Answer:

Explanation:
In this case we have to work with vectors. Firs of all we have to compute the angles between x axis and the r vector (which points the charges):

the electric field has two components Ex and Ey. By considering the sign of the charges we obtain that:

Hence, by replacing E1 and E2 we obtain:
![\vec{E}=[(68.67N/C)cos(90\°)-(16.52N/C)cos(17.10\°)]\hat{i}+[(68.67N/C)sin(90\°)-(16.52N/C)sin(17.10\°)]\hat{j}\\\\\vec{E}=(-15.78\hat{i}+63.81\hat{j})\frac{N}{C}](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%3D%5B%2868.67N%2FC%29cos%2890%5C%C2%B0%29-%2816.52N%2FC%29cos%2817.10%5C%C2%B0%29%5D%5Chat%7Bi%7D%2B%5B%2868.67N%2FC%29sin%2890%5C%C2%B0%29-%2816.52N%2FC%29sin%2817.10%5C%C2%B0%29%5D%5Chat%7Bj%7D%5C%5C%5C%5C%5Cvec%7BE%7D%3D%28-15.78%5Chat%7Bi%7D%2B63.81%5Chat%7Bj%7D%29%5Cfrac%7BN%7D%7BC%7D)
hope this helps!!
Answer:
It got transferred to kinetic energy
Explanation:
Ax<span>”= ot * k-> 00 A(i+2) == X:"40: hence lim </span>Ax<span>") = 0</span>
Answer:
0.737
Explanation:
= Refractive indices of liquid A
= Refractive indices of liquid B
= Refractive indices of liquid C
Consider the total internal reflection at interface of liquid A and liquid B
= Angle of incidence = 32.0
Using Snell's law for total internal reflection

Consider the total internal reflection at interface of liquid A and liquid C
= Angle of incidence = 46
Using Snell's law for total internal reflection

Ratio is hence given as
