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salantis [7]
3 years ago
12

An electron (e = 1.6 x 10-19 C) is traveling at 4.00 x 107 m/s due North in a horizontal plane through a point where the earth’s

magnetic field has a component to the North of 2.00 x 10-5 T and a downward (in to the earth) of 5.00 x 10-5 T. Calculate the magnetic force (magnitude and direction) on the electron and its acceleration. Me = 9.11 x 10-31 kg
Physics
1 answer:
Irina18 [472]3 years ago
4 0

Answer:

Explanation:

Force on the electron due to magnetic field 's north component will be zero because both velocity of electron and direction of magnetic field are same .

Force due to downward component of magnetic field

= B q v where B is magnetic field , q is charge moving and v is velocity of charge

F = 5 .00 x 10⁻⁵ x 1.6 x 10⁻¹⁹ x 4 x 10⁷

= 32 x 10⁻¹⁷ N

acceleration = F / m where m is mass of electron

= 32 x 10⁻¹⁷ / 9.11 x 10⁻³¹

= 3.5 x 10¹⁴ m/s²

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Describe the mechanical energy of a roller coaster car immediately before it begins traveling down a long track
GrogVix [38]
At the top of the hill, the cars possess a large quantity of potential energy. Potential energy - the energy of vertical position - is dependent upon the mass of the object and the height of the object. The car's large quantity of potential energy is due to the fact that they are elevated to a large height above the ground. As the cars descend the first drop they lose much of this potential energy in accord with their loss of height. The cars subsequently gain kinetic energy. Kinetic energy - the energy of motion - is dependent upon the mass of the object and the speed of the object. The train of coaster cars speeds up as they lose height. Thus, their original potential energy (due to their large height) is transformed into kinetic energy (revealed by their high speeds). As the ride continues, the train of cars are continuously losing and gaining height. Each gain in height corresponds to the loss of speed as kinetic energy (due to speed) is transformed into potential energy (due to height). Each loss in height corresponds to a gain of speed as potential energy (due to height) is transformed into kinetic energy (due to speed). A roller coaster ride also illustrates the work and energy relationship. The work done by external forces is capable of changing the total amount of mechanical energy from an initial value to some final value. The amount of work done by the external forces upon the object is equal to the amount of change in the total mechanical energy of the object. The relationship is often stated in the form of the following mathematical equation.

KEinitial + PEinitial + Wexternal = KEfinal + PEfinal

The left side of the equation includes the total mechanical energy (KEinitial + PEinitial) for the initial state of the object plus the work done on the object by external forces (Wexternal) while the right side of the equation includes the total mechanical energy (KEfinal + PEfinal) for the final state of the object.

Once a roller coaster has reached its initial summit and begins its descent through loops, turns and smaller hills, the only forces acting upon the coaster cars are the force of gravity, the normal force and dissipative forces such as air resistance. The force of gravity is an internal force and thus any work done by it does not change the total mechanical energy of the train of cars. The normal force of the track pushing up on the cars is an external force. However, it is at all times directed perpendicular to the motion of the cars and thus is incapable of doing any work upon the train of cars. Finally, the air resistance force is capable of doing work upon the cars and thus draining a small amount of energy from the total mechanical energy which the cars possess. However, due to the complexity of this force and its small contribution to the large quantity of energy possessed by the cars, it is often neglected. By neglecting the influence of air resistance, it can be said that the total mechanical energy of the train of cars is conserved during the ride. That is to say, the total amount of mechanical energy (kinetic plus potential) possessed by the cars is the same throughout the ride. Energy is neither gained nor lost, only transformed from kinetic energy to potential energy and vice versa.

The conservation of mechanical energy by the coaster car in the above animation can be studied using a calculator. At each point in the ride, the kinetic and potential energies can be calculated using the following equations.

<span> KE = 0.5 * mass * (speed)^2 PE = mass * g * height</span>

If the acceleration of gravity value of 9.8 m/s/s is used along with an estimated mass of the coaster car (say 500 kg), the kinetic energy and potential energy and total mechanical energy can be determined

5 0
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Which are characteristics of fungi? Check all that apply.
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they are eukaryotes that have cell walls, are heterotrophs that feed by absorbing their food, and use spores to reproduce.

3 0
2 years ago
How much work is done by an applied force to lift a 45 newton block 6.0 meters at a constant speed ?
AleksandrR [38]

Answer:

270Joues

Explanation:

Step one:

given data

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Required

The work done in moving the block 6m

Step two:

We know that the expression for the work done is

WD= force* distance

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7 0
2 years ago
A vector has a magnitude of 30m at an angle of 225 degrees with respect to the positive x-axis and has a magnitude of 13m.what a
Aneli [31]

Answer:

I'm pretty sure this is not a complete question. My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.

IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43

It could have any direction of

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θ = 45 ± 25.679...

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components of vector B would be

Bx = |B|cosθ

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Nikolay [14]

Answer: 20 m/s

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Ekinetic=(1/2)*(m*v^2)

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8 0
3 years ago
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