Answer:
346 m/s
4.34 s (2 d.p.)
4960 m
23.33 °C (2 d.p.)
Explanation:
Speed of sound in air
[tex]\boxed{v=331+0.6T}[/tex]
Where:
- v = velocity in m/s
- T = temperature in °C
Substitute the given temperature of 25°C into the formula and solve for v:
[tex]\implies v=\sf331+0.6(25)=346\; m/s[/tex]
As 1 km = 1000 m then:
⇒ 1.5 km = 1500 m
[tex]\boxed{\sf Time=\dfrac{Distance}{Speed}}[/tex]
Substitute the distance of 1500 m and the speed of 346 m/s into the formula and solve for time:
[tex]\implies \sf Time=\dfrac{1500}{346}=4.34\;s\;(2\:d.p.)[/tex]
[tex]\boxed{\sf Distance=Speed \times Time}[/tex]
10 seconds more would be:
[tex]\implies \sf \dfrac{1500}{346}+10=14.33526012...[/tex]
Substitute the speed of 346 m/s and the time into the formula and solve for distance:
[tex]\begin{aligned}\implies \sf Distance &= \sf Speed \times Time\\&= \sf 346 \times 14.33526012...\\&=\sf 4960\;m \end{aligned}[/tex]
To calculate what temperature the sound can travel at 345 m/s, substitute this speed into the Speed of sound in air formula and solve for T:
[tex]\begin{aligned} v&=331+0.6T\\\implies 345 & = 331+0.6T\\345 -331& = 331+0.6T-331\\14&=0.6T\\T&=\dfrac{14}{0.6}\\\implies T&=23.33^{\circ}\sf C\end{aligned}[/tex]
