If the separation distance is doubled, then the electric field decreases by a factor of 4.
<h3>What is the electric field strength?</h3>
We know that the electric field strength is known to depend on the magnitude of the charge and the distance of separation. We know that the electric field refers to the region in which the influence of a charge is felt. Recall that a charge is a specie that is positively or negatively charged. The charge on a specie must always be shown by its sign.
We know that the electric field is the region in space where the influence of a charge can be felt. If a charge is placed in the vicinity of another charge, the second charge would experience a force due to the presence of the first charge. This is because the second charge was brought into the electric field of the first charge.
Thus we know that;
E = Kq/r^2
Where;
E = electric field strength
q = magnitude of charge
r = distance of separation
Now;
E = 9.0* 10^9 * 3.052 * 10^-6/(8.22 * 10^-2)^2
E = 4 N/C
Given that the electric filed strength is inversely proportional to the distance of separation, when the distance between the charges is doubled, the electric field decreases by a factor of 4.
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An incompressible flow field F in a 3D cartesian grid with components u,v,w:
F = u + v + w
where u,v,w are functions of x,y,z
Must satisfy:
∇·F = du/dx + dv/dy + dw/dz = 0
We have a field F defined:
F = u+v+w, u = ax+byz, v = cy+dxz
du/dx = a, dv/dy = c
Recall ∇·F = 0:
∇·F = du/dx + dv/dy + dw/dz = 0
a + c + dw/dz = 0
dw/dz = -a-c
Solve for w by separation of variables:
w = ∫(-a-c)dz
w = -az - cz + f(x,y)
f(x,y) is some undetermined function of x and y
The question states that w is not a function of x and y, therefore f(x,y) = 0...
w = -az - cz
I would say that it would be A.) Been insulted. This is because you are angered by someone else. It does not necessarily mean that something has de-escelated, you've lost control of your car, or you've nothing to lose. Hopefully that helps. :)
Explanation:
It varies with altitude, but at sea level, it's 9.8 m/s².
To find the answer for this question, you simply need to divide 20 by 8, which is the speed he is traveling.
20 / 8 = 2.5
The football player will run 20 yards in 2 1/2 seconds.
Hope that helped! =)