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:
It's constant everywhere in its trajectory.
Explanation:
the projectile was launched with an initial velocity, the only acceleration that is affecting the projectile's velocity is gravity.
The acceleration of gravity is practically equal everywhere on earth, so during its trajectory, we have to take into consideration only the acceleration because of gravity.
This is only correct because the projectile was launched with an initial velocity and it's not accelerating from rest and then falls.
Answer:
2271.16N/C upward
Explanation:
The diagram well illustrate all the forces acting on the mass. The weight is acting downward and the force is acting upward in other to balance the weight.since the question says it is motionless, then indeed the forces are balanced.
First we determine the downward weight using
Hence for a mass of 3.82g 0r 0.00382kg we have the weight to be
To calculate the electric field,
Since the charge on the mass is negative, in order to generate upward force, there must be a like charge below it that is repelling it, Hebce we can conclude that the electric field lines are upward.
Hence the magnitude of the electric force is 2271.16N/C and the direction is upward
<span>B. velocity .................</span>
If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be .
<h3>What is an escape velocity?</h3>
The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.
The formula to calculate the escape velocity of earth is given below:-
Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-
.
Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be .
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