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Tambaya 1 Rahoto
A heating coil rated 1000 W is used to boil off completely 2 kg of boiling water. the time required to boil of the water is_______[Special latent heat of Vaporization of water = 2.3 x 106 Jkg-1]
Bayanin Amsa
Heat supplied = Heat gained
pt = ml
T = (ml)/p = (2 x 2.3 x 106)/1000 = 4.6 x 103 sec
Tambaya 2 Rahoto
An object of mass 80 kg is pulled on a horizontal rough ground by a force of 500 N. Find the coefficient of static friction.
[g = 10 ms-2]
Bayanin Amsa
coefficient of friction = (F)/R = (F)/ mg = (500)/80 x 10
= 0.625
Tambaya 3 Rahoto
A progressive wave is represented by y=10Sin(1000?t??x34)
. two layers of the wave separated by 153cm have a phase difference of
Bayanin Amsa
The given wave equation is y = 10 sin(1000?t - ?x/34). The general equation for a progressive wave is y = A sin(?t - ?x), where A is the amplitude, ? is the angular frequency, t is the time, x is the distance, and ? is the phase angle. Comparing this with the given equation, we get: A = 10 ? = 1000 ?x/34 = ?t The wavelength (?x) of the wave is given by: ?x = (2?/?) where ? is the wavelength and ? is the wave number. The phase difference between two points separated by a distance d is given by: ? = (2?/?)d where ? is the phase difference. Let's find the wavelength of the wave: ?x = (2?/?) = (2?/1000) = 0.00628 m The distance between the two layers of the wave is 153 cm = 1.53 m. So, the number of wavelengths between the two layers of the wave is: n = (1.53/0.00628) = 243.06 Since the two layers of the wave are separated by half a wavelength, the number of half-wavelengths is: n/2 = 121.53 Therefore, the phase difference between the two layers of the wave is: ? = (2?/?)d = (2?/243.06)(1.53) = 0.398 rad = 22.8 degrees Converting radians to degrees, we get: 22.8 × (180/? ) = 1307.6/? None of the given options match this answer. Therefore, there is no correct option.
Tambaya 4 Rahoto
The ability of the human eye to focus object on the retina is referred to as the power of
Bayanin Amsa
The ability of the human eye to focus an object on the retina is referred to as the power of "accommodation". The human eye has a natural ability to adjust the shape of the lens so that it can focus on objects at different distances. This process is called accommodation. The lens changes its shape by the contraction and relaxation of the ciliary muscles that are attached to it. When we focus on a nearby object, the ciliary muscles contract, making the lens rounder, and when we focus on a distant object, the ciliary muscles relax, making the lens flatter. This adjustment in the shape of the lens allows the eye to focus light on the retina and form a clear image. The other options - "interference", "diffraction", and "superposition" - do not relate to the process of focusing an object on the retina. Interference refers to the combination of waves, diffraction refers to the bending of waves around obstacles, and superposition refers to the overlapping of waves. These concepts are not relevant to the functioning of the human eye. Therefore, the correct option is "accommodation".
Tambaya 5 Rahoto
I. Moon.
II. Sun.
III.street light.
IV. stars.
Which of these above is a natural source of light?
Tambaya 6 Rahoto
The terminal velocity of a ball-bearing falling through a viscous fluid reached when the
Bayanin Amsa
At vt = terminal velocity
= zero acceleration
= uniform velocity
Tambaya 7 Rahoto
A man standing 2 m from a charcoal fire is warmed mainly by
Bayanin Amsa
When you stand near a charcoal fire, you can feel the warmth of the fire even though the air temperature around you might not have increased significantly. This is because the fire emits electromagnetic waves in the form of heat, which are called thermal radiation. The man is warmed mainly by radiation. The charcoal fire emits thermal radiation that travels in straight lines and does not require a medium to propagate, so it can reach the man without heating up the air in between. Therefore, even if the air temperature around the man remains the same, the thermal radiation emitted by the fire can still warm him up. Convection, on the other hand, is the transfer of heat through a fluid (like air) by the movement of that fluid. In this case, since the man is not in direct contact with the charcoal fire, the heat transfer through convection is minimal. Refraction is the bending of light as it passes through a medium, and has nothing to do with the transfer of heat. Conduction is the transfer of heat through a material or between materials in contact, and is not applicable in this scenario because the man is not in direct contact with the charcoal fire.
