Bruh how did i get 3 warnings and 3 brainliests removed
2 answers:
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By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.
Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .
now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.
charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.
Now , use vector for solve it.
vector = vector Fe + vector Fn,
⇒ || =
⇒ Fe = Fs = KqQ/(1m)² = KqQ
⇒ = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}
⇒ = √2KqQ
Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.
To learn more about positive charges here
#SPJ4
Answer:
The relationship is only between the coefficients A, E and J which is:
. The remaining coefficients can be anything without any constraints.
Explanation:
Given:
The three components of velocity is a velocity field are given as:
The fluid is incompressible.
We know that, for an incompressible fluid flow, the sum of the partial derivatives of each component relative to its direction is always 0. Therefore,
Now, let us find the partial derivative of each component.
Hence, the relationship between the coefficients is:
There is no such constraints on other coefficients. So, we can choose any value for the remaining coefficients B, C, D, F, G and H.