Reset the PhET simulation (using the button in the lower right) and set it up in the following manner: select Oscillate, select
No End, and use the parameters in parentheses by sliding the bars for Amplitude (1.00 cmcm), Frequency (1.40 HzHz ), Damping (none), and Tension (highest). Using the available Rulers, calculate the frequency of a photon that corresponds to the wavelength of the resulting wave. Assume the length with units (cmcm) of the ruler represents the real photon wavelength and that the speed of light is 3.00×108 m/s3.00×108 m/s.
1 answer:
Answer:
Check Explanation
Explanation:
This is a question that is as a result of an experimental procedure.
The Phet simulation is put on the settings given in the question;
- select Oscillate
- Select No End
- Use the parameters in parentheses by sliding the bars for Amplitude (1.00 cm), Frequency (1.40 Hz)
- Damping (none)
- Tension (highest)
I'll attach an interface of the Phet simulation to this solution.
Once all of these settings have been fixed, the simulation gives a wave pattern whose wavelength can be read from the ruler attached to the background of the simulation.
The wavelength is the distance from crest to crest or from trough to trough.
From the simulation example I have attached, the distance from crest to crest is from the green indicator on one crest to the green indicator on the next crest, that is about 5 to 5.1 cm
The velocity of a wave, v, is related to the frequency, f, and wavelength, λ of the wave through
v = fλ
For the photon, the velocity of the wave is the speed of light,
v = c = (3.00 × 10⁸) m/s
The wavelength computed from the simulation = λ = 5.1 cm = 0.051 m
c = fλ
frequency = (c/λ) = (3.00 × 10⁸) ÷ 0.051 = (5.88 × 10⁹) Hz
So, this step can be used to obtaim rhe required frequency of the photon, just follow these steps and use the calculation method too. You should be able to obtain the frequency of the photon in your experiment.
Hope this Helps!!!
You might be interested in
Answer:
h = v₀² / 2g , h = k/4g x²
Explanation:
In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches
Starting point. Lower compressed spring
Em₀ = K = ½ m v²
Final point. Highest on the path
= U = mg h
As or no friction the energy is conserved
Em₀ = Em_{f}
½ m v₀²² = m g h
h = v₀² / 2g
We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse
½ k x² = 2 g h
h = k/4g x²
156.8 Joules of energy is in the box's gravitational potential energy store
<u>Explanation</u>:
<em>Given:</em>
Mass of the box Dane is holding = 8 Kilograms
Height at which Dane is holding the box above the ground= 2 metres
<em>To Find:</em>
Gravitational potential energy in the box=?
<em>Solution:</em>
gravitational potential energy is the work done per mass on a object to move that object from one fixed location to to another location against gravity.Its unit is joules or J
Thus Gravitational potential energy is represented as,
where
is the gravitational potential energy
m is the mass
h is the height
g is the gravitational force( 9.8 )
Now substituting the given values,
A) Zero
The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.
B) -9.8 m/s^2
The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:
where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:
where the negative sign means it points downward.
C) 7.6 m/s
The x-component of the shot's velocity is given by:
where
is the initial velocity
is the angle of the shot
Substituting into the equation, we find
D) 9.3 m/s
The y-component of the shot's velocity is given by:
where
is the initial velocity
is the angle of the shot
Substituting into the equation, we find
E) 7.6 m/s
We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:
F) -11.1 m/s
The y-component of the velocity at time t is given by:
where
is the initial y-velocity
a = g = -9.8 m/s^2 is the vertical acceleration
t is the time
Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:
where the negative sign means the vertical velocity is now downward.
Since the tower base is square with a side length of 125 m,
Therefore,
Square root of 31250 = 176.776953 (Diameter) , so this is the diameter of the cylinder to enclose it, and radius, r = 88.38834765 m and height, h = 324 m.
The volume of cylinder,
Thus, the mass of the air in the cylinder,
Hence, the mass of the air in the cylinder is this more than the mass of the tower.