JAMB UTME - Physics - 1990
Ajụjụ 1 Ripọtì
The photocell works on the principle of the
Akọwa Nkọwa
The photocell is a device that converts light energy into electrical energy. It works on the principle of the emission of electrons by incident radiation. When light falls on the surface of a photocell, it knocks electrons out of the surface of the material. These electrons can be collected as a current if the photocell is connected to an external circuit. The current generated by the photocell can be used for various applications, including light sensors, solar panels, and even in some medical devices. The other options, such as voltaic cell, emission of protons by incident electrons, and photographic plate, are not directly related to the operation of a photocell.
Ajụjụ 2 Ripọtì
Which of the following statements are true of an insulated charged body carrying positive charge? i. it contains excess positive charges. ii. it creates an electric field. iii. it possesses potential energy. iv. it carries electric current.
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
For a short sighted person, light rays from a point on a very distant object is focused
Ajụjụ 5 Ripọtì
A stone of mass m kg is held h meters above the floor for 50s. The work done in joules over this period is
Akọwa Nkọwa
Work is the amount of energy transferred when a force acts on an object, causing it to move. The formula for work done is given as: work = force x distance x cos(theta) where force is the force applied on the object, distance is the distance moved by the object, and theta is the angle between the force and the direction of motion. In the given problem, a stone of mass m kg is held h meters above the floor for 50s. We need to find the work done in joules over this period. Since the stone is being held stationary, its distance moved is zero. Therefore, the work done on the stone is also zero. Therefore, the answer is O, which means zero. The option that represents zero is O.
Ajụjụ 6 Ripọtì
Which of the following is a vector?
Akọwa Nkọwa
A vector is a quantity that has both magnitude and direction. Among the given options, the only quantity that has both magnitude and direction is an electric field. Therefore, the electric field is the vector quantity.
Ajụjụ 7 Ripọtì
A body of mass 10kg rests on a rough inclined plane whose angle of tilt θ is variable. θ is gradually increased until the body starts to slide down the plane at 30o. The coefficient of limiting friction between the body and the plane
Akọwa Nkọwa
To find the coefficient of limiting friction between the body and the inclined plane, we can use the following steps: 1. Draw a free-body diagram of the body on the inclined plane. 2. Resolve the forces acting on the body along the incline and perpendicular to the incline. 3. Write down the equation for the limiting friction force, which is equal to the product of the coefficient of friction and the perpendicular force acting on the body. 4. Set the limiting friction force equal to the force acting down the incline, which is equal to the component of the weight of the body along the incline. 5. Solve for the coefficient of friction. Given that the body starts to slide down the plane at an angle of 30 degrees, we know that the force acting down the incline is equal to the component of the weight of the body along the incline. We can use trigonometry to find this component, which is equal to mg*sin(30), where m is the mass of the body and g is the acceleration due to gravity. The perpendicular force acting on the body is equal to mg*cos(30), and the limiting friction force is equal to the coefficient of friction times this perpendicular force. Therefore, we can write: mu*mg*cos(30) = mg*sin(30) Solving for the coefficient of friction, we get: mu = sin(30)/cos(30) = tan(30) = 1/sqrt(3) Therefore, the coefficient of limiting friction between the body and the inclined plane is 1/sqrt(3), which is approximately 0.577.
Ajụjụ 8 Ripọtì
An inclined plane which makes an angle of 30o with the horizontal has a velocity ratio of
Akọwa Nkọwa
The velocity ratio of an inclined plane is the ratio of the distance an object is moved along the plane to the vertical distance it is lifted. It is also equal to the inverse of the sine of the angle of inclination of the plane. In this case, the angle of inclination of the plane is 30 degrees. Therefore, the sine of the angle is 0.5 (sin 30 = 0.5). The velocity ratio is the inverse of the sine, which is 2 (1/sin 30 = 2). Therefore, the answer is 2.
Ajụjụ 9 Ripọtì
One of the properties of the earth's magnetic field is that the
Ajụjụ 11 Ripọtì
Which of the following statements? i. land and sea breezes are natural convection currents. ii. convection may occur in liquids or gasses but not in solids. iii. the vacuum in a thermos flask prevents heat loss due to convection only.
