The fraction of the engine power used to make the airplane climb is 83% to two significant figures.
The power generated by the engine is given by the equation P=W/t, where P is power, W is work, and t is time.
The work done in lifting the airplane can be calculated as W=mgh, where m is mass, g is the gravitational constant, and h is the altitude gained.
Therefore, the power used to make the airplane climb can be calculated as P=mgh/t, where t is the time taken to gain the altitude.
Since the rate of climb is given as 2.5 m/s, the time taken to gain the altitude can be calculated as t=h/2.5, where h is the altitude gained.
Substituting the values into the equation, the power used to make the airplane climb can be calculated as P=750*9.8*h/2.5, where h is the altitude gained.
Therefore, the fraction of the engine power used to make the airplane climb is
P/(80*1000)=750*9.8*h/(2.5*80*1000).
Finally, the fraction of the engine power used to make the airplane climb expressed as a percentage is
(750*9.8 h/(2.5*80*1000))*100=83%.
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the boundary between two adjacent air masses is called the boundary between two adjacent air masses is called an air strip. a front. entrainment. adiabatic.
A front is the line separating two contiguous air masses. Two air masses having differing characteristics, such as temperature, humidity, and density, collide and interact in a small area known as a front.
Cold fronts, warm fronts, stationary fronts, and occluded fronts are the four primary types of fronts. As a cold front enters a region, it pushes the warm air mass out of the way, resulting in a sharp drop in temperature and frequently severe weather. On the other hand, when a warm front approaches a region, it forces the cold air mass out of the way, causing the air to gradually warm up and increasing the likelihood of rain or drizzle. When two weak air masses collide, stationary fronts develop. enough to push the other out of the way, resulting in a stationary boundary between them. Occluded fronts occur when a cold front overtakes a warm front, lifting the warm air mass above the ground and forming a new boundary between the cold air behind the cold front and the cool air ahead of the warm front.
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a 67.97 kg student is standing atop a spring in an elevator that is accelerating upward at 3.66 m/s2. the spring constant is 2658 n/m. by how much is the spring compressed?
The spring is compressed by 0.70 m.
The 67.97 kg student standing atop a spring in an elevator that is accelerating upward at 3.66 m/s2 has a spring constant of 2658 n/m. The amount the spring is compressed can be calculated using the formula:
F = kx,
where F is the force, k is the spring constant, and x is the amount of compression.
Therefore, we can calculate the amount of compression in the spring as follows:
x = F/k = (67.97 kg * 3.66 m/s2)/2658 N/m = 0.70 m
Therefore, the spring is compressed by 0.70 m.
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a syringe containing an incompressible fluid is oriented vertically and the plunger slowly depressed. at which point is the kinetic energy the lowest?
The point at which the kinetic energy is lowest is 3 in the syringe containing an incompressible fluid that is vertically oriented and the plunger is slowly depressed.
The kinetic energy of an object is the energy it has due to its motion. When an object is in motion, it has kinetic energy. It is a scalar quantity that is proportional to the mass of the object and the square of its velocity. The formula for kinetic energy is given as follows:
KE = 1/2mv²
Where m is the mass of the object and v is its velocity.
Points 1 and 2 have higher kinetic energy because the incompressible fluid is still being compressed in the syringe. Point D is incorrect because the kinetic energy of the incompressible fluid is not the same at all three points. Point E is incorrect because enough information has been provided. Therefore, when a syringe containing an incompressible fluid is vertically oriented and the plunger is slowly depressed, the kinetic energy is lowest at point 3.
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isolated/insulated equipment grounding circuits must include how many equipment grounding conductors to meet the requirements of the nec?
Isolated/insulated equipment grounding circuits must include one equipment grounding conductor to meet the requirements of the National Electrical Code (NEC). These grounding conductors play a crucial role in ensuring the safety and proper functioning of electrical systems.
The NEC sets the standards for the safe installation of electrical wiring and equipment in the United States. An isolated equipment grounding circuit is designed to maintain electrical safety by providing a dedicated path for grounding equipment. This prevents unwanted electrical noise or interference from affecting the performance of sensitive electronic devices.
