A quantum system has three energy levels, so three wavelengths appear in its emission spectrum. the shortest observed wavelength
1 answer:
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Answer:
a)
b)
Explanation:
Part a
For this case we can begin finding the period like this:
Then we know that the centripetal acceleration is given by:
And the velocity is given by:
If we replace this into the acceleration we got:
And we can replace the values and we got:
Part b
For this case we want to find a value of k such that:
Where a = 9.74, so then we can solve for k like this:
The average speed between 0 h and 2.340 h is 6.97 Km/h
Average speed is defined as the total distance travelled divided by the total time taken to cover the distance.
With the above formula, we can obtain the average speed between 0 h and 2.340 h as illustrated below:
- Total time = 2.340 – 0 = 2.340 h
- Total distance = 16.3 – 0 = 16.3 Km
- Average speed =?
Learn more about average speed: brainly.com/question/24884027
Answer:
P = 1 (14,045 ± 0.03 ) k gm/s
Explanation:
In this exercise we are asked about the uncertainty of the momentum of the two carriages
Δ (Pₓ / Py) =?
Let's start by finding the momentum of each vehicle
car X
Pₓ = m vₓ
Pₓ = 2.34 2.5
Pₓ = 5.85 kg m
car Y
Py = 2,561 3.2
Py = 8,195 kgm
How do we calculate the absolute uncertainty at the two moments?
ΔPₓ = m Δv + v Δm
ΔPₓ = 2.34 0.01 + 2.561 0.01
ΔPₓ = 0.05 kg m
Δ = m Δv + v Δm
ΔP_{y} = 2,561 0.01+ 3.2 0.001
ΔP_{y} = 0.03 kg m
now we have the uncertainty of each moment
P = Pₓ /
ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²
ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²
ΔP = 0.006 + 0.0026
ΔP = 0.009 kg m
The result is
P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s