Chapter 2, Problem 12P

<a>

To determine

The ratio $\beta =\frac{v}{c}$, for speed $95\text{ km / h}$.

Answer to Problem 12P

The ratio $\beta =\frac{v}{c}$, for speed $95\text{ km / h}$ is $8.79\times {10}^{-8}$.

Explanation of Solution

Convert, speed kilometer per hour to meter per second. Thus

$\begin{array}{c}v=95\text{ km / h}\\ \text{=}95\text{ km / h}\times \left(\frac{1000\text{ m}}{1.0\text{ km}}\right)\left(\frac{1.0\text{ h}}{3600\text{ s}}\right)\\ =\frac{950}{36}\text{ m / s}\\ =26.38\text{ m / s}\end{array}$

Substitute $v=26.38\text{ m / s}$ in $\beta =\frac{v}{c}$, where, $c=3\times {10}^8\text{ m / s}$ is the speed of the light. Thus,

$\begin{array}{c}\beta =\frac{26.38\text{ m / s}}{3\times {10}^8\text{ m / s}}\\ =8.79\times {10}^{-8}\end{array}$

Conclusion:

Therefore, the ratio $\beta =\frac{v}{c}$, for $v=95\text{ km / h}$ is $8.79\times {10}^{-8}$.

<b>

To determine

The ratio $\beta =\frac{v}{c}$, for $v=240\text{ m / s}$.

Answer to Problem 12P

The ratio $\beta =\frac{v}{c}$, for $v=240\text{ m / s}$ is $8.0\times {10}^{-7}$.

Explanation of Solution

Substitute $v=240\text{ m / s}$ in $\beta =\frac{v}{c}$, where, $c=3\times {10}^8\text{ m / s}$ is the speed of the light. Thus,

$\begin{array}{c}\beta =\frac{240\text{ m / s}}{3\times {10}^8\text{ m / s}}\\ =\frac{24}{3}\times {10}^{-7}\\ \text{=8}\text{.0}\times {10}^{-7}\end{array}$

Conclusion:

Therefore, the ratio $\beta =\frac{v}{c}$, for $v=240\text{ m / s}$ is $8.0\times {10}^7$.

<c>

To determine

The ratio $\beta =\frac{v}{c}$, for an airplane travelling with 2.3 Mach.

Answer to Problem 12P

The ratio $\beta =\frac{v}{c}$, for an airplane travelling with 2.3 Mach is $\text{2}\text{.53}\times {10}^{-6}$.

Explanation of Solution

Velocity of the sound in the is $v_s\text{=330 m / s}$.

Write the formula for the conversion of the speed of the supersonic plane from Mach to meter per second. Thus,

$v=M{v}_s$, where, $v_s$ is the velocity of the sound in the air and M is the Mach number.

Substitute $v_s\text{=330 m / s}$ and $\text{M}=2.3$ in in $v=\text{M}{v}_s$ to  get v. Thus,

$\begin{array}{c}v=2.3\times 330\text{ m / s}\\ \text{=759 m / s}\end{array}$

Substitute $v=759\text{ m / s}$ in $\beta =\frac{v}{c}$, where, $c=3\times {10}^8\text{ m / s}$ is the speed of the light. Thus,

$\begin{array}{c}\beta =\frac{759\text{ m / s}}{3\times {10}^8\text{ m / s}}\\ =253\times {10}^{-8}\\ \text{=2}\text{.53}\times {10}^{-6}\end{array}$

Conclusion:

Therefore, the ratio $\beta =\frac{v}{c}$, for $v=2.3\text{ Mach}$ is $\text{2}\text{.53}\times {10}^{-6}$.

<d>

To determine

The ratio $\beta =\frac{v}{c}$, for $v=27000\text{ km / h}$.

Answer to Problem 12P

The ratio $\beta =\frac{v}{c}$, for $v=27000\text{ km / h}$ is $\text{2}\text{.5}\times {10}^{-5}$.

