WAEC SSCE - Physics - 1998
Ajụjụ 1 Ripọtì
which of the following pairs of physical quantities comprises vectors?
Akọwa Nkọwa
The physical quantities that have both magnitude and direction are called vectors. Among the given options, the pair of physical quantities that comprise vectors are "force and velocity". Both force and velocity have a direction and magnitude associated with them. The direction of force is along the line of action of the force and the direction of velocity is the direction of motion. Therefore, the correct answer is "force and velocity".
Ajụjụ 2 Ripọtì
How much heat is emitted when a body of mass 200g cools from 37oC to 31oC? (specific heat capacity of the body = 0.4Jg-1K-1,)
Akọwa Nkọwa
Ajụjụ 3 Ripọtì
The lower and upper fixed points of a mecury-in-glass thermometer are marked Y and 180mm respectively. On a particular the mecury meniscus in the thermometer rises to 45mm. If the corresponding reading on a scale is 10o, calculate the value of Y.
Ajụjụ 4 Ripọtì
It is always not possible to determine exactly and simultaneously the position and momentum of a particle. This statement is known as the
Akọwa Nkọwa
Heisenberg's uncertainty principle states that it is impossible to simultaneously determine the exact position and momentum of a particle. This principle is a fundamental concept in quantum mechanics, and it means that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa. The principle applies to all particles, including subatomic particles like electrons and photons. The uncertainty principle has important implications for the behavior of quantum systems and the interpretation of quantum mechanics. It is named after Werner Heisenberg, a German physicist who first described the principle in 1927.
Ajụjụ 6 Ripọtì
Two radioactive elements X and Y have half-lives of 100 and 50 years respectively. Samples of X and Y initially contain equal number of atoms. What is the ratio of the number of the remaining atoms of X to that of Y after 200 years?
Akọwa Nkọwa
Ajụjụ 7 Ripọtì
An object is protected with a velocity of 100m-1 at an angle of 60o to the vertical. Calculate the time taken by the object to reach the highest point (Take g as 10ms-2
Akọwa Nkọwa
The motion of the object can be divided into two components, horizontal and vertical. The horizontal component of the velocity will remain constant, while the vertical component will change due to the acceleration due to gravity acting in the downward direction. We can use the vertical component of the velocity to determine the time taken by the object to reach the highest point. The initial vertical velocity of the object can be calculated using the formula: v0sinθ = (100m/s)sin60o = 86.6m/s where v0 is the initial velocity, and θ is the angle of the velocity vector with the horizontal. At the highest point, the vertical component of the velocity will be zero. Using the formula: v = v0 + at where v is the final velocity, a is the acceleration due to gravity (-10m/s2), and t is the time taken to reach the highest point, we can calculate the time taken as follows: 0 = 86.6 - 10t t = 8.66s Therefore, the time taken by the object to reach the highest point is approximately 8.7 seconds. The closest option to this answer is (b) 8.7s.
Ajụjụ 9 Ripọtì
What is the image distance of an object placed at a distance of 2f from a converging lens of focal length f?
Akọwa Nkọwa
The image distance of an object placed at a distance of 2f from a converging lens of focal length f can be found using the lens formula: 1/f = 1/v - 1/u where f is the focal length of the lens, v is the image distance, and u is the object distance. Since the object is placed at a distance of 2f, u = 2f. Substituting these values in the lens formula, we get: 1/f = 1/v - 1/2f Multiplying both sides by 2f, we get: 2 - f/f = v Simplifying this equation, we get: v = 2f Therefore, the image distance of an object placed at a distance of 2f from a converging lens of focal length f is 2f.
Ajụjụ 10 Ripọtì
A proton of charge 1.6 x 1019C is projected into a uniform magnetic field of flux density with a constant speed of 1.6 x 106ms-1, calculate the magnitude of the force exerted on it by the field
Akọwa Nkọwa
Ajụjụ 11 Ripọtì
I the box is pulled from P to Q, which of the following expresses the velocity ratio of the inclined plane?
Akọwa Nkọwa
Ajụjụ 12 Ripọtì
The diagram below illustrates an a.c.source of 50V(r.m.s.), \(\frac{100}{\pi}\)Hz connected in series with an inductor of inductor of inductance L and a resistor of resistance R. The current in the circuit is 2A and the p.d across L and R are 30V and 40V respectively. Calculate the power factor of the circuit
Akọwa Nkọwa
Ajụjụ 13 Ripọtì
The images formed by diverging lenses are always
Akọwa Nkọwa
Diverging lenses are also known as concave lenses. These lenses are thinner at the centre and thicker at the edges. The images formed by diverging lenses are always diminished, virtual and erect. Diminished means that the image is smaller than the object. Virtual means that the image appears to be behind the lens and cannot be projected on a screen. Erect means that the image appears to be in the same orientation as the object. The reason for these characteristics is that the diverging lens spreads out the light rays that enter it, causing them to diverge or move apart. This creates an image that is smaller and appears to be located behind the lens.
Ajụjụ 14 Ripọtì
The intensity of light falling on the film in a camera depend on the i. brightness of the object. ii. diameter of the aperture stop. iii. speed of the shutter. Which of the statements is/are correct/
Akọwa Nkọwa
Ajụjụ 15 Ripọtì
A piece of metal is heated until it becomes red-hot, it is then quickly transferred into a beaker containing boiling water. What effect will this have on the boiling water?
Akọwa Nkọwa
Ajụjụ 16 Ripọtì
A cell of e.m.f. 1.5v is connected in series with a resistor of 3\(\Omega\). A high resistance voltmeter connected across the cell registers only 0.9v. Calculate the internal resistance of the cell
Akọwa Nkọwa
Ajụjụ 18 Ripọtì
The process of increasing the energy of an atom via inelastic collision with an electron is known as
Akọwa Nkọwa
The process of increasing the energy of an atom via inelastic collision with an electron is known as excitation. When an electron collides with an atom, it can transfer some of its energy to the atom. This results in the atom's electrons jumping to higher energy levels, or excited states. Excitation can occur through various processes, such as collisions with other particles or the absorption of light. In this case, the excitation occurs through an inelastic collision with an electron. The energy transferred to the atom can be used for various purposes, such as to emit light or to initiate chemical reactions.
Ajụjụ 19 Ripọtì
An astronomical telescope having an objective of focal length 100cm and an eyepiece of focal length 10cm is used in normal adjustment, calculate the separation of the lenses
Akọwa Nkọwa
The magnifying power of a telescope in normal adjustment is given by the ratio of the focal length of the objective to that of the eyepiece: Magnifying power, M = fo/fe where fo is the focal length of the objective and fe is the focal length of the eyepiece. The separation of the lenses in a telescope is the sum of the focal lengths of the lenses: separation of lenses = fo + fe Given that fo = 100cm and fe = 10cm, we can find the magnifying power and the separation of the lenses: M = fo/fe = 100cm/10cm = 10 separation of lenses = fo + fe = 100cm + 10cm = 110cm = 1.10m Therefore, the separation of the lenses is 1.10m. Answer:.
Ajụjụ 20 Ripọtì
Which of the following explains the concave meniscus of water in a clean glass tube? the
Akọwa Nkọwa
The concave meniscus of water in a clean glass tube is due to the adhesive forces between the water molecules and the glass tube being greater than the cohesive forces between the water molecules themselves. This causes the water to be attracted to the glass, resulting in the formation of a concave meniscus. Therefore, the correct option is "adhesion between water and glass molecules is greater than the cohesion between water molecules."
Ajụjụ 21 Ripọtì
An eclipse of the sun by the moon occurs when the sun, the moon and the earth are in a straight line and the
Akọwa Nkọwa
An eclipse of the sun by the moon occurs when the sun, the moon and the earth are in a straight line and the moon is between the sun and the earth. The moon casts its shadow on the earth, blocking the light from the sun, and this causes a temporary darkness on the earth. This phenomenon is known as a solar eclipse, and it occurs when the new moon comes between the sun and the earth in its monthly orbit.
