hung at one end of it. If the beam is 12 m long, determine the distance of a support from a 100 N body for it to balance horizontally.
(a) OBSERVATIONS
(i) Value of M0 correctly determined and recorded to at least 1 d.p in grammes
(ii) Value of m correctly determined and recorded to at least 1 d.p in grammes
(iii) Value of d correctly read and recorded to at least 1 d.p in cm
(iv) Five values of M correctly recorded in grammes
(v) Five values of Z = AK correctly read and recorded to at least1 d.p in cm and in trend.
Trend: As M increases Z decreases.
(vi) Five values of y = (Z - 5) correctly evaluated and recorded.
(vii) Five values of x = (d - Z) correctly evaluated and recorded
(vii) Five values of v = xy correctly evaluated and recorded to at least 3 d.p
(ix) Composite table showing at least M, Z,y, x and v.
(b)(i) Condition for a body to be in stable
- The algebraic sum of all external forces on the object is zero
OR
- The algebraic sum of all external torques from the external forces acting on the object is zero.
OR
- The sum of clockwise moments about a point in the object is equal to the sum of anticlockwise moment about the same point.
OR
- The algebraic sum of all moments about a point on the object is zero.
(ii) If the distance is y,
100y = 50(6 - y)
100y = 300 - 50y
150y = 300
y = 2m
Akọwa Nkọwa
(a) OBSERVATIONS
(i) Value of M0 correctly determined and recorded to at least 1 d.p in grammes
(ii) Value of m correctly determined and recorded to at least 1 d.p in grammes
(iii) Value of d correctly read and recorded to at least 1 d.p in cm
(iv) Five values of M correctly recorded in grammes
(v) Five values of Z = AK correctly read and recorded to at least1 d.p in cm and in trend.
Trend: As M increases Z decreases.
(vi) Five values of y = (Z - 5) correctly evaluated and recorded.
(vii) Five values of x = (d - Z) correctly evaluated and recorded
(vii) Five values of v = xy correctly evaluated and recorded to at least 3 d.p
(ix) Composite table showing at least M, Z,y, x and v.
(b)(i) Condition for a body to be in stable
- The algebraic sum of all external forces on the object is zero
OR
- The algebraic sum of all external torques from the external forces acting on the object is zero.
OR
- The sum of clockwise moments about a point in the object is equal to the sum of anticlockwise moment about the same point.
OR
- The algebraic sum of all moments about a point on the object is zero.
(ii) If the distance is y,
100y = 50(6 - y)
100y = 300 - 50y
150y = 300
y = 2m
Ajụjụ 59 Ripọtì
The effective potential energy, E, of a lunar satellite of mass, m, moving in an. elliptical orbit around the moon of mass, m, is given by
E = K22m1r2−Gm1m2r
where r is the distance of the satellite from the mooń and G is the universal gravitational constant of dimensions, M−1
L3
T22
.
Ďetermine the dimensions of the angular momentum, K, of the satellite using dimensional analysis.
Dimensions of the angular momentum, K =
[K]2[m1][r]2=[G][m1][m2][r]
[k]2ML2=M−1L3T−2M2L
[K] = (M2 L4T−2 )½
∴ [k] = ML2T−1
Akọwa Nkọwa
Dimensions of the angular momentum, K =
[K]2[m1][r]2=[G][m1][m2][r]
[k]2ML2=M−1L3T−2M2L
[K] = (M2 L4T−2 )½
∴ [k] = ML2T−1
Ajụjụ 60 Ripọtì
(a) State one condition each necessary for the characteristics each of the following occurrences:
(i) Constructive interference of waves.
(ii) Total internal reflection.
(iii) Production of beats.
(b) In a resonance tube experiment using a tuning fork of frequency 256 Hz, the first position of resonance was 35 cm, the next position was 100 cm. Calculate the velocity of sound in air from the experiment.
(c)(i) State the three classifications of musical instalments
(ii) Give one example each of the classifications stated in (c)(i).
(d) Calculate the critical angle for light traveling from glass to air. [refractive index of glass = 1.5].
