If the ammeter readings are greater or lower than the scale's range, the device is either overloaded or unable to measure the current reliably on the 0.5A scale.
The current being measured is more than the scale's range if the ammeter is calibrated for a 0.5 A scale and reads 2A. Alternatively put, If the ammeter is calibrated for 0.5 A and displays 0.2 A, Similarly, if the ammeter registers 0.002A, the current being measured is 0.002A or less and falls within the range of the scale. A reading of 0.02A on the ammeter indicates that the current being measured is 0.02A or less and falls within the range of the scale. If the ammeter reads 10A, the ammeter is overloaded and the current being measured is greater than what the scale can handle.
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you push a 15.0 kg box across the floor at a steady speed of 1.25 m/s for 16.0 s. the coefficient of kinetic friction between the box and the floor is 0.680. what is your power output?
The power output of pushing a 15.0 kg box across the floor at a steady speed of 1.25 m/s for 16.0 s, with a coefficient of kinetic friction of 0.680, can be calculated using the formula:
Power = Force x Velocity
Force = mass x acceleration due to friction
Acceleration due to friction = coefficient of kinetic friction x gravitational force
Power = (15.0 kg x 0.680 x 9.8 m/s2) x (1.25 m/s)
Power = 128.4 Watts
Therefore, the power output of pushing the box for 16.0 s is 128.4 Watts.
pls help 60 pts
screenshot, fill out, then screenshot . attach document to answer using the paperclip button
Answer:
Here you go, hope this helps you pass :)
Explanation:
You're welcome.
use the listen to a single source tab in sound waves to starts your investigation of sound. when you change the frequency, how does the visual model change?
When you change the frequency in the "Listen to a Single Source" tab of Sound Waves, the visual model will change in several ways.
What is sound?Sound is a type of energy that travels through a medium in waves. These waves propagate in the form of longitudinal or compression waves, and they vibrate the particles in the medium through which they travel. Sound waves can travel through solids, liquids, and gases, but they do so at different speeds.
The amplitude (height of the wave) will change based on the volume of the sound, and the wavelength (length of the wave) will change based on the frequency.
As the frequency increases, the wavelength will become shorter and the amplitude will become higher. As the frequency decreases, the wavelength will become longer and the amplitude will become lower.
The Listen to a Single Source tab in Sound Waves is a useful tool for beginning your investigation into sound. When you adjust the frequency, the visual model changes.
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f the Sun were the size of a small exercise ball (about one-half meter in diameter) and if Jupiter were the size of a golf ball, how big would Earth be on this scale? The size of a hot-air balloon, because Earth is larger than the Sun. The size of a golf ball, because Earth is about the same size as Jupiter. The size of a baseball, because Earth is larger than Jupiter. The size of a pea, because Earth is smaller than Jupiter.
Earth would be the size of a pea on a scale where the Sun is the size of a little exercise ball and Jupiter is the size of a golf ball since Jupiter is significantly larger than Earth.
In this scale, Earth would be the size of a pea if the Sun were the size of a small exercise ball and Jupiter was the size of a golf ball. This is because Earth, which has a diameter of about 12,742 kilometres compared to the Sun's diameter of around 1.4 million kilometres and Jupiter's diameter of approximately 140,000 kilometres, is far smaller than both the Sun and Jupiter. Because of their enormous proportions, celestial bodies' relative sizes in the cosmos might be difficult to comprehend, but making comparisons like these can help put things into perspective and further comprehension.
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mastering astronomy based on current evidence, how common are planetary systems? based on current evidence, how common are planetary systems? all stars should have planets orbiting around them. only 10 % of stars have planetary systems. about 70 % of stars have planetary systems. almost exactly half of stars have planetary systems.
Based on current evidence, planetary systems appear to be relatively common.
