Answer:
Explanation:
As given point p is equidistant from both the charges
It must be in the middle of both the charges
Assuming all 3 points lie on the same line
Electric Field due a charge q at a point ,distance r away
Where
- q is the charge
- r is the distance
-
is the permittivity of medium
Let electric field due to charge q be F1 and -q be F2
I is the distance of P from q and also from charge -q
⇒
F1
F2
⇒
F1+F2=
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Answer:
change in relative vorticity 0.0590
Explanation:
Given data
pressure = 1000 hPa
temperature lapse rate q1 = 3.1◦C per 50 hPa
pressure = 850 hPa
temperature lapse rate q2= -0.61◦C per 50 hPa
to find out
change in relative vorticity
solution
we will apply here formula that is
N = (g / potential temperature ) × (potential vertical temperature) × exp^1/2 ............................1
here we know g = 9.8 m/s
and q1 = potential temperature=3.3 degree celsius
potential vertical temperature gradient = 3.1 - 0.61 / 1000 -850
potential vertical temperature gradient = 0.0166 degree celsius/hpa
so
N = 9.8 / 2.75 × 0.0166 × exp^1/2
N = 0.0590
I’m pretty sure it’s lean mass but let another guy answer you before you use mine
Answer:
Explanation:
The two media must have differing index of refraction.
Index of refraction is an indication of how fast light can move through the media.
If a light wave approaches an interface at an angle and the media the light is moving into has a higher index of refraction meaning slower light speed in the media, then the part of the wave front hitting first gets slowed down sooner. As each section of the wave front crosses the interface, the whole wave front has changed direction toward the side with the first drag. Much like a column of marching soldiers executing a slight turn each at the precise moment to keep both columns and lined up in "military precision"
If the media the light is moving into has a smaller index of refraction, meaning higher light speed, then the side of the light beam hitting the interface first speeds up before the opposite side of the beam. This makes the angle leaving the interface much closer to the interface surface than the angle inside the higher index media. There becomes a point called the critical angle where the light cannot exit the higher index of refraction and the condition of total internal reflection exists. Think fiber optic cable which can transmit signals hundreds of miles without significant loss.
