Answer:
<em>The frequency of of the note = 131 Hz.</em>
Explanation:
<em>Frequency:</em><em> Frequency can be defined as the number of complete oscillation completed by a wave in one seconds. The S.I unit of frequency is Hertz ( Hz)</em>
v = λf ............................ Equation 1
Making f the subject of the equation,
f = v/λ .......................... Equation 2
Where v = Speed, λ = wavelength, f = frequency
<em>Given: v = 343 m/s, λ = 2.62 m.</em>
<em>Substituting these values into equation 2</em>
<em>f = 343/2.62</em>
<em>f = 131 Hz</em>
<em>Thus the frequency of of the note = 131 Hz.</em>
Period number is 1. It's group number also is 1. Hydrogen is a non-metal, yet happens to be on the metal side since it has that many valence electrons.
Explanation:
As per the problem,
![\Delta U = (V_{B} - V_{A})(-q) > 0](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20%28V_%7BB%7D%20-%20V_%7BA%7D%29%28-q%29%20%3E%200)
When q > 0 then -q is a negative charge . Since, change in potential energy (
) increases.
or,
> 0
or, ![V_{A} > V_{B}](https://tex.z-dn.net/?f=V_%7BA%7D%20%3E%20V_%7BB%7D)
Therefore, both positive and negative charge will move from
to
and as
so both of them move through a negative potential difference.
Thus, we can conclude that the true statements are as follows.
- The positively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).
- The negatively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).
Answer:
668 bright fringes
Explanation:
t = Thickness = 0.2 mm
= Wavelength = 600 nm
m = Number of fringes
We have the fringe width relation
![2t=\left(m+\frac{1}{2}\right)\lambda\\\Rightarrow m=\dfrac{2t}{\lambda}-\dfrac{1}{2}\\\Rightarrow m=\dfrac{2\times 0.2\times 10^{-3}}{600\times 10^{-9}}-\dfrac{1}{2}\\\Rightarrow m=666.166\approx 667](https://tex.z-dn.net/?f=2t%3D%5Cleft%28m%2B%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Clambda%5C%5C%5CRightarrow%20m%3D%5Cdfrac%7B2t%7D%7B%5Clambda%7D-%5Cdfrac%7B1%7D%7B2%7D%5C%5C%5CRightarrow%20m%3D%5Cdfrac%7B2%5Ctimes%200.2%5Ctimes%2010%5E%7B-3%7D%7D%7B600%5Ctimes%2010%5E%7B-9%7D%7D-%5Cdfrac%7B1%7D%7B2%7D%5C%5C%5CRightarrow%20m%3D666.166%5Capprox%20667)
So, total number of fringes will be including m = 0 is ![1+667=668](https://tex.z-dn.net/?f=1%2B667%3D668)