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
The core of a star must be at the temperature of 10,000,000 degrees Celsius for hydrogen fusion to begin.
True I hope this helps you out
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
