= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)
= -0.00225 m
New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
D
Which of the following descriptions apply to the Milky Way galaxy
2 answers:
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= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)
= -0.00225 m
New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
D
Answer:
The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
Explanation:
Given that,
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.
The speed of sound in air is 343 m/s.
To find,
The wavelength range for the corresponding frequency.
Solution,
The speed of sound is given by the following relation as :
Wavelength for f = 45 Hz is,
Wavelength for f = 375 Hz is,
So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.