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Swali 1 Ripoti
An object of mass 20kg slides down an inclined plane at an angle of 30o to the horizontal. The coefficient of static friction is?
[g = 10ms-2]
Swali 2 Ripoti
If a load of 1 kg stretches a cord by 1.2cm, what is the force constant of the cord? {g=10ms-2}
Maelezo ya Majibu
The force constant of a cord represents the force required to stretch it by a unit length. In this case, the cord stretches by 1.2 cm when a load of 1 kg is applied to it. We can use Hooke's Law to find the force constant of the cord. Hooke's Law states that the force required to stretch a spring or cord is proportional to the amount of stretch. Mathematically, it can be expressed as F = kx, where F is the force, k is the force constant, and x is the amount of stretch. In this case, we know that the load applied is 1 kg or 1*10^3 g. We can convert this to mass in kg by dividing it by the acceleration due to gravity, g, which is given as 10 m/s^2. Therefore, the mass of the load is: m = 1*10^3 g / (10 m/s^2) = 100 g = 0.1 kg The amount of stretch, x, is given as 1.2 cm or 0.012 m. Using Hooke's Law, we can write: F = kx The force, F, required to stretch the cord by 0.012 m is equal to the weight of the load, which is: F = mg = 0.1 kg * 10 m/s^2 = 1 N Therefore, we can solve for the force constant, k, as: k = F/x = 1 N / 0.012 m = 83.3 N/m Thus, the force constant of the cord is 83.3 N/m. Looking at the answer options, we can see that option B is the closest to our answer of 83.3 N/m. Therefore, the correct answer is option B, 833 Nm^-1.
Swali 3 Ripoti
If a freely suspended object is pulled to one side and released, it oscillates about the point of suspension because the
Maelezo ya Majibu
If a freely suspended object is pulled to one side and released, it oscillates about the point of suspension because of the property of the object called "restoring force". Restoring force refers to any force that tends to bring an object back to its equilibrium position when it is displaced. When an object is pulled to one side and released, it is displaced from its equilibrium position. The object then experiences a restoring force that pulls it back towards the equilibrium position. This restoring force is proportional to the displacement of the object from its equilibrium position. As a result, the object oscillates back and forth about its equilibrium position. The oscillation continues until all the energy transferred to the object during the initial displacement is dissipated by frictional forces and the object comes to rest. Therefore, the correct answer is option (A) acceleration is directly proportional to the displacement.
Swali 5 Ripoti
The equation of a wave traveling in a horizontal direction is expressed as y = 15 sin 2/5 (60t - x) what is its wave length?
Maelezo ya Majibu
Swali 6 Ripoti
If a reverse-biased voltage is applied across a p-n junction, the depletion layer width is
Maelezo ya Majibu
A p-n junction is a boundary between a p-type semiconductor and an n-type semiconductor. When a reverse-biased voltage is applied across a p-n junction, the negative terminal of the voltage source is connected to the p-type semiconductor, and the positive terminal is connected to the n-type semiconductor. This creates an electric field that opposes the flow of current through the junction. The electric field pushes the majority carriers away from the junction, widening the depletion layer. The depletion layer is the region around the p-n junction where there are no free charge carriers. Therefore, the correct answer is that the depletion layer width is increased when a reverse-biased voltage is applied across a p-n junction.
Swali 7 Ripoti
Calculate the total distance covered by a train before coming to rest if its initial speed is 30ms-1 with a constant retardation of 0.1m-2
Maelezo ya Majibu
To solve this problem, we can use the following kinematic equation: v^2 = u^2 + 2as where: - v is the final velocity (which is zero since the train comes to rest) - u is the initial velocity (which is 30 m/s) - a is the acceleration (which is the negative of the retardation, -0.1 m/s^2) - s is the distance covered Substituting the values into the equation, we get: 0^2 = (30 m/s)^2 + 2(-0.1 m/s^2)s Simplifying, we get: 0 = 900 - 0.2s 0.2s = 900 s = 4500 meters Therefore, the total distance covered by the train before coming to rest is 4500 meters. Explanation: When a train is moving with a certain velocity and then comes to rest due to a constant retardation, its total distance covered can be calculated using the above kinematic equation. This equation relates the initial velocity, final velocity, acceleration, and distance covered by the train. By substituting the given values into the equation and solving for the distance covered, we get the answer. In this case, the train starts with an initial velocity of 30 m/s and comes to rest with a constant retardation of 0.1 m/s^2. The total distance covered by the train before coming to rest is found to be 4500 meters.
Swali 8 Ripoti
Maelezo ya Majibu
To get an enlarged image when using a concave mirror to shave, a man can place his face between the principal focus and the pole of the mirror. A concave mirror has a reflective surface that curves inward like the inside of a bowl. When an object is placed in front of a concave mirror, the mirror reflects light rays that converge and meet at a point called the "focal point" or "principal focus". The distance between the mirror and the focal point is called the "focal length". When a man places his face between the principal focus and the pole of the concave mirror, the mirror forms a virtual, upright, and enlarged image of his face behind the mirror. The image is larger than the actual size of his face because the mirror bends and converges the light rays in such a way that they create a magnified image. If the man places his face at the principal focus or beyond, the mirror will form a real and inverted image of his face, which will be smaller than the actual size of his face. Therefore, the correct answer is option (C) between the principal focus and the pole.
Swali 9 Ripoti
A car starts from rest and moves with a uniform acceleration of 30ms-2 for 20s. Calculate the distance covered at the end of the motion
Maelezo ya Majibu
The distance covered at the end of the motion is 600 meters. The distance covered by an object with uniform acceleration can be calculated using the equation: distance = initial velocity x time + (1/2) x acceleration x time^2 In this case, the initial velocity is 0 (the car starts from rest), the acceleration is 30 m/s^2, and the time is 20 seconds. Plugging in these values, we get: distance = 0 x 20 + (1/2) x 30 x 20^2 distance = 0 + (1/2) x 30 x 400 distance = 600 meters So, the car covers a distance of 600 meters at the end of the 20 seconds of motion.
Swali 10 Ripoti
Two metals P and Q are heated through the same temperature difference. If the ratio of the linear expansivities of P to Q is 2:3 and the ratio of their
length is 3: 4 respectively, the ratio of the increase in lengths of P to Q is
Maelezo ya Majibu
The increase in length of a material due to heating is proportional to its linear expansivity and the change in temperature. The linear expansivity of a material is defined as the change in length per unit length per degree Celsius change in temperature. Let the increase in length of metal P be x, and the increase in length of metal Q be y. Let the temperature difference be ΔT. Then, we can write: x = αP ΔT Lp ...(1) y = αQ ΔT Lq ...(2) where αP and αQ are the linear expansivities of metals P and Q respectively, and Lp and Lq are their original lengths. We are given that the ratio of the linear expansivities of P to Q is 2:3. So, we can write: αP/αQ = 2/3 Cross-multiplying, we get: 3αP = 2αQ αP/αQ = 2/3 = 2k/3k ...(3) where k is a constant. We are also given that the ratio of the lengths of P to Q is 3:4. So, we can write: Lp/Lq = 3/4 Cross-multiplying, we get: 4Lp = 3Lq Lp/Lq = 3/4 = 3m/4m ...(4) where m is a constant. Substituting equations (3) and (4) into equations (1) and (2), we get: x = (2k/3k)αQ ΔT (3m) y = αQ ΔT (4m) Simplifying, we get: x/y = (2k/3k) (3/4) x/y = 1/2 Therefore, the ratio of the increase in lengths of P to Q is 1:2. Hence, the correct answer is 1:2.
Swali 11 Ripoti
Maelezo ya Majibu
White light is composed of a spectrum of colors that include red, orange, yellow, green, blue, indigo, and violet. These colors have different wavelengths and frequencies, which determine their position in the electromagnetic spectrum. When white light enters a medium like a prism, it slows down and bends, causing the different colors to refract at different angles. This process is called dispersion and results in the separation of white light into its component colors. The degree of refraction depends on the wavelength of the light, with shorter wavelengths refracting more than longer ones. As a result, the different colors of light are spread out, creating a spectrum of colors that is visible to the eye. Therefore, the answer to the question is: "Dispersion of white light is the ability of white light to separate into its component colors." In conclusion, the dispersion of white light is an important phenomenon in optics that allows us to see the different colors of the visible spectrum. It is caused by the different refractive indices of the colors of light, which leads to their separation when they pass through a medium like a prism.
