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

[firemouse]

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.

 

[ivybond]

#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

 

[quicknote]

I would suggest using the appropriate equations and units to solve for the jet speed and heat transfer equation. For the first problem, you can use the formula v = d/t, where v is the velocity (jet speed), d is the distance, and t is the time. In this case, you have two equations: v = d/6 and v = d/5, where d is the distance between city A and B. Since the distance is the same for both equations, you can set them equal to each other and solve for v. The velocity of the jet relative to the air would be the same in both directions, so the answer would be 6/5 times the 50 miles per hour headwind, or 60 miles per hour.

For the second problem, you are on the right track with your equation, but it's missing the thickness "t" in the denominator. The correct equation would be T2 = (H*t)/(A*(T1-T2)). To solve for T2, you can rearrange the equation and solve for T2 by isolating it on one side of the equation. It's important to make sure your units are consistent throughout the equation to get an accurate answer. I hope this helps.