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at location a, what are the directions of the electric fields contributed by the electron. calculate the magnitudes of the elect

Lisa [10]

We can use the equation E = k | Q | r 2 E = k | Q | r2 to find the magnitude of the electric field. The direction of the electric field is determined by the sign of the charge,

<h3>What is electric and magnetic field ?</h3>

With the use of electricity and other types of artificial and natural illumination, invisible energy fields known as electric and magnetic fields (EMFs) and radiation are created.

While the magnetic field is discernible by the force it exerts on other magnetic particles and moving electric charges, the electric field is actually the force per unit charge experienced by a non-moving point charge at any given location inside the field.

Learn more about Electromagnetic field here:

brainly.com/question/14372859

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4 0

1 year ago

A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate

lilavasa [31]

Take the stone's position at ground level to be the origin, and the downward direction to be negative. Then its position in the air $y$ at time $t$ is given by

$y=100\,\mathrm m-\dfrac g2t^2$

Let $d$ be the depth of the well. The stone hits the bottom of the well after 5.00 s, so that

$-d=100\,\mathrm m-\dfrac g2(5.00\,\mathrm s)^2\implies d=\boxed{22.6\,\mathrm m}$

7 0

3 years ago

Two tugboats are moving a barge. Tugboat A pushes on the barge with a force of 3000n. Tugboat B pulls with a force of 5000 Newto

kenny6666 [7]

The net force on the barge is 8000 N

Explanation:

In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.

In this problem, we have two forces:

The force of tugboat A, $F_A = 3000 N$ , acting in a certain direction
The force of tugboat B, $F_B = 5000 N$ , also acting in the same direction

Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get

$F=F_A+F_B = 3000 + 5000 = 8000 N$