Explanation:
The energy of the system before the collision must equal the energy after the collision.
After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is

Before the collision only the bullet has kinetic energy.
So we can now determine the speed of the bullet using

The work done for moving a charge is due to the change in potential between the points. That is the work done is given by the expression W = qΔV. Where q is the charge and ΔV is the change in potential between two points.
In case of equipotential lines the value of potential is same at all the points.
So the value of ΔV when moving along a equipotential line is zero.
So W = qΔV = q*0=0
So no work is needed to move charge along an equipotential line.
Work done = 0
Answer:
Heat transfer during the process = 0
Work done during the process = - 371.87 KJ
Explanation:
Initial pressure
= 0.02 bar
Initial temperature
= 200 K
Final pressure
= 0.14 bar
Gas constant for helium R = 2.077 
This is an isentropic polytropic process so temperature - pressure relationship is given by the following formula,
= ![[\frac{P_{2} }{P_{1} } ]^{\frac{\gamma - 1}{\gamma} }](https://tex.z-dn.net/?f=%5B%5Cfrac%7BP_%7B2%7D%20%7D%7BP_%7B1%7D%20%7D%20%5D%5E%7B%5Cfrac%7B%5Cgamma%20-%201%7D%7B%5Cgamma%7D%20%7D)
Put all the values in above formula we get,
⇒
= ![[\frac{0.14 }{0.02 } ]^{\frac{1.4 - 1}{1.4} }](https://tex.z-dn.net/?f=%5B%5Cfrac%7B0.14%20%7D%7B0.02%20%7D%20%5D%5E%7B%5Cfrac%7B1.4%20-%201%7D%7B1.4%7D%20%7D)
⇒
= 1.74
⇒
= 348.72 K
This is the final temperature of helium.
For isentropic polytropic process heat transfer to the system is zero.
⇒ ΔQ = 0
Work done W = m × (
-
) × 
⇒ W = 1 × ( 200 - 348.72 ) × 
⇒ W = 371.87 KJ
This is the work done in this process. here negative sign shows that work is done on the gas in the compression of gas.
Answer:
0.652 mA
Explanation:
According to Faraday's Law :



where ;
A = 



Induced current I = 
= 
= 
= 0.652 mA
Thus, the induced current in the loop of wire over this time = 0.652 A