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Ajụjụ 1 Ripọtì
A vibrator causes water ripples to travel across the surface of a tank. The wave travels 50cm in 2s and the distance between successive crests is 5cm. Calculate the frequency of the vibrator
Akọwa Nkọwa
The frequency of the vibrator can be calculated using the formula: frequency = speed / wavelength where speed is the speed of the wave, and wavelength is the distance between successive crests. In this case, we are given that the wave travels 50cm in 2s, which means the speed of the wave is: speed = distance / time = 50cm / 2s = 25cm/s We are also given that the distance between successive crests is 5cm, which is the wavelength. Therefore, the frequency of the vibrator is: frequency = speed / wavelength = 25cm/s / 5cm = 5Hz So the correct answer is 5Hz.
Ajụjụ 2 Ripọtì
Electrons were discovered by
Akọwa Nkọwa
Electrons were discovered by J.J. Thompson. In the late 19th century, he performed a series of experiments using cathode ray tubes, which are glass tubes containing low-pressure gas and electrodes. By applying high voltage, he observed a beam of negatively charged particles traveling from the negative electrode to the positive electrode. He concluded that these particles, which he called "corpuscles," were fundamental units of negative charge and later were renamed electrons. This discovery led to the development of the modern understanding of atomic structure and the electron's role in it.
Ajụjụ 3 Ripọtì
The part of the human eye that does similar work as the diaphragm of a camera lens is the
Akọwa Nkọwa
The part of the human eye that does similar work as the diaphragm of a camera lens is the iris. The iris is the colored part of the eye and is responsible for controlling the amount of light that enters the eye. Just like the diaphragm in a camera lens, the iris can adjust its size to allow more or less light into the eye. This helps to regulate the amount of light reaching the retina, which is responsible for sensing light and transmitting the image to the brain.
Ajụjụ 4 Ripọtì
When water is boiling, it
Akọwa Nkọwa
When water is boiling, it changes from a liquid state to a gaseous state called steam. This happens when the water is heated to its boiling point, which is when it reaches a temperature of 100 degrees Celsius (212 degrees Fahrenheit) at sea level. As the water is heated, it absorbs energy and the molecules start to move faster and faster, eventually reaching a point where they escape into the air as steam. The temperature of the water during boiling does not change, as all the energy is being used to break the bonds between the water molecules rather than increasing the temperature. Therefore, the options "gets hotter," "increase in mass," and "decreases in mass" are not correct when describing what happens when water is boiling.
Ajụjụ 5 Ripọtì
In the molecular explanation of conduction, heat is transferred by the
Akọwa Nkọwa
In the molecular explanation of conduction, heat is transferred by the Free electrons. In metals, free electrons move randomly and collide with other particles as they gain kinetic energy. These free electrons transfer the energy to the adjacent particles, which in turn gain kinetic energy and transmit it to other adjacent particles, thus transferring heat energy from one part of the material to another. This process of heat transfer by free electrons is called conduction. Therefore, the correct option is "Free electrons."
Ajụjụ 6 Ripọtì
Which of the following is consistent with Charles' law?
I
II
III
IV.
Akọwa Nkọwa
This is the correct graph. The graph is volume against 1/ temperature where temperature is in Celsius.
Ajụjụ 7 Ripọtì
One newton × One meter equals?
Akọwa Nkọwa
One newton times one meter is equal to one Joule. A newton is the unit of measurement for force, and a meter is the unit of measurement for distance. When force is applied over a distance, work is done, which is measured in Joules. Therefore, one newton multiplied by one meter results in one Joule of work done. The other options listed (one water, one ampere, one kilogram) are not correct units of measurement for this calculation.
Ajụjụ 8 Ripọtì
If a body moves with a constant speed and at the same time undergoes an acceleration, its motion is said to be
Akọwa Nkọwa
If a body moves with a constant speed and at the same time undergoes an acceleration, its motion is said to be rectilinear. When an object moves with constant speed, it means that it covers the same distance in equal time intervals. On the other hand, acceleration is the rate of change of velocity with time. If an object undergoes acceleration, its velocity changes with time. Therefore, if a body moves with constant speed and undergoes an acceleration, it means that its direction of motion changes while it covers equal distances in equal time intervals. This type of motion is called rectilinear motion, where the object moves in a straight line, but its velocity changes due to the acceleration. In contrast, circular motion is when an object moves in a circular path with a constant speed, while oscillatory motion is when an object moves back and forth around a fixed point. Rotational motion is when an object rotates around an axis. None of these descriptions fit the scenario of a body moving with constant speed and undergoing acceleration, so the answer is rectilinear motion.
