Answer:
The magnitude of the electric field is 5.75 N/C towards positive x- axis.
Explanation:
Given that,
Point charge at origin = 2 nC
Second charge = 5 nC
Distance at x axis = 8 m
We need to calculate the electric field at the point x = 2 m
Using formula of electric field

Put the value into the formula


The direction is toward positive x- axis.
Hence, The magnitude of the electric field is 5.75 N/C towards positive x- axis.
The chemical formular for water is H2O.
The H aspect of the formula stands for hydrogen gas and the subscript 2 which is attached to the H symbol signifies that two atoms of hydrogen are joined together, that is two atom of hydrogen are present.
The chemical formula of water indicates that, two atom of hydrogen react with one atom of oxygen to form one molecule of water.
In chemical formulae, subscripts are normally used to indicate the number of atoms that are present in a molecule.
I believe the best example of Newton's First Law of motion would be the example or illustration with the basketball player. An object will move in a straight line or a given direction at a constant speed unless or until another force acts upon the object, causing a change in speed and or direction.
Answer:
The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.
Explanation:
As data is incomplete here, so by seeing the complete question from the search the data is
vx_0=1.1 x 10^6
ax=0 As acceleration is zero in the horizontal axis so
Equation of motion in horizontal direction is given as


Now for the vertical distance
vy_o=0
than the equation of motion becomes

Now using this acceleration the value of electric field is calculated as

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.