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Tambaya 1 Rahoto
If the time of flight is 96seconds, calculate the horizontal range through the point of projection.
Bayanin Amsa
Time of flight, T = 96s
R = (Ucosθ) *time* T = 640 × 96 = 61,440m
Tambaya 2 Rahoto
In the molecular explanation of conduction, heat is transferred by the
Bayanin Amsa
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."
Tambaya 3 Rahoto
The pin-hole camera produces a less sharply defined image when the
Bayanin Amsa
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.
Tambaya 4 Rahoto
The limiting frictional force between two surfaces depends on
I. the normal reaction between the surfaces
II. the area of surface in contact
III. the relative velocity between the surfaces
IV. the nature of the surfaces
Bayanin Amsa
- Friction depends on the nature of the surfaces in contact
- Solid friction is independent of the area of the surfaces in contact and the relative velocity between the surfaces.
Tambaya 5 Rahoto
Which of the following characteristics of a wave is used in the measurement of the depth of the Sea?
Bayanin Amsa
Depth of sea can be measured by echo, a reflected sound waves.
Tambaya 6 Rahoto
Heat may be transferred by conduction, convention and radiation. By which of these methods does heat travel through vacuum?
Bayanin Amsa
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".
Tambaya 7 Rahoto
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
Bayanin Amsa
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.
Tambaya 8 Rahoto
In semi-conductor, the carriers of current at room temperature are
Bayanin Amsa
In a semiconductor, the carriers of current at room temperature are both electrons and holes. Semiconductors are materials with properties that are in between those of conductors (e.g. metals) and insulators (e.g. rubber). At room temperature, a semiconductor crystal contains both free electrons and positively charged vacancies called holes. When a voltage is applied across the semiconductor, the electrons move towards the positive end of the circuit and the holes move towards the negative end. This movement of charge carriers constitutes an electric current. In summary, both electrons and holes can carry current in a semiconductor at room temperature, making the correct answer.
Tambaya 9 Rahoto
The Earth's magnetic equator passes through Jos in Nigeria. At Jos, the
Bayanin Amsa
The Earth has a magnetic field that is generated by the movement of molten iron in its core. The magnetic field has different properties at different locations on the Earth's surface. The magnetic equator is an imaginary line on the Earth's surface where the inclination or tilt of the Earth's magnetic field is zero, meaning that the magnetic field lines are parallel to the Earth's surface. At Jos, Nigeria, the Earth's magnetic equator passes through, which means that the angle of inclination or dip of the Earth's magnetic field is zero. Therefore, the correct answer is that the angle of dip is zero. This means that a magnetic needle suspended by a thread or placed on a horizontal surface would remain horizontal and not point downwards or upwards, as it would at other locations on the Earth's surface. This makes Jos an important location for studying the Earth's magnetic field and for conducting experiments related to magnetism.
Tambaya 10 Rahoto
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 )
Bayanin Amsa
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.
Tambaya 11 Rahoto
A mass of 0.5kg is whirled in a vertical circle of radius 2m at a steady rate of 2 rev/s. Calculate the centripetal force
Bayanin Amsa
The centripetal force is the force that acts towards the center and keeps an object moving in a circular path. To calculate the centripetal force, we can use the following formula: f = m * v^2 / r where: - f = centripetal force - m = mass of the object (0.5 kg) - v = velocity of the object (2 rev/s * 2 * pi m/rev = 12.57 m/s) - r = radius of the circle (2 m) Plugging in the values, we get: f = 0.5 kg * 12.57 m/s^2 / 2 m f = 31.43 N Rounding to the nearest whole number, the centripetal force is 31 N. So, the closest answer from the options is 160N.
Tambaya 12 Rahoto
In the molecular explanation, heat is transferred by the
Bayanin Amsa
- Conduction is explained in terms of the free electrons
- Convection is explained in terms of the movement of the fluid involved
- Radiation is explained in terms of invisible electromagnetic waves.
Tambaya 13 Rahoto
A copper rod, 5m long when heated through 20c, expands by 1mm. If a second copper rod, 2.5m long is heated through 5c, by how much will it expand?