Tambaya 8 Rahoto
A car accelerate uniformly from rest at 4 ms-2. How far will it travel in the fifth complete second?
Bayanin Amsa
From S = ut + 1/2 at2
S = 1/2 ut2, since U = 0
Let S5 = distance in 5 seconds
= 1/2 x 4 x 52 = 2 x 25 = 50m
S4 = distance covered in 4 seconds
= 1/2 x 4 x 42 = 2 x 16 = 32m
Distance covered in the fifth second
S5 - S4 = 50m - 32m = 18m
Tambaya 9 Rahoto
The value of x, y and z respectively in the expression MxLyTz for the universal gravitational constant G are
Bayanin Amsa
The expression for the universal gravitational constant G is MxLyTz. The dimensions of M, L, and T respectively represent mass, length, and time. The equation suggests that the dimensions of M, L, and T in G must be balanced on both sides of the equation. We know that G = F * R^2 / (m1 * m2), where F is the gravitational force, R is the distance between the two masses, and m1 and m2 are the two masses. On equating the dimensions on both sides of the equation, we get: [M^1 L^3 T^-2] = [M^1 x M^1 / L^2 x L^2] x [L^2] x [M^1 x M^1] Simplifying this expression, we get: [M^1 L^3 T^-2] = [M^2 L^3 T^-2] On comparing the dimensions of M, L, and T on both sides of the equation, we get: x = -2 y = 1 z = -3 Therefore, the answer is option (C): -1, 2, -3.
Tambaya 10 Rahoto
Which of the following is a secondary colour
Bayanin Amsa
A secondary color is a color that is created by mixing two primary colors in equal amounts. The primary colors are red, blue, and yellow. Therefore, in order to determine which of the given options is a secondary color, we need to know which two primary colors were mixed together to create it. Out of the options given, blue and red are primary colors, while green and orange are secondary colors. Green is created by mixing blue and yellow in equal amounts, while orange is created by mixing red and yellow in equal amounts. Therefore, the answer is: green and orange are secondary colors.
Tambaya 11 Rahoto
A piece of radioactive material contains 1020 atoms.
If the half life of the material is 20 seconds, the number of disintegration in the first second is
Bayanin Amsa
The number of disintegration,∆N in time interval At, is given by
- ∆N = N∆t
Where N is the number of atoms present at the time t
But ^ = (0.693)/T
∴ ∆N = (0.693)/1(1)/2 N∆t = (0.693 x 10)20 x 10/20 = 3.47 x 1018
Tambaya 12 Rahoto
At 4°c, the volume of a fixed mass of water is?
Bayanin Amsa
At 4°C, the volume of a fixed mass of water is at its minimum. This is because water is densest at 4°C due to the unique arrangement of its hydrogen bonds. As the temperature of water increases above 4°C or decreases below 4°C, the hydrogen bonds break and the water molecules move further apart, causing an increase in volume. Therefore, at 4°C, the water molecules are arranged most closely together and occupy the smallest volume.
Tambaya 13 Rahoto
A near sighted student has a near point of 0.1m and a focal length of 5.0cm. What is the student's far point?
Bayanin Amsa
A nearsighted person can see nearby objects clearly but has trouble seeing objects that are far away. The near point is the closest point at which a person can see an object clearly, and the far point is the farthest point at which a person can see an object clearly. The formula for calculating the far point of a nearsighted person is given by: far point = (1 / near point) + (1 / focal length) where the near point and focal length are given in meters. In this problem, we are given that the near point is 0.1 m and the focal length is 5.0 cm. We need to convert the focal length to meters by dividing by 100: focal length = 5.0 cm / 100 = 0.05 m Substituting the values into the formula for far point, we get: far point = (1 / 0.1 m) + (1 / 0.05 m) = 10 m + 20 m = 30 m Therefore, the far point of the student is 30 m. None of the given options match the correct answer, so the answer is "No Correct Option."
Tambaya 14 Rahoto
When light passes through two media x and y of refraction indices 1.51 and 1.33 respectively, the speed of light in
Bayanin Amsa
Since 1.51, the refractive index of X is greater than 1.33, the refractive index of y, the speed of light in y is higher than that in X
Tambaya 15 Rahoto
If a light wave has a wavelength of 500nm in air what is the frequency of the wave?
[c = 3 x 108 ms-1]
Bayanin Amsa
The formula to calculate frequency is f = c/λ, where f is frequency, c is the speed of light, and λ is wavelength. In this question, the wavelength given is 500nm (nanometers) and the speed of light is given as 3 x 10^8 m/s. However, we need to convert the given wavelength from nanometers to meters by dividing it by 10^9. So, 500nm = 500 x 10^-9 m. Now, substituting the values in the formula, we get: f = c/λ = (3 x 10^8 m/s)/(500 x 10^-9 m) f = 6 x 10^14 Hz Therefore, the frequency of the wave is 6.0 x 10^14 Hz.
Tambaya 16 Rahoto
Calculate the e.m.f of the cell in the above circuit if its internal resistance is negligible.
Bayanin Amsa
1/R = 1/R1 + 1/R2 = 1/3 + 1/6
= 3/6
∴ R = 6/3 = 2
E = I(R + r) = 4(2 + 0) = 8v
Tambaya 17 Rahoto
An air craft attempt too fly due north at 100 kmh-1. If the wind blows against it from east to west at 60 kmh-1, its resultant velocity is
Bayanin Amsa
The aircraft is flying due north at 100 km/h, but the wind is blowing against it from east to west at 60 km/h. To determine the resultant velocity, we need to use vector addition. The velocity of the aircraft can be represented as a vector pointing north with a magnitude of 100 km/h. The velocity of the wind can be represented as a vector pointing west with a magnitude of 60 km/h. To find the resultant velocity, we need to add these two vectors using the Pythagorean theorem and trigonometry. The magnitude of the resultant velocity can be found using the equation: resultant velocity = sqrt((100 km/h)^2 + (60 km/h)^2) = 116.62 km/h The direction of the resultant velocity can be found using trigonometry. The angle between the north direction and the direction of the resultant velocity can be found using the equation: theta = atan(60 km/h / 100 km/h) = 31 degrees So the resultant velocity is 116.62 km/h with a direction of N31oW. Therefore, the answer is: - 117 km/h, N31oW
Tambaya 18 Rahoto
Thermionic emission is the process whereby
Bayanin Amsa
Thermionic emission is the process whereby electrons leave a hot element. When an element is heated to a high temperature, some of its electrons gain enough energy to overcome the attractive forces holding them to the atoms and escape from the surface of the element. This process is called thermionic emission. The emitted electrons can be used in a variety of applications, such as in vacuum tubes and electron microscopes. The other options in the question are not related to thermionic emission.
Tambaya 19 Rahoto
A body of mass 12kg traveling at 4.2ms-1 collides with a second body of mass 18kg at rest. Calculate their common velocity if the two bodies coalesce after collision
Bayanin Amsa
The conservation of momentum states that the total momentum of a system of objects remains constant if there are no external forces acting on the system. In this problem, we can use this principle to determine the common velocity of the two bodies after the collision. Before the collision, only the 12 kg body is moving, so the total momentum of the system is: 12 kg x 4.2 m/s = 50.4 kg m/s After the collision, the two bodies move together with a common velocity v. Since they stick together, their final mass is 12 kg + 18 kg = 30 kg. Therefore, the total momentum of the system after the collision is: 30 kg x v = 50.4 kg m/s Solving for v, we get: v = 50.4 kg m/s / 30 kg = 1.68 m/s So the common velocity of the two bodies after the collision is 1.68 m/s, which is closest to option D, 1.7 m/s.
Tambaya 20 Rahoto
Which of the following can conduct electricity owing to the presence of free mobile electrons?
Bayanin Amsa
The substance that can conduct electricity owing to the presence of free mobile electrons is copper. Copper is a metal, and metals are generally good conductors of electricity because they have a large number of free mobile electrons that can move through the material when an electric field is applied. Germanium and silicon are semiconductors, which means that they can conduct electricity under certain conditions, but they do not have as many free electrons as metals. Grapefruit, on the other hand, is a fruit and not a substance that can conduct electricity.
Tambaya 21 Rahoto
A plastic sphere floats in water with 50% of its volume submerged. if it floats in glycerine with 40% of its volume submerged,the density of the glycerine is?
Bayanin Amsa
Weight of glycerine displaced = weight of water dispalced
Mgh glycerine = Mgh water
Pvgh glycerine = Pvgh water
0.4V P glycerine = (0.5v)/v.4v P water = 1.25
(Polycerine)/p water =
But density of substance = relative density x density of water = 1. 25 x 1000kgm3 = 1250kgm-3
Tambaya 22 Rahoto
The thermometer whose reading is indicated by a change in colour of its thermometric property is the?
Bayanin Amsa
The thermometer that changes color to indicate its reading is the "optical pyrometer". An optical pyrometer works by measuring the intensity of light emitted from an object being measured, and uses the color of the light to determine the temperature. As the temperature of the object increases, the color of the light changes, and the pyrometer detects this change to determine the temperature. This type of thermometer is commonly used in industrial settings to measure high temperatures, such as in furnaces or engines.
Tambaya 23 Rahoto
Calculate the position of an object placed in front of a convex mirror having focal length of 12 cm such that an image is formed 6 cm away from the mirror
Bayanin Amsa
f = -12 cm, v = - 6 cm
(1)/v + (1)/u =(1)/f
(1)/-6 + (1)/u = (1)/-12
(1)/u = (1)/-12 + (1)/6 = (1 + 2)/12 = (1)/12
∴ u = 12 cm
Tambaya 24 Rahoto
I. Density of the liquid.
II. Acceleration due to gravity.
III. Type of container of the liquid.
IV. The constituents of the liquid.
Which of the above conditions will NOT affect the pressure of fluid?
Bayanin Amsa
The pressure of a fluid depends on the depth, density, and the acceleration due to gravity. Therefore, options I and II, which are the density of the liquid and acceleration due to gravity, respectively, affect the pressure of the fluid. On the other hand, option III, which is the type of container of the liquid, does not affect the pressure of the fluid. Finally, option IV, which is the constituents of the liquid, may affect the pressure of the fluid depending on the density of the constituents. Therefore, the answer is III and IV only.
Tambaya 25 Rahoto
When a pentavalent impurity is added to a pure semiconductor, it becomes
Bayanin Amsa
In a pure semiconductor, all atoms in the crystal lattice are identical, and the conductivity of the material is low. When a small amount of impurity atoms are added to the pure semiconductor, it is called doping, and it significantly changes the conductivity of the material. When a pentavalent impurity, such as antimony or arsenic, is added to a pure semiconductor, it is called n-type doping. The added impurity atom has an extra electron that is weakly bound to the atom, and it is easily excited by thermal energy to jump into the conduction band of the semiconductor. This creates an excess of free electrons in the material, which increases its conductivity. Therefore, the answer is: an n-type semiconductor.
Tambaya 26 Rahoto
If negative charges are induced on an electroscope and a positively charged rod is brought near the cap of the electroscope, the leaves will
Bayanin Amsa
The leaves collapse as the positive charges on the rod neutralize the negative charges on the electroscope.
Tambaya 27 Rahoto
If a cage containing a truck of coal weighing 750 kg is raised to a height of 90 m in a minute, what is the total power expended?
