Answer:
a) 
b) 
c) 
d) 
Explanation:
<u>Given equation of pressure variation:</u>
![\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]](https://tex.z-dn.net/?f=%5CDelta%20P%3D%20%281.78%5C%20Pa%29%5C%20sin%5C%20%5B%280.888%5C%20m%5E%7B-1%7D%29x-%28500%5C%20s%5E%7B-1%7D%29t%5D)
We have the standard equation of periodic oscillations:
<em>By comparing, we deduce:</em>
(a)
amplitude:
(b)
angular frequency:
∴Frequency of oscillations:
(c)
wavelength is given by:
(d)
Speed of the wave is gives by:
No, that's silly.
You've got your Pfund series where electrons fall down to the 5th level,
your Brackett series where they fall to the 4th level, and your Paschen
series where they fall to the 3rd level. All of those transitions ploop out
photons at Infrared wavelengths.
THEN next you get your Balmer series, where the electrons fall in
to the 2nd level. Most of those are at visible wavelengths, but even
a few of the Balmer transitions are in the Ultraviolet.
And then there's the Lyman series, where electrons fall all the way
down to the #1 level. Those are ALL in the ultraviolet.
<span>We can use an equation to find the gravitational force exerted on the HST.
F = GMm / r^2
G is the gravitational constant
M is the mass of the Earth
m is the mass of the HST
r is the distance to the center of the Earth
This force F provides the centripetal force for the HST to move in a circle. The equation we use for circular motion is:
F = mv^2 / r
m is the mass of the HST
v is the tangential speed
r is the distance to the center of the Earth
Now we can equate these two equations to find v.
mv^2 / r = GMm / r^2
v^2 = GM / r
v = sqrt{GM / r }
v = sqrt{(6.67 x 10^{-11})(5.97 x 10^{24}) / 6,949,000 m}
v = 7570 m/s which is equal to 7.570 km/s
HST's tangential speed is 7570 m/s or 7.570 km/s</span>
