The magnitude of the electric field at the third vertex of the triangle is determined as zero.
<h3>Electric field at the third vertex of the triangle </h3>
The electric field at the third vertex of the equilateral triangle due to the other charges placed on the first and second vertices is calculated as follows;
E = E(13) + E(23)
E = (kq₁)/r² + (kq₂)/r²
where;
- q1 is positive charge
- q2 is negative charge
E = (kq₁)/r² - (kq₂)/r²
E = 0
Thus, the magnitude of the electric field at the third vertex of the triangle is determined as zero.
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Answer:
Vx = 10.9 m/s , Vy = 15.6 m/s
Explanation:
Given velocity V= 19 m/s
the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°
θ = 55°
to Find Vx = ? and Vy= ?
Vx = V cos θ
Vx = 19 m/s × cos 55°
Vx = 10.9 m/s
Vx = V sin θ
Vy = 19 m/s × sin 55°
Vy = 15.6 m/s
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don't get your eyes off the road
don't get distracted
