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Answer:

0.000439077936334 m
Explanation:
q = Charge of electron = 
E = Electric field = 
= Permittivity of free space = 
d = Distance between plates = 2 cm (assumed)
m = Mass of electron = 
The beam consists of electrons which means it has negative charge this means the upper plates will be positive and the lower plate will be negative.
The direction is upper to lower lower plate.
= Permittivity of free space = 
Electric flux is given by

The charge per unit area on the plates is 
Deflection is given by

The deflection is 0.000439077936334 m
The process you're fishing for is "polarization", but that's a
misleading description.
Polarization doesn't do anything to change the light waves.
It simply filters out (absorbs, as with a polarizing filter) the
light waves that aren't vibrating in the desired plane, and
allows only those that are to pass.
The intensity of a light beam is always reduced after
polarizing it, because much (most) of the original light
has been removed.
A laser light source may be thought of as an exception,
since everything coming out of the laser is polarized.
To solve this problem we will proceed to use the equations given for the calculation of the resistance, in order to find the radius of the cable. Once the length is found we can find the number of turns of the solenoid and finally the net length of it
The resistance of the wire is

= Resistivity
L = Length
A = Cross-sectional Area
That can be also expressed as,

Rearranging the equation for the length of the wire we have



The number of turns of the solenoid is
Denominator is equal to the circumference of the loop


Finally the Length of he solenoid is

Where \phi is the diameter of wire



Therefore the length of the solenoid is 7.532m
Answer :
south
Explanation :
Given that,
Force = 200 N
Length = 2.00 m
Mass = 5.00 kg
We know that , the formula of the angular acceleration

We know the moment of inertia of rod

Now, the torque is

The angular acceleration



Negative sign represent of south direction.
Hence, this is the required solution.