Answer:
where E = electric field intensity
Explanation:
As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical
So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field
while weight of the ball is vertically downwards
so here we have
since string makes 30 degree angle with the vertical so we will have
where E = electric field intensity
Answer:
1.327363 m/s
0.00090243 m
Explanation:
u = Initial velocity
v = Final velocity
m = Mass of flea
Energy
The velocity of the flea when leaving the ground is 1.327363 m/s
The flea will travel 0.00090243 m upward
Hi there!
Acceleration = change in velocity / change in time = Δv/Δt
Thus:
a = (75 - 15)/4 = 60/4 = 15 mi/hr²
Answer:
Option (e) = The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere.
Explanation:
So, we are given the following set of infomation in the question given above;
=> "spherical Gaussian surface of radius R centered at the origin."
=> " A charge Q is placed inside the sphere."
So, the question is that if we are to maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located where?
The CORRECT option (e) that is " The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere." Is correct because of the reason given below;
REASON: because the charge is "covered" and the position is unknown, the flux will continue to be constant.
Also, the Equation that defines Gauss' law does not specify the position that the charge needs to be located, therefore it can be anywhere.
The answer ultraviolet rays just got it right on the test