Ajụjụ 23 Ripọtì
A rectangular block of wood floats in water with two-thirds of its volume immersed. When placed in another liquid, it floats with half of its volume immersed. Calculate the relative density of the liquid
Akọwa Nkọwa
When an object is immersed in a liquid, it displaces some volume of the liquid. According to Archimedes' principle, the weight of the liquid displaced is equal to the weight of the object. Since the object floats, the weight of the liquid displaced must be equal to the weight of the object. Let V be the volume of the rectangular block of wood and W its weight. Also, let the density of water be ρ1 and the density of the other liquid be ρ2. When the block is immersed in water, two-thirds of its volume is immersed. Therefore, the volume of water displaced is two-thirds of the volume of the block, which is 2V/3. According to Archimedes' principle, the weight of the water displaced is equal to the weight of the block, so: ρ1 x (2V/3) x g = W where g is the acceleration due to gravity. Rearranging, we get: ρ1 = W / [(2V/3) x g] Similarly, when the block is immersed in the other liquid, half of its volume is immersed. Therefore, the volume of the other liquid displaced is half of the volume of the block, which is V/2. According to Archimedes' principle, the weight of the other liquid displaced is equal to the weight of the block, so: ρ2 x (V/2) x g = W where g is the acceleration due to gravity. Rearranging, we get: ρ2 = W / [(V/2) x g] Dividing the second equation by the first equation gives: ρ2/ρ1 = [(V/2) x g] / [(2V/3) x g] Simplifying, we get: ρ2/ρ1 = 3/4 Therefore, the relative density of the liquid is 0.75. So the correct option is (E) 0.75.
Ajụjụ 24 Ripọtì
Air at temperature 527oC and pressure 30 atmospheres is admitted into the cylinder of an engine. Calculate the pressure of the gas when it has expanded to 5 times its volume and cooled to 127oC as it leaves the engine
Ajụjụ 25 Ripọtì
In the electrical method of magnetisation, the polarity of the magnet depends on the
Akọwa Nkọwa
The electrical method of magnetisation involves passing an electric current through a magnetic material to create a magnetic field. The polarity of the magnet that is produced depends on the direction of the current. The direction of the current can be determined by the direction of the flow of electrons in the wire, which is usually indicated by an arrow in a circuit diagram. Therefore, the correct answer is "direction of current."
Ajụjụ 26 Ripọtì
The direction of the magnetic field at a point in the vicinity of a bar magnetic is
Akọwa Nkọwa
Ajụjụ 27 Ripọtì
An object is heated from 30oC to 57oC. The increases in its temperature on the kelvin scale is
Ajụjụ 28 Ripọtì
A sonometer wire under a tension of N10N, produces a frequency of 250Hz when plucked. Keeping the length of the wire constant, the tension is adjusted to produce a new frequency of 350Hz. Calculate the new tension
Akọwa Nkọwa
The relationship between tension (T) and frequency (f) for a string fixed at both ends, such as a sonometer wire, is given by the equation: f = (1/2L)√(T/μ) where L is the length of the string, μ is the linear mass density of the string. Since the length of the wire is kept constant, we can equate the two expressions for frequency to find the relationship between the two tensions: (1/2L)√(T1/μ) = f1 = 250 Hz (1/2L)√(T2/μ) = f2 = 350 Hz Dividing the two equations gives: (f2/f1) = √(T2/T1) (350/250) = √(T2/N10N) Squaring both sides of the equation gives: (350/250)^2 = T2/N10N Solving for T2, we get: T2 = (350/250)^2 x N10N T2 = 19.6 N (to 2 significant figures) Therefore, the new tension required to produce a frequency of 350 Hz is 19.6 N. Answer option (B) is correct.
Ajụjụ 29 Ripọtì
If the cubic expansivity of brass between 27oC and 100oC is 5.7 x 10-5K-1, What is its linear expansivity?
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
Dry oxygen is trapped by a pellet of mercury in a uniform capillary tube which is sealed at one end. The length of the column of oxygen at 27oC is 50cm. if the pressure of the oxygen is constant, at what temperature will the length be 60cm?
Akọwa Nkọwa
This is a question about the relationship between the temperature and volume of a gas when pressure is held constant, known as Charles' Law. According to Charles' Law, the volume of a gas is directly proportional to its temperature when the pressure is constant. In this question, the length of the column of oxygen in the capillary tube is directly proportional to its volume. This means that if we increase the temperature of the oxygen while keeping the pressure constant, the length of the column of oxygen will increase proportionally. We can use the formula for Charles' Law, V1/T1 = V2/T2, to solve for the final temperature T2. Let V1 be the initial volume of the oxygen when the length of the column is 50cm, T1 be the initial temperature of the oxygen, V2 be the final volume of the oxygen when the length of the column is 60cm, and T2 be the final temperature of the oxygen that we want to find. We know that V1 = V2 since the pressure is constant, and we can substitute the given values into the formula to get: V1/T1 = V2/T2 V1 = 50 cm V2 = 60 cm T1 = 27°C + 273.15 = 300.15 K Solving for T2, we get: T2 = (V2/V1) x T1 = (60/50) x 300.15 = 360.18 K Converting back to Celsius, we get: T2 = 360.18 K - 273.15 = 87.03°C Therefore, the answer is option (D) 87.0oC.
Ajụjụ 31 Ripọtì
The average kinetic energy of the molecules of a perfect gas is directly proportional to the
Akọwa Nkọwa
Ajụjụ 32 Ripọtì
The diagram above illustrates a freely suspended bar magnet NS hanging from a point in a horizontal ceiling. The thread used for suspending the magnet and the axis of the magnet are as indicated on the diagram. The angle marked \(\theta\) is called the angle os
Akọwa Nkọwa
The angle marked theta in the diagram is called the "dip" of the magnet. The dip is the angle between the magnetic field lines of the Earth and a horizontal plane at the location of the magnet. The dip angle is important in navigation, especially for compasses, because it helps determine the latitude of the location.
Ajụjụ 33 Ripọtì
A lantern gives an image of 3m square of a slide 7.62cm square on a screen. If the screen is 10m from the projection lens of the lantern, calculate the focal length of the lens
Akọwa Nkọwa
Ajụjụ 34 Ripọtì
Calculate he wavelength of a note which is one octave lower than a note of 256Hz in a medium in which the speed of sound is 352ms-1
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
A converging lens of focal length 5cm is used as a magnifying glass by a man whose nearpoint is 35cm. Calculate the magnification given by the lens
Ajụjụ 36 Ripọtì
Use the following data to determine the length L of a wire when a fore of 30N is applied, assuming Hooke's law is obeyed
\(\begin{array}{c|c}\text{Force applied/N} & 0 & 5 & 10 & 30\\ \hline \text{length of wire/mm} & 500.0 & 500.0 & 501.0 & L \end{array}\)
Akọwa Nkọwa
Hooke's Law states that the force applied to a spring is directly proportional to the extension or compression of the spring, provided the elastic limit is not exceeded. This can be expressed mathematically as F=kx, where F is the force, k is the spring constant, and x is the extension or compression. To determine the length L of the wire when a force of 30N is applied, we need to first find the spring constant k. We can do this by calculating the extension x for each force applied, using the formula: x = (L - L0) where L0 is the original length of the wire (500.0 mm). From the data given, we can calculate the extensions as follows: For 0 N: x = (500.0 - 500.0) = 0 mm For 5 N: x = (500.0 - 500.0) = 0 mm For 10 N: x = (501.0 - 500.0) = 1.0 mm Using Hooke's Law, we can find the spring constant k for the wire: k = F/x = 10 N / 1.0 mm = 10 N/mm Now we can use Hooke's Law again to find the extension x for a force of 30 N: 30 N = 10 N/mm * x x = 3.0 mm Finally, we can find the length L of the wire when a force of 30 N is applied: L = L0 + x = 500.0 mm + 3.0 mm = 503.0 mm Therefore, the length of the wire when a force of 30N is applied is 503.0 mm. Answer is the correct answer.