Ajụjụ 12 Ripọtì
The pressure of a gas when cooled at constant volume will decrease because the molecules
Akọwa Nkọwa
When a gas is cooled at a constant volume, its molecules lose kinetic energy and move slower, which means they collide less frequently with the walls of the container. This decrease in collisions results in a decrease in pressure. Therefore, the correct answer is: "collide less frequently with the walls of the container." The other options are not applicable to this scenario. The average kinetic energy of the molecules remains the same because there is no change in the temperature, and the gas molecules do not break up into smaller molecules or decrease in number because the cooling process does not involve any chemical reaction or physical changes in the gas.
Ajụjụ 13 Ripọtì
The resistances of platinum wire at the ice and steam points are 0.75 ohm and 1.05 ohm respectively. Determine the temperature at which the resistance of the wire is 0.90 ohm
Akọwa Nkọwa
The resistance of a wire changes as its temperature changes. This change in resistance can be used to measure the temperature. The relationship between temperature and resistance is described by the temperature coefficient of resistance (TCR), which is different for different materials. For platinum, the TCR is approximately 0.00392 per degree Celsius. To determine the temperature at which the resistance of the wire is 0.90 ohm, we can use the following formula: Rt = R0 (1 + αt) where: Rt is the resistance of the wire at temperature t R0 is the resistance of the wire at the ice point (0°C or 273.15K) α is the TCR for platinum (0.00392/°C) t is the temperature of the wire in Celsius We can use this formula to solve for t by plugging in the values given in the problem: 0.90 = 0.75 (1 + 0.00392t) 0.90/0.75 = 1 + 0.00392t 1.2 = 1 + 0.00392t 0.2 = 0.00392t t = 0.2/0.00392 t ≈ 51 Therefore, the temperature at which the resistance of the wire is 0.90 ohm is approximately 51°C, which corresponds to answer choice (B) 50.0°C.
Ajụjụ 14 Ripọtì
A sonometer wire of length 100cm under a tension of 10N, has a frequency of 250Hz. Keeping the length of the wire constant, the tension is adjusted to produce a new frequency of 350Hz. The new tension is
Akọwa Nkọwa
The tension and frequency of a sonometer wire are directly proportional, meaning that an increase in tension results in an increase in frequency, and vice versa, as long as the length of the wire is kept constant. We can use the formula: f = (1/2L) * sqrt(T/μ) where: - f is the frequency of the wire - L is the length of the wire - T is the tension in the wire - μ is the mass per unit length of the wire If we assume that μ is constant and the length of the wire is 100cm, we can set up two equations for the two given frequencies: 250 = (1/2 * 100) * sqrt(10/T) 350 = (1/2 * 100) * sqrt(x/T) where x is the new tension we want to find. Simplifying the equations, we get: sqrt(10/T) = 5/2 sqrt(x/T) = 7/2 Squaring both sides of each equation, we get: 10/T = 25/4 x/T = 49/4 Multiplying both sides of each equation by T, we get: 10 = 25/4 * T x = 49/4 * T Solving for T in the first equation, we get: T = 10 * 4/25 = 1.6 N Substituting this value of T into the second equation, we get: x = 49/4 * 1.6 = 19.6 N Therefore, the new tension required to produce a frequency of 350 Hz is 19.6 N.
Ajụjụ 15 Ripọtì
Which of the following statements are correct about an object in equilibrium under parallel forces? i. the total clockwise moments of the forces about any point equals the total anti-clockwise moments about the same point. ii. the total forces in one direction equals the total forces in the opposite direction. iii. the resolved components along the x-axis equals the resolved components along the y-axis.
Akọwa Nkọwa
An object is said to be in equilibrium when it is not accelerating, which means the net force acting on it is zero. When an object is in equilibrium under parallel forces, the forces acting on it are parallel to each other. Now, let's consider each statement: i. The total clockwise moments of the forces about any point equals the total anti-clockwise moments about the same point: This statement is true. For an object to be in equilibrium under parallel forces, the sum of the moments of the forces on one side of the object must be equal to the sum of the moments of the forces on the other side. This ensures that there is no net moment causing rotation. ii. The total forces in one direction equals the total forces in the opposite direction: This statement is also true. For an object to be in equilibrium, the sum of the forces in one direction must be equal to the sum of the forces in the opposite direction. This ensures that there is no net force causing linear motion. iii. The resolved components along the x-axis equals the resolved components along the y-axis: This statement is not necessarily true. When an object is in equilibrium under parallel forces, the forces are acting in the same direction, so there are no resolved components along the x or y axis. This statement only applies when the forces are not parallel. Therefore, the correct statement is: i and ii only.
Ajụjụ 16 Ripọtì
40m3 of liquid is mixed with 60m3 of another liquid Q. If the density of P and Q are 1.00kgm-3 and 1.6kgm-3 respectively, what is the density of the mixture?
Akọwa Nkọwa
To solve this problem, we can use the formula: Density = Mass/Volume We are given the volume of each liquid, but we need to find their masses. We can find the mass of each liquid using the formula: Mass = Density x Volume For liquid P, Mass = 1.00 kgm^-3 x 40m^3 = 40 kg For liquid Q, Mass = 1.6 kgm^-3 x 60m^3 = 96 kg The total mass of the mixture is the sum of the masses of each liquid: 40 kg + 96 kg = 136 kg The total volume of the mixture is the sum of the volumes of each liquid: 40 m^3 + 60 m^3 = 100 m^3 Now we can calculate the density of the mixture using the formula: Density = Mass/Volume Density = 136 kg/100 m^3 = 1.36 kgm^-3 Therefore, the density of the mixture is 1.36 kgm^-3, which corresponds to the fourth option.
Ajụjụ 17 Ripọtì
A force of 5N act at a point Y on a rod XYZ as shown in the diagram above, if XY is 2m. What is the moment of the force about point X?
Akọwa Nkọwa
Ajụjụ 18 Ripọtì
Two horizontal forces, 10N and 8N and another force F1 inclined at 30 to the vertical acting as shown in the diagram above, keep the body P in equilibrium, The weight of the body is
Ajụjụ 19 Ripọtì
The pressure of a gas when cooled at constant volume will decrease because the molecules
Akọwa Nkọwa
When a gas is cooled at constant volume, its molecules lose kinetic energy and move slower. This means that they collide with the walls of the container less frequently, resulting in a decrease in pressure. The other options given are incorrect: - The number of molecules does not change, so is incorrect. - The average kinetic energy remains the same, so is incorrect. - The gas molecules do not break up into smaller molecules, so is incorrect.
Ajụjụ 20 Ripọtì
What is the average velocity of the sprinter whose velocity time graph is as shown above
Akọwa Nkọwa
The average velocity of the sprinter can be calculated by finding the slope of the line on the velocity-time graph. Since velocity is on the y-axis and time is on the x-axis, we need to find the change in velocity (y-axis) divided by the change in time (x-axis) over a certain interval. Looking at the graph, we can see that the sprinter's velocity changes from 0 m/s to 17 m/s over a time of 2 seconds. Therefore, the average velocity of the sprinter during this time interval would be: average velocity = change in velocity / change in time average velocity = (17 m/s - 0 m/s) / 2 s average velocity = 8.5 m/s So the correct option is: 8.5ms-1.
Ajụjụ 22 Ripọtì
The number of neutrons contained in the nucleus 23892U is
Akọwa Nkọwa
The number of neutrons contained in the nucleus of uranium-238 is 146.
Ajụjụ 23 Ripọtì
A 0 - 10mA galvanometer with a coil resistance of 30Ω can be converted to a 0 - 10A ammeter by using
Akọwa Nkọwa
To convert the galvanometer into an ammeter, a shunt resistor needs to be connected in parallel with the galvanometer. This allows only a fraction of the current to pass through the galvanometer, while the rest of the current flows through the shunt resistor. Using Ohm's Law, the resistance of the shunt resistor can be calculated. The formula for the shunt resistance is: Rs = G * Rg / (Imax - G) where Rs is the shunt resistance, Rg is the coil resistance of the galvanometer, Imax is the maximum current that the ammeter can measure (in this case, 10A), and G is the sensitivity of the galvanometer (in this case, 10mA). Substituting the values given in the question, we get: Rs = 0.01 * 30 / (10 - 0.01) = 0.03 Ω Therefore, the correct answer is a 0.03Ω shunt resistor.
Ajụjụ 25 Ripọtì
A substance has a half of 3mins. after 6mins. the count rate was observed to be 400. What was its count rate at zero time?
Akọwa Nkọwa
The count rate of a substance is directly proportional to the amount of the substance present. The half-life of a substance is the time it takes for half of the substance to decay. In this problem, we know that the half-life of the substance is 3 minutes. This means that after 3 minutes, half of the substance would have decayed, and after another 3 minutes (total of 6 minutes), another half of the remaining substance would have decayed, leaving only a quarter of the original substance. We also know that at 6 minutes, the count rate was observed to be 400. This count rate corresponds to a quarter of the original substance because half of the substance decayed in the first 3 minutes and another half decayed in the next 3 minutes. Therefore, the count rate at zero time (when all of the substance was present) would have been four times the count rate observed at 6 minutes, which is 4 times 400 = 1600. Therefore, the correct answer is 1600.
Ajụjụ 26 Ripọtì
A lamp is rated 240V, 60W. The resistance of the filament is
Akọwa Nkọwa
The resistance of the filament in the lamp can be found using Ohm's law. Ohm's law states that the voltage (V) across a conductor is proportional to the current (I) flowing through it, with the constant of proportionality being the resistance (R) of the conductor. Mathematically, V = IR, or R = V/I. Given that the lamp is rated at 240V and 60W, we can calculate the current flowing through the filament using the formula for power: P = IV. Rearranging this formula, we get I = P/V. Substituting the given values, we get I = 60W/240V = 0.25A. Now, using Ohm's law, we can calculate the resistance of the filament as R = V/I = 240V/0.25A = 960 ohms. Therefore, the answer is 960 ohms.
Ajụjụ 28 Ripọtì
A bar of initial length /0 is heated through a temperature change Δ
t to a new length I. The linear expansivity, a, of the bar is
Ajụjụ 30 Ripọtì
A boy looks at the image of an object in a plane mirror. He observes two images, a main bright one and the other faint. The observed images result from
Akọwa Nkọwa
When a boy looks at the image of an object in a plane mirror, he observes two images, a main bright one and the other faint. The observed images result from reflection only. The main bright image is the result of regular reflection, where light rays are reflected from the smooth surface of the mirror and the reflected rays maintain their parallel paths. The faint image is the result of irregular or diffused reflection, where light rays are reflected in different directions due to the unevenness of the mirror's surface, and some of these rays may reach the boy's eye, forming a faint and blurred image. Therefore, the correct option is "reflection only".
Ajụjụ 31 Ripọtì
Which of the following conditions will make water boil at a temperature of 100oC when the atmospheric pressure is 750mmHg?
Akọwa Nkọwa
Water boils when its vapor pressure equals the atmospheric pressure above it. The atmospheric pressure decreases with altitude. At sea level, the atmospheric pressure is about 760 mmHg, and pure water boils at 100°C. However, at high altitudes where the atmospheric pressure is lower than 760 mmHg, water boils at a lower temperature than 100°C. Therefore, to make water boil at a temperature of 100°C when the atmospheric pressure is 750 mmHg, the external pressure needs to be increased to match the vapor pressure of water at 100°C and 750 mmHg. Thus, the correct option is "increase the external pressure."
Ajụjụ 32 Ripọtì
What is the length of liquid column in a barometer tube that would support an atmospheric pressure of 102000Nm-2 if the density of the liquid is 2600kgm-3? [g = 10ms-2]
Akọwa Nkọwa
The pressure of the atmosphere can be measured by using a barometer, which works by balancing the weight of a column of liquid against the atmospheric pressure. The height of the liquid column is directly proportional to the atmospheric pressure. The formula for calculating the height of the liquid column is h = P/(ρg), where h is the height of the liquid column, P is the atmospheric pressure, ρ is the density of the liquid, and g is the acceleration due to gravity. Substituting the given values in the above formula, we get: h = P/(ρg) = 102000/(2600 x 10) = 3.92m Therefore, the length of the liquid column in the barometer tube that would support an atmospheric pressure of 102000Nm^-2 if the density of the liquid is 2600kgm^-3 is 3.92m. Hence, the correct option is (c).
