Answer:
The relationship between the initial stored energy
and the stored energy after the dielectric is inserted
is:
c) 
Explanation:
A parallel plate capacitor with
that is connected to a voltage source
holds a charge of
. Then we disconnect the voltage source and keep the charge
constant . If we insert a dielectric of
between the plates while we keep the charge constant, we found that the potential decreases as:

The capacitance is modified as:

The stored energy without the dielectric is
The stored energy after the dielectric is inserted is:

If we replace in the above equation the values of V and C we get that


Finally

Answer:
Explanation:
1760 yd/mi / 120 yd/field = 14⅔ fields/mi
Beaker would be most appropriate for measuring the approximate volume of a liquid.
Answer:
b
Explanation:
it compresses hot air turning into cool air almost like a reverse tornado