Answer:
The intensity of laser 2 is 4 times of the intensity of laser 1.
Explanation:
The intensity in terms of electric field is given by :
E is electric field
It means, 
In this problem, lasers 1 and 2 emit light of the same color, and the electric field in the beam of laser 1 is twice as strong as the e-field of laser 2.
Let E is electric field in the beam of laser 1 and E' is the electric field in the beam of laser 2. So,
We have,
E'=2E
So,
So, the intensity of laser 2 is 4 times of the intensity of laser 1.
Complete question:
A uniform electric field is created by two parallel plates separated by a
distance of 0.04 m. What is the magnitude of the electric field established
between the plates if the potential of the first plate is +40V and the second
one is -40V?
Answer:
The magnitude of the electric field established between the plates is 2,000 V/m
Explanation:
Given;
distance between two parallel plates, d = 0.04 m
potential between first and second plate, = +40V and -40V respectively
The magnitude of the electric field established between the plates is calculated as;
E = ΔV / d
where;
ΔV is change in potential between two parallel plates;
d is the distance between the plates
ΔV = V₁ -V₂
ΔV = 40 - (-40)
ΔV = 40 + 40
ΔV = 80 V
E = ΔV / d
E = 80 / 0.04
E = 2,000 V/m
Therefore, the magnitude of the electric field established between the plates is 2,000 V/m
Where is the rest of the sentence?
I'm not sure if a figure or some choices go along with this, but the closer to the sea floor the diver is, the lower the potential energy
Answer:
Newton's 2nd law think soo