Ajụjụ 33 Ripọtì
Which of the following is an essential physical property of the wires used for making fuses?
Akọwa Nkọwa
Ajụjụ 34 Ripọtì
To keep a vehicle moving at a constant speed v, requires power P, from the engine. The force provided by the engine is
Akọwa Nkọwa
The force provided by the engine is given by the formula P/v. This means that the force is directly proportional to the power provided by the engine, and inversely proportional to the velocity of the vehicle. In simpler terms, if the engine provides more power, the force will increase, and if the velocity of the vehicle increases, the force will decrease. So, to keep a vehicle moving at a constant speed v, the engine needs to provide a force equal to P/v.
Ajụjụ 35 Ripọtì
What must be the distance between an object and a converging lens of focal length 20cm to produce an erect image two times the object height?
Akọwa Nkọwa
This is a question related to optics and the use of converging lenses. When an object is placed in front of a converging lens, it forms an image on the other side of the lens. The distance between the lens and the object affects the properties of the image formed. The formula to determine the distance between the object and the lens is: 1/f = 1/di + 1/do where f is the focal length of the lens, di is the distance between the image and the lens, and do is the distance between the object and the lens. Since we want to produce an image that is two times the height of the object and the image is erect (upright), the image distance (di) is positive, and the magnification (M) is 2. Therefore: M = -di/do = 2 Solving for di, we get: di = -2do Substituting this value of di in the lens formula: 1/f = 1/di + 1/do 1/20 = 1/(-2do) + 1/do Multiplying both sides by -2do, we get: -2 + 20/do = -1 Solving for do, we get: do = 10cm Therefore, the distance between the object and the lens must be 10cm to produce an erect image that is two times the height of the object. The correct option is: 10cm.
Ajụjụ 36 Ripọtì
An organic pipe closed at one end is 80cm long. Determine the frequency of the fundamental note assuming that the speed of sound in air is 340ms-1
Akọwa Nkọwa
The frequency of the fundamental note of an organic pipe closed at one end can be calculated using the formula: f = v/4L Where: f = frequency of the fundamental note v = speed of sound in air L = length of the pipe Given: v = 340ms-1 L = 80cm = 0.8m (since the pipe is closed at one end, we need to consider only its effective length, which is half of its total length) Substituting the values in the formula, we get: f = (340ms-1)/(4 x 0.8m) = 106.25Hz Therefore, the frequency of the fundamental note is 106Hz (approximately), which is closest to option A. Explanation: When we blow into a pipe, the air inside the pipe starts to vibrate, producing sound waves. The sound waves that are produced depend on the length of the pipe, the speed of sound in air, and the boundary conditions of the pipe (whether it is open or closed at one or both ends). In the case of an organic pipe closed at one end, the fundamental frequency is the lowest frequency that can be produced, and it corresponds to the wavelength of twice the effective length of the pipe. Using the formula mentioned above, we can calculate the frequency of the fundamental note.
Ajụjụ 37 Ripọtì
Which of the following is a fundamental unit?
Akọwa Nkọwa
The fundamental unit among the options given is the "Second". A fundamental unit is a unit of measurement that is independent and cannot be derived from any other units. The second is a fundamental unit of time and is defined as the duration of 9,192,631,770 periods of the radiation corresponding to the transition between two hyperfine levels of the ground state of the caesium-133 atom. In contrast, the other units listed are derived units, which means they are combinations of fundamental units. For example, the Newton is a unit of force, which is derived from the fundamental units of mass, length, and time.
