Answer: D Although the total energy remains constant, nonrenewable fuels convert chemical energy into forms that are difficult or impossible to use again.
Explanation:
The first law of thermodynamics says that energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another.
Answer:
The maximum amount of work is 
Explanation:
From the question we are told that
The temperature of the environment is 
The volume of container A is 
Initially the number of moles is 
The volume of container B is 
At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as
![W = P_A V_A ln[ \frac{V_B}{V_A} ]](https://tex.z-dn.net/?f=W%20%3D%20%20P_A%20V_A%20%20ln%5B%20%5Cfrac%7BV_B%7D%7BV_A%7D%20%5D)
Now from the Ideal gas law
So substituting for 
in the equation above
![W = nRT ln [\frac{V_B}{V_A} ]](https://tex.z-dn.net/?f=W%20%3D%20%20nRT%20ln%20%5B%5Cfrac%7BV_B%7D%7BV_A%7D%20%5D)
Where R is the gas constant with a values of 
Substituting values we have that
![W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]](https://tex.z-dn.net/?f=W%20%3D%201.2%20%2A%20%288.314%29%20%2A%20%28280%29%20%2A%20ln%20%5B%5Cfrac%7B3.5%7D%7B2%7D%20%5D)
Answer:
31,360J
Explanation:
Gravitation potential energy (gpe) is calculated from the formula mgh.
That implies, gpe = mgh
Therefore substituting the values of m and h as given in the question, knowing in mine that the acceleration due to gravity( g) is 9.8 N/kg, will give 31,360J
Never forget to put your SI units, because even if your answer is numerical correct, it will be incorrect because it represents no physical quantity.
Answer:
Explanation:
E=(σ/ε0)
As noted by Dirac the field is the same no matter how far you are from the sheet. When your charge covers a conducting plane, as in your case, the field is, D/eo ,(D is charge density). Because the field inside the conductor (no matter how thin) is zero. The only time the field is, D/2eo, is when you have just a sheet of charge, by itself, not on a conducting plane."
V₁ = (1/g)₁ = Way₁ = 20(9.81)(0) = 0
V₂ (Vg)₂ = -WAy₂ = -20(9.81)(0.5) = -98.1J
The kinetic energy because the pool rotates about a fixed axis
W = VA/rA = VA/0.2 5VA
Mass momen of inertila about fixed axis which passes through point 0
I₀ = mko² = 50(0.280)² = 3.92 kg. m²
∴ The kinetic energy of the system is
T = 1/2 I₀w² + 1/2mAVA²
= 1/2(3.92)(5VA)² + 1/2 (20) VA² = 59VA²
Now that the system is at rest then T₁ = 0
Energy conservation is
T₁ +V₁ = T₂ + V₂
0+ 0 = 59VA² + (-98.1)
VA = 1.289 m/s
= 1.29 m/s
