In order to solve this problem, we will first need to find the electric field at the origin without the 3rd charge
E1 = (9x10^9)(13.4x10^-9)/(9.4x10^-2)^2 = 13648.7 V/m towards the negative y-axis
E2 = (9x10^9)(4.23x10^-9)/(4.99x10^-2)^2 = 15289.1 V/m towards the positive x-axis
The red arrow shows the direction of which the electric field points.
To make the electric field at the origin 0, we must find a location where q3 = the magnitude of q1 and q2
Etotal = sqrt(E1+E2) = 20494.97 V/m
E3 = 20494.97 = (9x10^9)(14.23x10^-9)/(d)^2
d = 0.079 m = 7.9 cm
The correct answer from the choices listed above is the first option. The statement that is true would be that c<span>ompound AB has chemical and physical properties that are completely different from those of A and B. They completely different substances with different properties.</span>
Answer:
As you know, the denser objects have more weight per unit of volume, this will mean that the force that pulls down these objects is a bit larger.
This will mean that the denser objects will always go to the bottom.
This clearly implies that the red liquid, the one with one of the smaller densities, can not be at the bottom.
There are some cases where a liquid with a small density may become a lot denser as the temperature or pressure changes, and in a case like that, we could see the red liquid at the bottom, but for this case, there is no mention of changes in the temperature nor in the pressure, so this can be discarded.
The only thing that makes sense is that the red part at the bottom is the base of the tube, and has nothing to do with the red liquid.