(e) The speed of sound in a medium at a temperature of 102 °C is 240 m s?1 . If the speed of sound in the medium is 3 10 m s?1 . Calculate its temperature
(a)
(i) One necessary condition for constructive interference of waves is that the waves have the same frequency, and the crests and troughs of the waves align with each other.
(ii) One necessary condition for total internal reflection is that the angle of incidence is greater than the critical angle for the boundary between two media, and the wave travels from a denser medium to a less dense medium.
(iii) One necessary condition for the production of beats is that two waves with slightly different frequencies interfere with each other, and their amplitudes vary periodically in time.
(b) The velocity of sound in air can be calculated as follows:
- The distance between the first and second position of resonance is 100 cm - 35 cm = 65 cm = 0.65 m.
- The wavelength of the sound wave is twice the distance between the first and second position of resonance, which is 2 x 0.65 m = 1.3 m.
- The frequency of the tuning fork is 256 Hz. - Using the equation v = fλ, where v is the velocity of sound, f is the frequency, and λ is the wavelength, we can calculate the velocity of sound as
v = 256 Hz x 1.3 m = 332.8 m/s.
(c) (i) The three classifications of musical instruments are:
- Stringed instruments
- Wind instruments
- Percussion instruments
(ii) Examples of each classification are:
- Stringed instruments: guitar, violin
- Wind instruments: flute, trumpet
- Percussion instruments: drums, xylophone
(d) The critical angle for light traveling from glass to air can be calculated as follows:
- The refractive index of glass is given as 1.5.
- Using the formula sin ?c = 1/n, where ?c is the critical angle and n is the refractive index, we can calculate the critical angle as sin θc = 1/1.5 = 0.67.
- Taking the inverse sine of 0.67, we can find the critical angle as θc = 42.3 degrees.
(e) The speed of sound in a medium is directly proportional to the square root of the temperature of the medium. Using this relationship, we can calculate the temperature of the medium as follows:
- Let T1 be the temperature of the medium where the speed of sound is 240 m/s, and T2 be the temperature of the medium where the speed of sound is 310 m/s.
- The ratio of the speeds of sound is 310/240 = 1.29.
- The ratio of the square roots of the temperatures is √(T2/T1) = 1.29. - Solving for T2, we get T2 = T1 x (1.29)^2 = T1 x 1.6641. - Substituting T1 = 102 + 273 = 375 K, we get T2 = 625 K. -
Therefore, the temperature of the medium where the speed of sound is 310 m/s is 625 - 273 = 352 °C.
Akọwa Nkọwa
(a)
(i) One necessary condition for constructive interference of waves is that the waves have the same frequency, and the crests and troughs of the waves align with each other.
(ii) One necessary condition for total internal reflection is that the angle of incidence is greater than the critical angle for the boundary between two media, and the wave travels from a denser medium to a less dense medium.
(iii) One necessary condition for the production of beats is that two waves with slightly different frequencies interfere with each other, and their amplitudes vary periodically in time.
(b) The velocity of sound in air can be calculated as follows:
- The distance between the first and second position of resonance is 100 cm - 35 cm = 65 cm = 0.65 m.
- The wavelength of the sound wave is twice the distance between the first and second position of resonance, which is 2 x 0.65 m = 1.3 m.
- The frequency of the tuning fork is 256 Hz. - Using the equation v = fλ, where v is the velocity of sound, f is the frequency, and λ is the wavelength, we can calculate the velocity of sound as
v = 256 Hz x 1.3 m = 332.8 m/s.
(c) (i) The three classifications of musical instruments are:
- Stringed instruments
- Wind instruments
- Percussion instruments
(ii) Examples of each classification are:
- Stringed instruments: guitar, violin
- Wind instruments: flute, trumpet
- Percussion instruments: drums, xylophone
(d) The critical angle for light traveling from glass to air can be calculated as follows:
- The refractive index of glass is given as 1.5.
- Using the formula sin ?c = 1/n, where ?c is the critical angle and n is the refractive index, we can calculate the critical angle as sin θc = 1/1.5 = 0.67.
- Taking the inverse sine of 0.67, we can find the critical angle as θc = 42.3 degrees.
(e) The speed of sound in a medium is directly proportional to the square root of the temperature of the medium. Using this relationship, we can calculate the temperature of the medium as follows:
- Let T1 be the temperature of the medium where the speed of sound is 240 m/s, and T2 be the temperature of the medium where the speed of sound is 310 m/s.
- The ratio of the speeds of sound is 310/240 = 1.29.
- The ratio of the square roots of the temperatures is √(T2/T1) = 1.29. - Solving for T2, we get T2 = T1 x (1.29)^2 = T1 x 1.6641. - Substituting T1 = 102 + 273 = 375 K, we get T2 = 625 K. -
Therefore, the temperature of the medium where the speed of sound is 310 m/s is 625 - 273 = 352 °C.
Ajụjụ 61 Ripọtì
You are provided with a loaded boiling tube with a centimeter scale fixed inside it, a transparent vessel filled with water, standard masses 2 g, 5g and 10 g, and a slide vernier caliper. Use the diagram above as a guide to perform the experiment.
(i) Use the slide vernier caliper to measure and record the external diameter, D, of the boiling tube.
(ii) Evaluated A = 0.25? D2 , where ? = 3.14.
(iii) Place the loaded boiling tube gently in the water in the transparent vessel such that it floats vertically.
(iv) Read and record the depth of immersion, y, from the zero mark of the scale fixed inside, the boiling tube.
(v) Add a mass, m = 2g, to the boiling tube. Read and record the new depth of immersion, y, from the zero mark of the scale.
(vi) Evaluate h = (y - y0 ), log h and log m.
(vii) Repeat the experiment for four other values of m = 5g, 7g, 10 g, and 12g. In each case, record y and evaluate h, log h, and log m.
(viii) Tabulate the results.
(ix) Plot a graph with log m on the vertical axis and log h on the horizontal axis starting both axes from the origin (0,0).
(b)(i) State in full the law on which the experiment in (a) is based.
(ii) A uniform cylindrical rod is 0.63 m long and it has a cross-sectional area of 0.1 m2 . Calculate the depth of immersion of the rod if it floats vertically in a liquid of relative density 1.26. [density of rod =720 kg m?3 , g = 10 m s?2 ].
(a) OBSERVATIONS
(i) Value of D correctly measure and recorded to at least 2 d.p in cm
(ii) Value of A = 0.25π D2 correctly evaluated
(iii) Value of y0 correctly read and recorded to at least 1 d.p in cm.
(iv) Five values of m correctly recorded in grammes.
(v) Five values of y correctly read and recorded to at least 1 d.p in cm and in trend.
(vi) Five values of h = (y - yo) correctly evaluated.
(vii) Five values of log h correctly evaluated to at least 3 d.p
(viii) Five values of log m correctly evaluated to 3d.
(ix) Composite table showing at least m, y, h, log h and log m.
(b)(i) The Law of Flotation;
A floating body displaces its own weight of fluid in which it floats.
(ii) Weight of rod = 0.63 x 0.1 x 720 x 10 = 453.6
Let y be the depth of immersion
∴ Weight of water displaced
= (0.1y) x 1.26 x 1000 x 10
1260y = 453.6 y=453.6=9.36
= 1260y = 453.6
y = 453.61260 = 9.36
Akọwa Nkọwa
(a) OBSERVATIONS
(i) Value of D correctly measure and recorded to at least 2 d.p in cm
(ii) Value of A = 0.25π D2 correctly evaluated
(iii) Value of y0 correctly read and recorded to at least 1 d.p in cm.
(iv) Five values of m correctly recorded in grammes.
(v) Five values of y correctly read and recorded to at least 1 d.p in cm and in trend.
(vi) Five values of h = (y - yo) correctly evaluated.
(vii) Five values of log h correctly evaluated to at least 3 d.p
(viii) Five values of log m correctly evaluated to 3d.
(ix) Composite table showing at least m, y, h, log h and log m.
(b)(i) The Law of Flotation;
A floating body displaces its own weight of fluid in which it floats.
(ii) Weight of rod = 0.63 x 0.1 x 720 x 10 = 453.6
Let y be the depth of immersion
∴ Weight of water displaced
= (0.1y) x 1.26 x 1000 x 10
1260y = 453.6 y=453.6=9.36
= 1260y = 453.6
y = 453.61260 = 9.36
Ajụjụ 62 Ripọtì
(a) What is a geostationary satellite?
