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:
W = (F1 - mg sin θ) L, W = -μ mg cos θ L
Explanation:
Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane
Y Axis
N -
=
N = W_{y}
X axis
F1 - fr - Wₓ = 0
fr = F1 - Wₓ
Let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
We substitute
fr = F1 - W sin θ
Work is defined by
W = F .dx
W = F dx cos θ
The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1
W = -fr x
W = (F1 - mg sin θ) L
Another way to calculate is
fr = μ N
fr = μ W cos θ
the work is
W = -μ mg cos θ L
If a theory is studied in let's say the 17th century, the theory has had many years to be studied and explained by many different people many different ways.
vf ^2 = kx^2/m = 56(0.75)^2 / 2.5 = 12.6
Therefore, v= 3.5 m/s.
Answer:0.6kw
Explanation:
Power=force×velocity
Power=20×30=600w
In kw it's going to be 600/1000=0.6kw