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rodikova [14]
3 years ago
6

What is the highest degrees above the horizon the moon ever gets during the year in the Yakima Valley ?

Physics
1 answer:
Ivahew [28]3 years ago
5 0

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.


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0.4029 \mathrm{m} / \mathrm{s}^{2} is the acceleration of the box.

<u>Explanation:</u>

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Mass of the box = 3.74 kg

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First, we need to find the net horizontal force acting on the box. With the given data, the equation can be formed as below.  Net horizontal force acting on the box (F) is given by

F=\left(4.20 \times \cos 50^{\circ}\right)+\left(2.25 \times \cos 122^{\circ}\right)

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8 0
3 years ago
A girl delivering newspapers covers her route by traveling 5.00 blocks west, 5.00 blocks north, and then 7.00 blocks east. What
FromTheMoon [43]

Answer:

P = 2i + 5j

Therefore she is 2 blocks east and 5 blocks north.

Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks

Angle = taninverse (5/2)

Angle = 68.2°

Explanation:

Given:

Let west be negative and east be positive x axis.

Let north be the positive y axis.

5.00blocks west = -5.00 i

5.00 blocks north = 5.00 j

7.00 blocks east = 7.00i

Addition of the vector form of hee position is;

P = -5i +7i -5j

P = 2i + 5j

Therefore she is 2 blocks east and 5 blocks north.

Resultant P = √(2^2 + 5^2) = √(4+25) = √29 = 5.4 blocks

Angle = taninverse (5/2)

Angle = 68.2°

3 0
3 years ago
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