91Ó°ÊÓ

The rate of continental drift is $V_c=1\mathrm{m}/y$

The drift speed of electron isrole="math" localid="1663066400474" $V_d=05\mathrm{mm}ls$

The highway speed limit is $V_h=55mi/h$

The speed of supersonic plane is role="math" localid="1663066459491" ${\mathrm{V}}_{\mathrm{s}}=110\mathrm{km}lh$

The rate of cosmological expansion is$u=3×{10}^{4}km/s$

The speed parameter is the ratio of speed of particle with speed of light. It describes the comparison of the particle with light particles.

(a)

The speed parameter of continental drift is given as:

$\mathrm{β}=\frac{V_c}{c}$

Here, c is the speed of light and its value is $3×{10}^{8}m/s$.

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{β}=\frac{\left(1\text{in}/\text{y}\right)\left(\frac{0.0254\text{m}}{1\text{in}}\right)\left(\frac{1\text{y}}{3.154×{10}^7\text{s}}\right)}{c\left(\frac{3×{10}^8\text{m}/\text{s}}{c}\right)} \\ =3×{10}^{-18} \end{gathered}$$

Therefore, the speed parameter of continental drift is $3×{10}^{-18}$.

(b)

The speed parameter of drift speed of electron is given as:

$\mathrm{β}=\frac{V_d}{c}$

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{β}=\frac{\left(0.5\text{mm}/\text{s}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{c\left(\frac{3×{10}^8\text{m}/\text{s}}{c}\right)} \\ =2×{10}^{-12} \end{gathered}$$

Therefore, the speed parameter of drift speed of electron is $2×{10}^{-12}$.

(c)

The speed parameter for highway speed limit is given as:

$\mathrm{β}=\frac{V_h}{c}$

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{β}=\frac{\left(55\text{mi}/\text{h}\right)\left(\frac{1600\text{m}}{1\text{mi}}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)}{c\left(\frac{3×{10}^8\text{m}/\text{s}}{c}\right)} \\ =8.2×{10}^{-8} \end{gathered}$$

Therefore, the speed parameter for highway speed limit is $8.2×{10}^{-8}$.

(d)

The speed parameter for root mean square speed of hydrogen molecule is given as:

$\mathrm{β}=\frac{V_r}{c}$

Here, $v_r$is the root mean square speed of hydrogen molecule at room temperature and its value is $2400m/s$.

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{β}=\frac{\left(2400\text{m}/\text{s}\right)}{c\left(\frac{3×{10}^8\text{m}/\text{s}}{c}\right)} \\ =8×{10}^{-6} \end{gathered}$$

Therefore, the speed parameter for root mean square speed of hydrogen molecule is $8×{10}^{-6}$.

The speed parameter for supersonic flying plane is given as:

$\mathrm{β}=\frac{V_s}{c}$

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{β}=\frac{\left(1200\text{km}/\text{h}\right)\left(\frac{1000\text{m}}{1\text{km}}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)}{c\left(\frac{3×{10}^8\text{m}/\text{s}}{c}\right)} \\ =1.1×{10}^{-6} \end{gathered}$$

Therefore, the speed parameter for supersonic flying plane is $1.1×{10}^{-6}$.

(f)

The speed parameter for projectile from Earth’s surface is given as:

$\mathrm{β}=\frac{V_e}{c}$

Here, $v_e$is the escape speed of projectile on Earth and its value is $11200m/s$.

Substitute all the values in the above equation.

role="math" localid="1663067724544" $$\begin{gathered} \mathrm{β}=\frac{\left(11200\text{m}/\text{s}\right)}{c\left(\frac{3×{10}^8\text{m}/\text{s}}{c}\right)} \\ =3.7×{10}^{-5} \end{gathered}$$

Therefore, the speed parameter for projectile from Earth’s surface is$3.7×{10}^{-5}$

(g)

The speed parameter for earth in its orbit around sun is given as:

$\mathrm{β}=\frac{V_o}{c}$

Here, $v_0$is the orbital speed of Earth around sun and its value is $297000m/s$.

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{β}=\frac{\left(297000\text{m}/\text{s}\right)}{c\left(\frac{3×{10}^8\text{m}/\text{s}}{c}\right)} \\ =9.9×{10}^{-5} \end{gathered}$$

Therefore, the speed parameter for earth in its orbit around sun is $9.9×{10}^{-5}$

(h)

The speed parameter for distant quasar is given as:

$\mathrm{β}=\frac{u}{c}$

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{β}=\frac{\left(3×{10}^4\text{km}/\text{s}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right)}{c\left(\frac{3×{10}^8\text{m}/\text{s}}{c}\right)} \\ =0.10 \end{gathered}$$

Therefore, the speed parameter for distant quasar is 0.10.