The air temperature over a lake decreases linearly with height after sunset, since air cools faster than water. The mean molar mass of air is M=28.8*10-3kg/mol.

The questions are:

If the temperature at the surface is 28.00 ∘C and the temperature at a height of 260.0 m is 8.000 ∘C how long does it take sound to rise 260.0 m directly upward? [Hint: Use the equation v=sqrt (γRT/M) and integrate.]

At a height of 260.0 m, how far does sound travel horizontally in this same time interval?

Question

Hudson Mcclain

Physics

28 February, 19:22

The air temperature over a lake decreases linearly with height after sunset, since air cools faster than water. The mean molar mass of air is M=28.8*10-3kg/mol.

The questions are:

If the temperature at the surface is 28.00 ∘C and the temperature at a height of 260.0 m is 8.000 ∘C how long does it take sound to rise 260.0 m directly upward? [Hint: Use the equation v=sqrt (γRT/M) and integrate.]

At a height of 260.0 m, how far does sound travel horizontally in this same time interval?

+4

Answers (2)

Know the Answer?

Aydan Johnson · Answer 1

Answer: t = 0.878s

Explanation: Dear big brain

since your temperature decreases linearly, you can assume that your velocity should behave linearly too. Here this isn't exactly the case (cfr. formula). But there's another way to prevent endlessly boring calculus. Use the principle of interpolation. (x/v_surface + x/v_top) / 2 = t.

This answer won't be exactly the same, but it's a quite good approx. You can use this eqation, hence you don't use large distances.

Sierra Warner · Answer 2

0.771 s

Explanation:

v = sqrt (γRT/M) = dx/dt

t = int (1/v, x=0 ... 260) = 260/sqrt (γRT/M)

In your case:

t = 260/sqrt ((1.4*8.31*281) / (28.8*10^-3)) = 0.771 s