(b) Name two types of Lasers.
(a) A geostationary satellite is an artificial satellite that orbits the Earth at the same rate as the Earth's rotation. This means that it stays fixed at a particular point in the sky over the Earth's equator. From the ground, it appears as though the satellite is stationary. (b) Two types of lasers are solid-state lasers and gas lasers. Solid-state lasers use a solid material as the active medium, while gas lasers use a gas or a mixture of gases as the active medium. Solid-state lasers are commonly used in industry, medicine, and research, while gas lasers are used for applications such as cutting, welding, and laser printers.
Akọwa Nkọwa
(a) A geostationary satellite is an artificial satellite that orbits the Earth at the same rate as the Earth's rotation. This means that it stays fixed at a particular point in the sky over the Earth's equator. From the ground, it appears as though the satellite is stationary. (b) Two types of lasers are solid-state lasers and gas lasers. Solid-state lasers use a solid material as the active medium, while gas lasers use a gas or a mixture of gases as the active medium. Solid-state lasers are commonly used in industry, medicine, and research, while gas lasers are used for applications such as cutting, welding, and laser printers.
Ajụjụ 63 Ripọtì
State:
(a) The S.I. unit of the intensity of a blackbody radiation.
(b) Two features of the intensity-wavelength graph of a perfect blackbody at different temperatures.
(a) S.l unite for the intensity of a blackbody radiation = Wm-² OR Js−1 m-²
(b) Two features of the graph
i. The shape of the graph is non-symmetric.
ii. Peaks of the intensity have lower wavelengths/Peaks decrease with wavelength.
iii. The line joining the peaks have a negative gradient.
Akọwa Nkọwa
(a) S.l unite for the intensity of a blackbody radiation = Wm-² OR Js−1 m-²
(b) Two features of the graph
i. The shape of the graph is non-symmetric.
ii. Peaks of the intensity have lower wavelengths/Peaks decrease with wavelength.
iii. The line joining the peaks have a negative gradient.
Ajụjụ 64 Ripọtì
State three differences between magnetic and non-magnetic materials.
Magnetic and non-magnetic materials are two distinct categories of materials. Here are three differences between these two: 1. Magnetic materials are attracted to a magnet, while non-magnetic materials are not. This means that if you hold a magnet near a magnetic material, it will be pulled towards the magnet, but the same won't happen with a non-magnetic material. 2. Magnetic materials can be magnetized, whereas non-magnetic materials cannot. Magnetization is the process of aligning the magnetic domains of a material, and this is only possible with magnetic materials. 3. Magnetic materials have magnetic domains, while non-magnetic materials do not. Magnetic domains are regions within a material where the magnetic moments of the atoms are aligned in the same direction, creating a magnetic field. In non-magnetic materials, the magnetic moments are randomly oriented, resulting in no net magnetic field. These are the three main differences between magnetic and non-magnetic materials.
Akọwa Nkọwa
Magnetic and non-magnetic materials are two distinct categories of materials. Here are three differences between these two: 1. Magnetic materials are attracted to a magnet, while non-magnetic materials are not. This means that if you hold a magnet near a magnetic material, it will be pulled towards the magnet, but the same won't happen with a non-magnetic material. 2. Magnetic materials can be magnetized, whereas non-magnetic materials cannot. Magnetization is the process of aligning the magnetic domains of a material, and this is only possible with magnetic materials. 3. Magnetic materials have magnetic domains, while non-magnetic materials do not. Magnetic domains are regions within a material where the magnetic moments of the atoms are aligned in the same direction, creating a magnetic field. In non-magnetic materials, the magnetic moments are randomly oriented, resulting in no net magnetic field. These are the three main differences between magnetic and non-magnetic materials.
Ajụjụ 65 Ripọtì
You are provided with a loaded boiling tube with a centimeter scale fixed inside it, a transparent vessel filled with water, standard masses 2 g, 5g and 10 g, and a slide vernier caliper. Use the diagram above as a guide to perform the experiment.
(i) Use the slide vernier caliper to measure and record the external diameter, D, of the boiling tube.
(ii) Evaluated A = 0.25? D2 , where ? = 3.14.