Swali 13 Ripoti
Clinical thermometers are examples of
Maelezo ya Majibu
Clinical thermometers are examples of mercury-in-glass thermometers. They consist of a thin, sealed glass tube that is filled with mercury. The glass tube has a narrow constriction in it, which prevents the mercury from flowing back down the tube once it has expanded due to a rise in temperature. When the thermometer is placed under the tongue or in the armpit, the heat from the body causes the mercury to expand and rise up the tube. The temperature can be read off a calibrated scale on the glass tube, which indicates the temperature based on the height of the mercury column. Mercury-in-glass thermometers are commonly used in clinical settings because they are accurate and have a fast response time. The mercury expands and contracts rapidly in response to temperature changes, allowing for quick and accurate temperature readings. Therefore, the correct answer is option (D) mercury-in-glass thermometer.
Swali 14 Ripoti
The current output from of an a.c. source is given as I = 10 sin to t. The d.c equivalent of the current is
Maelezo ya Majibu
The given equation of current, I = 10 sin ωt represents an alternating current (a.c.) which varies sinusoidally with time. The maximum value of this current is 10A, which occurs when sin ωt = 1. However, the question asks for the DC equivalent of this current, which is the constant value of current that would produce the same amount of heat in a resistor as the given a.c. current. The DC equivalent of an AC current is the root mean square (rms) value of the AC current, which is the square root of the mean of the square of the current over one cycle. For a sinusoidal current, the rms value is given by the formula I_rms = I_max/√2. Substituting the given maximum value of current (10A) into the formula for rms current gives: I_rms = 10A/√2 = 7.07A (approx) Therefore, the DC equivalent of the given AC current is approximately 7.1A, which is option (B) in the answer choices.
Swali 15 Ripoti
Maelezo ya Majibu
To find the upthrust on the balloon, we need to use the Archimedes' principle which states that an object submerged in a fluid experiences an upward force that is equal in magnitude to the weight of the fluid displaced. In this case, the balloon is not submerged but is filled with hydrogen which is lighter than air. Thus, the upthrust on the balloon is equal to the weight of the air displaced by the balloon. The weight of the air displaced by the balloon is given by the difference between the weight of the air that would fill the same volume as the balloon and the weight of the hydrogen that actually fills the balloon. The weight of air that would fill the same volume as the balloon is given by: mass of air = density of air x volume of balloon mass of air = 1.3 kg/m^3 x 300 m^3 mass of air = 390 kg The weight of the hydrogen that fills the balloon is given by: mass of hydrogen = density of hydrogen x volume of balloon mass of hydrogen = 0.09 kg/m^3 x 300 m^3 mass of hydrogen = 27 kg Therefore, the weight of the air displaced by the balloon is: weight of air = mass of air x acceleration due to gravity weight of air = 390 kg x 10 m/s^2 weight of air = 3900 N The upthrust on the balloon is equal to the weight of the air displaced, which is 3900 N. Therefore, the correct answer is "3900N." Explanation: The upthrust on an object in a fluid is equal to the weight of the fluid displaced by the object. In this case, the balloon filled with hydrogen displaces air, which has a known density. By calculating the weight of air displaced and subtracting the weight of the hydrogen filling the balloon, we can find the upthrust on the balloon. By using the given values and applying Archimedes' principle, we find that the upthrust on the balloon is 3900 N. Therefore, the correct answer is "3900N."
Swali 16 Ripoti
The fuse in an electric device is always connected to the
Maelezo ya Majibu
The fuse in an electric device is always connected to the live side of an electric supply. The live wire, also known as the "hot" wire, carries the electrical current from the power source to the device. The fuse is designed to protect the device and its components from damage in the event of an electrical surge or overloading. If the current becomes too high, the fuse will "blow," cutting off the flow of electricity and preventing the device from being damaged. By connecting the fuse to the live wire, it can protect the device from any electrical problems that may arise.
Swali 17 Ripoti
A load is pulled at a uniform speed along a horizontal floor by a rope at 45 degree floor. If the force in the rope is 1500N, what is the frictional force on the load ?