Ajụjụ 38 Ripọtì
Shadows and eclipses result from the
Akọwa Nkọwa
Shadows and eclipses result from the "rectilinear propagation of light." This means that light travels in a straight line, and when an opaque object is placed in the path of the light, it blocks the light rays and creates a shadow. Similarly, an eclipse occurs when an object such as the Moon passes between the Sun and the Earth, blocking the Sun's light and casting a shadow on the Earth. In both cases, it is the straight-line path of light that is responsible for the formation of shadows and eclipses. Therefore, the correct answer is "rectilinear propagation of light."
Ajụjụ 39 Ripọtì
Three cells each of e.m.f. 1.5v and an internal resistance of 1.0Ω are connected in parallel across a load resistance of 2.67Ω . Calculate the current in the load?
Akọwa Nkọwa
The circuit given consists of three cells connected in parallel across a load resistance of 2.67 Ω. Each cell has an emf of 1.5V and an internal resistance of 1.0 Ω. To calculate the current in the load, we can use the formula: I = V / R where I is the current in the load, V is the total voltage across the load, and R is the resistance of the load. First, we need to calculate the total voltage across the load. Since the cells are connected in parallel, the voltage across each cell is the same as the total voltage, which is: V = 1.5V Next, we can use the formula for calculating the equivalent resistance of cells connected in parallel, which is: 1/R = 1/R1 + 1/R2 + 1/R3 where R1, R2, and R3 are the resistances of each cell. Since all cells have the same resistance, we can simplify the equation to: 1/R = 3/1 R = 1/3 Ω The total resistance of the circuit, including the internal resistance of the cells and the load resistance, is: Rt = R + r where r is the internal resistance of each cell, and Rt is the total resistance. So, Rt = 1/3 + 1 = 4/3 Ω Now, we can use Ohm's law to calculate the current in the load: I = V / Rt I = 1.5V / (4/3) Ω I = 0.5A Therefore, the current in the load is 0.50A.
Ajụjụ 40 Ripọtì
A car moving with a speed of 90kmh-1 was brought uniformly to rest by the application of the brakes in 10s. How far did the car travel after the brakes were applied?
Akọwa Nkọwa
The question describes a car that is initially moving with a speed of 90 km/h, and then brought to a stop by the application of the brakes in 10 seconds. To find the distance traveled by the car during this time, we can use the formula: distance = (initial velocity * time) + (0.5 * acceleration * time^2) where initial velocity is the velocity of the car before braking, time is the duration of the braking, and acceleration is the deceleration of the car due to the braking. We can convert the initial velocity from km/h to m/s by multiplying it by 1000/3600, which gives 25 m/s. We know that the car was brought uniformly to rest, so its acceleration is simply its initial velocity divided by the time it took to come to a stop, or: acceleration = initial velocity / time Plugging in the given values, we get: acceleration = 25 / 10 = 2.5 m/s^2 Now we can plug in the values of initial velocity, time, and acceleration into the distance formula to get: distance = (25 * 10) + (0.5 * 2.5 * 10^2) = 125 + 125 = 250 meters Therefore, the car traveled a distance of 250 meters after the brakes were applied. So, the correct option is (c) 250m.
Ajụjụ 41 Ripọtì
A light wave of frequency 5 x 1014Hz moves through water which has a refractive index of 43 . Calculate the wavelength in water if the velocity of light in air is 3 x 108ms-1
Akọwa Nkọwa
The problem involves finding the wavelength of a light wave in water, given its frequency and the refractive index of water, and assuming that the velocity of light in air is known. First, we need to use the formula v = fλ, where v is the velocity of light, f is the frequency, and λ is the wavelength. We know that the velocity of light in air is 3 x 10^8 m/s, and the frequency of the light wave in water is 5 x 10^14 Hz. We need to find λ. Second, we can use the formula n = c/v, where n is the refractive index, c is the speed of light in a vacuum (which is the same as the speed of light in air), and v is the velocity of light in the medium (in this case, water). We know that the refractive index of water is 4/3. We can solve for v to get v = c/n. Now we can substitute the value of v into the formula v = fλ to get λ = v/f. We have already found v to be c/n, and we know f to be 5 x 10^14 Hz. Substituting these values, we get: λ = (3 x 10^8 m/s)/(4/3)(5 x 10^14 Hz) = 4.5 x 10^-7 m Therefore, the wavelength of the light wave in water is 4.5 x 10^-7 meters. In simple terms, we can say that the problem asks us to find the distance between two consecutive peaks or troughs of a light wave in water, given its frequency and the refractive index of water. To solve the problem, we use formulas that relate the velocity, frequency, wavelength, and refractive index of light in air and water.
