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Darina [25.2K]
3 years ago
15

A child has an ear canal that is 1.3 cm long. At what sound frequencies in the audible range will the child have increased heari

ng sensitivity?
Physics
1 answer:
ElenaW [278]3 years ago
7 0

Answer:

The answer to this is

6600 Hz to 19,800 Hz

Explanation:

The shape of the human ear is analogous to a closed ended pipe hence

we have λ = 4L or wavelength = 4 * length of the child ear

The frequency  c/λ  where c = speed of sound = 343 m/s

hence the child's audible range is multiples of 343/(4*0.13) =6600Hz

or 13200 Hz or 19,800 Hz

The generally quoted range of human hearing is 20 Hz to 20 kHz

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A spaceship travels toward the Earth at a speed of 0.97c. The occupants of the ship are standing with their torsos parallel to t
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Answer:

Explanation:

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L=\frac{.5}{\sqrt{1-(\frac{.97c}{c})^2 } }

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A compact car has a mass of 1310 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that
Nady [450]

Answer:

a) k= 3232.30 N / m,  b)  f = 4,410 Hz

Explanation:

In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.

The expression for the angular velocity is

          w = √k/m

the angular velocity is related to the period

          w = 2π / T

we substitute

          T = 2\pi  √m/ k

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           k = 4π² m / T²

           k = 4 π² 1310/2 2

           k = 12929.18 N / m

This is the equivalent constant of the short springs

           F1 + F2 + F3 + F4 = k_eq x

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           k = k_eq / 4

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In this case the plus is the mass of the vehicle plus the masses of the passengers

            m_total = 1360 + 4 70

            m_total = 1640 kg

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              w = 2pi f

we substitute

             2 pi f = Ra K / m

in this case the spring constant changes us

             k_eq = 12929.18 N / m

           

             f = 1 / 2π √ 12929.18 / 1640

             f = π / 2 2.80778

             f = 4,410 Hz

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