We are given an electromagnetic wave with a frequency of 5.09 x 10^14 Hz and travelling through a transparent medium. If the medium was vacuum, the speed of the wave would be equal to the speed of light. Otherwise, the main factor that would determine the speed of the wave is its wavelength.
Answer:
625000 N/ m
Explanation:
m= 20 kg
v= 30 m/s
x= 12 cm
k = ?
Here when the mass when hits at spring its speed is
Vi= 30 m/s
Finally it comes to rest after compressing for 12 cm
i-e Vf = 0 m/s
Distance= S= 12 cm = 0.12 m
using
2aS= Vf2 - Vi2
==> 2a ×0.12 = o- 30 × 30
==> a = 900 ÷ 0.24 = 3750 m/sec2
Now we know;
F = ma
F= -Kx
==> ma= -kx
==> 20 × 3750 = -K × 0.12
==> k = 625000 N/ m
Answer:
E/4
Explanation:
The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
Where;
E is the electric field
σ is the surface charge density
ε₀ is the electric constant.
Formula to calculate σ is;
σ = Q/A
Where;
Q is the total charge of the sheet
A is the sheet's area.
We are told the elastic sheet is a square with a side length as d, thus ;
A = d²
So;
σ = Q/d²
Putting Q/d² for σ in the electric field equation to obtain;
E = Q/(2ε₀d²)
Now, we can see that E is inversely proportional to the square of d i.e.
E ∝ 1/d²
The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.
From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;
E_new = E/4
2*3.5 = 7m/s
You multiply the acceleration per the time (they both are in seconds, otherwise, you should set them in the same units).
Answer:
The frequency of the photon decreases upon scattering
Explanation:
Here we note that when a photon is scattered by a charged particle, it is referred to as Compton scattering.
Compton scattering results in a reduction of the energy of the photon and hence an increase in the wavelength (from λ to λ') of the photon known as Compton effect.
Therefore, since the wavelength increases, we have from
λf = λ'f' = c
f = c/λ
Where:
f and f' = The frequency of the motion of the photon before and after the scattering
c = Speed of light (constant)
We have that the frequency, f, is inversely proportional to the wavelength, λ as follows;
f = c/λ
As λ = increases, and c is constant, f decreases, therefore, the frequency of the photon decreases upon scattering.