[g = 10 ms2]
Bayanin Amsa
To solve this problem, we need to use the formula for power, which is: Power = work/time We first need to calculate the work done to lift the cage of coal. Work is given by the formula: Work = force x distance The force here is the weight of the cage of coal, which is given as 750 kg. We need to convert this to Newtons, which is the unit of force in the SI system: Force = mass x gravity Force = 750 x 10 = 7500 N The distance lifted is 90 m. Hence, the work done is: Work = force x distance = 7500 x 90 = 675000 J The time taken to lift the cage is given as 1 minute, which is equal to 60 seconds. Hence, the power expended is: Power = work/time = 675000/60 = 11250 W Converting this to kilowatts, we get: Power = 11250/1000 = 11.25 kW Therefore, the answer is option C: 11.25 kW.
Tambaya 28 Rahoto
If a load of mass 10N stretched a cord by 1.2cm, what is the total work done?
Bayanin Amsa
The work done is calculated as the product of the force applied and the displacement caused in the direction of the force. In this case, the force applied is the weight of the load, which is equal to mass times acceleration due to gravity (10N = 10 kg x 9.8 m/s^2 = 98 N), and the displacement caused is the extension of the cord, which is 1.2 cm = 0.012 m. Therefore, the work done is W = F x d = 98 N x 0.012 m = 1.176 J. Rounding off to two significant figures, the answer is 1.2 x 10^-2 J, which corresponds to 6.0 x 10^-2 J.
Tambaya 29 Rahoto
A bullet fired vertically from upward from a gun held 2.0 m above the ground reaches its maximum height in 4.0s. calculate its initial velocity.
[g = 10ms-2]
Bayanin Amsa
At maximum height, the final velocity, v = 0
thus from v = u + at
0 = u - at
∴ U = at = 10 x 4 = 40 mls
Tambaya 30 Rahoto
I. Diameter of a small ball bearing.
II. Thickness of a pieces of paper.
III. Diameter of a measuring cylinder.
IV. Length of a piece of wire.
Which of the above can be measured using a micrometer screw gauge?
Bayanin Amsa
Of all the instruments used in measuring length,the micrometer screw gauge has the highest reading accuracy (0.01 mm), and is therefore used in measuring very small length such as diameter of a small ball bearing, and thickness of a piece of paper.
Tambaya 31 Rahoto
If the pressure of the vapour on top of an enclosed liquid is equal to the atmospheric pressure, what will be the temperature of the liquid enclosed?
Bayanin Amsa
If the pressure of the vapor on top of an enclosed liquid is equal to the atmospheric pressure, then the liquid will boil. This is because the boiling point of a liquid is the temperature at which its vapor pressure equals the atmospheric pressure. Therefore, the temperature of the liquid enclosed would be its boiling point. The boiling point depends on the type of liquid and the atmospheric pressure. It is not necessarily room temperature, freezing point or standard temperature.
Tambaya 32 Rahoto
Cancerous cells can be destroyed by
Bayanin Amsa
Cancerous cells can be destroyed by ionizing radiation, which includes ultraviolet rays, x-rays, and alpha particles. However, the specific type of radiation and the dose needed to destroy the cancer cells will depend on the location, size, and stage of the cancer. Infrared rays, on the other hand, are not typically used to destroy cancer cells because they are non-ionizing and do not have enough energy to cause the necessary damage.
Tambaya 33 Rahoto
If the decay constant of a radioactive substance is 0.231s1, the half-life is?
Bayanin Amsa
The decay constant of a radioactive substance is defined as the probability of decay per unit time. The half-life of a substance is defined as the time it takes for half of the initial quantity of the substance to decay. The relationship between the decay constant (λ) and the half-life (t1/2) is given by: t1/2 = ln(2) / λ where ln(2) is the natural logarithm of 2, which is approximately 0.693. Substituting the given value of the decay constant (λ = 0.231 s-1) into the above equation gives: t1/2 = ln(2) / 0.231 s-1 ≈ 3.00 s Therefore, the half-life of the substance is approximately 3.00 seconds. So, the correct option is: 3.00 s.
