Answer:
The y-component of the electric force on this charge is 
Explanation:
<u>Given:</u>
- Electric field in the region,

- Charge placed into the region,

where,
are the unit vectors along the positive x and y axes respectively.
The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

Thus, the y-component of the electric force on this charge is 
to your question is 54 cm
Answer:
You are doing this wrong. Make sure that if you have 2 rows and on the top it has pounds then on the second bottom row you also have pounds so that they cancel each other out.
Explanation:
Answer:
Your zenith is 43 N of 90 deg (equator)
Thus, your zenith is 90 - 43 = 47 deg
(At the N pole your zenith would be 0 deg from the N pole)