Bayanin Amsa
l1
= 5m, ΔT = 10c, l2
- l1
= 1mm
l1
= 2.5m, ΔT = 5c, l2
- l1
= ?
using | α | = | l2 - l1 l1 ΔT |
15(10) | = | l2 - l1 2.5(5) |
l2 | - | l2 | = | 2.5(5)5(10) | = | 14 | = | 0.25mm |
Tambaya 14 Rahoto
The diagram shows four positions of the bob of a simple pendulum. At which of these positions does the bob have maximum kinetic energy and minimum potential energy
Bayanin Amsa
At position 1, the bob of the simple pendulum has the maximum potential energy and zero kinetic energy. At position 4, the bob has the maximum kinetic energy and minimum potential energy. To understand this, we need to know that the energy of a simple pendulum is converted back and forth between kinetic energy and potential energy as it swings back and forth. When the bob is at its highest point (position 1), it has the maximum potential energy because it is farthest from the ground and has the most potential to move downward. At this point, the bob has zero kinetic energy because it is momentarily at rest. As the bob swings downward towards the equilibrium point, it gains speed and its potential energy is converted to kinetic energy. At the equilibrium point (position 2), the bob has equal amounts of kinetic and potential energy. As the bob continues to move downward, its potential energy decreases and its kinetic energy increases. At position 3, the bob has minimum potential energy and some amount of kinetic energy. At the lowest point of its swing (position 4), the bob has maximum kinetic energy because it is moving at its fastest speed. At this point, the bob has minimum potential energy because it is closest to the ground and has the least amount of potential to move downward. So, to summarize, the bob has maximum potential energy at position 1, equal amounts of kinetic and potential energy at position 2, minimum potential energy at position 3, and maximum kinetic energy at position 4.
Tambaya 15 Rahoto
A single force which produces the same effect as a set of forces acting together at a point is known as the
Bayanin Amsa
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.
Tambaya 16 Rahoto
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
Bayanin Amsa
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.
Tambaya 17 Rahoto
A well 1km deep is filled with a liquid of density 950kg/m3 and g = 10m/s2 , the pressure at the bottom of the well is
Bayanin Amsa
P = Pa + ρgh = (1.00 × 105
) + (950 × 10 × 1000)
P = 105
+ (95 × 105
) = 105
(1 + 95) = 96 × 105
P = 9.6 × 106
N/m2
Tambaya 18 Rahoto
If a body moves with a constant speed and at the same time undergoes an acceleration, its motion is said to be
Bayanin Amsa
If a body moves with a constant speed but at the same time undergoes an acceleration, its motion is called rectilinear motion. This means that the body moves in a straight line and its speed changes at a constant rate, causing an acceleration. It is different from oscillation, circular and rotational motions which involve changes in direction, as well as changes in speed.
Tambaya 19 Rahoto
When blue and green colours of light are mixed, the resultant colour is
Tambaya 21 Rahoto
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.
Bayanin Amsa
From two parts 20m apart
a = 10m, x = 6m, A = 5
V = ω√A2−X2
= 5√102−62
= 40m/s
Tambaya 22 Rahoto
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 )
Bayanin Amsa
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
Tambaya 23 Rahoto
During the transformation of matter from the solid to the liquid state, the heat supplied does not produce a temperature increase because
Bayanin Amsa
During the transformation of matter from the solid to the liquid state, the heat supplied does not produce a temperature increase because all the heat is used to break the bonds holding the molecules of the solid together
Tambaya 24 Rahoto
A supply of 400V is connected across capacitors of 3μf and 6μf in series. Calculate the charge
Bayanin Amsa
CT | = | C1 × C2 C1 + C2 |
= | 3 × 63 + 6 |
= 189
= 2μf
Q = CV
⇒ 2 × 10−6
× 400
⇒ 800 × 10−6
C = 8 × 10−4
C
Tambaya 25 Rahoto
A body was slightly displaced from its equilibrium position. Which one of the following is a condition for its stable equilibrium
Bayanin Amsa
The condition for stable equilibrium of a body that has been slightly displaced from its equilibrium position is "an increase in the potential energy of the body." When an object is at its equilibrium position, it has a minimum potential energy. When the object is displaced from its equilibrium position, it has a higher potential energy. For the object to be in stable equilibrium, it must be able to return to its equilibrium position after it has been displaced. If the potential energy of the object increases as it is displaced, it means that the equilibrium position is a point of stable equilibrium. This is because the object will experience a restoring force that will push it back towards its equilibrium position, as the potential energy decreases. Therefore, an increase in potential energy is a condition for a body to be in stable equilibrium after it has been slightly displaced from its equilibrium position. An increase in kinetic energy or height does not necessarily indicate stability, as it depends on the specific situation and other factors at play.
Tambaya 26 Rahoto
The momentum of a car moving at a constant speed in a circular track
Bayanin Amsa
Movement of an object in a circle with an acceleration towards its center is provided by change in velocity and centripetal force a α V α Fc
Tambaya 27 Rahoto
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
Bayanin Amsa
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.
Tambaya 28 Rahoto
Electrons were discovered by
Bayanin Amsa
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.
Tambaya 29 Rahoto
The pitch of a screw jack is 0.45cm and the arm is 60cm long. If the efficiency of the Jack is 75/π %, calculate the mechanical advantage.