Maelezo ya Majibu
F = mg cos 45 = 1500 x 0.708
Swali 18 Ripoti
Maelezo ya Majibu
V2=P1V1T2P2T2
V2=700×2000×273760×300
Swali 19 Ripoti
Which of the following particies CANNOT be deflected by both electric and magnetic fields?
Maelezo ya Majibu
Gamma rays cannot be deflected by both electric and magnetic fields. Gamma rays are high-energy electromagnetic radiation and do not have a charge, unlike alpha and beta particles. This means they are not affected by electric or magnetic fields and continue to travel in a straight line. On the other hand, alpha and beta particles, which are charged particles, can be deflected by electric and magnetic fields.
Swali 20 Ripoti
A particle carrying a charge of 1. 0 x 10-8C enters a magnetic field at 3.0x 10-2 ms -1 at right angles to the field. If the force on this particle is 1.8 x 10-8N, what is the magnitude of the field?
Maelezo ya Majibu
F = BQV
B = FQV
B = 1.8×10−81.0×10−8×3.0×10−2
Swali 21 Ripoti
The process of detecting a pin mistakenly swallowed by a child is X-ray
Maelezo ya Majibu
The process of detecting a pin mistakenly swallowed by a child is X-ray. X-ray is a diagnostic imaging tool that uses high-energy radiation to produce images of the internal structure of the body. In the case of a child swallowing a pin, an X-ray can be used to locate the position of the pin and determine if it is causing any harm. The X-ray produces a picture of the inside of the body, allowing medical professionals to see the location and position of the pin, and make any necessary medical decisions based on that information. So, X-ray is a diagnostic tool, not a therapy or a tool for crystallography or mammography.
Swali 22 Ripoti
A piece of radioactive material contains 1000 atmos. If its half -life is 20seconds, the time taken for 125 atoms to remain is
Maelezo ya Majibu
1000 → 500 → 250 → 125
20 + 20 + 20 = 60s
Swali 23 Ripoti
A glass block of thickness 10cm is placed on an object. If an observer views the object vertically, the displacement of the object is
Maelezo ya Majibu
Swali 24 Ripoti
Which of the following is NOT a factor that can increase the rate of evaporation of water in a lake?
Maelezo ya Majibu
Swali 25 Ripoti
An object of volume 1 m 3 and mass 2kg is totally immersed in a liquid of density 1 kgm-3 . Calculate its apparent weight.
Maelezo ya Majibu
The apparent weight of the object is the difference between the actual weight of the object and the buoyant force acting on it when it is immersed in a fluid. The buoyant force is equal to the weight of the fluid displaced by the object. In this case, the volume of the object is 1 m^3 and its mass is 2 kg, which means its density is 2 kg/m^3. The density of the liquid is given as 1 kg/m^3. Since the density of the object is greater than the density of the liquid, it will sink in the liquid. When the object is totally immersed in the liquid, it will displace a volume of liquid equal to its own volume. Therefore, the weight of the displaced liquid is given by the product of the volume and density of the liquid, which is 1 m^3 * 1 kg/m^3 = 1 kg. This is also the buoyant force acting on the object. The actual weight of the object is 2 kg * 9.81 m/s^2 = 19.62 N (where 9.81 m/s^2 is the acceleration due to gravity). Therefore, the apparent weight of the object is the difference between the actual weight and the buoyant force, which is 19.62 N - 1 kg * 9.81 m/s^2 = 9.81 N. Therefore, the apparent weight of the object is 9.81 N, which is closest to.
Swali 26 Ripoti
Maelezo ya Majibu
To find the height of the building, we can use the principle of similar triangles. The ratio of the height of the building to the distance from the pinhole to the building is equal to the ratio of the height of the image to the distance from the pinhole to the screen. So, we can write the following equation: (height of building) / (distance from pinhole to building) = (height of image) / (distance from pinhole to screen) Plugging in the given values, we get: (height of building) / 300m = 2.5cm / 5cm Solving for the height of the building, we get: (height of building) = (2.5cm / 5cm) * 300m Since 1 meter is equal to 100 cm, we can convert the answer to meters: (height of building) = (2.5 / 5) * 300 * (100 cm/m) = 150m So, the height of the building is 150 meters.
