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?
To solve these problems, we need to use the formula for sound speed in air and the speed of light.
(a) To find the distance of the explosion neglecting the time taken for light to reach the physicist, we can use the formula for sound speed in air:
Speed of sound (v) = distance (d) / time (t)
Given that the time (t) is 0.800 s, we need to find the distance (d). The speed of sound in air depends on the air temperature, which is given as 22.0°C.
The speed of sound in air at 22.0°C is approximately 343 m/s.
We can rearrange the formula to solve for distance:
d = v * t
So, plugging in the values, we have:
d = 343 m/s * 0.800 s = 274.4 meters.
Therefore, the explosion is approximately 274.4 meters away.
(b) To calculate the distance taking into account the speed of light, we need to subtract the time it takes for light to travel from the initial distance we calculated. The speed of light is approximately 299,792,458 m/s.
So, the time it takes for light to reach the physicist is:
Time of light (tl) = d / c
where c is the speed of light.
Therefore, the distance (d) would be:
d' = d - c * tl
Substituting the values:
d' = 274.4 m - (299,792,458 m/s * 0.800 s) = 22.87 meters.
Therefore, if we account for the speed of light, the explosion would be calculated to be approximately 22.87 meters further away.
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.
(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