Ajụjụ 9 Ripọtì
Which of the following statements are correct of the production and propagation of waves?
I. vibration produces waves
II. waves transmit energy along the medium
III. the medium through which the wave travels does not travel with the wave
IV. waves do not require any medium for transmission
Akọwa Nkọwa
The correct statement is: I and II and III only. Explanation: - Statement I is correct because the production of waves involves some kind of disturbance that creates a vibration in the medium, which then propagates as a wave. - Statement II is correct because waves carry energy along the medium as they propagate. This is why waves can be used to transmit information or power over long distances. - Statement III is correct because the medium through which a wave travels does not move with the wave. Instead, the wave passes through the medium, causing it to oscillate or vibrate, but not to move along with the wave. - Statement IV is incorrect because most waves require a medium through which to propagate. For example, sound waves require air, water waves require water, and seismic waves require the Earth's crust. There are some types of waves, such as electromagnetic waves, that can propagate through a vacuum, but this is not true for all waves.
Ajụjụ 11 Ripọtì
Which of the following is/are the limitations to the Rutherford's atomic models?
I. It is applicable when energy is radiated as electrons are revolving
II. It is applicable when energy is radiated in a continuous mode
III. It is applicable to an atom with only one electron in the other shell
Akọwa Nkọwa
Rutherford assumed that (I) energy is radiated when electrons are revolving (II) energy is radiated in a continuous mode. These are limitations of Rutherford's model
Ajụjụ 12 Ripọtì
Heat may be transferred by conduction, convention and radiation. By which of these methods does heat travel through vacuum?
Akọwa Nkọwa
Heat can be transferred by conduction, convection, and radiation. Conduction is the transfer of heat through a material by the movement of heat-carrying particles, such as atoms or molecules, from one part of the material to another. This method of heat transfer is not possible in a vacuum, as there are no particles present to carry heat. Convection is the transfer of heat by the movement of a fluid, such as air or water. This method of heat transfer is also not possible in a vacuum, as there are no fluids present to carry heat. Radiation is the transfer of heat through electromagnetic waves, such as light or infrared radiation. This method of heat transfer does not require any material or fluid medium, and can therefore occur in a vacuum. Therefore, the answer is "Radiation only".
Ajụjụ 13 Ripọtì
A siren having a ring of 200 hole makes 132 rev/min. A jet of air is directed on the set of holes. Calculate the frequency and wavelength in air of the note produced (take v = 350m/s)
Akọwa Nkọwa
n = 200, S = 132 rev/min, v = 350m/s2
| f | = | ns | = | 200 | × | 132 | revmin | × | 1min60s | = | 440Hz |
| λ | = | vf | = | 350440 | = | 0.875m |
Ajụjụ 14 Ripọtì
If the attraction of the sun is suddenly ceased, the earth would continue to move in a straight line making a tangent with the original orbit. This statement is derived from Neutron's
Akọwa Nkọwa
The correct answer is the First law of motion. The First law of motion, also known as the law of inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. In this case, the earth is moving in its orbit around the sun because of the force of gravity between the two objects. If the force of gravity suddenly ceased, the earth would no longer be acted upon by an external force and would continue to move in a straight line, making a tangent with its original orbit. This idea is attributed to Sir Isaac Newton, who developed the laws of motion and the law of universal gravitation. However, the specific statement mentioned in the question is derived from the First law of motion.
Ajụjụ 15 Ripọtì
The mass of a nucleus is the
Akọwa Nkọwa
The mass of a nucleus is the total number of its protons and neutrons. The protons and neutrons are the subatomic particles that make up the nucleus of an atom. The mass of an atom is mostly concentrated in its nucleus, and the electrons orbiting the nucleus have a much smaller mass. Therefore, the mass of an atom is mostly determined by the number of protons and neutrons in its nucleus. The number of protons determines the element, and the number of neutrons can vary, resulting in isotopes of that element.
Ajụjụ 17 Ripọtì
The angular dispersion of a prism depends on
Akọwa Nkọwa
Dispersion is due to different refractive indices speeds and wavelengths.
Ajụjụ 18 Ripọtì
The diagram shows a uniform meter rule AB which balances horizontally at the 90cm mark when a mass of 0.2kg is suspended from B. Calculate the mass of the meter rule.
