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Rudiy27
3 years ago
15

Focus groups

Physics
1 answer:
Montano1993 [528]3 years ago
4 0
I think the answer is a
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What is the highest degrees above the horizon the moon ever gets during the year in the Yakima Valley ?
Ivahew [28]

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.


5 0
2 years ago
What is refraction? <br><br>this my in sta id-:akhilrawat9453​
My name is Ann [436]

Explanation:

ij jdjcjxjjdjnndnxnsmxnjxjebxnc

6 0
2 years ago
Read 2 more answers
Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0
Alisiya [41]

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

            y’= R sin 35

           

The modulus vector can be found using the Pythagorean theorem

            R² = x² + y²

            R = 450 m

we calculate

            x ’= 450 cos 35

            x ’= 368.61 m

            y ’= 450 sin 35

            y ’= 258.11 m

4 0
2 years ago
How many cycles have been completed after 3 seconds on the graph​
Reptile [31]

Answer:

3 Cycles

Explanation:

Every cycle on a cosine function are marked by the completion of 5 key points when moving along the x-axis:

(5, 0, -5, 0, 5)

This cycle has a frequency of 2, occurring once every second, thus answering our question easily.

6 0
3 years ago
Which of these tools is used to measure temperature?
Feliz [49]

Answer:

D a thermometer

Explanation: It measures and track Celcius and Feirinheit.

3 0
2 years ago
Read 2 more answers
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