Swali 27 Ripoti
The magnification of an object 2cm tall when placed 10cm in front of a plane mirror is
Maelezo ya Majibu
When an object is placed in front of a plane mirror, its image appears to be behind the mirror, at the same distance from the mirror as the object is in front of it. The image appears to be a virtual image that is the same size as the object, but reversed from left to right. In this case, the object is 2cm tall and is placed 10cm in front of the mirror. The image will appear to be 10cm behind the mirror, which is the same distance as the object is in front of the mirror. The image will also be the same size as the object, which is 2cm tall. To find the magnification, we need to divide the height of the image by the height of the object: Magnification = height of image / height of object In this case, the height of the image is 2cm and the height of the object is also 2cm, so: Magnification = 2cm / 2cm = 1.0 Therefore, the magnification of the object when placed 10cm in front of a plane mirror is 1.0. This means that the image appears to be the same size as the object.
Swali 28 Ripoti
I. Small size.
II. Low power requirement
III. Not easily damaged by high temperature
IV. Highly durable. Which of the above are the advantages of semiconductors?
Maelezo ya Majibu
Semiconductors are materials with electrical conductivity between conductors (like metals) and insulators (like ceramics). They are used extensively in electronic devices like transistors, diodes, and integrated circuits due to their unique properties, which include: I. Small size: Semiconductors can be fabricated into very small sizes, allowing for the production of compact electronic devices. II. Low power requirement: Semiconductors require very little power to operate, making them suitable for use in low-power devices like mobile phones and laptops. III. Not easily damaged by high temperature: Semiconductors can withstand high temperatures without damage, making them suitable for use in high-temperature applications like automotive and aerospace electronics. IV. Highly durable: Semiconductors are very durable and can withstand a lot of wear and tear, making them suitable for use in industrial and military applications. Therefore, the answer to the question is: "The advantages of semiconductors are I, II, III, and IV." In conclusion, semiconductors are a crucial component of modern electronics due to their unique properties. They allow for the production of small, low-power, and durable electronic devices that can withstand high temperatures, making them suitable for a wide range of applications.
Swali 29 Ripoti
A rocket is fired from the earth's surface to a distant planet. By Newton's law of universal gravitation, the force F will
Maelezo ya Majibu
Swali 30 Ripoti
In order to remove the error of parallax when talking measurements with a metre rule, the eye should be focused
Maelezo ya Majibu
When taking measurements with a meter rule, it is important to avoid parallax error. Parallax error occurs when the eye is not positioned correctly in relation to the meter rule markings, causing an incorrect reading to be taken. To avoid this error, the eye should be positioned directly above the marking being measured, looking vertically downwards at it. This ensures that the eye is perpendicular to the markings and there is no angle between the eye and the markings. Therefore, the correct answer is "vertically downwards on the markings." Focusing the eye slantingly towards the right or left on the markings, or vertically upwards, would introduce an angle between the eye and the markings, resulting in a parallax error.
Swali 31 Ripoti
The pressure at any point in a liquid at rest depends only on the
Maelezo ya Majibu
The pressure at any point in a liquid at rest depends only on the depth and the density of the liquid. This is known as Pascal's principle or Pascal's law. According to this principle, the pressure at any point in a fluid is the same in all directions and is transmitted uniformly throughout the fluid. The pressure at a point in a liquid at rest is directly proportional to the depth of the point below the surface of the liquid. This is because the weight of the liquid above the point exerts a force on the liquid at the point, which results in a pressure that is directly proportional to the depth of the point. The pressure at a point in a liquid at rest is also directly proportional to the density of the liquid. This is because the weight of a given volume of liquid is proportional to its density, and the pressure at a point in the liquid is determined by the weight of the liquid above the point. Therefore, the pressure at any point in a liquid at rest depends only on the depth and the density of the liquid, and not on the mass, volume, quantity, surface area or viscosity of the liquid.