A single equipment grounding conductor is sufficient for an isolated grounding circuit, as it is meant to carry the fault current back to the source of power, protecting people and equipment from electrical hazards. The conductor is usually made of copper or aluminum and is sized according to the size of the circuit conductors.
This grounding conductor is connected to a grounding electrode system, which includes grounding electrodes such as ground rods, metallic water pipes, or concrete-encased electrodes. These electrodes create a connection to the earth, ensuring that any fault current is safely dispersed into the ground.
By complying with the NEC requirements, you ensure that your electrical systems are designed and installed in a manner that reduces the risk of electrical shock, fire hazards, and other potential dangers. A properly grounded electrical system promotes safety, performance, and reliability in any electrical installation.
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numerade a constant 10-n horizontal force is applied to a 20-kg cart at rest on a level floor. if friction is negligible, what is the speed of the cart when it has been pushed 8.0 m?
The speed of the cart is 4.0 m/s.
step by step explanation:
Force F is 10-N,
Distance d is 8.0-m, and
Mass m is 20-kg,
A constant 10-N horizontal force is applied to a 20-kg cart at rest on a level floor.
If friction is negligible, then the speed of the cart when it has been pushed 8.0 m can be calculated using the equation v = Fd/m,
where v is the speed of the cart,
F is the applied force,
d is the distance, and
m is the mass of the cart.
so the speed of the cart is v = (10 N × 8.0 m) / 20 kg = 4.0 m/s.
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when einstein's theory of gravity (general relativity) gained acceptance, it demonstrated that newton's theory had been?a. wrongb. incompletec. really only guess
When Einstein's theory of gravity (general relativity) gained acceptance, it demonstrated that Newton's theory had been (b) incomplete.
Newton's theory of gravity is a law that governs the behavior of objects. The formula [tex]F = \frac {G m_1 m_2}{ d^2}[/tex] explains the force of gravity between two objects, where F is the force of gravity, G is the universal gravitational constant, m1 is the mass of one object, m2 is the mass of another object, and d is the distance between the centers of the two objects. This formula shows that gravity decreases as distance increases.
Einstein's theory of gravity (general relativity): It is a theoretical framework proposed by Albert Einstein in 1915. It combines special relativity and Newton's law of universal gravitation. General relativity is based on the notion that gravitation is not a force acting between two masses but rather a curvature of spacetime created by the presence of massive objects. It differs from Newton's law of universal gravitation, which states that gravitation is caused by an attractive force acting between two masses.
When Einstein's theory of gravity (general relativity) gained acceptance, it demonstrated that Newton's theory had been incomplete. Therefore the correct answer is b.
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which of the following will increase the capacitance of a parallel-plate capacitor? (there could bemore than one correct choice.) a) an increase in the charge on the platesb) an increase in the potential difference
D) introduce a dielectric material between the plates, and E) decrease the separation between the plates will increase the capacitance of a parallel-plate capacitor.
The capacitance of a parallel-plate capacitor is given by the formula:
C = εA/d
where C is the capacitance, ε is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.
From this formula, we can see that the capacitance is directly proportional to the area of the plates and the permittivity of free space, and inversely proportional to the distance between the plates. Therefore, the following changes will increase the capacitance of a parallel-plate capacitor:
D) Introduce a dielectric material between the plates: A dielectric material has a higher permittivity than air, which increases the capacitance of the capacitor.
E) Decrease the separation between the plates: A decrease in the distance between the plates increases the capacitance of the capacitor.
Therefore, the correct choices are D) introduce a dielectric material between the plates, and E) decrease the separation between the plates.
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a 5 kg toy train car is connected to a 3 kg toy train car. the 3 kg car is given an external force of 16 n. what is the tension in the rope connecting the cars?
A 5 kg toy train car is connected to a 3 kg toy train car. the 3 kg car is given an external force of 16 n. the tension in the rope connecting the two cars is 29 N.
The tension in the rope connecting two toy train cars A toy train car with a mass of 5 kg is connected to a toy train car with a mass of 3 kg. An external force of 16 N is applied to the 3 kg car.
