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A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.800 s....Asked by Justin
A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.800 s.
(a) How far (m)away is the explosion if air temperature is 22.0°C, neglecting the time taken for light to reach the physicist?
(b) How much further (m)away would the explosion be calculated to be if the speed of light is taken into account?
(a) How far (m)away is the explosion if air temperature is 22.0°C, neglecting the time taken for light to reach the physicist?
(b) How much further (m)away would the explosion be calculated to be if the speed of light is taken into account?
Answers
Answered by
Steve
look up the speed of sound in 22° air.
distance = speed * time
The distance is reduced by a small amount if you have to allow for the fact the the light did not arrive at the instant of the explosion, but a bit later.
distance = speed * time
The distance is reduced by a small amount if you have to allow for the fact the the light did not arrive at the instant of the explosion, but a bit later.
Answered by
Elena
(a)
v=331.3 + 0.606•t =331.3 + 0.606•22 =344.6 m/s
s=vt=344.6•0.8=275.7 m
v=331.3 + 0.606•t =331.3 + 0.606•22 =344.6 m/s
s=vt=344.6•0.8=275.7 m
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