Answer:
I_FWHW = 3.2 μW / m²
Explanation:
In the analysis of optics and electricity a very useful magnitude is the width at half height (FWHW) and the intensity at this height, which is given by
I_FWHW = I₀ / 2
corresponds to the width of the line for this intensity.
In this case they give the maximum intensity for which
I_FWHW = 6.2 / 2
I_FWHW = 3.2 μW / m²
You do not give more data in your exercise, but the most interesting calculation is to find the angle values for which you have this intensity since it is this range is 50% of the energy of the system, have I write the equation for this calculation
I = Io cos² x₁ (sin x / x)²
x₁ = π d sin θ /λ
x = π b sin θ /λ
where d is the separation of the slits and b the width of each slit
Radiation is the only one that works across the vacuum of space
Here, you need to apply second equation of Kinematics
S = ut + 1/2 at²
s = at²/2 [ As initial velocity (u) = 0 ]
Now, a = 9.8 m/s²
t = 2.7 s
Substitute their values in the formula:
s = 9.8 × (2.7)² / 2
s = 35.721 m
After rounding-off to the nearest hundredth value, it would be 36 m
Finally, your answer would be option B.
Hope this helps!
Upside down if its further than 1 focal point you have seen this with a spoon and enlarged right side up if closer than 1 focal point
Answer:
So the acceleration of the child will be 
Explanation:
We have given angular speed of the child 
Radius r = 4.65 m
Angular acceleration 
We know that linear velocity is given by 
We know that radial acceleration is given by 
Tangential acceleration is given by
So total acceleration will be 
