Complete Question
One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 2 cm across, and you estimate that the distance from the window shade to the wall is about 5 m.
Required:
Estimate the diameter of the pinhole.
Answer:
The diameter is 
Explanation:
From the question we are told that
The central maxima is 
The distance from the window shade is 
The average wavelength of the sun is mathematically evaluated as

Generally the visible light spectrum has a wavelength range between 400 nm to 700 nm
So the initial wavelength of the sun is 
and the final wavelength is 
Substituting this into the above equation


The diameter is evaluated as

substituting values


I believe that the mechanical energy would transform from starting out as kinetic, the reaching the top it would be potential, then go back to kinetic as it is falling back down.
I'm not 100% sure that this is right but if I had to take a guess this is what I would say.
Answer:
The frequency of oscillation of the simple pendulum is 0.49 Hz.
Explanation:
Given that,
Mass of the simple pendulum, m = 0.35 kg
Length of the string to which it is attached, l = 1 m
We need to find the frequency of oscillation. The frequency of oscillation of the simple pendulum is given by :

So, the frequency of oscillation of the simple pendulum is 0.49 Hz. Hence, this is the required solution.
Answer:
Zero or +2
Explanation:
The noble gases already have a avplete outermost shell. They are the least reactive elements of earth?
Their normal oxidation number is zero but some have been shown to be reactive.