What is current evidence?Current evidence suggests that regular physical activity can improve physical and mental health, reduce the risk of developing chronic diseases, and promote well-being. Studies have found that regular exercise can reduce stress, improve mood, improve sleep quality, and increase energy levels. Exercise can also reduce the risk of developing heart disease, stroke, diabetes, and certain types of cancer. Additionally, regular physical activity can help to maintain a healthy weight, reduce the risk of osteoporosis, and improve cognitive function.
Studies using data from the Kepler Space Telescope suggest that around 70% of stars have at least one planetary system, with some estimates as high as 90%. This means that for every 10 stars, 7 of them could potentially have planetary systems orbiting them. This suggests that planetary systems are relatively common in the Universe.
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I need to figure out the missing boxes and work out the power, work done or time taken
To calculate the missing values in the table, we can use the formulas:
Power = Work Done / Time TakenWork Done = Power x Time TakenTime Taken = Work Done / PowerUsing these formulas, we can fill in the missing values:
Power(W) Work Done(J) Time Taken(s)
For a,. 3000 90000 30
For b,. 20 100 5
For c,. 50 100 2
For d. 700 245 0.35
For e. 25 1875 75
For f. 50000 500000 10
For g. 150 450 3
For h. 0.5 10 20
What is power?
Power is the rate at which work is done or energy is transferred. Mathematically, power is defined as the amount of work done or energy transferred per unit time. The unit of power is the watt (W), which is equal to one joule of work per second.
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To calculate the missing values in the table, we can use the formulas:
Power = Work Done / Time Taken
Work Done = Power x Time Taken
Time Taken = Work Done / Power
Using these formulas, we can fill in the missing values:
Power(W) Work Done(J) Time Taken(s)
For a,. 3000 90000 30
For b,. 20 100 5
For c,. 50 100 2
For d. 700 245 0.35
For e. 25 1875 75
For f. 50000 500000 10
For g. 150 450 3
For h. 0.5 10 20
What is power?
Power is the pace at which labour or energy is done or transferred. Power is described mathematically as the quantity of labour done or energy moved per unit time. The watt (W) is the measure of electricity, and it equals one joule of work per second.
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Materials that allow electricity to flow are called A) Insulators B) Conductors C) Electron flows D) Stimulators
B) Conductors, Materials known as conductors allow electricity to pass through them with little or no resistance. Metals like aluminum, silver, and copper are among the examples.
Materials known as conductors have a low resistance to electricity passing through them. They are able to move freely through the material because they have free electrons that are not strongly bonded to the atomic nucleus. Due to their abundance of free electrons, metals like copper, aluminium, silver, and gold are the most popular conductors. A material's conductivity—expressed as electrical conductivity or its opposite, resistivity—determines its capacity to carry electricity. Electrical wires, power transmission lines, electronic components, and circuit boards are just a few of the many uses for conductors. They also have a variety of industrial uses and are found in medical equipment like pacemakers and EEG machines.
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A train is moving up a steep grade at constant velocity (see following figure) when its caboose breaks loose and starts rolling freely along the track. After 5.0 s, the caboose is 30 m behind the train. What is the acceleration of the caboose?
The velocity of the caboose is constant, so the acceleration is zero. Therefore, the caboose's acceleration is 0 m/s².
Acceleration is the rate at which the velocity of an object changes over time. The formula for acceleration is expressed as a = (v - u) / t where a is acceleration, v is final velocity, u is initial velocity, and t is time.
The velocity of the train and the caboose is the same. The caboose breaks loose and starts rolling freely along the track. Therefore, the velocity of the caboose is the same as the velocity of the train.
Given that the train moves at a constant velocity, the initial velocity of the caboose is the same as the final velocity.
Using the formula above, the acceleration of the caboose is calculated as follows:
a = (v - u) / ta
= (0 - 0) / 5.0
a = 0 m/s²
Therefore, the acceleration of the caboose is 0 m/s². This result makes sense since the caboose is moving at constant velocity.
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11- An object of mass 0.2 kg moves in a circular path. If it makes 3/4 revolution in 0.3 s and its displacement is 6 m, calculate: a) The radius of the circular path ?
