Answer:
Intensity = 11.56W/m²
The energy flowing through the given area is 4.55 J
Explanation:
The expression for the intensity of the electromagnetic wave is,
Here,
is the permittivity of the free space,
is the electric field amplitude and
c is the speed of the light.
substitute
⁸m/s for c
8.85×10 −12 C²
/N⋅m² for 
and 93.3 V/m for 
The expression for the energy is,
E = I×A×t
Here, I is the intensity of the electromagnetic wave,
A is the area, and
t is the time.
Substitute
11.56W/m² for I
0.0287m ² for A
13.7s for t
The energy flowing through the given area is 4.55 J
What’s the question? Is it true or false?
Kinetic Energy I’m not 100% shure tho
Answer:
the lowest possible frequency of the emitted tone is 404.79 Hz
Explanation:
Given the data in the question;
S₁ ← 5.50 m → L
↑
2.20 m
↓
S₂
We know that, the condition for destructive interference is;
Δr = ( 2m + 
) × λ
where m = 0, 1, 2, 3 .......
Path difference between the two sound waves from the two speakers is;
Δr = √( 5.50² + 2.20² ) - 5.50
Δr = 5.92368 - 5.50
Δr = 0.42368 m
v = f × λ
f = ( 2m + )v / Δr
m = 0, 1, 2, 3, ....
Now, for the lowest possible frequency, let m be 0
so
f = ( 0 + )v / Δr
f = (v) / Δr
we know that speed of sound in air v = 343 m/s
so we substitute
f = (343) / 0.42368
f = 171.5 / 0.42368
f = 404.79 Hz
Therefore, the lowest possible frequency of the emitted tone is 404.79 Hz