Bayanin Amsa
P = 0.45cm, L = 60cm, Eff = 75/π%
VR | (Screw | system) | = | 2πrP | = | 2πLP |
M.A | = | Eff% × VR100 | = | 75π | × | 1100 | × | 2π × 600.45 | = | 75 × 800300 | = | 200 |
Tambaya 30 Rahoto
Any line or section taken through an advancing wave in which all the particles are in the same phase is called the
Bayanin Amsa
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.
Tambaya 31 Rahoto
During the transformation of matter from the solid to the liquid state, the heat supplied does not produce temperature increase because
Bayanin Amsa
When a solid is heated to its melting point, the heat supplied is used to overcome the intermolecular forces holding the molecules in a fixed position, resulting in the breaking of these bonds. As a result, the solid transforms into a liquid without any change in temperature. This is because the heat energy supplied is used in breaking the bonds between molecules rather than increasing the kinetic energy of the molecules, which is what causes an increase in temperature. Therefore, the correct option is: "all the heat is used to break the bonds holding the molecules of the solid together."
Tambaya 33 Rahoto
The statement 'Heat lost by the hot body equals that gained by the cold one' is assumed when determining specific that heat capacity by the method of mixtures. Which of the following validates the assumption?
I. Lagging the Calorimeter
II. Ensuring that only S.I units are used
III. Weighing the calorimeter, the lid and the stirrer.
Bayanin Amsa
The assumption 'Heat lost by the hot body equals that gained by the cold one' is based on the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one system to another. Thus, to validate this assumption, it's important to have a well-designed and insulated calorimeter so that as little heat as possible is lost to the environment. This is accomplished by lagging the calorimeter (Option I). Additionally, using the correct units (Option II) helps ensure that the energy transfer is accurately calculated and reported. Weighing the calorimeter, the lid, and the stirrer (Option III) is important for accurately measuring the amount of heat transferred, but by itself is not enough to validate the assumption. Therefore, the correct answer is "I and III only".
Tambaya 34 Rahoto
A train has an initial velocity of 44m/s and an acceleration of -4m/s2 . Calculate its velocity after 10 seconds
Bayanin Amsa
The velocity of the train after 10 seconds can be calculated using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we get: v = 44 m/s + (-4 m/s^2) x 10 s v = 44 m/s - 40 m/s v = 4 m/s Therefore, the velocity of the train after 10 seconds is 4m/s. Answer option D is correct. Explanation: The train has an initial velocity of 44 m/s and an acceleration of -4 m/s^2. The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which means that the train is slowing down. After 10 seconds, the train's velocity decreases by 40 m/s (4 m/s^2 x 10 s) to reach a final velocity of 4 m/s.
Tambaya 35 Rahoto
Radio waves belongs to the class of ware whose velocity is about
Bayanin Amsa
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.
Tambaya 36 Rahoto
The mass of a nucleus is the
Bayanin Amsa
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.
Tambaya 37 Rahoto
An object is acted upon by a system of parallel three causing the object to be in state equilibrium. Which of the following statement is not correct
Bayanin Amsa
all the parallel forces must be equal in magnitude and direction
Tambaya 38 Rahoto
Which of the following statements is/are correct for a freely falling body?
I. the total is entirely kinetic
II. the ratio of potential energy to kinetic energy is constant
III. the sum of potential and kinetic energy is constant
Bayanin Amsa
The correct answer is "III only". A freely falling body is one that is falling under the influence of gravity and experiences no other force or constraint. In this situation, the total energy of the body is conserved, meaning that the sum of its potential and kinetic energy remains constant. The potential energy of a body is directly proportional to its height above the ground, and its kinetic energy is directly proportional to its velocity. As the body falls, its potential energy decreases and its kinetic energy increases, but the total energy remains constant. Statement III is correct because the sum of potential and kinetic energy is indeed constant for a freely falling body. Statement I is incorrect because the body has both potential and kinetic energy, so the total energy is not entirely kinetic. Statement II is incorrect because the ratio of potential energy to kinetic energy is not constant for a freely falling body, as both are changing as the body falls.
Tambaya 39 Rahoto
Which of the following statement about the electromagnet shown above is correct?
Bayanin Amsa
A - B = S - N.
Also, starting end of the current is south while terminating end is North.
Tambaya 40 Rahoto
A mixture of blue and red pigment when illuminated by white light will appear
Bayanin Amsa
A mixture of blue and red pigment when illuminated by white light will appear purple. This is because when white light shines on a surface, it contains all the colors of the visible spectrum. When blue and red pigments are mixed together, they absorb all the other colors in the spectrum except for blue and red. Therefore, when white light shines on this mixture, the blue pigment absorbs all the colors except blue, while the red pigment absorbs all the colors except red. The result of this is that the blue and red pigments reflect only blue and red light, which then combines to form purple. Therefore, the mixture of blue and red pigments appears purple when illuminated by white light.
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