Answer:
Frequency, f = 15 Hz
Explanation:
We have,
Speed of an ocean wave is 45 m/s
Wavelength of a wave is 3 m
It is required to find the frequency of an ocean wave.
Speed of a wave, 
, f = frequency of ocean wave
So, the frequency of an ocean wave is 15 Hz.
Answer:
what is the question
Explanation: plz give the brainliest
When a wave is too steep to support itself, the wave front collapses therefore creating a break.
<h3>What is a Wave?</h3>
This is defined as the propagation of disturbance from one place to another in an organized manner.
In situations where the wave is too steep to support itself there is a break in the wavefront which advances up the shoreline.
Read more about Wave here brainly.com/question/6069116
Answer:
t< 75 nm
Explanation:
A soap bubble is a thin film where when the beam enters the film it has a 180º phase change due to the refractive index and the wavelength changes between
λ = λ₀ / n
In the case of constructive interference in the curve of the spherical film it is
2 nt = (m + ½) λ₀
Where t is the thickness of the film and n the refractive index that does not indicate that we use that of water n = 1.33, m is an integer. The thickness of the film for the first interference (m = 0) is
t = λ₀ / 4 n
A thickness less than this gives destructive interference.
Let's look for the thickness for the visible spectrum
Violet light λ₀ = 400 nm = 400 10⁻⁹ m
t₁ = 400 10⁻⁹ / 4 1.33
t₁ = 75.2 10-9 m
Red light λ₀ = 700 nm = 700 10⁻⁹ m
t₂ = 700 10⁻⁹ / 4 1.33
t₂ = 131.6 10⁻⁹ m
Therefore, for all wavelengths to have destructive interference, the thickness must be less than 75 10⁻⁹ m = 75 nm
b) a film like eta is very thin, it is achieved when gravity thins the pomp, but any movement or burst of air breaks it,
Answer:
The answer to the question is
The distance d, which locates the point where the light strikes the bottom is 29.345 m from the spotlight.
Explanation:
To solve the question we note that Snell's law states that
The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction
n₁sinθ₁ = n₂sinθ₂
y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and
θ₁ = 14.511 °
n₁ = 1.0003 = refractive index of air
n₂ = 1.33 = refractive index of water
Therefore sinθ₂ = 
= 
= 0.1885 and θ₂ = 10.86 °
Since the water depth is 4.0 m we have tanθ₂ = 
or x₂ = 
=
= 20.845 m
d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.
