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.
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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
at time
is given by

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

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,
, acting in a certain direction - The force of tugboat B,
, 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

and the direction is the same as the direction of the two forces.
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Answer:

Explanation:
Path difference due to a transparent slab is given as

here we know that

now total shift in the bright fringe is given as

Also we know that the fringe width of maximum intensity is given as

now we have

now the shift is given as

given that the shift is

here we have

now plug in all values in it



Answer:
1.87 A
Explanation:
τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s
d = diameter of copper wire = 2 mm = 2 x 10⁻³ m
Area of cross-section of copper wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
E = magnitude of electric field = 0.01 V/m
e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
m = mass of electron = 9.1 x 10⁻³¹ kg
n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³
= magnitude of current
magnitude of current is given as


= 1.87 A