The minimum frequency is

while the maximum frequency is

Using the relationship between frequency f of a wave, wavelength

and the speed of the wave v, we can find what wavelength these frequencies correspond to:


So, the wavelengths of the radio waves of the problem are within the range 188-545 m.
<u>Given</u><u>:</u>
- initial velocity, u = 200 m/s
- Final velocity, v = 300 m/s
<u>To</u><u> </u><u>be</u><u> </u><u>calculated</u><u>:</u>
Calculate the acceleration of given object ?
<u>Formula</u><u> </u><u>used</u><u>:</u>
Acceleration = v - u / t
<u>Solution</u><u>:</u>
We know that,
Acceleration = v - u / t
☆ Substituting the values in the above formula,we get
Acceleration ⇒ 300 - 200 / 20
⇒ 100/20
⇒ 5 m/s²
The magnitude of the current in wire 3 is (I₃)= 0.33A
<h3>How to calculate the value of the magnitude of the current in wire 3 ?</h3>
To calculate the magnitude of the current in wire 3 we are using the Kirchhoff’s current law,
I₁ + I₂ + I₃ = 0
Where we are given,
I₁ = current in wire 1
=0.40 A.
I₂ = current in wire 2
= -0.73 A.
We have to calculate the magnitude of the current in wire 3, I₃
Now we put the known values in above equation, we get,
I₁ + I₂ + I₃ = 0
Or, I₃ = -.(I₁ + I₂)
Or, I₃ = -.(0.40 - 0.73)
Or, I₃ = 0.33 A
From the above calculation, we can conclude that the current in wire 3 is I₃ = 0.33 A
Learn more about current:
brainly.com/question/25537936
#SPJ4