Answer:
Explanation:
As we know that the orbital speed of the satellite is given as
also we know that
time period of the revolution is given as
now from above equation we know that
so we will have
now plug in all data in this equation
Answer:
-0.038 N
Explanation:
F=K Q1 Q2/r^2 by COULOMB'S LAW
F= 9×10^9×1×10^-5×-1.5×10^-5/(6)^2
F= -0.038 N
Electromagnetic radiation
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