Ajụjụ 37 Ripọtì
The gravitational force of the moon is one-sixth that of the earth. If a body weighs 6.0N on the moon. Calculate its weight on the earth
Akọwa Nkọwa
Weight is the force experienced by a body due to gravity. It is given by the product of mass and gravitational acceleration, which is 9.8m/s2 on earth and 1.6m/s2 on the moon. Let's assume the mass of the body to be m kg. On the moon, weight, Wm = mgm = 6.0N, where gm = 1.6m/s2 is the gravitational acceleration on the moon. On the earth, weight, We = mge, where ge = 9.8m/s2 is the gravitational acceleration on the earth. We are required to find We, given that the gravitational force of the moon is one-sixth that of the earth, which means the gravitational acceleration on the moon is one-sixth that of the earth. So, gm = (1/6)ge Substituting this value of gm in the equation for weight on the moon, we get: Wm = mgm = m(1/6)ge = (1/6)(mge) Hence, weight of the body on earth, We = 6Wm = 6*(1/6)(mge) = mge Therefore, We = 6.0N * 9.8m/s2 = 58.8N Hence, the weight of the body on earth is 58.8N. Therefore, the answer is 36.0N.
Ajụjụ 38 Ripọtì
A radioactive nuclide of proton number X emits a \(\beta\)-particle to form a new nuclide to proton number Y. The correct equation relating X and Y is
Ajụjụ 39 Ripọtì
When light from a source is sent through a gas. i. certain wavelengths are absorbed. ii. a dark line is left. Which of the following is/are correct?
Ajụjụ 40 Ripọtì
The inside of a vacuum flak is usually coated with silver to reduce heat loss by
Akọwa Nkọwa
The correct answer is **radiation**. Heat can be transferred through vacuum by radiation, which is the emission of electromagnetic waves from a body due to its temperature. A vacuum flask consists of two walls of glass with a vacuum in between them, and this vacuum is a poor conductor of heat, which reduces heat loss by conduction and convection. However, the inner wall of a vacuum flask can still lose heat by radiation. Hence, to reduce heat loss by radiation, the inside of a vacuum flask is usually coated with a reflective surface, typically silver, which reflects heat back into the flask, keeping the contents warmer for longer.
Ajụjụ 41 Ripọtì
A piece of brass of mass 170kg has its temperature raised from 0oC to 30oC. Calculate its increase in volume, given the density of brass at 0oC as 8.5 x 103kgm3 and its cubic expansivity as 5.7 x 10-1k1
Akọwa Nkọwa
When a material is heated, it expands. The increase in volume can be calculated using the following formula: ΔV = VαΔT Where ΔV is the change in volume, V is the original volume, α is the coefficient of linear or cubic expansivity depending on the type of expansion, and ΔT is the change in temperature. For a cubic expansivity, the formula can be written as: ΔV = VβΔT Where Vβ is the coefficient of cubic expansivity. Using the given values, the increase in volume of the brass can be calculated as: ΔV = VβΔT = Vβ(Tf - Ti) Where Tf is the final temperature and Ti is the initial temperature. The initial volume of the brass can be calculated using its mass and density: V = m/ρ = 170/8.5 x 10^3 = 0.02 m^3 Substituting the values, we get: ΔV = Vβ(Tf - Ti) = 0.02 x 5.7 x 10^-3 x (30 - 0) = 3.42 x 10^-5 m^3 Therefore, the increase in volume of the brass is 3.42 x 10^-5 m^3. The correct option is (a) 3.4 x 10^-5 m^3.
Ajụjụ 42 Ripọtì
Calculate the resistance of the filament of lamp rated at 240V 60W
Akọwa Nkọwa
The power rating of the lamp is given as 60W, and the voltage rating is given as 240V. We can use the formula for power, which is P = V^2/R, to calculate the resistance of the filament. First, we rearrange the formula to solve for resistance R: R = V^2 / P Plugging in the given values, we get: R = (240V)^2 / 60W Simplifying, we get: R = 960\(\Omega\) Therefore, the resistance of the filament of the lamp is 960\(\Omega\).
