Answer:
a) E = 1.47 × 10^5 N/C
b) south
Explanation:
The magnitude of an electric field can be defined mathematically as;
E = F/q ........1
Where,
E = magnitude of the electric field
F = electric force
q = charge on the proton
Given;
F = 2.36 × 10^-14 N
Note that charge on a proton is known as Qp = 1.602 × 10^-19 C
q = 1.602 × 10^-19 C
Substituting into equation 1, we have;
E = 2.36 × 10^-14 N/1.602 × 10^-19 C
E = 1.47 × 10^5 N/C
b) The direction of the electric field;
From equation 1
E = F/q ........1
since both electric field and electric force are vector quantity and q is a positive charge (constant), then both the electric field and electric force would be parallel to each other. Therefore the electric field is directed to the south also.
(When a vector is multiplied by a positive constant the direction remains the same)
The normal force of the force given is calculated through the equation,
Fn = F(sin θ)
where Fn is the normal force, F is the force, and θ is the angle.
Fn = (25 N)(sin 60°) = 21.65 N
The x-component of the force applied is,
Fx = (25 N)(cos 60°) = 12.5 N
The value of the coefficient of static friction is calculated through the equation,
F = μFn
μ = Fx / Fn = 12.5 N / 21.65 N = 0.577
Answer: 61.2N
F=kq1q2/r^2
F=8.98E9(1.6E-19)(1.6E-19)/(1.93E-15)^2
F=61.2N
Explanation:
<h2>Yes!</h2>
<h3>In physics, constant velocity occurs when there is no net force acting on the object causing it to accelerate. In terms of airplane flight, the two main forces influencing its velocity forward are drag and thrust. At a constant altitude, when the force of thrust equals the opposing force of drag, then the airplane will experience uniform motion in one direction. This can be further explained by Newton’s First Law. </h3>
It pushes the currents to opposite sides