Akọwa Nkọwa
Mr
(90 - 50) = 0.2(100 - 90)
40Mr
= 0.2 × 10
Mr
= 240
= 0.05kg
Ajụjụ 19 Ripọtì
The earth's gravitational field intensity at its surface is about
(G = 6.7 × 10−11 Nm2 /kg2 , mass of the earth is 6 × 1024 kg, radius of the earth is 6.4 × 106 m, g on the earth = 9.8m/s2 )
Akọwa Nkọwa
The earth's gravitational field intensity at its surface can be calculated using the formula: g = G * M / r^2 where G is the gravitational constant, M is the mass of the earth, r is the radius of the earth, and g is the gravitational field intensity at the surface of the earth. Substituting the given values, we get: g = (6.7 × 10^-11 Nm^2/kg^2) * (6 × 10^24 kg) / (6.4 × 10^6 m)^2 g = 9.8 N/kg (approx.) Therefore, the answer is 9.8N/kg.
Ajụjụ 20 Ripọtì
A metal rod has a length of 100cm at 200oC . At what temperature will its length be 99.4cm. If the linear expansivity of the material of the rod is 2 × 10−5C−1
Akọwa Nkọwa
The linear expansivity of a material describes how its length changes with temperature. If the linear expansivity is given as 2 × 10^-5/°C, this means that for every 1°C change in temperature, the length of the material will change by 2 × 10^-5 times its original length. Given that the rod has a length of 100 cm at 200°C, we can use this information to find its length at a different temperature. If we let L be the length of the rod at temperature T, we can write the relationship as follows: L = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) To find the temperature at which the rod will have a length of 99.4 cm, we can set L equal to 99.4 cm and solve for T: 99.4 cm = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) 99.4 cm / 100 cm = 1 + 2 × 10^-5 * (T - 200°C) 0.994 = 1 + 2 × 10^-5 * (T - 200°C) -0.006 = 2 × 10^-5 * (T - 200°C) -0.006 / 2 × 10^-5 = T - 200°C -0.006 / (2 × 10^-5) = T - 200°C -0.006 / (2 × 10^-5) + 200°C = T So the temperature at which the rod will have a length of 99.4 cm is approximately equal to -0.006 / (2 × 10^-5) + 200°C, or -100°C. Therefore, the answer is -100°C.
Ajụjụ 21 Ripọtì
The diagram above represents the stress-strain graph of a loaded wire. Which of these statements is correct?
Akọwa Nkọwa
- I is the elastic limit
- the end of the constant part J is the yield point
- L is the break point.
Ajụjụ 22 Ripọtì
A straight wire 15cm long, carrying a current of 6.0A is in a uniform field of 0.40T. What is the force on the wire when it is at right angle to the field
Akọwa Nkọwa
The force on a current-carrying wire in a uniform magnetic field can be calculated using the equation: F = BILsinθ where F is the force in Newtons, B is the magnetic field strength in Tesla, I is the current in Amperes, L is the length of the wire in meters, and θ is the angle between the wire and the magnetic field. In this problem, the wire is 15cm long (0.15m), carrying a current of 6.0A, and the magnetic field is 0.40T. The angle between the wire and the magnetic field is 90 degrees (since the wire is at right angles to the field). Substituting the given values into the equation, we get: F = (0.40T)(6.0A)(0.15m)sin90 sin90 = 1, so we can simplify the equation to: F = (0.40T)(6.0A)(0.15m) F = 0.36N Therefore, the force on the wire is 0.36N. Answer option C is the correct answer.
Ajụjụ 24 Ripọtì
Workdone on an object to bring it to a certain point in space is called
Akọwa Nkọwa
The work done on an object to bring it to a certain point in space is called "Potential Energy". Potential energy is a form of energy that an object possesses due to its position relative to other objects. When an object is lifted or moved to a higher point against gravity, work is done on it, and this work is stored as potential energy. The potential energy of an object is directly proportional to its height and mass. It can be converted into other forms of energy, such as kinetic energy, when the object is released or allowed to move freely. Therefore, potential energy is a type of stored energy that an object has due to its position, and it can be released to do work.
Ajụjụ 25 Ripọtì
The equilibrium position of objects in any field corresponds to situation of
Akọwa Nkọwa
The equilibrium position of an object in any field corresponds to the situation of minimum potential energy. This means that at the equilibrium position, the object has the lowest possible potential energy within the field. In other words, the forces acting on the object are balanced, and the object is not being pushed or pulled in any direction. Therefore, the object will remain at rest at the equilibrium position unless it is acted upon by an external force. Of the options given, the correct answer is "minimum potential energy".
Ajụjụ 26 Ripọtì
Which of the following readings cannot be determined with a meter rule?
