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:
72
Explanation:
The displacement of an object can be found from the velocity of the object by integrating the expression for the velocity.
In this problem, the velocity of the sport car is given by the expression
In order to find the expression for the position of the car, we integrate this expression. We find:
where C is an arbitrary constant.
Here we want to find the displacement after 3 seconds. The position at t = 0 is
While the position after t = 3 s is
Therefore, the displacement of the car in 3 seconds is
Answer:
Sir Richard Branson's personal dilemma is that he is concerned about the environment and climate change, but he has made his fortune with an airline industry that contributes to the greenhouse gases.
brainliest plz
I agree with the other comment
Answer:
B= 55.6×10^(-7) Tesla
Explanation:
B= μoI/(2πr)
B: magnetic field strength
μo: permeability of free space and is equal to 4π×10^(-7) T.m/A
r: distance from the wire
I : current in the wire
B= (4π×10^(-7)×125)/(2π×4.5)
B= 55.6×10^(-7) Tesla
