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:
if i'm not mistaken that's either a plug in cord for a certain device, or what i am assuming to be a usb software cord.
Explanation:
Answer:
Explanation:
la frecuencia = ω/2π, nada cambio
v(max) = ωA → ω2Α = 2ωA duplicara velocidad máxima
a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima
la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía
I believe it's magnetic stripping. I hope that helps
Answer:
Explanation:
The strength of the electric field produced by a charge Q is given by
where
Q is the charge
r is the distance from the charge
k is the Coulomb's constant
In this problem, the electric field that can be detected by the fish is
and the fish can detect the electric field at a distance of
Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:
