let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
Radial acceleration is given by

where

then

Now

Using the relation


Putting into rpm
When I went through with the math, the answer I came upon was:
<span>6.67 X 10^14 </span>
<span>Here is how I did it: First of all we need to know the equation. </span>
<span>c=nu X lamda </span>
<span>(speed of light) = (frequency)(wavelength) </span>
<span>(3.0 X 10^8 m/s) = (frequency)(450nm) </span>
<span>We want the answer in meters so we need to convert 450nm to meters. </span>
<span>450nm= 4.5 X 10^ -7 m </span>
<span>(3.0 X 10^8 m/s) = (frequency)(4.5 X 10^ -7 m) </span>
<span>Divide the speed of light by the wavelength. </span>
<span>(3.0 X 10^8m/s) / (4.5 X 10^ -7m) =6.67 X 10^ 14 per second or s- </span>
<span>Answer: 6.67 X 10^14 s- hope this helps</span>