Answer :
Answer:
(a) [tex]T=37.5 ^oC[/tex]
(b) [tex]x=70.09\ m[/tex]
Explanation:
Speed of Sound
The sound travels in an open space at a predictable speed which mainly depends on the air temperature. It can be calculated with an approximate formula in terms of the temperature T in degrees Celsius.
[tex]v_s=331.3+0.606T[/tex]
The speed comes in m/s
If we have determined the speed of sound, it's possible to estimate the distance to a large obstacle that could cause a returning sound (echo). The formula for the distance is
[tex]x=v_st[/tex]
Where [tex]v_s[/tex] is the speed of sound and t is half the time we hear our echo, because we can measure the forth and back times, assumed to be equal.
(b) We know the air is at -10^oC, so the speed of sound is
[tex]v_s=331.3+0.606(-10)[/tex]
[tex]v_s=325.24\ m/s[/tex]
The echo was heard 0.431 seconds later, so t=0.431 / 2= 0.2155 seconds
The distance to the rear wall of the gym is
[tex]x=(325.24)(0.2155)[/tex]
[tex]x=70.09\ m[/tex]
(b) On a summer day, the experiment was repeated and the echo was heard at t = 0.396 seconds. Since we already know the distance to the wall, we can estimate the temperature of the air, by solving the equation for T
[tex]v_s=331.3+0.606T[/tex]
[tex]\displaystyle T=\frac{v_s-331.3}{0.606}[/tex]
We need to compute the new value of [tex]v_s[/tex], with
[tex]\displaystyle v_s=\frac{x}{t}[/tex]
We know [tex]x=70.09\ m, t=0.396/2=0.198\ sec[/tex]
[tex]\displaystyle v_s=\frac{70.09}{0.198}[/tex]
[tex]\displaystyle v_s=354\ m/s[/tex]
Now we can find T
[tex]\displaystyle T=\frac{354-331.3}{0.606}[/tex]
[tex]\displaystyle T=\frac{22.7}{0.606}[/tex]
[tex]T=37.5 ^oC[/tex]