We can use the kinematic equations for circular motion to solve this problem. The equation we will use is:
θ = (s / r)
where θ is the angle of rotation (in radians), s is the displacement, and r is the radius of the circular path.
In this case, the object makes 3/4 revolution, which is equivalent to an angle of rotation of:
θ = (3/4) x 2π radians = 1.5π radians
The displacement is given as 6 m.
Therefore, we can rearrange the equation to solve for the radius:r = s / θ
r = 6 m / (1.5π radians)
r ≈ 1.273 m
Therefore, the radius of the circular path is approximately 1.273 meters.
What is a path ?
A path is a route or way that someone or something follows to reach a destination or achieve a particular goal. It can be physical or abstract, and can refer to a variety of contexts such as travel, communication, decision-making, or personal development.
A physical sense, a path can refer to a trail, road, or sidewalk that people use to walk, cycle, or drive to a particular location. In computing, a path refers to the sequence of directories and that lead to a specific file or folder.
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During a baseball game, the sound of the bat hitting the ball can be heard in most parts of the stadium. That sound is weaker at greater distances. What is the cause of this phenomenon?(1 point)
The sound waves are spread out over a large area.
The sound waves are blocked by people in the stadium.
The sound waves can only travel through certain materials.
The sound waves slow down as they move away from the bat.
The cause of the phenomenon is because the sound waves are spread out over a large area. Option 1 is correct.
How do sound waves propagate in this case?In this case, the sound waves are spread out over a large area. This is due to the fact that sound waves propagate outward from their source in all directions, creating a spherical wavefront that expands as it moves away from the source. As the wavefront expands, the same amount of sound energy is distributed over a larger and larger area, causing the sound intensity to decrease with distance from the source. This is known as the inverse square law, which states that the intensity of a sound wave decreases proportionally to the square of the distance from the source. Therefore, the farther away you are from the source of the sound, the weaker it will be.
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the bottom of the tank is 30 feet in diameter and groundwater level is 10 ft above the bottom of the tank. how much force is being exerceted on the bottom of the tank?
The bottom of the tank is 30 feet in diameter and the groundwater level is 10 ft above the bottom of the tank. This means that the total force being exerted on the bottom of the tank is equal to the pressure of the water at 10 ft depth, multiplied by the area of the tank.
The pressure of water at 10 ft depth can be calculated using the formula: Pressure = Density of Water x Gravity x Depth
Therefore, the force exerted on the bottom of the tank is:
Force = Density of Water x Gravity x Depth x Area of the TankWhere:
Density of Water is 62.4 lb/ft³Gravity is 32.2 ft/s2, Depth is 10 ftArea of the Tank is 706.5 ft²Therefore, the total force exerted on the bottom of the tank is: Force = 62.4 x 32.2 x 10 x 706.5 = 1358561.2 lbf.
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If the magnetic field steadily decreases from B to zero during a time interval t , what is the magnitude E of the induced emf?
Express your answer in terms of x,y ,t , and B .
If the magnetic field steadily decreases from B to zero during a time interval t, the magnitude E of the induced emf is given by the formula; E = (Bx-y/t), where B is the magnetic field, x, and y are constants.
An induced emf is the voltage generated across a conductor when it moves through a magnetic field. It is also induced when there is a change in the magnetic field passing through a conductor.
The emf generated in a coil of wire is equal to the rate of change of magnetic flux through the coil. Magnetic flux is given by the formula: φ=B*A,
where - B is the magnetic field strength and
- A is the area of the coil.
If the magnetic field steadily decreases from B to zero during a time interval t, the change in magnetic flux is given by the formula: Δφ=B*A = B*ΔA, where ΔA is the change in area over time Δt.
The induced emf E is given by the formula: E = (-N * Δφ)/Δt
Where N is the number of turns in the coil. If the magnetic field is steadily decreasing, then ΔB/Δt is constant, and the induced emf E is given by the formula: E = (-N * B * ΔA/Δt) = (-N * B * x*y/t) = (Bx-y/t), where x and y are constants.
