The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
<h3>Electric field on the master charge</h3>
E = kq/r²
where;
- q is magnitude of master charge
- r is distance of separation
- k is Coulomb's constant
E = (9 x 10⁹ x 0.63)/(0.75²)
E = 1.008 x 10¹⁰ N/C
<h3>Force on the test charge</h3>
F = Eq
where;
- E is electric field
- q is the test charge
F = (1.008 x 10¹⁰) x (0.5)
F = 5.04 x 10⁹ N
Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
Learn more about electric field here: brainly.com/question/14372859
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Answer:
The energy lost by the atoms is given off as an electromagnetic wave. ... even if it's not very intense, will always cause electrons to be emitted.
Explanation:
The acceleration of this car is equal to 5
.
<u>Given the following data:</u>
- Initial velocity = 0 m/s (assuming it's starting from rest).
To determine the acceleration of this car:
<h3>How to calculate acceleration.</h3>
In Science, the acceleration of an object is calculated by subtracting the initial velocity from its final velocity and dividing by the time.
Mathematically, acceleration is given by this formula:

<u>Where:</u>
- U is the initial velocity.
- is the time measured in seconds.
Substituting the given parameters into the formula, we have;

Acceleration, a = 5 
Read more on acceleration here: brainly.com/question/24728358
Answer:
B)
Explanation:
That the time period of which they stop.