This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3
<span><u><em>This begins with the sun’s radiation,
which is absorbed differently on the earth’s surface. The earth's
surface is heated differently because of scenarios like cloud cover,
mountains, valleys, water bodies, vegetation and desert lands.</em></u>
<u><em></em></u>
<u><em>Good luck! </em></u></span><u><em></em></u>
It would be 1. B 2. A 3. A
Answer:
about 602 milliseconds
Explanation:
The motion can be approximated by the equation ...
y = -4.9t^2 -22.8t +15.5
where t is the time since the arrow was released, and y is the distance above the ground.
When y=0, the arrow has hit the ground.
Using the quadratic formula, we find ...
t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))
= (22.8 ± √823.64)/(-9.8)
The positive solution is ...
t ≈ 0.60195193
It takes about 602 milliseconds for the arrow to reach the ground.
Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector

.
The positive y-axis = the northern direction, with unit vector

.
The airplane flies at 340 km/h at 12° east of north. Its velocity vector is

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
![\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D_%7B2%7D%20%3D40%28cos%2834%5E%7Bo%7D%29%5Chat%7Bi%7D%20-%20sin%2824%5E%7Bo%7D%29%5D%5Chat%7Bj%7D%29%20%3D%2033.1615%5Chat%7Bi%7D%20-22.3677%5Chat%7Bj%7D%29)
The plane's actual velocity is the vector sum of the two velocities. It is

The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h
The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°
Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.