Ajụjụ 43 Ripọtì
a pilot records the atmospheric pressure outside his plane as 63cm of Hg while a ground observer records a reading of 75cm of Hg with his barometer. Assuming that the density of air is constant, calculate the height of the plane above the ground. (take the relative densities of air and mecury as 0.00136 and 13.6 respectively)
Akọwa Nkọwa
When the pilot recorded the atmospheric pressure outside the plane as 63cm of Hg, it means that the pressure due to the atmosphere is balancing the pressure exerted by a column of mercury 63cm high in a barometer tube. Likewise, when the ground observer recorded a reading of 75cm of Hg, it means that the pressure due to the atmosphere is balancing the pressure exerted by a column of mercury 75cm high in a barometer tube. The pressure due to the atmosphere at any point is proportional to the height of the column of air above that point. This implies that the difference in the atmospheric pressure between the pilot's location and the ground observer's location is directly proportional to the height difference between the two points. Using the given data, the pressure difference between the two points is 75 - 63 = 12 cm of Hg. Let h be the height of the plane above the ground, then by the principle of hydrostatics, the pressure difference is also given by: h = (pressure difference) x (density of mercury/density of air) x (gravitational acceleration) Substituting the given values and solving for h, we have: h = 12 x (13.6/0.00136) x (9.81) = 1200m Therefore, the height of the plane above the ground is 1200m. Answer: 1200m
Ajụjụ 44 Ripọtì
An electromagnetic wave of frequency 5.0 x 1014Hz, is incident on the surface of water of refractive index \(\frac{4}{3}\). Taking the speed of the wave in air as 3.0 x 108ms-1, calculate the wavelength of the wave in water.
Akọwa Nkọwa
The formula for calculating the wavelength of a wave is given by: \begin{equation*} \lambda = \frac{v}{f} \end{equation*} where - \(\lambda\) is the wavelength of the wave - v is the velocity of the wave in the medium - f is the frequency of the wave. The velocity of the electromagnetic wave in air is given as 3.0 x 108ms-1. The refractive index of water is given as \(\frac{4}{3}\). The refractive index is the ratio of the speed of light in a vacuum to the speed of light in a medium. Hence the speed of the wave in water is given as: \begin{equation*} v_{water} = \frac{v_{air}}{n} = \frac{3.0\times10^8}{4/3} = 2.25\times10^8ms^{-1} \end{equation*} Substituting the values of v and f in the formula above, we have: \begin{equation*} \lambda = \frac{v}{f} = \frac{2.25\times10^8}{5.0\times10^{14}} = 4.5\times10^{-7}m \end{equation*} Therefore, the wavelength of the electromagnetic wave in water is 4.5 x 10-7m. Hence the correct option is: 4.5 x 10-7m.
Ajụjụ 45 Ripọtì
Lenz's law of electromagnet states that
Akọwa Nkọwa
Lenz's law of electromagnetism states that the direction of an induced current in a conductor is always such as to oppose the change that produced it. In other words, when there is a change in magnetic flux through a circuit, it produces an induced electromotive force (emf) which in turn generates a current. This induced current flows in such a direction that it opposes the change that produced it. This law is a consequence of the conservation of energy principle, as the induced current creates a magnetic field that opposes the change in the magnetic field that caused the current to be induced.
Ajụjụ 46 Ripọtì
A waterfall is 1.260m high. Calculate the change in temperature of a quantity of water that falls from the top to the bottom of the waterfall. (Neglect heat losses to the surroundings, take g as 10m-2 and specific heat capacity of water as 4200Jkg-1K-1)
Akọwa Nkọwa
Ajụjụ 47 Ripọtì
If an ebonite rod is rubbed with fur
Akọwa Nkọwa
When an ebonite rod is rubbed with fur, electrons are transferred from the fur to the ebonite rod. Electrons are negatively charged, so when electrons move from one object to another, it leaves the object with a positive charge, and the other with a negative charge. Therefore, the ebonite rod will be negatively charged while the fur will be positively charged. Thus, the correct answer is: "the ebonite rod will be negatively charged while the fur will be positively charged".