Explanation of Solution

Convert, speed kilometer per hour to meter per second. Thus

$\begin{array}{c}v=27000\text{ km / h}\\ \text{=}27000\text{ km / h}\times \left(\frac{1000\text{ m}}{1.0\text{ km}}\right)\left(\frac{1.0\text{ h}}{3600\text{ s}}\right)\\ =\frac{270\times {10}^3}{36}\text{ m / s}\\ =7.5\times {10}^3\text{ m / s}\end{array}$

Substitute $v=7.5\times {10}^3\text{ m / s}$ in $\beta =\frac{v}{c}$, where, $c=3\times {10}^8\text{ m / s}$ is the speed of the light. Thus,

$\begin{array}{c}\beta =\frac{7.5\times {10}^3\text{ m / s}}{3\times {10}^8\text{ m / s}}\\ =\frac{7.5}{3}\times {10}^{-5}\\ =\text{2}\text{.5}\times {10}^{-5}\end{array}$

Conclusion:

Therefore, the ratio $\beta =\frac{v}{c}$, for $v=27000\text{ km / h}$ is $=\text{2}\text{.5}\times {10}^{-5}$.

<e>

To determine

The ratio $\beta =\frac{v}{c}$, when an electron covering a distance 25 cm in 2 ns.

Answer to Problem 12P

The ratio $\beta =\frac{v}{c}$, when an electron covering a distance 25 cm in 2 ns is $0.42$.

Explanation of Solution

Write the equation to calculate the speed of the electron. Thus

$v=\frac{d}{t}$, where, d is the distance covered in time t.

Distance covered by the electron is 25 cm. Convert this distance in meter, thus,

$\begin{array}{c}25\text{ cm}=25.0\text{ cm}\times \frac{{\text{10}}^{-2}\text{ m}}{1.0\text{ cm}}\\ =25.0\times {10}^{-2}\text{ m}\end{array}$

Time taken by the electron to cover 25 cm is 2 ns. Convert this time in second, thus,

$\begin{array}{c}2\text{ ns}=2\text{ ns}\times \frac{{10}^{-9}\text{ s}}{1\text{ ns}}\\ =2\times {10}^{-9}\text{ s}\end{array}$

Substitute $d=25.0\times {10}^{-2}\text{ m}$ and $t=2\times {10}^{-9}\text{ s}$ in $v=\frac{d}{t}$ to calculate the v. Thus,

$\begin{array}{c}v=\frac{25.0\times {10}^{-2}\text{ m}}{2\times {10}^{-9}\text{ s}}\\ =12.5\times {10}^7\text{ m / s}\\ =1.25\times {10}^8\text{ m / s}\end{array}$

Substitute $v=1.25\times {10}^8\text{ m / s}$ in $\beta =\frac{v}{c}$, where, $c=3\times {10}^8\text{ m / s}$ is the speed of the light. Thus,

$\begin{array}{c}\beta =\frac{1.25\times {10}^8\text{ m / s}}{3\times {10}^8\text{ m / s}}\\ =0.416\\ =0.42\end{array}$

Conclusion:

Therefore, the ratio $\beta =\frac{v}{c}$, for $v=1.25\times {10}^8\text{ m / s}$ is $0.42$.

<f>

To determine

The ratio $\beta =\frac{v}{c}$, when an electron covering a distance ${10}^{-14}\text{ m}$ in $3.5\times {10}^{-23}\text{ s}$.

Answer to Problem 12P

The ratio $\beta =\frac{v}{c}$, when an electron covering a distance ${10}^{-14}\text{ m}$ in $3.5\times {10}^{-23}\text{ s}$ is $0.95$

Explanation of Solution

Write the equation to calculate the Speed of the electron. Thus

$v=\frac{d}{t}$, where, d is the distance covered in time t.

Distance covered by the electron is ${10}^{-14}\text{ m}$ and time taken to cover this distance is $3.5\times {10}^{-23}\text{ s}$.

Substitute $d={10}^{-14}\text{ m}$ and $t=3.5\times {10}^{-23}\text{ s}$ in $v=\frac{d}{t}$ to calculate the v. Thus,

$\begin{array}{c}v=\frac{{10}^{-14}\text{ m}}{3.5\times {10}^{-23}\text{ s}}\\ =0.285\times {10}^{-14+23}\text{ m / s}\\ =0.285\times {10}^9\text{ m / s}\\ =2.86\times {10}^8\text{ m / s}\end{array}$

Substitute $v=2.86\times {10}^8\text{ m / s}$ in $\beta =\frac{v}{c}$, where, $c=3\times {10}^8\text{ m / s}$ is the speed of the light. Thus,

$\begin{array}{c}\beta =\frac{2.86\times {10}^8\text{ m / s}}{3\times {10}^8\text{ m / s}}\\ =0.95\end{array}$

Conclusion:

Therefore, the ratio $\beta =\frac{v}{c}$, for $v=2.86\times {10}^8\text{ m / s}$ is $0.95$.