Answer:
The work required to move this charge is 0.657 J
Explanation:
Given;
magnitude of charge, q = 4.4 x 10⁻⁶ C
Electric field strength, E = 3.9 x 10⁵ N/C
distance moved by the charge, d = 50 cm = 0.5m
angle of the path, θ = 40°
Work done is given as;
W = Fd
W = FdCosθ
where;
F is the force on the charge;
According the coulomb's law;
F = Eq
F = 3.9 x 10⁵ x 4.4 x 10⁻⁶ = 1.716 N
W = FdCosθ
W = 1.716 x 0.5 x Cos40
W = 0.657 J
Therefore, the work required to move this charge is 0.657 J
Answer:
Explanation:
1.) What is the net force in the horizontal (x) direction?
Fnet = 8 - 3 = 5 N left
2.) what is the acceleration in the horizontal (x) direction?
a = Fnet/m = 5/2 = 2.5 m/s² left
Answer:
D. absolute magnitude and apparent magnitude
