Question 607998: The attenuation of a 900 MHz mobile phone signal by a concrete wall reinforced with steel mesh is described by,
I(t)=1.2e^(-0.138t)
where t is the thickness of the wall in cm, and I(t) is the intensity
of signal, in mW/m2, measured on the interior of the wall.
(a) What is the signal intensity incident on the exterior of the
wall?
(a) Answer - I think is t=0
900=1.2e^1.2^-0.138*0
900=1.2*1
900/1.2=1.2*1/1.2
750=1
750/1=0
750
(b) If the wall is 25 cm thick, what intensity will be measured
inside? Round your answer to 2 decimal places.
(b) Answer - I think is
750=1.2e^-0.138*0.25
750=1.2e^-0.0345
750/1.2*0.966088=1.2*0.966088/1.2*0.966088
646.9389952
Approx =647
(c) What thickness of wall will drop the intensity to half of the
exterior value? Round your answer to 2 decimal places.
(c) Not sure a bit lost
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! you need to watch what values you substitute where...
(a) I(0) = 1.2 e^[-0.138(0)] = 1.2 e^0 = 1.2
(b) I(25) = 1.2 e^[-0.138(25)]
(c) 1/2 = e^[-0.138(x)] ___ ln(1/2) = -0.138 x
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