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stira [4]
2 years ago
8

In a long, straight, vertical lightning stroke, electrons move downward and positive ions move upward and constitute a current o

f magnitude 20.0kA . At a location 50.0m east of the middle of the stroke, a free electron drifts through the air toward the west with a speed of 300 m/s . (e) If it does not collide with any obstacles, how many revolutions will the electron complete during the 60.0-μs duration of the lightning stroke?
Physics
1 answer:
uranmaximum [27]2 years ago
7 0

The number of revolutions the electron completes in 60.0-μs of the strike is 134.

A magnetic field, a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

Electrons go downward and positive ions move upward in a long, straight, vertical lightning stroke, creating a current of magnitude I = 20.0 kA.

A free electron travels through the air at a speed of v = 300 m/s at a place r = 50.0 m east of the stroke's center.

Let the magnetic field be B, and F be the magnetic force.

Counterclockwise horizontal arcs of field lines are produced by the upward lightning current.

We have, B = 8 × 10⁻⁵ T and;

The mass of an electron is, m = 9.11 × 10⁻³¹ kg

The time interval is Δt = 60 μs = 60 × 10⁻⁶

The angular frequency is given as:

ω = qB /m = 2πN / Δt

Where the number of revolutions is N.

So,

N = qBΔt /2πm

N = (l.60 × l0⁻¹⁹)(8 × l0⁻⁵)(60 × 10⁻⁶) / 2π(9.11 × 10⁻³¹ kg)

N = 134 revolutions

Learn more about current here:

brainly.com/question/1100341

#SPJ4

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Difference between of echo and reverberation​
KiRa [710]

Answer:

A difference between of echo and reverberation​  is described below in details.

Explanation:

Here's a piece of immediate information: An echo is an individual consideration of a soundwave off a horizon exterior. Reverberation is the consideration of sound waves generated by the superposition of the before-mentioned echoes. ... A reverberation can happen when a sound wave is displayed off a nearby covering.

5 0
3 years ago
Equipotential surfaces a) make an angle of 45 degrees with the electric field. b) are parallel to the electric field. c) are per
Schach [20]

Answer:

Option c) are perpendicular to the electric field

Explanation:

Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90^{\circ} to the equipotential surface.

Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.

Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.

7 0
3 years ago
Help
photoshop1234 [79]

Answer:

It conserves both energy and momentum in the collision at the same time. By design, when the balls collide the strings that hold them up are vertical (assuming balls are only swung from one side).

Explanation:

Hope This Helps!!

7 0
3 years ago
A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t
elena55 [62]

Answer:

The first part can be solved via conservation of energy.

mgh = mg2R + K\\K = mg(h-2R)

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}

where N = 0 because we are looking for the case where the car loses contact.

mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

7 0
3 years ago
A bulb converts 9J of energy in each second from an input of 100J of energy. Calculate efficiency of this light bulb
77julia77 [94]
Efficiency is calculated through dividing the actual mechanical advantage by the hypothetical mechanical advantage:

- the actual mechanical advantage is 9J because that's how much work the light bulb doing

- the hypo. mechanical advantage is 100J. Ideally, in a perfect world, the light bulb can convert 100J input into 100J output, but do to resistance and other factors it is not possible.

\frac{9}{100}  = .09
change the decimal to a percentage:

.09 = 9\%
the light bulb had 9% efficiency
8 0
4 years ago
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