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.
Final Answer is - 197.53948
No matter what direction you throw it, or with what speed, its acceleration is immediately 9.8 m/s^2 downward as soon as you release it from your hand, and it doesn't change until the ball hits something.
Answer: 4.17m
Explanation:
The observer at C will hear a sound on no sound upon whether the interference is constructive or destructive.
If the listeners hears sounds it is caled constructive interference but if he hears no sound its called destructive interference.
So
d2 - d1 = (n *lamba)/ 2
Where n=1,3,5
lamda=v/f =349/62.8
lamda=5.56m
d2= d1 + nlamda/2
d2= 1 + 5.56/2
d2= 3.78m
X'= 1 cos 60= 0.5m
Y= 1 sin60= 0.866m
X"^2 + Y^2 =d2^2
X" =√(y^2 - d2^2)
X"=√(3.78^2 - 0.886^2)
X"= 3.67m
So therefore the closest that speaker A can be to speaker B so the listener does not hear any sound is X' + X"= 0.5 + 3.67
4.17m
Answer:
the final speed of the rain is 541 m/s.
Explanation:
Given;
acceleration due to gravity, g = 9.81 m/s²
height of fall of the rain, h = 9,000 m
time of the rain fall, t = 1.5 minutes = 90 s
Determine the initial velocity of the rain, as follows;
The final speed of the rain is calculated as;
Therefore, the final speed of the rain is 541 m/s.