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A tetherball is on a 2.1 m string which makes an angle of 56 degree with the vertical as it moves around the pole in a horizontal plane. If the mass of the ball is 1.3 kg, what is the ball's speed?
The tetherball's speed can be calculated as 4.23m/s.
The question asks for the speed of a tetherball on a 2.1 m string which is making an angle of 56 degrees with the vertical as it moves around the pole in a horizontal plane. The mass of the ball is 1.3 kg.
We can use the equation v2 = 2gLsin θ to solve for the speed. Where v is the speed of the ball, g is the acceleration due to gravity (9.8 m/s2), L is the length of the string (2.1 m), and θ is the angle (56 degrees).
Using this equation, we can solve for the speed of the tetherball: v = 4.23 m/s
Hence, the tetherball's speed can be calculated as 4.23m/s
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an appliance manufacturer has redesigned its original electrical fan. it has made the new fan more energy efficient. this means that the new fan
An appliance manufacturer that has redesigned its original electrical fan to make it more energy efficient means that the new fan will consume less electricity for the same amount of air circulation compared to the original fan.
This can result in several benefits such as: Lower electricity bills: As the new fan is more energy efficient, it will consume less electricity, which can lead to lower electricity bills for the consumer. Reduced environmental impact: By consuming less electricity, the new fan can reduce the overall demand for electricity, which can help to reduce the environmental impact associated with electricity generation. Longer lifespan: Energy-efficient fans tend to have a longer lifespan compared to less efficient models as they put less stress on the electrical components, which can result in fewer malfunctions and a longer lifespan. Improved performance: While the new fan consumes less electricity, it may still be capable of producing the same or better airflow compared to the original fan due to improvements in the design and technology used. Overall, making the new fan more energy efficient can result in several benefits for the consumer and the environment.
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True or False. when a system can go from state 1 to state 2 by several different processes, the change in the internal energy of the system will be the same for all processes.
The given statement: "When a system can go from state 1 to state 2 by several different processes, the change in the internal energy of the system will be the same for all processes." is False.
It is incorrect to claim that if a system may shift from state 1 to state 2 through various processes, the shift in the internal energy of the system will be the same for all processes. This is due to the fact that internal energy is a state function that is dependent solely on the initial and final states of the system, and not on the path by which the system was transformed from one state to another. As a result, the alteration in the internal energy of a system moving from one state to another is not influenced by the process used to achieve the transition. So, it can be concluded that the given statement is False.
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Suppose two protostars form at the same time, one with a mass of 0.5 msun and the other with a mass of 15 msun . Which of the following statements are true?
[Select ALL answers that are true in alphabetical order]
A) The 10MSun protostar will have a smaller change in surface temperature during this phase than the 0.5MSun protostar.
B) The 10MSun protostar will reach the main sequence cooler and fainter than the 0.5MSun protostar.
C) The 10MSun star will end its main-sequence life before the 0.5MSun star even completes its protostar stage.
D) The 10MSun protostar will have a smaller change in luminosity during the sequence shown than the 0.5MSun protostar.
E) The 10MSun protostar will be much more luminous than the 0.5MSun protostar.
The two protostars form at the same time, the 10MSun star will end its main-sequence life before the 0.5MSun star even completes its protostar stage, and 10MSun will be much more luminous than the 0.5MSun protostar. Thus, options C and E are correct.
More massive stars have shorter lifetimes because they burn through their nuclear fuel more quickly.
Therefore, the 10MSun star will end its main-sequence life before the 0.5MSun star even completes its protostar stage. Thus, option C is correct.
The luminosity of a star depends on its mass.
More massive stars have higher luminosities than less massive stars. Therefore, option E is correct.
Thus, options C and E are correct.
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waves that have traveled out of their area of origin and have become uniform and symmetrical are known as .
The waves that have traveled out of their area of origin and have become uniform and symmetrical are known as spherical waves.
