Find Math Help: Solving for Jet Speed and Heat Transfer Equation

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SUMMARY

The discussion focuses on solving two mathematical problems: determining the speed of a jet relative to the air and deriving the heat transfer equation. For the jet problem, the equation (x-50)6 = (x+50)5 leads to the conclusion that the jet's speed is 550 miles/hour. In the heat transfer equation, the correct expression for T2 is confirmed as T2 = (Ht / A) + T1, validating the user's solution.

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1) A jet took 6h to fly against headwinds of 50 miles/hour from city A to B. It took 5H on the return trip when the winds became tailwinds of 50miles/hour. How fast does the jet travel relative to the air?


2) The amount of heat H which passes through wall "t" thick is proportional to the product of the area "A" of a wall T2 - T1, the differnce in temperature of the surfaces of the wall and is inversly proportional to "t". Express the equation and solve for T2.


For #2 I got to: T2 = ((HT) / A) + T1

Is that right? thanks.

Thanks a lot guys, been havinga lot of trouble with these.
 
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#1.
A distance one way is the same as on a way back (AB = BA).
[distance] = [rate]*[time]
(x-50)6 = (x+50)5
x=550.

#2.
You got it right! Why doubt?
Just use "t" instead of "T":
T2 = Ht / A + T1
 

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