A bat emits a chirping sound of frequency 67.0 Hz while moth hunting. The chirp lasts for 1.0 ms and then is silent. The beginning of the echo returns just after the outgoing chirp is finished - how close to the moth is the bat? Assume a cool night with air temperature of 10 degrees .
To calculate how close the bat is to the moth based on the echo's delay, we need to consider the speed of sound in air. The speed of sound in air can be calculated using the formula:
v = 331.4 + 0.6 * T
where:
v is the speed of sound in meters per second (m/s)
T is the air temperature in degrees Celsius (°C)
In this case, the air temperature is given as 10 degrees Celsius. Plugging in the value into the formula, we get:
v = 331.4 + 0.6 * 10 = 331.4 + 6 = 337.4 m/s
Now, we need to calculate the time it takes for the sound to travel from the bat to the moth and back to the bat based on the frequency and duration of the chirp.
The time it takes for the sound to travel one way can be calculated using the formula:
t = d / v
where:
t is the time in seconds (s)
d is the distance in meters (m)
v is the speed of sound in air (m/s)
Since the sound travels from the bat to the moth and back to the bat, the total time it takes can be calculated as:
2t = 1.0 ms = 1.0 * 10^-3 s
Now we can rearrange the formula to find the distance, d:
d = v * t
Substituting the values we have:
d = 337.4 m/s * (1.0 * 10^-3) s = 0.3374 m
Therefore, the bat is approximately 0.3374 meters (or 33.74 centimeters) away from the moth.