Swali 32 Ripoti
Calculate the temperature change when 500J of heat is supplied to 100g of water
Maelezo ya Majibu
To calculate the temperature change when a certain amount of heat is supplied to a substance, we can use the following formula: Q = m * c * ΔT where Q is the amount of heat supplied, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature. In this problem, we are given that 500J of heat is supplied to 100g of water. We know that the specific heat capacity of water is 4.18 J/(g·°C). So, we can substitute the values into the formula and solve for ΔT: Q = m * c * ΔT 500J = 100g * 4.18 J/(g·°C) * ΔT ΔT = 500J / (100g * 4.18 J/(g·°C)) ΔT = 1.2°C Therefore, the temperature change when 500J of heat is supplied to 100g of water is 1.2°C. In conclusion, the temperature change of a substance when heat is supplied to it depends on the specific heat capacity of the substance, the mass of the substance, and the amount of heat supplied. The formula for calculating the temperature change can be used to solve problems involving heat and temperature change.
Swali 33 Ripoti
A newly charged 12V accumulator can easily start a car whereas eight new dry cells in series with a effective e.m.f. of 12V cannot start the same car because
Maelezo ya Majibu
Swali 34 Ripoti
A conductor of length 1m moves with a velocity of 50ms-1 at an angle of 30o to the direction of a uniform magnetic field of flux density 1.5Wbm-2. What is the e.m.f induced in the conductor?
Maelezo ya Majibu
The electromotive force (e.m.f) induced in a conductor moving in a magnetic field can be calculated using Faraday's law of electromagnetic induction, which states that the e.m.f induced in a conductor is equal to the rate of change of the magnetic flux through the conductor. The magnetic flux can be calculated as the product of the magnetic field strength and the area of the conductor that is perpendicular to the magnetic field. In this case, the magnetic field strength is given as 1.5 Wb/m^2, the length of the conductor is 1m, and the angle between the velocity of the conductor and the direction of the magnetic field is 30 degrees. The area of the conductor that is perpendicular to the magnetic field can be calculated as the product of the length of the conductor and the component of its velocity that is perpendicular to the magnetic field. The component of velocity that is perpendicular to the magnetic field can be calculated as the product of the velocity and the sine of the angle between the velocity and the magnetic field direction. Thus, the e.m.f induced in the conductor can be calculated as follows: e.m.f = rate of change of magnetic flux = magnetic field strength * area of conductor perpendicular to magnetic field = 1.5 Wb/m^2 * (1m) * (50 m/s * sin(30 degrees)) = 37.5V Therefore, the e.m.f induced in the conductor is 37.5V.
Swali 35 Ripoti
The characteristic of a vibration that determines its intensity is the
Maelezo ya Majibu
The characteristic of a vibration that determines its intensity, or how strong or weak it is, is the amplitude. The amplitude is the maximum displacement of a point on a vibrating body from its rest position. Simply put, it's the height of the vibration wave. The higher the amplitude, the stronger the vibration, and the more energy it carries. In contrast, a low amplitude vibration is weaker and carries less energy.
Swali 36 Ripoti
I. Rays of light travel from a less dense medium to a denser medium. II. The Angle of incidence is greater than critical angle. III. Rays of light travel from a denser medium to a less dense medium. Which of the statements above are conditions for total internal reflection to occur?
Swali 37 Ripoti
After reflection from the concave mirror, rays of light from the sun converges
Maelezo ya Majibu
A concave mirror is a spherical mirror with a reflecting surface that is curved inward. When parallel rays of light fall on a concave mirror, they are reflected in a way that depends on the distance between the mirror and the object, as well as the curvature of the mirror. If the object is located beyond the focal point of the concave mirror, the reflected rays will converge at a point between the focal point and the radius of curvature of the mirror. This point is known as a real image. However, if the object is located at a distance less than the focal length from the mirror, the reflected rays will diverge and not form a real image. Instead, they will form a virtual image that is located behind the mirror. In the case of the question, if rays of light from the sun fall on a concave mirror, and the mirror is positioned such that the sun's rays are parallel to its axis, then the reflected rays will converge at the focal point of the mirror. This is because the focal point is the point at which parallel rays of light, such as those from the sun, converge after reflection from the concave mirror. Therefore, the answer to the question is: "After reflection from the concave mirror, rays of light from the sun converge at the focus." In conclusion, the position of the object and the distance between the object and the concave mirror determine the nature of the image formed by the reflected rays of light. When the object is located at a distance greater than the focal length, a real image is formed that is inverted and smaller than the object. When the object is located at a distance less than the focal length, a virtual image is formed that is upright and larger than the object.
