Answer:
1.11
Explanation:
The index of the medium can be calculated using below formula
V= c/ n ............eqn(1)
Where V= velocity of the light is reduced to while traveling through the second medium= 2.7 x 10^8 m/s
n= index of the medium
c= speed of light= 3 x 10^8 m/s
Substitute for the values in eqn(1)
2.7 x 10^8 = (3 x 10^8 m/s)/ n
Making " n" subject of the formula, we have
n= (3 x 10^8 )/(2.7 x 10^8)
n= 1.11
Hence, the index of the medium is 1.11
<span>Yes, there are! r1 and r2 are numbers. The volume of the hollow shell is 4 π 3 ( r 3 1 − r 3 2 ) 4π3(r13−r23). Now multiply by ρ to get the mass.</span>
The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:
![W=q \Delta V](https://tex.z-dn.net/?f=W%3Dq%20%5CDelta%20V)
where q is the charge and
![\Delta V](https://tex.z-dn.net/?f=%5CDelta%20V)
is the potential difference.
In our problem, the work done is W=39 J while the potential difference of the battery is
![\Delta V = 9.0 V](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%209.0%20V)
, so we can find the charge transferred by the battery:
Data:
![f_{2} = 42 Hz](https://tex.z-dn.net/?f=f_%7B2%7D%20%3D%2042%20Hz)
n (Wave node)
V (Wave belly)
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
![f_{n} = \frac{nV}{2L}](https://tex.z-dn.net/?f=f_%7Bn%7D%20%3D%20%20%5Cfrac%7BnV%7D%7B2L%7D%20)
Wire 2 → 2º Harmonic → n = 2
![f_{n} = \frac{nV}{2L}](https://tex.z-dn.net/?f=f_%7Bn%7D%20%3D%20%5Cfrac%7BnV%7D%7B2L%7D%20)
![f_{2} = \frac{2V}{2L} ](https://tex.z-dn.net/?f=f_%7B2%7D%20%3D%20%5Cfrac%7B2V%7D%7B2L%7D%20%0A)
![2V = f_{2} *2L](https://tex.z-dn.net/?f=2V%20%3D%20%20f_%7B2%7D%20%2A2L)
![V = \frac{ f_{2}*2L }{2}](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cfrac%7B%20f_%7B2%7D%2A2L%20%7D%7B2%7D%20)
![V = \frac{42*2L}{2}](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cfrac%7B42%2A2L%7D%7B2%7D%20)
![V = \frac{84L}{2}](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cfrac%7B84L%7D%7B2%7D%20)
![V = 42L](https://tex.z-dn.net/?f=V%20%3D%2042L)
Wire 1 → 1º Harmonic or Fundamental rope → n = 1
![f_{n} = \frac{nV}{2L}](https://tex.z-dn.net/?f=f_%7Bn%7D%20%3D%20%5Cfrac%7BnV%7D%7B2L%7D%20)
![f_{1} = \frac{1V}{2L}](https://tex.z-dn.net/?f=f_%7B1%7D%20%3D%20%5Cfrac%7B1V%7D%7B2L%7D%20)
![f_{1} = \frac{V}{2L}](https://tex.z-dn.net/?f=f_%7B1%7D%20%3D%20%20%5Cfrac%7BV%7D%7B2L%7D%20)
If, We have:
V = 42L
Soon:
![f_{1} = \frac{V}{2L}](https://tex.z-dn.net/?f=f_%7B1%7D%20%3D%20%5Cfrac%7BV%7D%7B2L%7D%20)
![f_{1} = \frac{42L}{2L}](https://tex.z-dn.net/?f=f_%7B1%7D%20%3D%20%5Cfrac%7B42L%7D%7B2L%7D%20)
![\boxed{f_{1} = 21 Hz}](https://tex.z-dn.net/?f=%5Cboxed%7Bf_%7B1%7D%20%3D%2021%20Hz%7D)
Answer:
<span>The fundamental frequency of the string:
</span>
21 Hz