Tension in the rope between the two toy cars is what we need to calculate. According to Newton’s 2nd law, force equals mass multiplied by acceleration. If the two cars are moving in the same direction with the same acceleration, the tension in the rope can be calculated as follows:
Force acting on the two cars is the external force that is applied on the 3 kg car which is equal to 16 N. In this case, both cars will have the same acceleration.
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Two pieces of clay are thrown towards each other. The blue clay has a mass of 2 kg and is traveling at 1.5 m/s east. The red clay has a mass of 1.5 kg and is
traveling at 2.5 m/s west (negative velocity). They stick together after they collide. What is the final velocity of the combined clay pieces after the collision? East
is considered positive direction.
Explanation:
To solve this problem, we can use the law of conservation of momentum, which states that the total momentum of a system is conserved in the absence of external forces.
Before the collision, the momentum of the blue clay is:
momentum of blue clay = mass of blue clay * velocity of blue clay
= 2 kg * 1.5 m/s = 3 kg*m/s to the east (positive)
Before the collision, the momentum of the red clay is:
momentum of red clay = mass of red clay * velocity of red clay
= 1.5 kg * (-2.5 m/s) = -3.75 kg*m/s to the west (negative)
The total momentum before the collision is:
total momentum before collision = momentum of blue clay + momentum of red clay
= 3 kgm/s - 3.75 kgm/s = -0.75 kg*m/s to the west (negative)
After the collision, the two clays stick together and move as one combined object. Let's assume that the final velocity of the combined clay pieces after the collision is v.
By the law of conservation of momentum, the total momentum after the collision is equal to the total momentum before the collision:
total momentum after collision = total momentum before collision
= -0.75 kg*m/s
The combined mass of the two clays after the collision is:
combined mass = mass of blue clay + mass of red clay
= 2 kg + 1.5 kg = 3.5 kg
Therefore, the final velocity of the combined clay pieces after the collision is:
v = total momentum after collision / combined mass
= (-0.75 kg*m/s) / 3.5 kg
= -0.214 m/s to the west (negative)
Since the negative velocity indicates a direction to the west, the final velocity of the combined clay pieces after the collision is 0.214 m/s to the west.
A 750-kg roller coaster car drops from rest at a height of 90.0 m along a frictionless track. If the coefficient of kinetic friction due to braking along a horizontal track at the end of the ride is 0.720, over what distance does the car need to brake to come to a complete stop?
mass than air) at the same temperature. how does this affect the normal-mode frequencies of the pipe?
When a pipe is filled with a liquid of higher density than air, the frequency of the normal modes of the pipe decreases.
This is due to the fact that the speed of sound is proportional to the square root of the ratio of the bulk modulus to the density of the medium in which it travels.
When the density of the medium inside the pipe increases, the velocity of sound decreases, causing the frequency of the normal modes to decrease.
he wavelength of the sound waves inside the pipe is shortened due to the increase in density, resulting in a lower frequency of the normal modes.
The frequency of the normal modes of a pipe is influenced by a variety of factors, including the diameter and length of the pipe, as well as the speed of sound in the medium inside the pipe. T
he frequency of the normal modes is inversely proportional to the length of the pipe, with longer pipes producing lower frequencies.
In the case of a pipe filled with a liquid of higher density than air, the frequency of the normal modes would be lower than if it were filled with air.
This is because the speed of sound in the liquid would be lower than in air, resulting in a decrease in the frequency of the normal modes.
When a pipe is filled with a liquid of higher density than air, the frequency of the normal modes of the pipe decreases.
This is due to the fact that the speed of sound in the liquid is lower than in air, resulting in a decrease in the frequency of the normal modes.
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when a toolbox weighing 5 newtons is resting on the ground next to a sawhorse, how much potential energy does it have?
The potential energy of a toolbox weighing 5 newtons is zero.
The potential energy of a toolbox weighing 5 newtons depends on its height relative to the ground.
Potential energy (PE) is equal to the mass of the object (m) multiplied by the acceleration due to gravity (g) multiplied by its height (h): PE = mgh.