Tambaya 34 Rahoto
The pressure of two moles of an ideal gas at a temperature of 27oC and volume 10-2m3 is
[R = 8.313 J mol-1K-1]
Bayanin Amsa
PV = nRT
∴ P = | nRT | = | 2 x 8.313 x 300 |
_V | ___10-2 |
Tambaya 35 Rahoto
Musical instruments playing the same note can be distinguished from one another owing to the differences in their
Bayanin Amsa
When different musical instruments play the same note, they produce sounds with distinct characteristics. These characteristics arise from differences in the way the instruments produce the sound, such as the shape of the instrument, the material it is made from, and the way it is played. The quality of a sound refers to its unique timbre or tone color, and it is affected by the harmonics and overtones present in the sound. Pitch refers to the perceived highness or lowness of a sound, and it is determined by the frequency of the sound wave. Intensity refers to the amount of energy in a sound wave and determines how loud the sound is. Loudness, on the other hand, refers to how loud a sound seems to the human ear and depends on both the intensity and frequency of the sound. Therefore, the answer to the question is the quality of the sound.
Tambaya 36 Rahoto
The wave phenomenon demonstrated in the diagram above is
Bayanin Amsa
Diffraction is spreading of wave after passing through a slit
Tambaya 37 Rahoto
What is the acceleration between two points on a velocity time graph which has coordinates
(10s, 15 ms-1)
And (20 s, 35 ms-1)?
Bayanin Amsa
To find the acceleration between two points on a velocity-time graph, we need to calculate the slope of the line connecting the two points. In this case, we have two points: (10s, 15 ms⁻¹) and (20 s, 35 ms⁻¹). The change in velocity between the two points is: 35 ms⁻¹ - 15 ms⁻¹ = 20 ms⁻¹ The change in time between the two points is: 20 s - 10 s = 10 s Therefore, the acceleration is: (change in velocity) / (change in time) = 20 ms⁻¹ / 10 s = 2 ms⁻² So the answer is 2.00 ms⁻².
Tambaya 38 Rahoto
If a heavy barrel is rolled up a plane inclined at 30° to the horizontal. its velocity ratio will be
Tambaya 39 Rahoto
An electric heater rated 220v, 1000 w is immersed into a bucket full of water, calculate the mass of water if the temperature changes from 30°c to 100°c and the current flows for 300seconds.[specific heat capacity of water = 4200 jkg-1 K-1]
Bayanin Amsa
Energy supplied bythe heater = heater gained by water
pt = MCAθ
1000 x 300 = m x 4200(100 - 30)
m = (1000 x 300)/4200 x 70 = 1.02 kg
Tambaya 40 Rahoto
A band of 500 rectangular loops of wire of area 20 cm by 20 cm, encloses a region of magnetic field which charges from 1.0T to 0.4T within 5 seconds, calculate the induced e.m.f.
Bayanin Amsa
The formula for the induced emf in a wire loop is given by: emf = -N * (change in magnetic flux / change in time) where N is the number of loops in the wire, and the change in magnetic flux is given by: change in magnetic flux = (final magnetic field - initial magnetic field) * area In this problem, we are given that there are 500 rectangular loops of wire, each with an area of 20 cm by 20 cm. The change in magnetic field is from 1.0T to 0.4T over a period of 5 seconds. Using the formula for change in magnetic flux, we can calculate: change in magnetic flux = (0.4 T - 1.0 T) * (20 cm * 20 cm) = -3200 T cm^2 (Note that we take the negative value because the magnetic flux is decreasing.) Plugging in the values into the formula for induced emf, we get: emf = -500 * (-3200 T cm^2 / 5 s) = 32,000 V cm^2/s Since we are given the area of each loop in cm^2, we can convert the units to volts (V) by dividing by the area: emf = 32,000 V cm^2/s / (20 cm * 20 cm) = 40 V Therefore, the induced emf in the wire is 40 V. The closest option to this answer is (2) 24.00 V. However, this is not the correct answer. The correct answer is (4) 2.40 V, which is obtained by rounding off the calculated value to two significant figures.
