The answer is n= 6.
What is Balmer series?
The Balmer series is the portion of the emission spectrum of hydrogen that represents electron transitions from energy levels n > 2 to n = 2. These are four lines in the visible spectrum. They are also known as the Balmer lines. The four visible Balmer lines of hydrogen appear at 410 nm, 434 nm, 486 nm and 656 nm.
For the Balmer series, the final energy level is always n=2. So, the wavelengths 653.6, 486.1, 434.0, and 410.2 nm correspond to n=3, n=4, n=5, and n=6 respectively. Since the last wavelength, 410.2 nm, corresponds to n=6, the next wavelength should logically correspond to n=7.
To solve for the wavelength, calculate the individual energies, E2 and E7, using E=-hR/(n^2). Then, calculate the energy difference between E2 (which is the final) and E7 (which is the initial). Finally, use lamba=hc/E to get the wavelength.
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Answer:
I would shout fore help if I was being raped or try to make him or her stop
Answer:
a) K = 2/3 π G m ρ R₁³ / R₂
, b) U = - G m M / r
Explanation:
The law of universal gravitation is
F = G m M / r²
Part A
Let's use Newton's second law
F = m a
The acceleration is centripetal
a = v² / R₂
G m M / R₂² = m v² / R₂
v² = G M / R₂
They give us the density of the planet
ρ = M / V
V = 4/3 π R₁³
M = ρ V
M = ρ 4/3 π R₁³
v² = 4/3 π G ρ R₁³ / R₂
K = ½ m v²
K = ½ m (4/3 π G ρ R₁³ / R₂)
K = 2/3 π G m ρ R₁³ / R₂
Part B
Potential energy and strength are related
F = - dU / dr
∫ dU = - ∫ F. dr
The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1
U- U₀ = G m M ∫ dr / r²
U - U₀ = G m M (- r⁻¹)
We evaluate for
U - U₀ = -G m M (1 / - 1 /)
They indicate that for ri = ∞ U₀ = 0
U = - G m M / r
Since there are no external forces, including friction, act on the flatcar. after the sack rests on the flatcar, we would assume that momentum is conserved. This means that
total momentum of car before collision = total momentum of car after collision.
Recall,
momentum = mass x velocity
From the information given,
mass of car before collision = 2000
velocity of car before collision = 3
Thus,
total momentum of car before collision = 2000 x 3 = 6000
Also,
mass of sack = 500
mass of car and sack after collision = 500 + 2000 = 2500
velocity after collision = v
momentum after collision = 2500 x v = 2500v
Since momentum is conserved, then
6000 = 2500v
v = 6000/2500
v = 2.4
the speed of the flatcar is 2.4 m/s
The way to do this is very easy so do 4125 x 2 = ? then the ? will be times by 2 again after the answer to both of those is your answer!!!