This would be called the "Escape Velocity "
Answer:
T= 1.71×10^{-3} sec= 1.71 mili sec
t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec
Explanation:
First of all we write equation for current oscillation in LC circuits. Note, the maximum current (I_0)= 5.5 mA is the amplitude of this function. Then, we continue to solve for the angular frequency(ω). Afterwards, we calculate the time period T. qo = maximum charge on capacitor. = 1.5× 10 ^− 6 C
a) I(t) = -ωqosin(ωt+φ)
⇒Io= ωqo
⇒ω= Io/qo
also we know that T= 2π/ω
⇒T= 
now putting the values we get
= 
= 1.71×10^{-3} sec
b) note that the time
it takes the capacitor to from uncharge to fully charged is one fourth of the period . That is


t_{fc}= 4.281×10^{-4} sec or 0.4281 mili sec
Pushing, pulling is the answer
Apply the combined gas law
PV/T = const.
P = pressure, V = volume, T = temperature, PV/T must stay constant.
Initial PVT values:
P = 1atm, V = 8.0L, T = 20.0°C = 293.15K
Final PVT values:
P = ?, V = 1.0L, T = 10.0°C = 283.15K
Set the PV/T expression for the initial and final PVT values equal to each other and solve for the final P:
1(8.0)/293.15 = P(1.0)/283.15
P = 7.7atm