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Elina [12.6K]
3 years ago
5

The human ear canal is about 2.8 cm long. If it is regarded as a tube that is open at one end and closed at the eardrum, what is

the fundamental frequency around which we would expect hearing to be most sensitive?[3.1 kHz]
Physics
1 answer:
Diano4ka-milaya [45]3 years ago
8 0

Answer:

f = 3.1 kHz

Explanation:

given,

length of human canal =2.8 cm = 0.028 m

speed of sound = 343 m/s

fundamental frequency  = ?

The fundamental frequency of a tube with one open end and one closed end is,

f = \dfrac{v}{4L}

f = \dfrac{343}{4\times 0.028}

f = \dfrac{343}{0.112}

       f = 3062.5 Hz

       f = 3.1 kHz

hence, the fundamental frequency is equal to f = 3.1 kHz

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Answer:

As of right now the techology has not been invented to time travel

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Explanation:

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According to the law of universal gravitation, any two objects are attracted to each other. The strength of the gravitational fo
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A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of
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Answer:

7.74m/s

Explanation:

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Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

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8 0
3 years ago
Read 2 more answers
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14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

       Charge = Current \times Time

As two different current is passing at two different times, the net charge will be the different in current.  So,

        \text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

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Here k = 9 * 10^9, q is the charge and R is the radius. As q=2 \times 10^{-6} \times t and R =17 cm = 0.17 m, then the voltage will be

        V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}

The time is required to find to reach the voltage of 1500 V, so

1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}

\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}

So, 14 ms is required to reach the potential of 1500 V.

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