Swali 39 Ripoti
The quantity of heat energy required to melt completely 1 kg of ice at - 30oC is
[latent heat of fusion = 3.5 x 10 5Jkg -1, specific heat capacity of ice = 2.1x10 3Jkg -1K -1]
Maelezo ya Majibu
When a substance changes its phase, it either absorbs or releases heat energy. The amount of heat energy required to change the phase of a substance is known as the latent heat of fusion. In this case, we have to calculate the quantity of heat energy required to melt completely 1 kg of ice at -30°C. To do this, we need to use the following formula: Q = m Lf Where Q is the quantity of heat energy, m is the mass of the substance, and Lf is the latent heat of fusion. We are given that the mass of the ice is 1 kg, and the latent heat of fusion of ice is 3.5 x 10^5 Jkg^-1. So, substituting these values into the formula, we get: Q = 1 kg x 3.5 x 10^5 Jkg^-1 Q = 3.5 x 10^5 J This means that the quantity of heat energy required to melt completely 1 kg of ice at -30°C is 3.5 x 10^5 J. However, we need to be careful in calculating the quantity of heat energy required to lower the temperature of the ice from 0°C to -30°C before it starts melting. To do this, we need to use the following formula: Q = m c ΔT Where Q is the quantity of heat energy, m is the mass of the substance, c is the specific heat capacity of ice, and ΔT is the change in temperature. We are given that the mass of the ice is 1 kg, the specific heat capacity of ice is 2.1 x 10^3 Jkg^-1K^-1, and the change in temperature is 30°C. So, substituting these values into the formula, we get: Q = 1 kg x 2.1 x 10^3 Jkg^-1K^-1 x 30 K Q = 6.3 x 10^4 J Therefore, the total quantity of heat energy required to melt completely 1 kg of ice at -30°C is the sum of the quantity of heat energy required to lower the temperature of the ice from 0°C to -30°C and the quantity of heat energy required to melt the ice. Total Q = 6.3 x 10^4 J + 3.5 x 10^5 J Total Q = 4.13 x 10^5 J So, the correct answer is 4.13 x 10^5 J.
Swali 40 Ripoti
A block and tackle is used to raise a load of 250N through a vertical distance of 30m. What is the efficiency of the system if the work done against friction is 1500J?
Maelezo ya Majibu
A block and tackle is a system of pulleys that allows a person to lift a heavy load with less effort than it would take to lift the load directly. The efficiency of the block and tackle system is the ratio of the work output to the work input, expressed as a percentage. In this problem, the load being lifted is 250N, and it is being lifted through a vertical distance of 30m. The work done against friction is 1500J. We need to calculate the efficiency of the system. The work output of the system is the work done on the load, which is the product of the force applied to the load and the distance the load is lifted. In this case, the work output is: Work output = Force x Distance = 250N x 30m = 7500J The work input of the system is the work done by the person who is operating the block and tackle system. This includes the work done to lift the load as well as the work done against friction. We can calculate the work input as: Work input = Work output + Work against friction = 7500J + 1500J = 9000J Now we can calculate the efficiency of the system: Efficiency = (Work output / Work input) x 100% Efficiency = (7500J / 9000J) x 100% Efficiency = 83.3% Therefore, the efficiency of the system is 83.3%.
Swali 41 Ripoti
I. It is a rapid, constant and irregular motion of tiny particles. II. It gives evidence that tiny particles of matter called molecules exist. III. It takes place only in gases. IV. It gives evidence that molecules are in a constant state of random motion. which of the combinations above is correct about Brownian notion?
Maelezo ya Majibu
Swali 42 Ripoti
Maelezo ya Majibu
The angular velocity of an object moving in a circular path is the rate at which the object is rotating around the center of the circle. It is usually measured in radians per second (rad/s). To find the angular velocity of the object in this question, we can use the formula: angular velocity = speed / radius where speed is the speed of the object and radius is the radius of the circular path. Substituting the values given in the question, we get: angular velocity = 1 m/s / 0.5 m which simplifies to: angular velocity = 2 rad/s Therefore, the answer is 2 rads-1.
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