Answer:
The answer is choice A.
Explanation:
Assuming you are in a situation with a gravitational field. You can divide the motion of the bullet into two components. One horizontal and the other in the vertical.

EQUATOR IS THE LINE THAT DIVIDES EARTH INTO TWO HEMISPHERES .
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;

where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

Therefore, the mutual force between the two point charges is 319.64 N
<h3>
Answer:</h3>
49 N
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of the brick as 3 kg
- The coefficient of friction as 0.6
We are required to determine the force that must be applied by the woman so the brick does not fall.
- We need to importantly note that;
- For the brick not to fall the, the force due to gravity is equal to the friction force acting on the brick.
- That is; Friction force = Mg
But; Friction force = μ F
Therefore;
μ F = mg
0.6 F = 3 × 9.8
0.6 F = 29.4
F = 49 N
Therefore, she must use a force of 49 N