Although the video is not found here, the sentence makes reference to the transmission spectrum of colored filters.
<h3>What is the transmission spectrum?</h3>
The transmission spectrum indicates the light portion having a given wavelength that can be passed through a filter.
This spectrum (transmission spectrum) depends on the physical separation of the particles that form the filter.
In conclusion, although the video is not found here, the sentence makes reference to the transmission spectrum of colored filters.
Learn more about the transmission spectrum here:
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Answer:

Explanation:
The heaviside function is defined as:

so we see that the Heaviside function "switches on" when
, and remains switched on when 
If we want our heaviside function to switch on when
, we need the argument to the heaviside function to be 0 when 
Thus we define a function f:

The
term inside the heaviside function makes sure to displace the function 5 units to the right.
Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. (
when
, so it becomes just a 1, which we can safely ignore.)
Therefore our final result is:

I have made a sketch for you, and added it as attachment.
171.0798 M/S
In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.
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Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so
= A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s