The frequency of the human ear canal is 2.92 kHz.
Explanation:
As the ear canal is like a tube with open at one end, the wavelength of sound passing through this tube will propagate 4 times its length of the tube. So wavelength of the sound wave will be equal to four times the length of the tube. Then the frequency can be easily determined by finding the ratio of velocity of sound to wavelength. As the velocity of sound is given as 339 m/s, then the wavelength of the sound wave propagating through the ear canal is
Wavelength=4*Length of the ear canal
As length of the ear canal is given as 2.9 cm, it should be converted into meter as follows:
Then the frequency is determined as
f=c/λ=339/0.116=2922 Hz=2.92 kHz.
So, the frequency of the human ear canal is 2.92 kHz.
Answer:
1. the one with the raito
2. the one that stubbed their toe
Explanation:
Meselson and Stahl
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The classic experiment that supported the semiconservative model of dna replication was performed by Matthew Meselson and Franklin W. Stahl. In this model, the two strands of DNA unwind from each other, and each acts as a template for synthesis of a new, complementary strand. This results in two DNA molecules with one original strand and one new strand. They used E. coli bacteria as a model system.
The exact location of the 'subtropical-jet stream' is located at the North phase of 30 Degrees and the reason behind this is due to the variation of air which lies on the region of mid-altitude and warmer equatorial air. the correct answer would be Cool and Warm air masses meeting near the equator.
Answer:
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Explanation:
From the question we are told that:
The Electric field of strength direction =Right
The Velocity of The First Electron=V_0
The Velocity of The Second Electron=V_0
Therefore
Generally, the equation for the Horizontal Displacement of electron is mathematically given by
Where
Acceleration is given as
And
Time
Therefore horizontal displacement towards the left is
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