Akọwa Nkọwa
Meter rule has a reading accuracy of 0.5mm or 0.05cm, thus measurement is M ± 0.05cm i.e 2.00, 2.05, 2.50, 2.55 etc.
The reading that cannot be read is 2.56cm.
Ajụjụ 27 Ripọtì
A rectangular solid black has length 10cm, breadth 5cm and height 2cm. If it lies on a horizontal surface, and has density 100kg/m3 , calculate the pressure it exerts on the surface.
Akọwa Nkọwa
To calculate the pressure that the rectangular solid exerts on the surface, we need to use the formula for pressure: Pressure = Force / Area In this case, the force is the weight of the rectangular solid, which we can calculate using the formula: Weight = Mass x Gravity The mass of the rectangular solid can be calculated using its density and volume: Mass = Density x Volume The volume of the rectangular solid is simply its length x breadth x height: Volume = Length x Breadth x Height = 10 cm x 5 cm x 2 cm = 100 cm3 We need to convert this volume to cubic meters to use the density given in kg/m3: Volume = 100 cm3 = 0.0001 m3 Now we can calculate the mass: Mass = Density x Volume = 100 kg/m3 x 0.0001 m3 = 0.01 kg The gravity is the acceleration due to gravity, which we can assume to be 9.81 m/s2. Therefore, the weight is: Weight = Mass x Gravity = 0.01 kg x 9.81 m/s2 = 0.0981 N Now we can use this weight to calculate the pressure on the surface. The surface area in contact with the rectangular solid is simply its length x breadth: Area = Length x Breadth = 10 cm x 5 cm = 50 cm2 We need to convert this area to square meters: Area = 50 cm2 = 0.005 m2 Therefore, the pressure is: Pressure = Force / Area = 0.0981 N / 0.005 m2 = 19.62 N/m2 We can convert this to units of N/cm2 or N/mm2 if desired. This is equivalent to: Pressure = 0.1962 N/cm2 = 0.0001962 N/mm2 So the pressure that the rectangular solid exerts on the surface is 19.62 N/m2, which is approximately 20 N/m2. Therefore, the answer is 200 N/m2.
Ajụjụ 28 Ripọtì
A boy pushes a 500kg box along a floor with a force of 2000N. If the velocity of the box is uniform, the co-efficient of friction between the box and the floor is
Akọwa Nkọwa
The coefficient of friction is a measure of the amount of friction between two surfaces. It is represented by the symbol "μ" and is a dimensionless quantity. The coefficient of friction between two surfaces depends on the nature of the surfaces in contact and the force pressing them together. In this problem, the boy is pushing the box with a force of 2000N. If the box is moving with a uniform velocity, then the force of friction acting on the box is equal and opposite to the pushing force applied by the boy. We can calculate the force of friction using the formula: frictional force = coefficient of friction x normal force where the normal force is the force exerted by the floor on the box in a direction perpendicular to the floor. Since the box is not moving up or down, the normal force is equal to the weight of the box. The weight of the box can be calculated using the formula: weight = mass x gravity where mass is the mass of the box and gravity is the acceleration due to gravity (9.8 m/s^2). So, the weight of the box is: weight = 500 kg x 9.8 m/s^2 = 4900 N The force of friction is equal to the pushing force of 2000N, so we can set these two equal to each other and solve for the coefficient of friction: frictional force = 2000N coefficient of friction x normal force = 2000N coefficient of friction x 4900N = 2000N coefficient of friction = 2000N / 4900N = 0.408 So, the coefficient of friction between the box and the floor is approximately 0.4. Therefore, the correct answer is 0.4.
Ajụjụ 30 Ripọtì
The volume of 0.354g of helium at 273°C and 114cm of mercury pressure is 2667cm3 . Calculate the volume
Akọwa Nkọwa
m = 0.354g, T1
= 273°C = 273 + 273 = 576K
P1
= 114cmHg, V1
= 2667cm3
at STP
T2
= 273K, P2
= 76cmHg, V2
= ?
| P1 V1 T1 | = | P2 V2 T1 |
| V2 | = | 114 × 2667 × 27376 × 576 | = | 2000.25cm3 |
Ajụjụ 31 Ripọtì
In which of the points labelled A, B, C, D and E on the conductor shown would electric charge tend to concentrate most
Akọwa Nkọwa
- Charge are mostly concentrated at the outermost part of a hollow conductor
- Charge are also mostly concentrated at the pointed ends or places with high density point.
Ajụjụ 33 Ripọtì
Which of the following statement about the electromagnet shown above is correct?
Akọwa Nkọwa
A - B = S - N.
