The speed of sound in air in fact depends on the temperature of the air. At 0 o

Question

The speed of sound in air in fact depends on the temperature of the air. At 0 oC, the speed of sound in air is 331.5 m/s. For every Celsius degree above 0 oC, the speed increases by 0.6 m/s. For every Celsius degree below 0 oC, the speed decreases by 0.6 m/s.

PLEASE, SHOW ME HOW TO DO IT.

1) what is the speed of sound outside on a cold day when the tempretature is -17 oC?

Answer in m/s

2) The ultrasonic motion detector that you have used in the lab sends a very high frequency sound (above the range of human hearing) and essentially measures the time for the sound to bounce off the nearest object and return. The detector is calibrated for sound traveling at 342.5 m/s. (This assumes the temperature of the room is 20oC. ), you measure a distance of 6 m with the detector. How long does it take the signal to leave the detector and return if the detector measures 6m?

answer in s

3) you take this detector outside a very cold winter day when the temperature is -17oC. While outside, you measure a distance between an object and the detector to be of 6m. You know this cannont be the case because the temperature at which the detector was calibrated is not the outside temperature. What is the actual distance between the object and the detector?

answer in m

Explanation / Answer

v=sqrt(RT/M)

dv/v= 1/2 * dT/T

for 0 degrees C = 273 K

dT = 1K

T=273 K

v=331.5

==> dv= (1/2)*(1/273)*331.5 = 0.606 m/s

==> Change in v = 0.606 m/s for every 1K chage in temperature

a) V2/V1= sqt(T2/T1)

==> V/331.5= sqrt[(273-17)/273]

==> V = 321.01 m/s

b) t=2*6/v= 12/342.5 = 0.035 s

c) s2/s1=v2/v1=sqrt(T2/T1)

==> s=sqrt[(273-17)/(273+20)]*6 = 5.608 m

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