Ajụjụ 48 Ripọtì
If \(\theta\) is equal to \(\alpha\) at the time the box is just about to slide down the plank, the coefficient of static friction between the plank and box is
Akọwa Nkọwa
When the box is just about to slide down the plank, the force of static friction acting on the box is equal and opposite to the component of the weight of the box parallel to the surface of the plank. This component is given by mg sin\(\alpha\), where m is the mass of the box, g is the acceleration due to gravity and \(\alpha\) is the angle of inclination of the plank. Since \(\theta\) is equal to \(\alpha\), the angle of inclination of the plank is also \(\theta\). Therefore, the coefficient of static friction \(\mu\) is given by: \(\mu = \frac{\text{Force of friction}}{\text{Normal force}} = \frac{mg \sin \theta}{mg \cos \theta} = \tan \theta = \tan \alpha\) Hence, the correct option is tan\(\alpha\).
Ajụjụ 49 Ripọtì
As the end Q of the plank is raised, the component of W, normal to plank, will
Akọwa Nkọwa
As the end Q of the plank is raised, the component of W, normal to the plank will decrease. This is because the weight, W, acting vertically downwards on the plank can be resolved into two components: one perpendicular to the plank and the other parallel to the plank. The component perpendicular to the plank is Wcosθ, where θ is the angle between the weight vector and the normal to the plank. As Q is raised, the angle θ increases, causing the value of cosθ to decrease, hence decreasing the component of W perpendicular to the plank.
Ajụjụ 50 Ripọtì
A 90-W immersion heater is used to supply energy for five minutes. The energy supplied is used to completely melt 180g of a solid at is melting point. Neglecting energy losses to the surroundings, calculate the specific latent heat of fusion of the solid
Akọwa Nkọwa
Ajụjụ 51 Ripọtì
The phenomenon which occurs when light changes direction as it passes from one medium to another is called
Akọwa Nkọwa
The phenomenon which occurs when light changes direction as it passes from one medium to another is called refraction. Refraction is the bending of light as it passes from one medium to another, such as from air to water. The amount of bending depends on the angle at which the light strikes the surface, and the difference in the refractive indices of the two media. This phenomenon is the reason why objects can appear distorted when viewed through a curved surface like a lens, and it is also why a straw in a glass of water appears to bend at the water's surface.
Ajụjụ 52 Ripọtì
The diagram below illustrates an a.c.source of 50V(r.m.s.), \(\frac{100}{\pi}\)Hz connected in series with an inductor of inductor of inductance L and a resistor of resistance R. The current in the circuit is 2A and the p.d across L and R are 30V and 40V respectively. Calculate the average power dissipated in the circuit
Akọwa Nkọwa
Ajụjụ 53 Ripọtì
The frictional effect between the layers of a moving fluid is called
Akọwa Nkọwa
The frictional effect between the layers of a moving fluid is called viscosity. Viscosity is a measure of a fluid's resistance to flow, which arises due to the friction between the layers of the fluid as they move past each other. A highly viscous fluid has a thick and sticky consistency and resists flowing, while a less viscous fluid flows more easily. Viscosity is an important property of fluids, and it has many practical applications in areas such as engineering, chemistry, and biology.
Ajụjụ 54 Ripọtì
the shortest length of the air column in a resonance tube at resonance is 0.12m and the next resonance length is 0.37m. calculate the frequency of vibration given that the speed of sound in air is 340m-1
Akọwa Nkọwa
The difference between the first and second resonance lengths is equal to half the wavelength of the sound wave. Hence, the wavelength is: λ = 2 × (0.37m – 0.12m) = 0.5m The speed of sound is given as 340m/s. We can use the formula: v = fλ where v is the velocity of sound, f is the frequency of vibration, and λ is the wavelength of the sound wave. Substituting the values we get: 340 = f × 0.5 Solving for f, we get: f = 680 Hz Therefore, the frequency of vibration is 680Hz. So, the answer is 680Hz.
Ajụjụ 55 Ripọtì
A piece of stone attached to one end pf string is whirled round in a horizontal circle. When the string is suddenly cut, the stone will
Akọwa Nkọwa
When an object is whirled round in a horizontal circle, it moves with a constant speed and acceleration towards the centre of the circle. This acceleration is known as the centripetal acceleration and is provided by the tension in the string. When the string is suddenly cut, the centripetal force acting on the stone is removed, and it will fly off in a direction tangential to the circular path. Therefore, the correct option is "fly off in a direction tangential to the circular path."