When a wave is generated from a point source, it initially spreads out in all directions as a series of expanding spheres. These waves are called spherical waves because the wavefronts form concentric spheres around the source. As the waves travel further away from the source, they begin to merge and overlap, and the wavefronts become more uniform and symmetrical.
Spherical waves are a fundamental concept in wave physics and are used to describe a wide range of physical phenomena, including sound waves, electromagnetic waves, and water waves. They are particularly useful in situations where the wave source is small and the distance from the source is large compared to the wavelength of the wave.
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Discuss three applications of the effects of surface tension.
Earthquake _____________ measures shaking at a given location (and so is strongly influenced by _____________) while _____________ is a measurement of the total energy released.Intensity; substrate; magnitude
Earthquake intensity measures shaking at a given location and so is strongly influenced by the substrate, while earthquake magnitude is a measurement of the total energy released.
What is earthquake ?
An earthquake is a natural phenomenon that occurs when the Earth's tectonic plates, which are large pieces of the Earth's crust, move and shift against each other. These movements cause vibrations, or seismic waves, which can be felt on the Earth's surface as shaking or ground motion.
Earthquakes can range in size from small tremors that are barely felt to massive quakes that can cause widespread damage and loss of life. The strength of an earthquake is measured on the Richter scale, which assigns a numerical value to the amplitude of the seismic waves generated by the quake.
Earthquake intensity is a measure of the strength of shaking that is felt at a particular location, and it is influenced by a variety of factors, including the distance from the earthquake epicenter, the local geology and substrate, and the magnitude of the earthquake. Intensity is usually measured on the Modified Mercalli Intensity Scale, which assigns a numerical value to the observed effects of shaking at a given location.
Earthquake magnitude, on the other hand, is a measure of the total amount of energy released by an earthquake, and it is based on the seismic waves that are generated by the earthquake. Magnitude is usually measured on the Richter scale, which assigns a numerical value to the amplitude of the seismic waves as recorded by a seismograph.
In summary, earthquake intensity measures shaking at a particular location and is influenced by the substrate, while earthquake magnitude measures the total energy released by an earthquake and is based on the seismic waves generated by the earthquake.
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what is the net torque when you are holding a 10 lb weight 1 ft from your elbow (producing a clockwise torque) and your biceps is generating a 40 lb force with a moment arm of 6 in. (producing a counterclockwise torque)?
The net torque when you are holding a 10 lb weight 1 ft from your elbow (producing a clockwise torque) and your biceps are generating a 40 lb force with a moment arm of 6 in (producing a counterclockwise torque) can be calculated using the formula below:τ = r x F where τ is the torque, r is the moment arm, and F is the force. In this scenario, the clockwise torque produced by the weight is τ1 = (-1 ft) x (10 lb) = -10 lb-ft (negative because it is clockwise), and the counterclockwise torque produced by the biceps is τ2 = (0.5 ft) x (40 lb) = 20 lb-ft (positive because it is counterclockwise).To find the net torque, we must add the two torques together:
τnet = τ1 + τ2τnet = (-10 lb-ft) + (20 lb-ft)
τnet = 10 lb-ft
Therefore, the net torque is 10 lb-ft.
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Water is flowing in a circular pipe varying cross-sectional area, and at all points, the water completely fills the pipe.a) At one point in the pipe the radius is 0.150 m. What is the speed of the water at this point if the water is flowing into this pipe at a steady rate of 1.20 m3/s?b) At a second point in the pipe the water speed is 2.90 m/s. What is the radius of the pipe at this point?
