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
External force.
Air resistance, friction, grass rubbing on a rolling ball, etc.
If there is no external force on a moving object, it keeps going.
Forever !
Answer:
Explanation:
Given data
Mass m=67.0 kg
Final Speed vf=8.00 m/s
Initial Speed vi=2.00 m/s
Distance d=25.0 m
Force F=30.0 N
From work-energy theorem we know that the work done equals the change in kinetic energy
W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²
And
So
and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N
So