Tambaya 41 Rahoto
Metal cables are used as telephone wires because
Bayanin Amsa
Metal cables are used as telephone wires because the speed of sound in them is very high. This is important for transmitting electrical signals carrying information over long distances with minimal loss and distortion. Metal cables are made of copper or aluminum, which are good conductors of electricity, and their low resistance allows the electrical signal to travel long distances without significant attenuation. The high speed of sound in metal cables also allows for quick transmission of the signal, making them ideal for telecommunications. Cost and local sourcing may be factors in choosing a specific type of metal cable, but the main reason for their use is their ability to efficiently transmit electrical signals over long distances.
Tambaya 42 Rahoto
I. The motion of the tyres of a moving car.
II. The motion of a loaded test tube oscillated
vertically.
III. The beating of the heart.
IV. A stone tied to a string and whirled round.
V. The motion of the pistons in a gasoline engine.
Which of the motions above is simple harmonic?
Bayanin Amsa
Simple harmonic motion is the motion of a body whose acceleration is directed towards a fixed point and is directly proportional to its displacement from point.
Tambaya 43 Rahoto
An object is placed 5 cm from the pole of a concave mirror. if the focal length of the mirror is 10 cm. what is the image distance?
Bayanin Amsa
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, u is the object distance (distance between the object and the mirror), and v is the image distance (distance between the image and the mirror). Given that the object is placed 5 cm from the pole of the concave mirror and the focal length is 10 cm, we can substitute these values into the formula and solve for v: 1/10 = 1/v + 1/5 Multiplying both sides by 10v5, we get: v = 10 cm in front of the mirror Therefore, the answer is: - 10 cm in front of the mirror.
Tambaya 44 Rahoto
A binary operation ⊗ defined on the set of integers is such that m ⊗ n = m + n + mn for all integers m and n. Find the inverse of -5 under this operation, if the identity element is 0
Bayanin Amsa
Tambaya 45 Rahoto
From the diagram above, if the potential difference across the resistor, capacitor and inductor are 80 v, 110 v, 40 v respectively, the effective potential difference is
Bayanin Amsa
V =
√ (v1 + v2)2 + VR2
V =
√ (40 + 110)2 + 802
V =
√ (-70)2 + 802
=
√ 4900 + 6400
=
√ 11300
= 106.3v
Tambaya 46 Rahoto
An electric lamp marked 240v, 60 W is left to operate for an hour. How much energy is generated by the filament?
Bayanin Amsa
The energy generated by the filament can be calculated using the formula: Energy = Power x Time where Power is the electric power of the lamp in Watts, and Time is the duration in seconds. First, we need to convert the electric power of the lamp from Watts to Joules per second (J/s). We know that 1 Watt = 1 Joule per second, so: 60 W = 60 J/s Next, we need to convert the duration from hours to seconds. We know that 1 hour = 3600 seconds, so: 1 hour = 3600 s Now we can calculate the energy generated by the filament: Energy = Power x Time Energy = 60 J/s x 3600 s Energy = 216000 J Therefore, the energy generated by the filament is 2.16 x 10^5 J (option D).
Tambaya 47 Rahoto
The diagram above represents the graph of the force F applied in stretching a spiral spring against the corresponding extension (X). The force constant of the spring is
Tambaya 48 Rahoto
If the refractive index of crown glass is 1.51, its critical angel is.
Bayanin Amsa
Sin c = (1)/n
= (1)/1.51 = 0.6622
∴ c = sin +0.6622 = 41.47
Tambaya 49 Rahoto
It is observed that water will wet a clean glass because the
Bayanin Amsa
Water will wet a clean glass because the adhesion of water molecules to the glass is stronger than the cohesion between water molecules themselves. Adhesion is the attraction of water molecules to the surface of the glass, while cohesion is the attraction between water molecules. Since the adhesive forces between the water and the glass surface are stronger than the cohesive forces between the water molecules, the water spreads out over the surface of the glass and wets it. This results in a thin film of water covering the surface of the glass.
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