The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.
Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima. The latitude in the middle
of that intersection is 46.585° North. <u>That's</u> the number we need.
Here's how I would do it:
-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.
-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°.
-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.
This sets the limit of the highest in the sky that the moon can ever appear.
90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .
That doesn't happen regularly. It would depend on everything coming
together at the same time ... the moon happens to be at the point in its
orbit that's 5.14° above ==> (the point on the ecliptic that's 23.5° above
the celestial equator).
Depending on the time of year, that can be any time of the day or night.
The most striking combination is at midnight, within a day or two of the
Winter solstice, when the moon happens to be full.
In general, the Full Moon closest to the Winter solstice is going to be
the moon highest in the sky. Then it's going to be somewhere near
67° above the horizon at midnight.
A large geographical region having a distinct biodiversity<span> of </span>flora<span> and </span>fauna.<span>
</span>
Cadences.
These cadences are the resulting tensions that chords release from their resting points. This movement is classified from a unstable chord progression to a stable one. Thank you for your question. Please don't hesitate to ask in Brainly your queries.
Answer:
Part a)

Part b)


Part c)


Explanation:
Part a)
frequency of light will not change with change in medium but it will depend on the source only
so here frequency of light will remain same in both water and glass and it will be same as that in air



Part b)
As we know that the refractive index of water is given as

so the wavelength in the water medium is given as



Similarly the refractive index of glass is given as

so the wavelength in the glass medium is given as



Part c)
Speed of the wave in water is given as



Speed of the wave in glass is given as


