Respuesta :
Answer:
a) 137.6 m
b) 137.632 m
Explanation:
T = Temperature of air = 22.0 °C
v = speed of sound at temperature "T"
speed of sound at temperature "T" is given as
[tex]v = 331\sqrt{1 + \frac{T}{273}}[/tex]
[tex]v = 331\sqrt{1 + \frac{22}{273}}[/tex]
[tex]v = 344.08 [/tex] m/s
a)
t = time taken for the sound to reach the physicist = 0.400 s
d = distance of the location of explosion
Distance of the location of explosion is given as
d = v t
d = (344.08) (0.400)
d = 137.6 m
b)
d = distance of the location of explosion
v = speed of sound = 344.08 m/s
c = speed of light = 3 x 10⁸ m/s
[tex]t_{s}[/tex] = time taken by sound to reach the physicist = [tex]\frac{d}{v}[/tex] = [tex]\frac{d}{344.08}[/tex]
[tex]t_{L}[/tex] = time taken by light to reach the physicist = [tex]\frac{d}{c}[/tex] = [tex]\frac{d}{3\times 10^{8}}[/tex]
t = time taken for the sound to reach the physicist = 0.400 s
time taken for the sound to reach the physicist is given as
t = [tex]t_{s}[/tex] - [tex]t_{L}[/tex]
0.400 = [tex]\frac{d}{344.08}[/tex] - [tex]\frac{d}{3\times 10^{8}}[/tex]
d = 137.632 m
The distance to the explosion (in m) taking the speed of light into account is 137.6316 meters.
Given to us
times lag between seeing an explosion and hearing its sound = 0.400 s.
Air temperature = 22.0 °C
What is the speed of the sound?
We know that the speed of the sound at a temperature T is given as,
[tex]v =331\sqrt{1+\dfrac{T}{273}}[/tex]
where v is the speed of the sound and T is the air temperature in °C,
Substitute the values,
[tex]v =331\sqrt{1+\dfrac{22}{273}}[/tex]
[tex]v = 344.0786\rm\ m/sec[/tex]
What is the distance of the explosion?
We know that speed can be written as,
[tex]\rm speed = \dfrac{Distance}{Time}[/tex]
Substitute the values,
[tex]\rm 344.0786 = \dfrac{Distance}{0.400\ s}[/tex]
Distance, d = 137.6314 meters
Thus, the distance of the explosion is 137.6314 meters.
What is the distance to the explosion (in m) taking the speed of light into account?
We know the following information,
- speed of the sound, v = 344.0786 m\s
- speed of the light, c = [tex]3 \times 10^8[/tex] m/s
The time taken by the light to reach the physicist,
[tex]\rm time = \dfrac{distance}{speed}[/tex]
Substitute the values
[tex]\rm t_c= \dfrac{s}{3 \times 10^8}[/tex]
The time taken by the sound to reach the physicist,
[tex]\rm time = \dfrac{distance}{speed}[/tex]
Substitute the values
[tex]\rm t_s= \dfrac{s}{344.0786}[/tex]
We know that the lag between seeing an explosion and hearing its sound is 0.400 s. therefore,
[tex]t = t_s - t_c[/tex]
substitute the values,
[tex]0.400 =\dfrac{s}{344.0786} -\dfrac{s}{3 \times 10^8}[/tex]
s = 137.6316 meters
Hence, the distance to the explosion (in m) taking the speed of light into account is 137.6316 meters.
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