(iii) Place the loaded boiling tube gently in the water in the transparent vessel such that it floats vertically.
(iv) Read and record the depth of immersion, y, from the zero mark of the scale fixed inside, the boiling tube.
(v) Add a mass, m = 2g, to the boiling tube. Read and record the new depth of immersion, y, from the zero mark of the scale.
(vi) Evaluate h = (y - y0 ), log h and log m.
(vii) Repeat the experiment for four other values of m = 5g, 7g, 10 g, and 12g. In each case, record y and evaluate h, log h, and log m.
(viii) Tabulate the results.
(ix) Plot a graph with log m on the vertical axis and log h on the horizontal axis starting both axes from the origin (0,0).
(b)(i) State in full the law on which the experiment in (a) is based.
(ii) A uniform cylindrical rod is 0.63 m long and it has a cross-sectional area of 0.1 m2 . Calculate the depth of immersion of the rod if it floats vertically in a liquid of relative density 1.26. [density of rod =720 kg m?3 , g = 10 m s?2 ].
The law on which this experiment is based is the Archimedes' principle, which states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces. The buoyant force arises because of the difference in pressure acting on the top and bottom surfaces of the immersed object. This principle helps in determining the density of the liquid in which the object is immersed.
To calculate the depth of immersion of a uniform cylindrical rod, we can use the Archimedes' principle by considering the weight of the liquid displaced by the rod, which is equal to the weight of the rod. Given the length, cross-sectional area, density of the rod, and the relative density of the liquid, we can determine the depth of immersion of the rod.
(a) OBSERVATIONS
(i) Value of D correctly measure and recorded to at least 2 d.p in cm
(ii) Value of A = 0.25? D2 correctly evaluated
(iii) Value of y0 correctly read and recorded to at least 1 d.p in cm.
(iv) Five values of m correctly recorded in grammes.
(v) Five values of y correctly read and recorded to at least 1 d.p in cm and in trend.
(vi) Five values of h = (y - yo) correctly evaluated.
(vii) Five values of log h correctly evaluated to at least 3 d.p
(viii) Five values of log m correctly evaluated to 3d.
(ix) Composite table showing at least m, y, h, log h and log m.
(b)(i) The Law of Flotation;
A floating body displaces its own weight of fluid in which it floats.
(ii) Weight of rod = 0.63 x 0.1 x 720 x 10 = 453.6
Let y be the depth of immersion
? Weight of water displaced
= (0.1y) x 1.26 x 1000 x 10
1260y = 453.6 y=453.6=9.36
= 1260y = 453.6
y = 453.61260 = 9.36
Akọwa Nkọwa
The law on which this experiment is based is the Archimedes' principle, which states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces. The buoyant force arises because of the difference in pressure acting on the top and bottom surfaces of the immersed object. This principle helps in determining the density of the liquid in which the object is immersed.
To calculate the depth of immersion of a uniform cylindrical rod, we can use the Archimedes' principle by considering the weight of the liquid displaced by the rod, which is equal to the weight of the rod. Given the length, cross-sectional area, density of the rod, and the relative density of the liquid, we can determine the depth of immersion of the rod.
(a) OBSERVATIONS
(i) Value of D correctly measure and recorded to at least 2 d.p in cm
(ii) Value of A = 0.25? D2 correctly evaluated
(iii) Value of y0 correctly read and recorded to at least 1 d.p in cm.
(iv) Five values of m correctly recorded in grammes.
(v) Five values of y correctly read and recorded to at least 1 d.p in cm and in trend.
(vi) Five values of h = (y - yo) correctly evaluated.
(vii) Five values of log h correctly evaluated to at least 3 d.p
(viii) Five values of log m correctly evaluated to 3d.
(ix) Composite table showing at least m, y, h, log h and log m.
(b)(i) The Law of Flotation;
A floating body displaces its own weight of fluid in which it floats.
(ii) Weight of rod = 0.63 x 0.1 x 720 x 10 = 453.6
Let y be the depth of immersion
? Weight of water displaced
= (0.1y) x 1.26 x 1000 x 10
1260y = 453.6 y=453.6=9.36
= 1260y = 453.6
y = 453.61260 = 9.36