Also, starting end of the current is south while terminating end is North.
Ajụjụ 34 Ripọtì
Radio waves belongs to the class of ware whose velocity is about
Akọwa Nkọwa
Radio waves belong to the class of waves whose velocity is approximately 3 x 10^8 m/s. This velocity is commonly denoted as the speed of light, which is the speed at which all electromagnetic waves, including radio waves, travel in a vacuum. This constant velocity is one of the fundamental principles of physics and is important in understanding the behavior and properties of light and other electromagnetic waves. The speed of light is incredibly fast, and it's difficult for us to imagine just how fast it is. To put it into perspective, light can travel around the Earth's equator almost 7.5 times in just one second. This high speed is essential for radio communication, as it enables radio waves to travel long distances in a short amount of time, allowing us to communicate with people and devices far away from us.
Ajụjụ 35 Ripọtì
A body moves in SHM between two point 20m on the straight line Joining the points. If the angular speed of the body is 5 rad/s. Calculate its speed when it is 6m from the center of the motion.
Akọwa Nkọwa
From two parts 20m apart
a = 10m, x = 6m, A = 5
V = ω√A2−X2
= 5√102−62
= 40m/s
Ajụjụ 36 Ripọtì
The height at which the atmosphere cases to exist is about 80km. If the atmospheric pressure on the ground level is 760mmHg, the pressure at a height of 20km above the ground level is
(ρm = 13.6g/cm3 ρ = 0.00013g/cm3 )
Akọwa Nkọwa
ρm
hm
= ρa
ha
13.68(760 - p) × 10−3
= 13 × 10−5
(20 × 103
)
| 760 | - | p | = | 13 × 10−5 × 20 × 103 13.68 × 10−3 | = | 19.00 | × | 101 |
760 - p = 190
p = 760 - 190 = 570mmHg
Ajụjụ 37 Ripọtì
The pin-hole camera produces a less sharply defined image when the
Akọwa Nkọwa
The pin-hole camera produces a less sharply defined image when the pin-hole is larger. A pin-hole camera works by allowing light to pass through a small hole (the pin-hole) and project an inverted image of the outside world onto a screen or surface located behind the hole. The smaller the pin-hole, the sharper the resulting image, as light passing through a smaller hole produces less diffraction or spreading out of the light. When the pin-hole is larger, more light enters the camera, but the light rays also become more scattered, resulting in a less well-defined image. This is because the larger opening allows more light rays to enter at different angles, creating a wider range of paths that the light can take as it travels through the camera and onto the screen. As a result, the image is less clear and less defined, with less sharp edges and more blurring. is the correct answer because it correctly identifies the effect of a larger pin-hole on the image produced by the pin-hole camera. less illumination, would actually produce a dimmer image, but it would not affect the sharpness or definition of the image. the distance of the screen from the pin-hole, and the distance of the object from the pin-hole, would affect the size of the image and the scale of the objects, but they would not affect the sharpness or definition of the image.
Ajụjụ 38 Ripọtì
Any line or section taken through an advancing wave in which all the particles are in the same phase is called the
Akọwa Nkọwa
The answer is: wave front. A wave front is any imaginary line or surface that connects all points of a wave that are in the same phase, meaning they are at the same point in their cycle. In other words, it is a line or surface that separates the points of a wave that are in-phase from those that are out-of-phase. For example, consider the ripples on the surface of a pond when a stone is thrown in. The wave fronts are the concentric circles that emanate from the point where the stone entered the water. All points along a given circle are in-phase, meaning the water molecules at those points are at the same point in their oscillation cycle. In summary, a wave front is a line or surface that separates points in a wave that are in-phase from those that are out-of-phase.
Ajụjụ 39 Ripọtì
A single force which produces the same effect as a set of forces acting together at a point is known as the
Akọwa Nkọwa
The single force which produces the same effect as a set of forces acting together at a point is known as the "resultant". In other words, the resultant is the net force that results from combining all the individual forces acting on an object. It represents the combined effect of all the forces acting on the object and is the force that would produce the same motion as the original set of forces acting together. Therefore, when solving problems in physics, it is often useful to find the resultant force in order to determine the overall effect of multiple forces on an object.
Ajụjụ 40 Ripọtì
When a girl moves towards a plane mirror at a speed of 4.0m/s, the distance between the girl and her image reduces a speed of
Akọwa Nkọwa
| v | = | dt | or | v | α | d |
d = x, v = 4m/s
d = 2x, v = ? (girl and image)
| v | = | 2 × 4x | = | 8 | ms |
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