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(A) The relationship between wave speed $(v)$,frequency $(n)$,and wavelength $(\lambda)$ is given by the formula: $v = n \lambda$. Rearranging for wav

Vedclass · Accepted Answer

$4 	imes 10^{-4} \, m$

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(A) The relationship between wave speed $(v)$,frequency $(n)$,and wavelength $(\lambda)$ is given by the formula: $v = n \lambda$. Rearranging for wavelength,we get: $\lambda = \frac{v}{n}$. Given: Frequency $(n)$ = $4.2 \, MHz = 4.2 	imes 10^6 \, Hz$. Speed of sound $(v)$ = $1.7 \, km/s = 1.7 	imes 10^3 \, m/s$. Substituting the values: $\lambda = \frac{1.7 	imes 10^3}{4.2 	imes 10^6} = \frac{1.7}{4.2} 	imes 10^{-3} \approx 0.4047 	imes 10^{-3} \, m = 4.047 	imes 10^{-4} \, m$. Rounding to the nearest value,we get $\lambda \approx 4 	imes 10^{-4} \, m$.

$A$ hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is $4.2 \, MHz$. The speed of sound in a tissue is $1.7 \, km/s$. The wavelength of sound in the tissue is close to:

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