/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 The power of sound from the spea... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 77

The power of sound from the speaker of a radio is \(20 \mathrm{~mW}\). By turning the knob of volume control, the power of sound is increased to \(400 \mathrm{~mW}\). The power increase in \(\mathrm{dB}\) as compared to the original power is \(\left(\log _{10} 2=0.3\right)\) (A) \(1.3 \mathrm{~dB}\) (B) \(3.1 \mathrm{~dB}\) (C) \(13 \mathrm{~dB}\) (D) \(30.1 \mathrm{~dB}\)

Short Answer

Expert verified
The correct answer is (C) \(13 \mathrm{~dB}\).

Step by step solution

01

Understand the dB formula and given values

The formula to calculate the increase in power in dB is given by: \(P_{dB} = 10 \times \log_{10}\left(\frac{P_{2}}{P_{1}}\right)\) Where \(P_{1}\) is the initial power, \(P_{2}\) is the final power, and \(P_{dB}\) is the power increase in dB. In this problem, the initial power, \(P_{1}\), is 20 mW, and the final power, \(P_{2}\), is 400 mW.
02

Substitute the given values in dB formula

Now, we need to substitute the values of \(P_{1}\) and \(P_{2}\) in the formula: \(P_{dB} = 10 \times \log_{10}\left(\frac{400}{20}\right)\)
03

Simplify the expression

Now, we need to simplify the expression on the right side of the equation as follows: \(P_{dB} = 10 \times \log_{10}(20)\)
04

Calculate the power increase in dB

Now, we just need to calculate the value of the power increase in dB: \(P_{dB} = 10 \times \log_{10}(20)\) \(P_{dB} = 10 \times 1.30103\) \(P_{dB} = 13.0103\) The power increase in dB is approximately 13 dB. So, the correct answer is: (C) \(13 \mathrm{~dB}\)

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Most popular questions from this chapter

An open glass tube is immersed in mercury in such a way that a length of \(8 \mathrm{~cm}\) extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional \(46 \mathrm{~cm}\). What will be length of the air column above mercury in the tube now? (Atmospheric pressure \(=76 \mathrm{~cm}\) of \(\mathrm{Hg}\) ) (A) \(16 \mathrm{~cm}\) (B) \(22 \mathrm{~cm}\) (C) \(38 \mathrm{~cm}\) (D) \(6 \mathrm{~cm}\)

In forced oscillation of a particle the amplitude is maximum for a frequency \(\omega_{1}\) of the force while the energy is maximum for a frequency \(\omega_{2}\) of the force; then (A) \(\omega_{1}<\omega_{2}\) when damping is small and \(\omega_{1}>\omega_{2}\) when damping is large (B) \(\omega_{1}>\omega_{2}\) (C) \(\omega_{1}=\omega_{2}\) (D) \(\omega_{1}<\omega_{2}\)

A particle of mass \(m\) is attached to a spring (of spring constant \(k\) ) and has a natural angular frequency \(\omega_{0}\). An external force \(F(t)\) proportional to \(\cos \omega t\left(\omega \neq \omega_{0}\right)\) is applied to the oscillator. The time displacement of the oscillator will be proportional to \(\quad\) (A) \(\frac{1}{m\left(\omega_{0}^{2}+\omega^{2}\right)}\) (B) \(\frac{1}{m\left(\omega_{0}^{2}-\omega^{2}\right)}\) (C) \(\frac{m}{\omega_{0}^{2}-\omega^{2}}\) (D) \(\frac{m}{\left(\omega_{0}^{2}+\omega^{2}\right)}\)

A wave disturbance in a medium is described by \(y(x, t)=0.02 \cos \left(50 \pi t+\frac{\pi}{2}\right) \cos (10 \pi x)\), where \(x\) and \(y\) are in meter and \(t\) is in second. Then (A) First node occurs at \(x=0.15 \mathrm{~m}\) (B) First anti-node occurs at \(x=0.3 \mathrm{~m}\) (C) The speed of interfering waves is \(5.0 \mathrm{~m} / \mathrm{s}\) (D) The wavelength is \(0.2 \mathrm{~m}\)

The total energy of a particle, executing simple harmonic motion is (A) independent of \(x\) (B) \(\propto x^{2}\) (C) \(\propto x\) \((\mathrm{D}) \propto x^{1 / 2}\) where \(x\) is the displacement from the mean position.
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