Therefore, the potential energy of the toolbox is equal to 5*9.8*h (where h is the height of the toolbox above the ground).
Assuming that the toolbox is resting on the ground, it has zero potential energy since its height is zero. If the toolbox is lifted above the ground, however, then it will have a greater potential energy.
For example, if the toolbox is lifted to a height of 10 meters above the ground, then it will have a potential energy of 490 joules (5*9.8*10).
The potential energy of the toolbox when it is placed next to the sawhorse, the height of the sawhorse needs to be taken into consideration.
If the sawhorse is higher than the ground, then the toolbox will have a greater potential energy since it will be located at a greater height above the ground.
If the sawhorse is lower than the ground, then the toolbox will have a lesser potential energy than when it is resting on the ground.
The potential energy of a toolbox weighing 5 newtons when placed next to a sawhorse depends on the height of the sawhorse relative to the ground.
If the sawhorse is higher than the ground, then the toolbox will have a greater potential energy, and if it is lower than the ground, then the toolbox will have a lesser potential energy.
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if the retina is 1.7 cm from the lens in the eye, how large is the image on the retina of a person of height 1.8 m standing 8.0 m away?
The image size on the retina of a person of height 1.8 m standing 8.0 m away is: 0.094 cm.
The size of the image on the retina of a person of height 1.8 m standing 8.0 m away is determined by the size of the object, the distance between the object and the lens, and the distance between the lens and the retina.
The image size on the retina is inversely proportional to the distance between the object and the lens and is directly proportional to the distance between the lens and the retina. In this case, the object is 1.8 m away and the lens is 1.7 cm from the retina.
Therefore, the image size on the retina is (1.7 cm/1.8 m) times 8.0 m, or 0.094 cm. This means that the image size on the retina of a person of height 1.8 m standing 8.0 m away is 0.094 cm.
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an object is located in water 30 cm from the vertex of a convex surface made of plexiglas with a radius of curvature of 80 cm. where does the image form by refraction and what is its magnification? nwater
The image formed by refraction is at a distance of 120cm behind the lens and its magnification is 4.
The image formation by refraction and magnification of an object in water 30 cm from the vertex of a convex surface made of plexiglass with a radius of curvature of 80 cm can be calculated using the following steps:
1. Determine the object's distance from the lens. Object distance (u) = -30 cm. (negative sign as per the convention of the mirror)
2. Determine the focal length of the lens using the formula:
f = R/2 where, f = focal length of the lens, R = radius of curvature of the lens.
So, f = 80/2 = 40 cm.
3. Use the mirror formula to determine the image distance from the surface:
1/f = 1/v + 1/u where,v = image distance from the surface.
Substituting the given values (with proper sign convention), we get:
(-1/40) = 1/v + (-1/30)
Solving for v, we get:
v = 120 cm.
5. Use the magnification formula to determine the magnification of the image:
m = -(v/u)
where,m = magnification, v = 120 cm, u = 30 cm
Therefore,m = -(120/-30) = 4
Therefore, the image will form at a distance of 120 cm from the lens on the water side of the lens and is magnified by a factor of 4.
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jasmin, a cyclist, accelerates from rest. after 8 s, the wheels have made 3 revolutions. (a) what is the angular acceleration of the wheels? (b) what is the angular velocity of the wheels after 8 s?
a. The angular acceleration of the wheels is 0.2945 rad/s². b. The angular velocity of the wheels after 8 seconds is 2.3560 rad/s.
