Since the electron dropped from an energy level i to the ground state by emitting a single photon, this photon has an energy of 1.41 × 10⁻¹⁸ Joules.
<h3>How to calculate the photon energy?</h3>
In order to determine the photon energy of an electron, you should apply Planck-Einstein's equation.
Mathematically, the Planck-Einstein equation can be calculated by using this formula:
E = hf
<u>Where:</u>
In this scenario, this photon has an energy of 1.41 × 10⁻¹⁸ Joules because the electron dropped from an energy level i to the ground state by emitting a single photon.
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Answer
given,
Pressure on the top wing = 265 m/s
speed of underneath wings = 234 m/s
mass of the airplane = 7.2 × 10³ kg
density of air = 1.29 kg/m³
using Bernoulli's equation
Applying newtons second law
2 Δ P x A - mg = 0
A = 3.53 m²
The final speed of an airplane is v = 92.95 m/s
The rate of change of position of an object in any direction is known as speed i.e. in other word, Speed is measured as the ratio of distance to the time in which the distance was covered.
Solution-
Here given,
Acceleration a= 10.8 m/s2 .
Displacement (s)= 400m
Then to find final speed of airplane v=?
Therefore from equation of motion can be written as,
v²=u²+ 2as
where, u is initial speed, v is final speed ,a is acceleration and s is displacement of the airplane. Therefore by putting the value of a & s in above equation and (u =0) i.e. the initial speed of airplane is zero.
v²= 2×10.8 m/s²×400m
v²=8640m/s
v=92.95m/s
hence the final speed of airplane v =92.95m/s
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