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.
Learn more about electric field here: brainly.com/question/14372859
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Answer: a and d
Explanation: A.) the power lines themselves
B.) the wooden pole that supports the lines
C.) the rubber soles on the worker’s boots
D.) the metal tools the worker uses
E.) the wooden ladder leaning against the lines
Hello!
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?
Data:
For a spring (or an elastic), the elastic potential energy is calculated by the following expression:
Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.
Solving:
Answer:
The displacement of the spring = 0.8 m
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I Hope this helps, greetings ... Dexteright02! =)
Answer:
a = change in v / change in time
= (5.2 - 11) / 3.1
= -1.87 m/s^2
Explanation:
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