Calculation:
a. The formula for angular acceleration is: α = (ω2 - ω1) / (t2 - t1) Whereα is angular acceleration, ω2 is final angular velocity, ω1 is initial angular velocity, t2 is final time, t1 is initial time. To calculate the angular acceleration, we can use the formula:α = (ω2 - ω1) / (t2 - t1)
The initial angular velocity of the wheels is zero since Jasmin starts from rest, soω1 = 0. We know that the wheels make 3 revolutions after 8 seconds, so the final angular velocity can be calculated as follows: ω2 = (3 revolutions / 8 s) x (2π radians / 1 revolution) = 2.3562 rad/s
Therefore,α = (2.3562 rad/s - 0 rad/s) / (8 s - 0 s) = 0.2945 rad/s². The angular acceleration of the wheels is 0.2945 rad/s².
b. To calculate the angular velocity of the wheels after 8 seconds, we can use the formula:ω = ω1 + αtWhereω is angular velocity,ω1 is initial angular velocity,α is angular acceleration, t is time. The initial angular velocity of the wheels is zero since Jasmin starts from rest, so ω1 = 0
We have already calculated the angular acceleration to be 0.2945 rad/s², and we know that the time is 8 seconds, soω = ω1 + αt = 0 + (0.2945 rad/s²) x (8 s) = 2.3560 rad/s. Therefore, the angular velocity of the wheels after 8 seconds is 2.3560 rad/s.
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when rotating the platform, the hanging mass should be removed from the platform. question 2 options: true false
The given statement, while the platform is rotating, the hanging mass remains attached to the test mass and is not removed from the platform is true, if the purpose of the experiment or test is to determine the effect of the hanging mass on the rotation or stability of the platform.
In this case, the hanging mass must remain attached to the test mass during the rotation to observe the behavior of the system under the specified conditions. If the purpose of the experiment or test is to study the effect of the hanging mass on the platform's rotation or stability, the hanging mass must remain attached to the test mass during the rotation. This is because the presence of the hanging mass affects the overall weight and center of gravity of the system. Removing the hanging mass would alter the system's behavior and prevent accurate observations of the phenomenon under investigation. Therefore, if the experiment requires the hanging mass to be present, it must remain attached to the test mass while the platform is rotating.
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--The complete question is, While the platform is rotating, the hanging mass remains attached to the test mass and is not removed from the platform. State true/false.--
pry on the power steering reservoir to adjust the tension of the power steering belt. true or false?
The statement "pry on the power steering reservoir to adjust the tension of the power steering belt" is: false.
The tension of the power steering belt is adjusted by adjusting the position of the power steering pump. There is a tension adjustment bolt on the power steering pump that is used to adjust the tension of the power steering belt. The adjustment bolt should be turned clockwise or counterclockwise to adjust the tension of the belt.
A belt tension gauge may be used to ensure that the belt is properly tensioned. A pry bar should not be used on the power steering reservoir to adjust the tension of the power steering belt. This could cause damage to the reservoir or other components of the power steering system. The reservoir should be inspected for damage or leaks, but it should not be used to adjust the tension of the belt.
In summary, the tension of the power steering belt should be adjusted by adjusting the position of the power steering pump, not by prying on the power steering reservoir.
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the tilt of the moon's orbit shifts a little bit each year, which changes the dates eclipses occur each year. this shift of the orbital plane is called
The shift of the Moon's orbit is called precession and it occurs over a period of 18.6 years.
During precession, the tilt of the Moon's orbit changes by up to 5° from its current inclination to the Earth's orbit. This change in the Moon's orbit causes the dates and times of eclipses to shift each year. Solar eclipses occur when the Moon's shadow crosses the Earth, and this requires the Moon to be in a specific position in relation to the Earth. Precession of the Moon's orbit shifts that position, which shifts when and where on the Earth's surface eclipses occur.
Precession also affects the visibility of lunar eclipses. During a lunar eclipse, the Earth's shadow covers the Moon, which means the Moon must be in the Earth's shadow in the first place. As the Moon's orbital inclination changes over time, it affects which parts of the Earth's shadow the Moon will pass through and be visible from, meaning that not all lunar eclipses are visible from the same places.
Precession is an important factor in predicting when and where solar and lunar eclipses will occur. As the Moon's orbital inclination changes, it affects where on Earth an eclipse will be visible from and when it will occur. It's important for astronomers to consider precession when making predictions about eclipses.
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you are standing on a scale in an elevator. suddenly you notice your weight increases. what do you conclude?
When standing on a scale in an elevator, if one notices an increase in their weight, it means that: the elevator is accelerating upwards.