Ajụjụ 42 Ripọtì
What precaution should a manufacturer take to ensure that energy loss in a transformer is minimized?
Akọwa Nkọwa
Ajụjụ 43 Ripọtì
1kg of copper is transferred quickly from boiling water to a block of ice. Calculate the mass of ice melted, neglecting heat loss. [specific heat capacity of copper 400Jkg-1K-1 and latent heat of fusion of ice 333 x 103Jkg-1]
Akọwa Nkọwa
When 1kg of copper is transferred from boiling water to ice, it will first cool down from 100°C to 0°C, giving off heat to the surroundings, and then it will transfer heat to the ice, melting some of it. To calculate the mass of ice melted, we need to use the specific heat capacity of copper and the latent heat of fusion of ice. The amount of heat lost by the copper as it cools from 100°C to 0°C can be calculated as: Q1 = mCΔT where m is the mass of copper, C is its specific heat capacity, and ΔT is the change in temperature. ΔT = (100°C - 0°C) = 100°C Substituting the values, we get: Q1 = (1kg)(400Jkg-1K-1)(100°C) = 40,000J This heat is absorbed by the ice to melt it. The amount of heat required to melt a given mass of ice can be calculated as: Q2 = mL where m is the mass of ice and L is the latent heat of fusion of ice. L = 333 x 10³ Jkg-1 Substituting the values, we get: Q2 = (m)(333 x 10³ Jkg-1) Since the heat lost by the copper (Q1) is equal to the heat gained by the ice (Q2), we can equate the two equations: Q1 = Q2 mCΔT = mL Substituting the values, we get: (1kg)(400Jkg-1K-1)(100°C) = m(333 x 10³ Jkg-1) Solving for m, we get: m = 1.2 kg or 1200g Therefore, the mass of ice melted is 1200g, which is equivalent to 1.2 kg. Answer: 120g
Ajụjụ 44 Ripọtì
When light is incident on an object which is magenta in colour, which of the following colour would be absorbed?
Akọwa Nkọwa
Ajụjụ 45 Ripọtì
Electrical power is transmitted at a high voltage rather than low voltage because the amount of energy loss is reduced due to
Akọwa Nkọwa
Ajụjụ 46 Ripọtì
In a resonance tube experiment, the effective length of the air column for the first resonance is 20cm when set into vibration by a tuning fork of frequency 480Hz. Neglecting and effect, the velocity of sound in air is
Akọwa Nkọwa
λ = 4(L + C) ; where C = o
λ = 4 x 20 = 80cm
V = λ F
= 80 x 10-2 x 480 = 384ms-1
Ajụjụ 47 Ripọtì
An object which is 3cm high is placed vertically 10cm in front of a concave mirror. If this object produces an image 40cm from the mirror, the height of the image is
Akọwa Nkọwa
To solve this problem, we can use the mirror formula: 1/f = 1/v + 1/u where f is the focal length of the mirror, v is the distance of the image from the mirror, and u is the distance of the object from the mirror. We are given u = 10cm and v = 40cm. To find f, we can rearrange the formula: 1/f = 1/v + 1/u 1/f = 1/40 + 1/10 1/f = 0.03 f = 33.33cm Now, we can use the magnification formula: m = -v/u where m is the magnification of the image. Plugging in the values, we get: m = -v/u m = -40/10 m = -4 The negative sign indicates that the image is inverted. Finally, we can use the magnification formula to find the height of the image: m = h'/h where h' is the height of the image and h is the height of the object. Plugging in the values, we get: m = h'/h -4 = h'/3 h' = -12cm Again, the negative sign indicates that the image is inverted. To find the absolute value of the height, we ignore the sign: | h' | = 12cm Therefore, the height of the image is 12cm. Answer is correct.