The speed of water at the point with a radius of 0.150 m is 16.97 m/s while the radius of the pipe at the point where the water speed is 2.90 m/s is 0.0682 m.
a) To find the speed of the water at a point of a circular pipe where the radius is 0.150 m if the water is flowing into this pipe at a steady rate of 1.20 m³/s, we'll use the equation;
Q = A₁V₁ = A₂V₂ Where Q = Flow rate (m³/s)A₁ = Cross-sectional area at one point (m²)V₁ = Velocity of water at one point (m/s)A₂ = Cross-sectional area at a second point (m²)V₂ = Velocity of water at the second point (m/s)At one point in the pipe, the radius is 0.150 m.Therefore, the cross-sectional area, A₁ is given by:
A₁ = πr₁² = π (0.150 m)² = 0.0707 m²Given that the water is flowing into the pipe at a steady rate of 1.20 m³/s, we can write;Q = A₁V₁1.20 m³/s = 0.0707 m² V₁V₁ = 1.20/0.0707V₁ = 16.97 m/s.Therefore, the speed of water at the point with a radius of 0.150 m is 16.97 m/s.
b) To find the radius of the pipe at a point where the water speed is 2.90 m/s, we'll use the same equation as in part (a);Q = A₁V₁ = A₂V₂At a second point in the pipe, the water speed is 2.90 m/s.Given that the water completely fills the pipe, we know that the volume flow rate, Q will remain constant at 1.20 m³/s.So, we have:
Q = A₁V₁ = A₂V₂We know that A₁ = πr₁²So, Q = πr₁²V₁Also, we know that A₂ = πr₂²So, Q = πr₂²V₂Since the volume flow rate is constant, we can equate both equations,πr₁²V₁ = πr₂²V₂Dividing both sides of the equation by π, we have;r₁²V₁ = r₂²V₂But we are interested in finding the radius of the pipe at the second point, r₂.So, we can express r₁ in terms of r₂ using the relationship between the cross-sectional areas;
A₁ = A₂r₁² = (A₂/A₁)²r₂²r₁ = r₂ (A₂/A₁)^(1/2).We know that A₁ = πr₁²We can find A₂ using the fact that the water completely fills the pipe;
A₁V₁ = A₂V₂πr₁²V₁ = A₂V₂π(0.150 m)²(16.97 m/s) = A₂(2.90 m/s)A₂ = π(0.150 m)²(16.97 m/s)/(2.90 m/s)A₂ = 0.0707 m²
So,r₂ = r₁(A₂/A₁)^(1/2)r₂ = 0.150 m × (0.0707 m²/π)/(0.0150 m²)^(1/2)r₂ = 0.0682 m. Therefore, the radius of the pipe at the point where the water speed is 2.90 m/s is 0.0682 m.
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what diameter should the nichrome wire be in order for the electric field strength to be the same in both wires?
The diameter of the nichrome wire should be 0.71 times the diameter of the copper wire in order for the electric field strength to be the same in both wires.
To determine the diameter ratio of nichrome wire to copper wire, we can use the equation for electric field strength (E = V / d), where V is the potential difference and d is the distance between the wires.
Since the potential difference between the wires is the same for both wires, we can set the electric field strength for the two wires equal to each other: E_nichrome = E_copper
Using the equation for electric field strength, we can rewrite this as: V_nichrome / d_nichrome = V_copper / d_copper
We know that the diameters of the wires are inversely proportional to their respective distances from each other, so we can rewrite this equation as: V_nichrome / (k * d_copper) = V_copper / d_copper, where k is the ratio of the diameter of the nichrome wire to the diameter of the copper wire.
Solving for k, we get: k = (V_nichrome / V_copper)^(1/2)
Substituting the given values, we get: k = (60 / 40)^(1/2) = 0.866
Therefore, the diameter of the nichrome wire should be 0.71 times the diameter of the copper wire (since 1 / 0.866 = 1.155, and 1.155 * diameter of copper wire = diameter of nichrome wire).
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a -pound elevator is suspended by a -foot cable that weighs lb/ft. how much work is required to raise the elevator from the basement to the third floor, a distance of ft?
The work required to raise the elevator from the basement to the third floor is equal to the force times the distance moved. The work required is equal to (-pounds)(ft) = -foot-pounds.
We have to calculate the work required to raise the elevator from the basement to the third floor using the given data. In order to calculate the work, we need to determine the force required to lift the elevator.