Ajụjụ 56 Ripọtì
a ball of mass 5.0kg hits a smooth vertical wall normally with a speed of 2m-1 and rebounds with the same speed. Determine the impulse experienced by the ball
Akọwa Nkọwa
The impulse experienced by an object is equal to the change in momentum of the object. Momentum is defined as the product of mass and velocity, i.e., p=mv. When the ball hits the wall, it experiences an impulse that causes a change in its momentum. Initially, the ball is moving towards the wall with a velocity of 2m/s, and the velocity changes direction after the collision, so the final velocity is -2m/s. The change in velocity, Δv = (-2) - (2) = -4m/s. Using the formula for impulse, J = Δp = mΔv, we can calculate the impulse experienced by the ball: J = 5.0kg × (-4m/s) = -20.0kgm/s. Therefore, the impulse experienced by the ball is 20.0 kgm/s. Note that the negative sign indicates that the impulse is in the opposite direction to the initial velocity of the ball.
Ajụjụ 57 Ripọtì
The diagram above represents a section of a pair of vernier calipers. The reading on the instrument is
Ajụjụ 58 Ripọtì
(a) Distinguish between heat and temperature.
(b) State two physical properties of substances which may be used to measure temperature.
(c) State two reasons why mercury is preferred to alcohol as a thermometric liquid.
(d)(i) Describe how a mercury-in-glass thermometer is calibrated.
(ii) State two precautions necessary to ensure an accurate result.
(e) Explain how land and sea breezes occur.
Akọwa Nkọwa
None
Ajụjụ 59 Ripọtì
(a)(i) Describe, with the aid of a circuit diagram, an experiment to measure the resistance of a wire given an ammeter of low resistance, a battery, a key, a rheostat, va high-resistance voltmeter and some connecting wires.
(ii) State two precautions necessary to obtain an accurate result.
(b) Using the experimental result and any necessary measurements, explain how the resistivity of the wire may be determined.
(c) Two cells, each of e.m.f. 2V and internal resistance 0.552, are connected in series. They are made to supply current to a combination of three resistors, one of resistance 20 connected in series to a parallel combination of two other resistors each of resistance 3Q. Draw the circuit diagram and calculate the:
(i) current in the circuit
(ii) potential difference across the parallel combination of the resistors
(iii) lost volts of the battery.
Akọwa Nkọwa
None
Ajụjụ 60 Ripọtì
(a) Explain the term uniform acceleration
(b)(i) Sketch and describe the velocity-time graph for the motion of a ball from the time it is projected vertically upwards until it returns to the point of projection.
(ii) Neglecting air resistance and using ycur sketch, explain how the acceleration of free fall due to gravity g, and the maximum height attained when the ball is projected vertically upwards can be determined.
(c) A stone is projected vertically upwards with a velocity of 20ms\(^{-1}\). Two seconds later, a second stone is similarly projected with the same velocity. When the two stones meet, the second one is rising at a velocity of 10ms\(^{-1}\). Neglecting air resistance, calculate the:
(i) length of time the second stone is in motion before they meet,
(ii) velocity of the first stone when they meet (Take g as 10ms\(^{-2}\))
Ajụjụ 61 Ripọtì
(a) Explain the term photoelectric effect.
(b)
The diagram above represents a photocell with its associated electric circuit. Identify each of the physical quantities
represented by the letters A, B, C, D, E and F
(c) What factor determines the: (i) current produced by the photocell
(ii) maximum kinetic energy of the photoelectrons?
(d) State one similarity and one difference between photoemission and evaporation
(e) Name two methods by which a beam of free electrons may be produced other than photoemission
(f) State two applications of photoelectric effect.
(g) A light wavelength 5.0 x 10\(^{-7}\) m is incident on metal resulting in photoemission of electrons. If the work function of the metal is 3.04 x 10\(^{-19}\)J, calculate the:
(i) frequency of the light
(ii) energy of the incident photon,
(iii) maximum kinetic energy of the photoelectrons (Speed of light = 3.00 x 108ms\(^{-1}\); Planck's constant = 6.6 x 10\(^{-34}\)Js).
