This is somewhat misleading, and I encountered the same question in my homework. An electric field strength of 1*10^5 N/C is provided, along with a drag force of 7.25*10^-11 N, and the critical detail is that it maintains a constant velocity, indicating that the particle is in equilibrium and not accelerating.
<span>To solve, utilize F=(K*Q1*Q2)/r^2 </span>
<span>You'll want to equate F with the drag force, where the electric field strength translates to (K*Q2)/r^2; substituting the values results in </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
The appropriate choice is C.
In physics, the law of gravity helps us understand how gravity varies with height. As altitude increases, so too does the experience of gravity. Changes in altitude also result in variations in weight, though these differences are not particularly significant. Consequently, weighing metals at different heights shows negligible variance as the impact of gravity remains constant across them.
We start by finding the angle of inclination with the sine function,
sin θ = 1 m / 4 m
θ = 14.48°
Next, we compute the work done by the movers using the following formula:
W = Fnet * d
We need to first determine Fnet. It is the weight force minus the frictional force.
Fnet = m g sinθ – μ m g cosθ
Fnet = 1,500 sin14.48 – 0.2 * 1,500 * cos14.48
Fnet = 84.526 N
The work done is therefore:
W = 84.526 N * 4 m
<span>W = 338.10 J</span>
