Answer:
here w means work done as we can see that it is the formula for work done where f is force and d is distance travelled. Hope it helped :-)
Explanation:
Answer:
a) 
b) 
c) 
Explanation:
Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:


Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.
When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.
The new surface charge density can be calculated as follows:

c) The electric field outside the cylinder can be found by Gauss' Law:

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

Answer:
Zero.
Explanation:
An adiabatic process is one in which there is no exchange of heat energy. Therefore, in an adiabatic process, heat is neither added to the system not it is removed from the system.
The work done by the gas on the environment is 20 J. This energy is equal to the change in internal energy for an adiabatic process.
Therefore, for an ideal gas to undergo an adiabatic process in which it expands and does 20 J of work on its environment, the heat exchange is zero.
All the planets is the answer