The half empty will fall quicker
Please leave a thanks <3
Answer:
(a) the observed frequency is 200 Hz
(b) the observed frequency is 188 Hz.
Explanation:
speed of the truck, Vs = 27 m/s
frequency of the truck as it approaches, Fs = 185 Hz
(a) Apply Doppler effect to determine the frequency you will hear.
As the truck approaches you, the observed frequency will be higher than the source frequency because of decrease in distance.
![F_s = F_o [\frac{V}{V_S + V} ]](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7BV%7D%7BV_S%20%2B%20V%7D%20%5D)
Where;
Fo is the observed frequency which is the frequency you will hear.
V is speed of sound in air
![F_s = F_o [\frac{V}{V_S + V} ]\\\\185 = F_o [\frac{340}{27 + 340} ]\\\\185 = F_o (0.926)\\\\F_o = \frac{185}{0.926}\\\\F_o = 199.78 \ Hz](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7BV%7D%7BV_S%20%2B%20V%7D%20%5D%5C%5C%5C%5C185%20%3D%20F_o%20%5B%5Cfrac%7B340%7D%7B27%20%2B%20340%7D%20%5D%5C%5C%5C%5C185%20%3D%20F_o%20%280.926%29%5C%5C%5C%5CF_o%20%3D%20%5Cfrac%7B185%7D%7B0.926%7D%5C%5C%5C%5CF_o%20%3D%20199.78%20%5C%20Hz)
(b) Apply the following formula for a moving observer and a moving source;
](https://tex.z-dn.net/?f=F_o%20%3D%20F_s%5B%5Cfrac%7BV-V_o%7D%7BV%7D%20%5D%28%5Cfrac%7BV%7D%7BV-V_S%7D%20%29)
The observed frequency is negative since you are driving away from the truck and the source frequency is also negative since it is driving towards you.
\\\\F_o = 185[\frac{340-22}{340} ](\frac{340}{340-27} )\\\\F_o = 185(0.9353)(1.0863)\\\\F_o = 188 \ Hz](https://tex.z-dn.net/?f=F_o%20%3D%20F_s%5B%5Cfrac%7BV-V_o%7D%7BV%7D%20%5D%28%5Cfrac%7BV%7D%7BV-V_S%7D%20%29%5C%5C%5C%5CF_o%20%3D%20185%5B%5Cfrac%7B340-22%7D%7B340%7D%20%5D%28%5Cfrac%7B340%7D%7B340-27%7D%20%29%5C%5C%5C%5CF_o%20%3D%20185%280.9353%29%281.0863%29%5C%5C%5C%5CF_o%20%3D%20188%20%5C%20Hz)
Conductor, a conductive materail readily accepts the flow of electrons.
Answer:
The answer is 138.5
Explanation:
STEP 1:
The inductance per unit length of a coaxial transmission line is
L′=L<em>/ </em>I
=Ø/H
=μoI/2π In (b/a)
In this a is the radius of inner conductor
b is the radius of outer conductor
I is the coaxial transmission
μ is the magnetic permeability
Since the transmission of the charge exists in air, the value of the relative permeability is μr= I and permeability of free space is μo= 4π x 10-7 H/m . So the magnetic permeability will be
μ = μoμ r
μ =μ o(I) 4π x 10-7 H/m
L′= μoI/2π In (b/a)
= (4π x 10-7 ) (2)/2π In (10/5)
=2.77 x 10-7 H
STEP 2:
Obtain the magnetic energy stores in the magnetic field H of a volume of the coaxial transmission line containing a material with permeability μ, by using the formula given below:
Wm= 1/2 LI^2
= 1/2 (2.77x 10^-7 I^2
= 138.5 X 10^-9 I^2 J
Now we will simplify the equation
Wm= 185.5<em>I</em>^2 nJ
So, the magnetic energy stored in insulating medium is 185.5<em>I</em>^2 nJ
<span>because electrical charge builds up otherwise know as static electricity</span>
