I think the answer may be the letter B
W = ∫ (x from 0.1 to +oo) F dx
= ∫ (x from 0.1 to +oo) A e^(-kx) dx
= A/k x [ - e^(-kx) ](between 0.1 and +oo)
= A/k x [ 0 + e^(-k * 0.1) ]
<span>
= A/k x e^(-k/10) </span>
Answer:
1. Hydrogen
2. Helium
Explanation:
Nuclear fusion is when two atoms of Hydrogen join together to form one Helium atom.
To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion
From the definition we know that the frequency can be expressed as
![f = \frac{v}{\lambda}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7Bv%7D%7B%5Clambda%7D)
Where,
![v = Velocity \rightarrow 20m/s](https://tex.z-dn.net/?f=v%20%3D%20Velocity%20%5Crightarrow%2020m%2Fs)
![\lambda = Wavelength \rightarrow 35*10^{-2}m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20Wavelength%20%5Crightarrow%2035%2A10%5E%7B-2%7Dm)
Therefore the frequency would be given as
![f = \frac{20}{35*10^{-2}}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B20%7D%7B35%2A10%5E%7B-2%7D%7D)
![f = 57.14Hz](https://tex.z-dn.net/?f=f%20%3D%2057.14Hz)
The frequency is directly proportional to the angular velocity therefore
![\omega = 2\pi f](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20f)
![\omega = 2\pi *57.14](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20%2A57.14)
![\omega = 359.03rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20359.03rad%2Fs)
Now the maximum speed from the simple harmonic movement is given by
![V_{max} = A\omega](https://tex.z-dn.net/?f=V_%7Bmax%7D%20%3D%20A%5Comega)
Where
A = Amplitude
Then replacing,
![V_{max} = (1*10^{-2})(359.03)](https://tex.z-dn.net/?f=V_%7Bmax%7D%20%3D%20%281%2A10%5E%7B-2%7D%29%28359.03%29)
![V_{max} = 3.59m/s](https://tex.z-dn.net/?f=V_%7Bmax%7D%20%3D%203.59m%2Fs)
Therefore the maximum speed of a point on the string is 3.59m/s