The energy required by the excitation of the line is:
ΔE = hν = hc / λ
where:
ΔE = energy difference
h = Planck constant
ν = line frequency
c = speed of light
λ = line wavelength
The energy difference must be supplied by the electron, supposing it transfers all its kinetic energy to excite the line:
Therefore,
And solving for v we get:
Plugging in numbers (after trasforing into the correct SI units of measurement):
=9.4 · 10⁵ m/s
Hence, the electron must have a speed of
9.4 · 10<span>
⁵ m/s in order to excite the <span>492nm</span> line.</span>
Answer:
(a) Position Vectors V₁= -2î km, V₂=5î km
(b) Displacement Δx=7 km
Explanation:
Given data
Distance=2 km west at t=0
Distance=5 km east at t=6 min
Positive x is the east direction
To find
(a)Car position vector at given times
(b)Displacement between 0 to 6.0 min
Solution
For Part (a) car position vector at given times
At t=0 the distance=2 km west so conclude that x₁=-2 because it is in negative side So vector V₁
V₁= -2î km
At t=6.0 the distance=5 km east so conclude that x₂=5 because it is in positive side So vector V₂
V₂=5î km
For (b) displacement between 0 to 6.0 min
According to following mathematical law we can conclude that
Δx=x₂-x₁
Δx=5-(-2)km
Δx=7 km
ANOTHER RUNNING DOG
Explanation:
In the given question it is to find a suitable reference point to describe the motion of dog. Here I could suggest that it is better to compare the dog with another running dog to create the relative speed difference to get a reliable motion variation.
Because the motion of dog is in the linear with respect to the another dog and to the acceleration produced by the dog in the required interval is easy to calculate with respect to another dog which is already in motion.
Hence, I suggest that Motion of dog can be analysed better by analyse the motion variation of dog with another dog running.
THE ANSWER IS A USING CAMERA IMAGING AND WIRELESS DEVICE HOPE THIS HELPS SOMEONE.