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
Height of baby carriage from ground = 21m
Mass of carriage with baby = 1.5 kg
The carriage has potential energy by virtue of its height.
Potential energy = mgh = 1.5×10×21 = 315 J
Hence, potential energy of the carriage is 315 Joule.
Answer:
The workdone is 
Explanation:
From the question we are told that
The potential difference is 
Generally the charge on 
is 
Generally the workdone is mathematically represented as
=> 
=>
Answer:
550 kg
Explanation:
mass = E / gh
= 33000/60
=550
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