Answer:
i think that bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad but actually isnt bad because their good at being bad but they are actually bad, that makes me say, its A
Answer:
Explanation:
Explanation:
Answer:
47 mW
Explanation:
The average value of the Poynting vector, S = 0.939 W/m² = Intensity of wave, I
S = I S
Also, I = P/A where P = Et, P = power of electromagnetic wave, E = energy of electromagnetic wave in time t and t = time = 1 min = 60 s and A = area = lb since the electromagnetic waves falls on area equal to that of a rectangle.
So, S = Et/A
E = SA/t
= Slb/t
= 0.939 W/m² × 1.5 m × 2.0 m/60 s
= 2.817 W/60 s
= 0.047 W
= 47 mW
So, 47 mW of electromagnetic energy falls on the area in 1.0 minute.
<span>A full moon is at its brightest, and here is no disk to be seen. New moons are barely visable.</span>
Answer:
A) I = Io 0.578, B) he light that leaves the polarized is completely polarized, being perpendicular to the axis of the second filter
Explanation:
A) Light passing through a polarizer must comply with the / bad law
I = Io cos2 tea
Where is at the angle of the polarizer and incident light
I = Io cos2 45
I = Io 0.578
Therefore the beam intensity is 0.578 of the incident intensity
.B) the light that leaves the polarized is completely polarized, being perpendicular to the axis of the second filter
