Answer:
Incident ray
Explanation:
"To describe the reflection of light, we will use the following terminology. The incoming light ray is called the incident ray. The light ray moving away from the surface is the reflected ray. The most important characteristic of these rays is their angles in relation to the reflecting surface."
https://www.siyavula.com/read/science/grade-11/geometrical-optics/05-geometrical-optics-03
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
Answer:
-6N
Explanation:
The force to the east is acting in the positive x-direction therefore it is positive. The force to the east is in the negative x-direction therefore it is negative. The net force is just the sum of the two so 3-9=-6
Answer:
The smallest diameter is 
Explanation:
From the question we are told that
The resolution of the telescope is 
The wavelength is 
From the question we are told that

So 
Therefore


Now 
So 
=> 

The smallest diameter is mathematically represented as

substituting values


1.) appearance
2.)texture
3.)color
4.)melting point
5.)odor