<h2>
Answer:When electrons absorb or emit quantized units of energy in the form of photons.</h2>
Explanation:
When a electron is collided with a photon with exactly the same energy it would require to get to any of the farther orbits,electron transition takes place to an orbit depending on the energy of the photon.
When electrons emit exactly the same amount energy that is difference between the current energy level and the new level,then the electron makes a transition to the new level.
Answer:
![y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x](https://tex.z-dn.net/?f=y_n%28x%29%20%3DC_n%20%5Csin%20%5Csqrt%7B%5Cfrac%7B%5Crho%7D%7BT%7D%20%7D%20w_nx%3DC_n%20%5Csin%20%5Csqrt%7B%5Cfrac%7B%5Crho%7D%7BT%7D%20%7D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Crho%7D%20%7D%20%5Cfrac%7Bn%20%5Cpi%7D%7BL%7D%20x)
![y_n(x) = C_n \sin \frac{n \pi x}{L}](https://tex.z-dn.net/?f=y_n%28x%29%20%3D%20C_n%20%5Csin%20%5Cfrac%7Bn%20%5Cpi%20x%7D%7BL%7D)
Explanation:
The given differential equation is
and y(0) = 0, y(L) =0
where T and ρ are constants
The given rewrite as
![\frac{d^2y}{dx^2} + \frac{\rho w^2}{T} y=0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%20%2B%20%5Cfrac%7B%5Crho%20w%5E2%7D%7BT%7D%20y%3D0)
auxiliary equation is
![m^2+ \frac{\rho w^2}{T} =0\\\\m= \pm\sqrt{\frac{\rho}{T} } wi](https://tex.z-dn.net/?f=m%5E2%2B%20%5Cfrac%7B%5Crho%20w%5E2%7D%7BT%7D%20%3D0%5C%5C%5C%5Cm%3D%20%5Cpm%5Csqrt%7B%5Cfrac%7B%5Crho%7D%7BT%7D%20%7D%20wi)
Solution of this de is
![y(x)=C_1 \cos \sqrt{\frac{\rho}{t} } wx + C_2 \sin \sqrt{\frac{\rho}{T} } wx](https://tex.z-dn.net/?f=y%28x%29%3DC_1%20%5Ccos%20%5Csqrt%7B%5Cfrac%7B%5Crho%7D%7Bt%7D%20%7D%20wx%20%2B%20C_2%20%5Csin%20%5Csqrt%7B%5Cfrac%7B%5Crho%7D%7BT%7D%20%7D%20wx)
y(0)=0 ⇒ C₁ = 0
![y(x) = C_2 \sin \sqrt{\frac{\rho}{T} } wx](https://tex.z-dn.net/?f=y%28x%29%20%3D%20C_2%20%5Csin%20%5Csqrt%7B%5Cfrac%7B%5Crho%7D%7BT%7D%20%7D%20wx)
y(L) = 0 ⇒
![C_2 \sin \sqrt{\frac{\rho}{T} } wL=0](https://tex.z-dn.net/?f=C_2%20%5Csin%20%5Csqrt%7B%5Cfrac%7B%5Crho%7D%7BT%7D%20%7D%20wL%3D0)
we need non zero solution
⇒ C₂ ≠ 0 and
![\sin \sqrt{\frac{\rho}{T} } wL=0](https://tex.z-dn.net/?f=%5Csin%20%5Csqrt%7B%5Cfrac%7B%5Crho%7D%7BT%7D%20%7D%20wL%3D0)
![\sin \sqrt{\frac{\rho}{T} } wL=0 \rightarrow \sqrt{\frac{\rho}{T} } wL=n \pi](https://tex.z-dn.net/?f=%5Csin%20%5Csqrt%7B%5Cfrac%7B%5Crho%7D%7BT%7D%20%7D%20wL%3D0%20%5Crightarrow%20%5Csqrt%7B%5Cfrac%7B%5Crho%7D%7BT%7D%20%7D%20wL%3Dn%20%5Cpi)
![w_n = \sqrt{\frac{T}{\rho} } \frac{n \pi}{L}](https://tex.z-dn.net/?f=w_n%20%3D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Crho%7D%20%7D%20%5Cfrac%7Bn%20%5Cpi%7D%7BL%7D)
solution corresponding these
values
![y_n(x) =C_n \sin \sqrt{\frac{\rho}{T} } w_nx=C_n \sin \sqrt{\frac{\rho}{T} } \sqrt{\frac{T}{\rho} } \frac{n \pi}{L} x](https://tex.z-dn.net/?f=y_n%28x%29%20%3DC_n%20%5Csin%20%5Csqrt%7B%5Cfrac%7B%5Crho%7D%7BT%7D%20%7D%20w_nx%3DC_n%20%5Csin%20%5Csqrt%7B%5Cfrac%7B%5Crho%7D%7BT%7D%20%7D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Crho%7D%20%7D%20%5Cfrac%7Bn%20%5Cpi%7D%7BL%7D%20x)
![y_n(x) = C_n \sin \frac{n \pi x}{L}](https://tex.z-dn.net/?f=y_n%28x%29%20%3D%20C_n%20%5Csin%20%5Cfrac%7Bn%20%5Cpi%20x%7D%7BL%7D)
An animal doesnt have a chloroplast
The dimension of soccer field in ft is 377.315 ft long and 277.885 ft in width.
The dimension of soccer field in ft is 4527.78 in long and 3334.62 in in width.
As it is given that 1 m =3.281 ft
Therefore the length of the soccer field in ft is
115*3.281=377.315 ft
And width of the field is 85*3.281=277.885 ft
We know that 1 ft =12 in
Therefore the length of the soccer field in in is
377.315*12=4527.78 in
And width of the field is 277.885 *12=3334.62 in