Answer:
fundamental frequency in helium = 729.8 Hz
Explanation:
Fundamental frequency of an ope tube/pipe = v/2L
where v is velocity of sound in air = 340 m/s; λ is wave length of wave = 2L ; L is length of the pipe
To find the length of the pipe,
frequency = velocity of sound / 2L
272 = 340 / 2 L
L = 0.625 m
If the pipe is filled with helium at the same temperature, the velocity of sound will change as well as the frequency of note produced since velocity is directly proportional to frequency of sound.
Also, the velocity of sound is inversely proportional to square root of molar mass of gas; v ∝ 1/√m
v₁/v₂ = √m₂/m₁
v₁ = velocity of sound in air, v₂ = velocity of sound in helium, m₁ = molar mass of air, m₂ = molar mass of helium
340 / v = √4 / 28.8
v₂ = 340 / 0. 3727
v₂ = 912.26 m /s
fundamental frequency in helium = v₂ / 2L
fundamental frequency in helium = 912.26 / (2 x 0.625)
fundamental frequency in helium = 729.8 Hz
The answer is
D) Betsy is correct since the line is the same on both graphed and both objects are traveling at a speed of 6.25 meters per second.
19.927 moles are in 1.20 times atoms of phosphorus.
<h3>What are moles?</h3>
A mole is defined as 6.02214076 × of some chemical unit, be it atoms, molecules, ions, or others.
1 mole of any substance contain Avogadro's number of molecules so we can calculate the number of moles by dividing the provided number of atoms over Avogadro's number to obtain the number of moles .
Moles=
Moles= 1.20 X atoms ÷ 6.022 X
= 19.927
Hence, 19.927 moles are in 1.20 times atoms of phosphorus.
Learn more about moles here:
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