Answer: 16.22 m/s^2
Explanation: g= GM/r^2 G= (6.67x 10^-11) M= 1.66(6x 10^24) r=(6400x 10^3) so
((6.67x10^-11)(1.66x 6x 10^24))/ (6400x10^3)^2 = 16.22 m/s^2
Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A
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.