This can be done using the equation: F = W + w x L
where, F is the force required to lift the elevator, W is the weight of the elevator, w is the weight of the cable per foot, and L is the length of the cable. F = 1500 + (0.4 × 75) = 1500 + 30 = 1530 pounds.
Therefore, the force required to lift the elevator is 1530 pounds. The work done to raise the elevator is given by the equation: W = F × d
where, W is the work done, F is the force required to lift the elevator, and d is the distance traveled by the elevator.
W = 1530 × 30 = 45,900 foot-pounds.
Therefore, the work required to raise the elevator from the basement to the third floor is 45,900 foot-pounds.
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what is the log2(100)? it or use a calculator. (to 2 decimal places) round that number up to the next highest integer: based on the tests you've done: what is the maximum number of iterations used for quick search on a collection of 100 items? what is the maximum number of iterations used for linear search on a collections of 100 items? what is the bigo of quick search? hint: n
Log2(100) is 6.64 to 2 decimal places. Rounded up to the next highest integer, it is 7. Based on the tests done, the maximum number of iterations used for quick search on a collection of 100 items is 7 iterations.
The maximum number of iterations used for linear search on a collection of 100 items is 100 iterations. The big O of quick search is O(n log n).
log2(100) is the same as x in 2^x=100
This means 2^6.64=100
To solve for 6.64, take the logarithm of both sides.
log(2^6.64)=log(100)
6.64log(2)=log(100)
6.64=2log(10)+2log(5)
log(2)+log(1.25)+log(10)=6.64
log(2)+log(1.25)=6.64-log(10)
log(2)+log(1.25)=2.02
log(2x1.25)=2.02x2^(log(2)+log(1.25))=2.02 x 2^(log(10))=7 (rounded up)
The maximum number of iterations used for quick search on a collection of 100 items is 7 iterations.
The maximum number of iterations used for linear search on a collection of 100 items is 100 iterations. In big O notation, the time complexity of linear search is O(n).
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a train moves from a train station at 30meter per second after 5 seconds its cover the distance of 100 m and the acceleration is ten meters per second square find the speed
The speed of a train that moves from a train station at 30 meters per second after 5 seconds and covers a distance of 100 m with an acceleration of ten meters per second square would be 80 m/s.
Speed of a trainWe can use the equation of motion to solve for the final velocity of the train:
v = u + at
where:
v = final velocityu = initial velocity = 30 m/sa = acceleration = 10 m/s^2t = time = 5 sSubstituting the values, we get:
v = 30 + 10(5)
v = 80 m/s
Therefore, the speed of the train after 5 seconds is 80 m/s.
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Imagine sitting on a merry-go-round and riding along as it spins. Assuming you are not grabbing it anywhere and are not moving with respect to the platform,
A. static friction (directed inwards) causes you to accelerate.
B. you are not accelerating because you aren't moving on the platform.
C. static friction (directed outwards) causes you to accelerate.
D. sliding friction makes you accelerate inwards.
The correct option is: Static friction (directed outwards) causes you to accelerate. (Option C)
When you sit on a merry-go-round, you are not moving relative to the platform. Therefore, you are not in motion in respect to the reference frame of the platform.
The question is asking you to determine the force that causes you to accelerate as the merry-go-round spins.
Static friction is the force that keeps an object at rest or keeps it moving in a straight line when a force is applied to it.
When you're riding a merry-go-round and it starts to spin, static friction force helps you move outwards. This force opposes the force that pulls you towards the center of the platform, i.e., centripetal force.
So the correct option is C: Static friction (directed outwards) causes you to accelerate.
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which consumes more energy, a 1.2 kw hair dryer used for 10 min or a 10 w night light left on for 24 hr
A 1.2 kw hair dryer used for 10 min consumes more energy than a 10 w night light left on for 24 hr
To determine which device consumes more energy, you need to calculate the energy consumed by each device. The formula for calculating energy is: Energy = Power x Time
where power is measured in watts (W) and time is measured in hours (h).
