Answer:
F = 236063.6N
Explanation:
Please see attachment below.
Answer:
Interface
Explanation:
This is a classic example of Interface technology.
An interface allows different software packages to communicate without re-entering data.
Here in this case also systems are able to communicate with one another without duplicating data entry. For example, practice management software and another for their electronic health record.
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.
Answer:
1. 0.45 s.
2. 4.41 m/s
Explanation:
From the question given above, the following data were obtained:
Height (h) = 1 m
Time (t) =?
Velocity (v) =?
1. Determination of the time taken for the pencil to hit the floor.
Height (h) = 1 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1 = ½ × 9.8 × t²
1 = 4.9 × t²
Divide both side by 4.8
t² = 1/4.9
Take the square root of both side
t = √(1/4.9)
t = 0.45 s.
Thus, it will take 0.45 s for the pencil to hit the floor.
2. Determination of the velocity with which the pencil hit the floor.
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 0.45 s.
Final velocity (v) =?
v = u + gt
v = 0 + (9.8 × 0.45)
v = 0 + 4.41
v = 4.41 m/s
Thus, the pencil hit the floor with a velocity of 4.41 m/s
