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>
Answer:
Q stay the same
Explanation:
Charging of capacitor is done by battery . If battery is disconnected , charging will stop . There will not be any discharging as plates are separate . So pulling the plates apart will not affect the charge lying on the capacitor . It will decrease its capacity and increase its potential , keeping its charge constant.
Answer: B, increase the rate at which he turns the crank
Explanation: i got it right
Answer:
work = 1275.3 J
Explanation:
work = (force)(distance)cosø ------- force = ma
=(mass*acceleration)(distance)cosø
=(20*9.81)(6.5)cos0
=1275.3J
nite that the angle of cosine is the difference between the angle of force and the distance. in this case, the force and the distance are in the same direction. :)
Your answer is cod!!!
Hope this helped!!
~xox Melis