For this problem, we would be using the formula: Vf^2 = Vi^2 + 2ad
where:
Vf = 400m/s
Vi = 300m/s
a = ?
d = 4.0km
= 4000m
400^2 = 300^2 + 2a4000
a = [ 160000 - 90000 ] / 8000
a = 8.75m/s^2
rounding it off to 2 significant figures, will give us 8.8 m/s^2.
For an object moving in a path that's a circle or a part of one,
the centripetal force acts in the direction toward the center of
the circle. That direction is perpendicular to the way the object
is moving.
Answer:
Light includes ALL of these answers: Radio/Microwaves. Visible light and X-rays/Gamma rays.
Answer:
The magnitude of the electric field is 1.124 X 10⁷ N/C
Explanation:
Magnitude of electric field is given as;

where;
E is the magnitude of the electric field, N/C
q is the point charge, C
k is coulomb's constant, Nm²/C²
r is the distance of the point charge, m
Given;
q = 5mC = 5×10⁻³ C
r = 2m
k = 8.99 × 10⁹ Nm²/C²
Substitute these values and solve for magnitude of electric field E

Therefore, the magnitude of the electric field is 1.124 X 10⁷ N/C