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:
4 m/s^2
Explanation:
The acceleration is defined as: Δv/Δt (the difference of the velocity over a time period in which happens that difference).
Remember that a difference is calculated by subtracting the initial value of a physical quantity from its final value.
In our case:
Δv = Vfinal - Vinitial = 36m/s - 0 m/s = 36m/s
Δt = 9s
a = Δv/Δt = 36m/s / 9s = 4m/s^2
The sun orbits the eth at 2kilogram per sec
Answer:
This link was diagram
Explanation:
https://doubtnut.app.link/FnsNC80Dccb
Newton’s Law: F = MA
A = F/M (change equation)
12.6 N/ 2.4 kg = 5.25
Answer: acceleration is 5.25 m/s^2