This is due to the fact that the scale underfoot has to counter the upward acceleration of the elevator, which causes the weight measured on the scale to increase. The scale measures the normal force, which is the weight being exerted on the scale, which is equal to the mass of the individual multiplied by the gravitational acceleration on the surface of the earth.
This can be represented by the formula: W = mg,
where W is the weight, m is the mass of the object and g is the gravitational acceleration.
When the elevator is stationary or moving at a constant velocity, the gravitational acceleration is the same as the normal force and the weight of the individual remains constant. However, when the elevator begins to accelerate upwards, the normal force exerted by the scale must increase to counter the upward acceleration of the elevator.
This causes an increase in weight measured on the scale. Therefore, if one notices an increase in their weight while standing on a scale in an elevator, it indicates that the elevator is accelerating upwards.
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a box is given a push so that it slides across the floor. how far will it go, given that the coefficient of kinetic friction is 0.11 and the push imparts an initial speed of 3.8 m/s ?
The box will slide a distance of 6.96 m before coming to a stop due to the force of kinetic friction.
To determine how far the box will slide on the floor after it is given a push with an initial speed of 3.8 m/s, we need to use the equations of motion for constant acceleration. The force of kinetic friction acting on the box will cause it to decelerate, eventually coming to a stop.
The distance traveled by the box can be found using the equation:
d = [tex](v_i^2 - v_f^2) / (2 * a)[/tex]
where d is the distance traveled, v_i is the initial speed, v_f is the final speed (which is zero since the box comes to a stop), and a is the deceleration caused by the force of kinetic friction.
The deceleration can be found using the equation:
a = -F[tex]_friction / m[/tex]
where Ffriction is the force of kinetic friction and m is the mass of the box.
Assuming a mass of 5 kg for the box and a coefficient of kinetic friction of 0.11, the force of kinetic friction can be found using the equation:
F_friction = friction coefficient * F_normal
where F_normal is the normal force (equal to the weight of the box) and the friction coefficient is a dimensionless quantity that depends on the nature of the contact surface.
The weight of the box is:
Fweight = m * g
where g is the acceleration due to gravity (9.81 m/s²).
Therefore, the force of kinetic friction is:
F_friction = (0.11) * (5 kg * 9.81 m/s²) = 5.40 N
Using the equation for deceleration, we get:
a = -Ffriction / m = -(5.40 N) / (5 kg) = -1.08 m/s²
Finally, we can use the equation for distance traveled to find the distance the box will slide:
d = [tex](v_i^2 - v_f^2) / (2 * a)[/tex] =[tex](3.8 m/s)^2 / (2 * 1.08 m/s^2)[/tex] = 6.96 m
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a boat moves through the water with two forces acting on it. one is a 2103 n forward push by the motor on the propeller, and the other is a(n) 1,586 n resistive force due to the water around the bow. what is the acceleration of the 1193 kg boat?
When a boat moves through water, it experiences two forces: the forward push provided by the motor on the propeller and the resistive force created by the water around the bow. The acceleration of the 1193 kg boat is 0.404 m/s².
The net force acting on the boat can be calculated by subtracting the resistive force from the forward force:
F_net = F_forward - F_resistiveF_net = 2103 N - 1586 NF_net = 517 N
The acceleration of the boat can be calculated using the formula: a = F_net/m Where F_net is the net force acting on the boat, and m is the mass of the boat. Substituting the values we know, a = 517 N / 1193 kg a = 0.404 m/s²Therefore, the acceleration of the 1193 kg boat is 0.404 m/s².
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a physics student wants to construct a model of an electric motor. The student creates a chart for the steps in the process. the chart shows what the student recorded so far:
what is the correct ordering of these cards to model how an electric motor works?
An electric motor is a device that transforms electrical energy into mechanical energy. Most electric motors create force in the form of torque imparted to the motor's shaft by interacting between the magnetic field of the motor and electric current in a wire winding.
How does an electric motor work?Electric motors generate motion by transferring electrical energy to mechanical energy. The interaction of a magnetic field and winding alternating (AC) or direct (DC) current generates force within the motor.
