The range of frequencies of visible light in a vacuum is mathematically given as
Fmin=4.19*10^14Hz to Fmax=1*10^15Hz
<h3>What is the range of frequencies of visible light in a vacuum?</h3>
Question Parameters:
The wavelengths of visible light vary from about 300 nm to 700 nm.
Generally, the equation for the frequency is mathematically given as
F=C/\lambda
Therefore
For Fmax

Fmax=1*10^15Hz
Where

Fmin=4.19*10^14Hz
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Answer:
Vf = 23 m/s
Explanation:
First we need to find the distance covered by the motorcycle 2 when it passes motorcycle 1. Using the uniform speed equation for motorcycle 1:
s₁ = v₁t₁
where,
s₁ = distance covered by motorcycle 1 = ?
v₁ = speed of motorcycle 1 = 6.5 m/s
t₁ = time = 10 s
Therefore,
s₁ = (6.5 m/s)(10 s)
s₁ = 65 m
Now, for the distance covered by motorcycle 2 at the meeting point. Since, the motorcycle started 50 m ahead of motorcycle 2. Therefore,
s₂ = s₁ + 50 m
s₂ = 65 m + 50 m
s₂ = 115 m
Now, using second equation of motion for motorcycle 2:
s₂ = Vi t + (1/2)at²
where,
Vi = initial velocity of motorcycle 2 = 0 m/s
Therefore,
115 m = (0 m/s)(10 s) + (1/2)(a)(10 s)²
a = 230 m/100 s²
a = 2.3 m/s²
Now, using 1st equation of motion:
Vf = Vi + at
Vf = 0 m/s + (2.3 m/s²)(10 s)
Vf = 23 m/s
The compression curve would be theoretically given for a system of bodies in which the spring applies the force (Although in the same way the following process can be extrapolated to any system, depending on the type of Force to consider) For a spring mass system, the strength is given by Hooke's law as

Where,
K = Spring constant
x = Displacement
If we integrate based on distance we would have

This integral represents the area under the Force Curve based on each distance segment traveled.



This is the same formula that represents the elastic potential energy of a body. Therefore the correct answer is D.
5.625 hours and it is 450 divided by 80
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