Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0
Explanation:
- a) First of all find the distance between the two charges;
- x = 0, y = 0.30 and x = 0.40 m, y = 0
hence, the force F = 2Kq1q2cosθ /r²...............equation 1
but cosθ = y/r = 0.3/0.5
cosθ = 0.6
plugging back to equation 1;
F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2
F = 540 x 10^-3
Magnitude of Force = 0.54N
b) Direction is at angle 90
Answer:
Depending on which hemisphere it is, like western to eastern, It would most likely get stuck at the center. You would also have to put more things into thought like acceleration, velocity, and speed.
BUT since the question asked "would it pop out the other side?", I'm assuming it's talking about northern to southern hemisphere. so in that case it would pop out the other side since gravity makes things go downwards.
v = x/t
v = average velocity, x = displacement, t = elapsed time
Given values:
x = 6km south, t = 60min
Plug in and solve for v:
v = 6/60
v = 0.1km/min south
Answer:
Explanation:
Speed = distance / time
Velocity = displacement / time
So ,
Speed = 50 km / 0.5 hr = 100 km/h
Velocity = 40 km / 0.5hr = 80 km/h
Answer:
check image
Explanation:
For any question related to newons law of motion first draw the free body diagram(FBD),