Converting 10 minutes into hours, you get: 10 minutes = 10/60 hours = 0.167 hours
Energy consumed by the 1.2 kW hair dryer in 10 minutes: Energy = Power x Time
Energy = 1.2 kW x 0.167 hours
Energy = 0.2 kWh
Therefore, the energy consumed by the 1.2 kW hair dryer in 10 minutes is 0.2 kWh.
Converting 24 hours into minutes, you get: 24 hours = 24 x 60 minutes = 1440 minutes
Energy consumed by the 10 W night light in 24 hours:
Energy = Power x Time
Energy = 10 W x 24 hours
Energy = 240 Wh
Therefore, the energy consumed by the 10 W night light in 24 hours is 240 Wh.
Comparing the two values, we see that 1.2 kW hair dryer used for 10 min consumes more energy than a 10 W night light left on for 24 hours.
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A 200 g air-track glider is attached to a spring. The glider is pushed 10.0 cm against the spring, then released. A student with a stopwatch finds that 10 oscillations take 12.0 s. What is the spring constant? A 200 g ball is tied to a string. It is pulled to an angle of 8.00degree and released to swing as a pendulum. A student with a stopwatch finds that 10 oscillations take 12.0 s. How long is the string?
The spring constant of spring is 39.9 N/m and the length of the string is about 47.5 meters.
What is spring constant?Mass of the air-track glider (m) = 200 g = 0.2 kg
Displacement of the air-track glider (x) = 10.0 cm = 0.1 m
Number of oscillations (n) = 10
Time taken for 10 oscillations (t) = 12.0 s
T = 2π√(m/k)
where, T is the time period of oscillation. Substituting the given values, we get:
12 s = 2π√(0.2 / k)
Solving for k, we get:
The spring constant is 39.9 N/m.
Mass of the ball (m) = 200 g = 0.2 kg
Angle of displacement (θ) = 8.00°
Number of oscillations (n) = 10
Time taken for 10 oscillations (t) = 12.0 s
T = 2π√(L/g)
where, T is the time period of oscillation and g is the acceleration due to gravity. Substituting the given values, we get:
12 s = 2π√(L/9.8)
Solving for L, we get:
L = (12/2π)² × 9.8 = 47.5 m
Therefore, the length of the string is 47.5 meters.
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Two soccer balls (ball A and ball B) are kicked down the field. Ball A is kicked farther than ball B,
but ball B reaches a higher height than ball A. Which of the balls spent more time in the air?
(Come up with your answer and then discuss among your group until there is a consensus.)
Two soccer balls were kicked down the field, Ball A and Ball B. Ball A was kicked farther than Ball B, but Ball B reached a higher height than Ball A. It can be determined that the ball that spent more time in the air was Ball B.
Why Ball B spent more time in the air than Ball A?
Ball B was kicked into the air at a higher angle, meaning it travelled upwards for a longer amount of time. The ball's horizontal velocity would have been lower than Ball A's, causing it to travel a shorter distance horizontally.
However, the additional time Ball B spent travelling upwards and falling back down would have compensated for the shorter horizontal distance travelled, allowing it to remain in the air for longer than Ball A.
Ball A's flight time would have been shorter than Ball B's flight time because of its high horizontal velocity. Because Ball B had a higher initial upward velocity, it travelled higher in the air and took longer to fall back down, resulting in a longer flight time. As a result, Ball B spent more time in the air than Ball A.
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19. Which statement correctly describes a short circuit?
O A. Having a part with very low resistance keeps the circuit cooler, preventing
damage caused by overheating.
O B. Almost no current flows through a part of a circuit that has very low resistance
because it goes through the load.
O C. Almost all the current flows through a part of a parallel circuit with very low
resistance, instead of through the rest of the circuit.
O D. Having a part with very low resistance keeps circuit breakers and fuses from
opening the circuit
Answer:
.
C. Almost all the current flows through a part of a parallel circuit with very low resistance, instead of through the rest of the circuit.