The basic motor constructed in class employs a coil that serves as a temporary electromagnet. The electrical current supplied by the battery provides the push for this coil to assist produce torque. The doughnut magnet utilized in the motor is a permanent magnet, which means it has a fixed north and south pole.
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Mercury has a mass of 3.29E23 kg and a radius of 2.44E6 m. Venus has a mass of 4.87E24 kg and a radius of 6.05E6 m. The gravitational field near the surface of Mercury is? N/kg. The gravitational field near the surface of Venus is? N/kg.
Gravitational field near the surface of Mercury is approximately 3.7 N/kg
Gravitational field near the surface of Venus is approximately 8.87 N/kg.
Gravitational field near the surface of Mercury and Venus, we can use the formula:
gravitational field (g) = (G * M) / R^2
where G is the gravitational constant (6.67430 × 10^(-11) m^3 kg^(-1) s^(-2)), M is the mass of the planet, and R is the radius of the planet.
For Mercury:
M = 3.29E23 kg
R = 2.44E6 m
g = (6.67430 × 10^(-11) m^3 kg^(-1) s^(-2) * 3.29E23 kg) / (2.44E6 m)^2
g ≈ 3.7 N/kg
For Venus:
M = 4.87E24 kg
R = 6.05E6 m
g = (6.67430 × 10^(-11) m^3 kg^(-1) s^(-2) * 4.87E24 kg) / (6.05E6 m)^2
g ≈ 8.87 N/kg
So, the gravitational field near the surface of Mercury is approximately 3.7 N/kg, and the gravitational field near the surface of Venus is approximately 8.87 N/kg.
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engineers at the university of texas at austin are developing an advanced locomotive propulsion system that uses a gas turbine and perhaps the largest highspeed flywheel in the world in terms of the energy it can store. the flywheel can store 4.8 * 108 j of energy when operating at its maximum rotational speed of 15,000 rpm. at that rate, the perimeter of the rotor moves at approximately 1,000 m/s. determine the radius of the flywheel and its rotational inertia.
The radius of the flywheel and its rotational inertia will be 0.64m and 389kgm² respectively.
What is the rotational inertia?Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to rotational motion. It is similar to the concept of mass in linear motion. Just as mass is a measure of an object's resistance to linear motion, the moment of inertia is a measure of an object's resistance to rotational motion.
The moment of inertia of an object depends on its shape and mass distribution. Objects with more mass distributed farther from the axis of rotation have a higher moment of inertia than objects with the same mass but a more compact distribution of mass. The moment of inertia is measured in units of kilograms square meters (kg m²) in the SI system.
The radius will be:
= 1000 / 15000(2πrad / 60)
= 0.64m
The inertia will be:
= 2(4.8 × 10^8) / 100 (2π/60)
= 389kgm²
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at an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. people sit in seats facing the axis, their backs against the outer wall. at one instant the outer wall moves at a speed of 3.13 m/s, and an 83.6 kg person feels a 578 n force pressing against his back. what is the radius of a chamber
The radius of the cylindrical chamber in the amusement park ride can be calculated using formula for centripetal force and the given values for velocity and force. In this case, the radius is approximately 14.3 meters.
The person is seated facing the axis, with their back against outer wall. At a given instant, the outer wall moves at a speed of 3.13 m/s, and the person feels a 578 N force pressing against their back.
To determine the radius of the chamber, we can use formula for centripetal force, [tex]Fc = (mv^2) / r[/tex].
Rearranging the formula to solve for r, we get r = (mv^2) / Fc. Substituting the given values, we get[tex]r = (83.6 kg * (3.13 m/s)^2) / 578 N,[/tex] which simplifies to [tex]r = 14.3 m[/tex]. Therefore, the radius of the cylindrical chamber is approximately [tex]14.3 meters[/tex].
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logs sometimes float vertically in a lake because one end has become water-logged and denser than the other. what is the average density of a uniform-diameter log that floats with 20.0% of its length above water?
Uneven-diameter logs that float with 20.0% of their length above water have an average density of 0.8g/cm3. The density is the proportion of weight to capacity.
