Answer:
first one is b 2nd one is a 3rd is c and the 4th one is c also
Explanation: have a nice day
Answer:
z = 0.8 (approx)
Explanation:
given,
Amplitude of 1 GHz incident wave in air = 20 V/m
Water has,
μr = 1
at 1 GHz, r = 80 and σ = 1 S/m.
depth of water when amplitude is down to 1 μV/m
Intrinsic impedance of air = 120 π Ω
Intrinsic impedance of water = 
Using equation to solve the problem

E(z) is the amplitude under water at z depth
E_o is the amplitude of wave on the surface of water
z is the depth under water



now ,


taking ln both side
21.07 x z = 16.81
z = 0.797
z = 0.8 (approx)
Answer:
Long question good luck:)
Explanation:
Answer:
Engular velocity: 
Linear velocity: 
The time it takes:

Explanation:
The magnitude of the centripetal acceleration can be related to the angular velocity and radius as:
(1)
Solving for w:
(2)
Replacing a=9,8m/s2 and r=6,375,000m:
(3)
And the angular velocity relates to the linear velocity:

The perimeter of the orbit is:

The time it takes:
