The answer is C. You divide 4000 kg/s by 115 kg.
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.
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>