An item it's far less compact that liquid may be supported up liquid water, and hence it floats. More dense objects can sink when submerged in water. Less dense logs float whereas more thick logs sink. A body can change its condition of rest or motion by the application of force
Instead of obliquely reading from either the side, read the scale stick straight from of the end of both the log. → The diameter of a log is only ever calculated within the bark. Employ a log measuring rod to determine the log's small end's "diameter from within bark," also known as "d.i.b."
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a strong lightning bolt transfers an electric charge of about 21 c to earth (or vice versa). how many electrons are transferred? avo
Approximately [tex]1.31 \times 10^{20}[/tex] electrons are transferred in a strong lightning bolt carrying an electric charge of 21 C.
The electric charge of one electron is equal to [tex]-1.602 \times 10^{-19}[/tex] Coulombs (C). Therefore, we can calculate the number of electrons transferred by dividing the total charge transferred by the charge of a single electron:
Number of electrons = Total charge transferred / Charge of a single electron
Number of electrons = [tex]\frac{21 C }{-1.602 \times 10^{-19} C}[/tex]
The number of electrons ≈ [tex]1.31 \times 10^{20} electrons[/tex]
Hence the number of electrons transferred during the lightning bolt is [tex]1.31 \times 10^{20}[/tex].
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1. Which graphic organizer best compares the energy inputs and outputs of cellular respiration and photosynthesis? Chemical energy. Cellular respiration. Thermal energy. Photosynthesis. Light energy. Light energy. Photosynthesis. Chemical energy. Cellular respiration. Chemical energy. Chemical energy. Photosynthesis. Thermal energy. Cellular respiration. Chemical energy. Light energy. Cellular respiration. Chemical energy. Photosynthesis. Thermal energy.
The best sources of energy are light, photosynthesis, chemicals, cellular respiration, and chemicals. the energy inputs and outputs of photosynthesis and cellular respiration are compared.
What exactly does energy mean?The capability to do tasks is how energy is most frequently defined. In other words, energy is a property of everything that can accomplish work. Making or producing change is another name for conducting work in the context of energy. With every act of labor, energy is either changed or transferred.
Why does energy exist?Energy is a word that denotes a property of matter & non-matter fields; it is not a substance in and of itself. For instance, it is argued that matter has kinetic energy when it moves quickly. Potential energy can take many different forms.
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a 6 mf capacitor, a 10 mf capacitor, and a 16 mf capacitor are connected in parallel. what is their equivalent capacitance?
The equivalent capacitance of a 6 mF capacitor, a 10 mF capacitor, and a 16 mF capacitor connected in parallel is: 32 mF
This is because when capacitors are connected in parallel, their total capacitance is equal to the sum of their individual capacitances. The formula for calculating the equivalent capacitance (C) of capacitors connected in parallel is: C = C1 + C2 + C3 + ... In this example, C = 6 mF + 10 mF + 16 mF = 32 mF.
Capacitors are electrical components that store energy in the form of an electric field between two conductors (plates). When capacitors are connected in parallel, the electric field between the plates of each capacitor is the same, but the overall capacitance is increased due to the combined plate area of all the capacitors.
This increase in plate area is why the equivalent capacitance of the three capacitors in this example is 32 mF, which is larger than any of the individual capacitances.
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you are holding a box of books with flat hands. if you press harder, what happens to the friction force applied by your hands onto the sides of the box?
When you hold a box of books with flat hands and press harder, the friction force applied by your hands onto the sides of the box will increase.
The force causes motion because if an object is at rest, it remains at rest until acted upon by a force. If the object is in motion, it remains in motion unless acted upon by a force to slow it down, speed it up, or change its direction. So, we have to look at the direction of the force and the motion to understand how the force will affect it. In general, the frictional force opposes motion.
The force of friction is proportional to the force pressing the two surfaces together. In this case, the force pressing the box onto your hands will be greater if you press harder, resulting in a greater frictional force applied by your hands onto the sides of the box, according to Coulomb's laws. Therefore, the friction force will increase when you press harder.
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