An empty capacitor is capable of storing 1.0 × 10-4 j of energy when connected to a certain battery. if the distance between the plates is halved and then filled with a dielectric (κ = 4.3), how much energy could this modified capacitor store when connected to the same battery?

Explanation:

Energy stored in a capacitor

= 1/2 C₁V²

capacity of a capacitor

c = εK A / d

k is dielectric and d is distance between plates .

When  the distance between the plates is halved and then filled with a dielectric (κ = 4.3)

capacity becomes 4.3  x 2 times

New capacity

C₂ = 8.6 C₁

Energy of modified capacitor

1/2 C₂ V²= 1/2 x 8.6 c x V²

Energy becomes

8.6 times.

Energy stored = 8.6 x 10⁻⁴ J

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day

there are no components of a physical goal mentioned on the list of choices that you've provided.