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Chapter 17: Q.55PE (page 630)

The factor of \({\rm{1}}{{\rm{0}}^{{\rm{ - 12}}}}\)in the range of intensities to which the ear can respond, from threshold to that causing damage after brief exposure, is truly remarkable. If you could measure distances over the same range with a single instrument and the smallest distance you could measure was\(1\;{\rm{mm}}\), what would the largest be?

Short Answer

Expert verified

The largest distance is \(1{0^9}\;m\).

Step by step solution

01

Given Data

The factor is\(1{0^{ - 12}}\).

The smallest distance is\(d = 1\;mm = 1{0^{ - 3}}\;m\).

02

Concept

The expression for the factor of intensity is given by,

\(f = \frac{d}{D}\)

Here \(f\) is the factor of intensity, \(d\) is the smallest distance that can be measured, \(D\) is the possible largest distance.

03

Calculation of the largest distance  

The factor of intensity for the smallest distance\({\rm{d}}\)and the largest distance\({\rm{D}}\)is,

\(\frac{d}{D}\)

Plugging the values,

\(\begin{aligned}{}1{0^{ - 12}} &= \frac{{1{0^{ - 3}}}}{D}\\D &= \frac{{1{0^{ - 3}}}}{{1{0^{ - 12}}}}\\D &= 1{0^9}\;m\end{aligned}\)

Therefore the largest distance is \({10^9}\,{\rm{m}}\).

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Most popular questions from this chapter

In the clinical use of ultrasound, transducers are always coupled to the skin by a thin layer of gel or oil, replacing the air that would otherwise exist between the transducer and the skin. (a) Using the values of acoustic impedance given inTable 17.5calculate the intensity reflection coefficient between transducer material and air. (b) Calculate the intensity reflection coefficient between transducer material and gel (assuming for this problem that its acoustic impedance is identical to that of water). (c) Based on the results of your calculations, explain why the gel is used.

An 8-hour exposure to a sound intensity level of\(90.0\;{\rm{dB}}\)may cause hearing damage. What energy in joules falls on a\(0.800\;{\rm{cm}}\)diameter eardrum so exposed?

Based on the graph inFigure\({\bf{17}}.{\bf{36}}\), what is the threshold of hearing

in decibels for frequencies of\({\bf{60}},{\rm{ }}{\bf{400}},{\rm{ }}{\bf{1000}},{\rm{ }}{\bf{4000}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{15}},{\bf{000}}{\rm{ }}{\bf{Hz}}\)? Note

that many AC electrical appliances produce 60 Hz, music is commonly

\({\bf{400}}{\rm{ }}{\bf{Hz}}\), a reference frequency is\({\bf{1000}}{\rm{ }}{\bf{Hz}}\), your maximum sensitivity

is near\({\bf{4000}}{\rm{ }}{\bf{Hz}}\), and many older TVs produce a\({\bf{15}},{\bf{750}}{\rm{ }}{\bf{Hz}}\)whine.

It is more difficult to obtain a high-resolution ultrasound image in the abdominal region of someone who is overweight than for someone who has a slight build. Explain why this statement is accurate.

If a woman needs an amplification of \({\rm{5 \times 1}}{{\rm{0}}^{{\rm{12}}}}{\rm{ W}} \cdot {{\rm{m}}^{{\rm{ - 2}}}}\) times the threshold intensity to enable her to hear at all frequencies, what is her overall hearing loss in dB? Note that smaller amplification is appropriate for more intense sounds to avoid further damage to her hearing from levels above\({\rm{90 dB}}\).
See all solutions

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