Answer:
True The grid with more slits gives more angle separation increases
True. The grating with 10 slits produces better-defined (narrower) peaks
Explanation:
Such a system can be seen as a diffraction network in this case with different number of lines per unit length, the expression for the constructive interference of a diffraction network is
d sin θ = m λ
where d is the distance between slits or lines, m the order of diffraction and λ the wavelength.
For network with 5 slits
d = 1/5 = 0.2
For the network with 10 slits
d = 1/10 = 0.1
let's calculate the separation (teat) for each one
θ = sin⁻¹ (m λ / d)
for 5 slits
θ₅ = sin⁻¹ (m λ 5)
for 10 slits
θ₁₀ = sin⁻¹ (m λ 10)
we can appreciate that for more slits the angle increases
the intensity of a series of slits is
I = I₀ sin²2 (N d/2) / sin² d/2)
when there are more slits (N) the peaks have greater intensity and are more acute (half width decreases)
let's analyze the claims
False
True The grid with more slits gives more angle separation increases
False
True The expression for the intensity of the diffraction peaks the intensity of the peaks increases with the number of slits as well as their spectral width decreases
False
Answer:
a) 
For this case we know the following values:
So then if we replace we got:
b) 
With 
And replacing we have:
And then the scattered wavelength is given by:
And the energy of the scattered photon is given by:
c) 
Explanation
Part a
For this case we can use the Compton shift equation given by:
For this case we know the following values:
So then if we replace we got:
Part b
For this cas we can calculate the wavelength of the phton with this formula:
With
And replacing we have:
And then the scattered wavelength is given by:
And the energy of the scattered photon is given by:
Part c
For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:
Answer:
<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>
Explanation:
The Non conservative force is defined as a force which do not store energy or get he energy dissipate the energy from the system as the system progress with the motion.
Given are
<em> mass of the student 73 kg</em>
<em> height of water glide 11.8 m</em>
<em> work done as -5.5*10³ J</em>
Have to find speed at which the student goes down the glide.
According to<em> Law of Conservation of energy</em>,
K.E =P.E+Work Done
mv²/2=mgh +W
Rearranging the above eqn for v
v = √2(gh+W/m)
Substituting values,
V = 12.48 m/s.
<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>
D chest tightness
I experienced these the past few days...
but hoped I helped ^_^
