There are multiple reasons for this. First of all, water is available in almost every place on the Earth. It doesn't pollute the air, doesn't cause health use and is easily handle.
Other factor is the fact that water has a really high specific heat. This means that water, and more specifically steam, can aborb and transport more energy. A lower heat capacity would imply the need to boil more of the liquid to obtain the same amount of energy. This combine with the fact that water expands at a large rate when boiling, combine with everything mentioned previously, and you get a liquid with all the characteristics that a efficient turbine requires to work.
Answer:
the final kinetic energy is 0.9eV
Explanation:
To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:
you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is
![E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV](https://tex.z-dn.net/?f=E_%7Bn_2-n_1%7D%3D-13.6eV%5B%5Cfrac%7B1%7D%7Bn_2%5E2%7D-%5Cfrac%7B1%7D%7Bn_1%5E2%7D%5D%5C%5C%5C%5CE_%7B2-1%7D%3D-13.6eV%5B%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B1%7D%5D%3D-10.2eV)
-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:
Answer:
E/4
Explanation:
The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
Where;
E is the electric field
σ is the surface charge density
ε₀ is the electric constant.
Formula to calculate σ is;
σ = Q/A
Where;
Q is the total charge of the sheet
A is the sheet's area.
We are told the elastic sheet is a square with a side length as d, thus ;
A = d²
So;
σ = Q/d²
Putting Q/d² for σ in the electric field equation to obtain;
E = Q/(2ε₀d²)
Now, we can see that E is inversely proportional to the square of d i.e.
E ∝ 1/d²
The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.
From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;
E_new = E/4
