Answer:
![\theta = 20.98 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2020.98%20degree)
Explanation:
As we know that the speed of the sound is given as
![v = 332 + 0.6 t](https://tex.z-dn.net/?f=v%20%3D%20332%20%2B%200.6%20t)
now at t = 273 k = 0 degree
![v = 332 m/s](https://tex.z-dn.net/?f=v%20%3D%20332%20m%2Fs)
so we have
![a sin\theta = N\lambda](https://tex.z-dn.net/?f=a%20sin%5Ctheta%20%3D%20N%5Clambda)
![a sin\theta = N(\frac{v_1}{f})](https://tex.z-dn.net/?f=a%20sin%5Ctheta%20%3D%20N%28%5Cfrac%7Bv_1%7D%7Bf%7D%29)
now when temperature is changed to 313 K we have
![t = 313 - 273 = 40 degree](https://tex.z-dn.net/?f=t%20%3D%20313%20-%20273%20%3D%2040%20degree)
now we have
![v = 332 + (0.6)(40)](https://tex.z-dn.net/?f=v%20%3D%20332%20%2B%20%280.6%29%2840%29)
![v_2 = 356 m/s](https://tex.z-dn.net/?f=v_2%20%3D%20356%20m%2Fs)
![a sin\theta' = N(\frac{v_2}{f})](https://tex.z-dn.net/?f=a%20sin%5Ctheta%27%20%3D%20N%28%5Cfrac%7Bv_2%7D%7Bf%7D%29)
now from two equations we have
![\frac{sin19.5}{sin\theta} = \frac{332}{356}](https://tex.z-dn.net/?f=%5Cfrac%7Bsin19.5%7D%7Bsin%5Ctheta%7D%20%3D%20%5Cfrac%7B332%7D%7B356%7D)
so we have
![sin\theta = 0.358](https://tex.z-dn.net/?f=sin%5Ctheta%20%3D%200.358)
![\theta = 20.98 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2020.98%20degree)
Answer:
![\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%2B%5Cfrac%7B%5Cbeta%7D%7Bm%7D%5Cfrac%7Bdx%7D%7Bdt%7D%2B%5Cfrac%7Bk%7D%7Bm%7Dx%3D0)
Explanation:
let
be the mass attached, let
be the spring constant and let
be the positive damping constant.
-By Newton's second law:
![m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}](https://tex.z-dn.net/?f=m%5Cfrac%7Bd%5E2x%7D%7Bdt%5E2%7D%3D-kx-%5Cbeta%20%5Cfrac%7Bdx%7D%7Bdt%7D)
where
is the displacement from equilibrium position. The equation can be transformed into:
shich is the equation of motion.
Answer:
The Resultant Induced Emf in coil is 4∈.
Explanation:
Given that,
A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.
To find :-
find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).
So,
Emf induced in the coil represented by formula
∈ =
...................(1)
Where:
.
{ B is magnetic field }
{A is cross-sectional area}
.
No. of turns in coil.
.
Rate change of induced Emf.
Here,
Considering the case :-
&
Putting these value in the equation (1) and finding the new emf induced (∈1)
∈1 =![-N1\times\frac{d\phi1}{dt}](https://tex.z-dn.net/?f=-N1%5Ctimes%5Cfrac%7Bd%5Cphi1%7D%7Bdt%7D)
∈1 =![-2N\times2\frac{d\phi}{dt}](https://tex.z-dn.net/?f=-2N%5Ctimes2%5Cfrac%7Bd%5Cphi%7D%7Bdt%7D)
∈1 =![4 [-N\times\frac{d\phi}{dt}]](https://tex.z-dn.net/?f=4%20%5B-N%5Ctimes%5Cfrac%7Bd%5Cphi%7D%7Bdt%7D%5D)
∈1 = 4∈ ...............{from Equation (1)}
Hence,
The Resultant Induced Emf in coil is 4∈.