Ask question

Ask question

All categories

aleksandr82 [10.1K]

2 years ago

8

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op

en at only one end. The other end is closed by the eardrum. A typical length for the auditory canal in an adult is about 3.1 cm. The speed of sound is 336 m/s. What is the fundamental frequency of the canal

1 answer:

Juliette [100K]2 years ago

7 0

Answer:

The fundamental frequency of can is 2.7 kHz.

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

$f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz$

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

You might be interested in

What is the current if a charge of 25 coulombs passes in 5 seconds?

Jobisdone [24]

Answer: 5 amps

Explanation: I = Q/t

8 0

3 years ago

A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9

11111nata11111 [884]

Answer:

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

$m_{1} v_{1}=m_{2} v_{2}$

$\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }$

$\mathrm{v}_{1} \text { velocity before the collision }$

$\mathrm{v}_{2} \text { velocity after the collision }$

$\mathrm{m}_{1}=600 \mathrm{kg}$

$\mathrm{m}_{2}=900 \mathrm{kg}$

$\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}$

$600 \times 5=900 \times v_{2}$

$3000=900 \times v_{2}$

$\mathrm{v}_{2}=\frac{3000}{900}$

$\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

5 0

3 years ago

After exercise your heart rate is

harina [27]

Your heart rate is higher than normal.

7 0

3 years ago

Read 2 more answers

Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin

Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

$\mu_o=4\pi \times 10^{-7}\ T-m/A$

Resistivity, $\rho=1.72\times 10^{-8}\ \Omega-m$

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

$R=\rho \dfrac{l}{A}$

$R=\rho \dfrac{l}{\pi r^2}$

$r=\sqrt{\dfrac{\rho l}{R\pi}}$

$r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}$

r = 0.00199 m

or

$r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m$

The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{2\pi rB}{\mu_o}$

$I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}$

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0

2 years ago

To see why an MRI utilizes iron to increase the magnetic field created by a coil, calculate the current needed in a 400-loop-per

gtnhenbr [62]

Answer:

B = 4.059 x 10¹⁵ T

Explanation:

Given,

Number of loop, N = 400

radius of loop, r = 0.65 x 10⁻¹⁵ m

Current, I = 1.05 x 10⁴ A

Magnetic field at the center of the loop

$B = \dfrac{\mu_0NI}{2R}$

$B = \dfrac{4\pi\times 10^{-7}\times 400 \times 1.05 \times 10^4}{2\times 0.65\times 10^{-15}}$

B = 4.059 x 10¹⁵ T

7 0

3 years ago