the potential at the center of curvature of the arc = v = Q ∕ (4πε∘a) or 
ATQ,
We have density of charge,
λ = 
Where L is the rod's length, in this case the semicircle's length L = πr
Q is the charge on the rod
The potential created at the center by an differential element of charge is:
where k is the coulomb's constant
r is the distance from dQ to center of the circle
v = ∫ 
, Where a = radius, k = 1 / 4πε∘
v = or Q ∕ (4πε∘a)
To learn more about potential from the given link
brainly.com/question/25923373
#SPJ4
Answer:
77.35 m / s
Ф = -17° from + X axis or 343° from + X axis
Explanation:
v1 = 75 m/s 25° east of north
v2 = 100 m/s 25° east of south
Write the velocities in vector form ,we get
Now add the velocity vectors to get the resultant of the velocities.
magnitude of resultant velocity is 
= 77.35 m / s
The direction is Ф from X axis
Ф = -17° from + X axis or 343° from + X axis
The transmission of light waves is usually done through cornea of the eyes, then move through another opening which is regarded as pupil before it will get to the retina.
- Light waves can be regarded as moving energy which contains microscopic particles known as photons.
- The vision of the eye can be completed through the light wave passing through the components of the eyes and this process goes thus;
- Light will move through the (cornea) which is situated at the front area of the eyes into lens.
- Then both the cornea and the lens give room for the focusing of the light rays to the retina which is situated at the back of the eye .
- Then through the help of the cells in the retina, the light will be absorbed and then be converted to electrochemical impulses and then transfer it to the brain as well as optic nerve.
Therefore, light wave are form of tiny microscopic particles.
brainly.com/question/19734585?referrer=searchResults
Do 25-14 and you will get your answer
