<span>0.0001 km / year or 10^-5 km/year just take 50 km and divide it by 5 million</span>
Because 'acceleration' does NOT mean 'speeding up'.
It means ANY change in motion ... speeding up, slowing down,
or changing DIRECTION.
When traveling a roundabout, or any curved path, the direction
is constantly changing even if the speed is constant, so there is
constant acceleration going on.
Answer:
Explanation:
Given an RL circuit
A voltage source of.
V = 108V
A resistor of resistance
R = 1.1-kΩ = 1100 Ω
And inductor of inductance
L = 34 H
After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor
A. Time the inductor current will reduce to 12% of it's initial current
Let the initial charge current be Io
Then, final current is
I = 12% of Io
I = 0.12Io
I / Io = 0.12
The current in an inductor RL circuit is given as
I = Io ( 1—exp(-t/τ)
Where τ is time constant and it is given as
τ = L/R = 34/1100 = 0.03091A
So,
I = Io ( 1—exp(-t/τ))
I / Io = ( 1—exp(-t/τ))
Where I/Io = 0.12
0.12 = 1—exp(-t/τ)
0.12 — 1 = —exp(-t/τ)
-0.88 = -exp(-t/0.03091)
0.88 = exp(-t/0.03091)
Take In of both sides
In(0.88) = In(exp(-t/0.03091)
-0.12783 = -t/0.030901
t = -0.12783 × 0.030901
t = 3.95 × 10^-3 seconds
t = 3.95 ms
B. Energy stored in inductor is given as
U = ½Li²
So, the current at this time t = 3.95ms
I = Io ( 1—exp(-t/τ))
Where Io = V/R
Io = 108/1100 = 0.0982 A
Now,
I = Io ( 1—exp(-t/τ))
I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))
I = 0.0982(1—exp(-0.12783)
I = 0.0982 × 0.12
I = 0.01178
I = 11.78mA
Therefore,
U = ½Li²
U = ½ × 34 × 0.01178²
U = 2.36 × 10^-3 J
U = 2.36 mJ
Actually, the real reason it appears yellow-white is because of the temperature of the star. the stars that are hotter tend to be more blue and the stars that aren’t as hot tend to be more red/orange/yellow.
