Answer:
The air-water interface is an example of<em> </em>boundary. The <u><em>transmitted</em></u><em> </em> portion of the initial wave energy is way smaller than the <u><em>reflected</em></u><em> </em> portion. This makes the <u><em>boundary</em></u> wave hard to hear.
When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can <u><em>travel directly to your ear</em></u>.
Explanation:
The air-to-water sound wave transmission is inhibited because more of reflection than transmission of the wave occurs at the boundary. In the end, only about 30% of the sound wave eventually reaches underwater. For sound generated underwater, all the wave energy is transmitted directly to the observer. Sound wave travel faster in water than in air because, the molecules of water are more densely packed together, and hence can easily transmit their vibration to their neighboring molecules, when compared to air.
.....When astronomers developed ways to measure distances, they could plot, this angle from north, this many light years, so relative to us that star is here. Plot many stars this way and you will see the shape of the galaxy.... i think this is a response right?
Answer:
thickness t = 528.433 nm
Explanation:
given data
wavelength λ1 = 477.1 nm
wavelength λ2 = 668.0 nm
n = 1.58
solution
we know for constructive interference condition will be
2 × t × μ = (m1+0.5) × λ1 ....................1
2 × t × μ = (m2+0.5) × λ2 ....................2
so we can say from equation 1 and 2
(m1+0.5) × λ1 = (m2+0.5) × λ2
so
..............3
put here value and we get
= 1.4
...................4
so we here from equation 4
m1+0.5 = 7
m1 = 3 .................5
m2+0.5 = 4
m2 = 2 .................6
so now put value in equation 1
2 × t × μ = (m1+0.5) × λ1
2 × t × 1.58 = (3+0.5) × 477.1
solve it we get
thickness t = 528.433 nm
Answer:
F = π/4 ρ d² v²
Explanation:
Force is mass times acceleration:
F = ma
Acceleration is change in velocity over change in time:
F = m Δv / Δt
Since there's no rebound, Δv = v.
F = m v / Δt
Mass is density times volume:
F = ρ V v / Δt
Volume per time is flow rate:
F = ρ Q v
Flow rate is velocity times cross sectional area:
F = ρ (v A) v
F = ρ A v²
Area of a circle is pi times radius squared, or pi/4 times diameter squared:
F = ρ (π/4 d²) v²
F = π/4 ρ d² v²