91影视

1. Mass of the block varies between $7.9脳{10}^7\text{鈥塳驳}<m<1.7脳{10}^8\text{鈥塳驳}$.
2. Height from which the block is dropped,$h=500\text{鈥尘}$.
3. Impact lasted for time,$螖t=0.5\text{鈥塻}$.
4. Depth of vertical cylinder,$d=5.0\text{鈥尘}$.
5. Distance of grounded triggered seismographs,$r=200\text{鈥塳尘}$ .

We know the height from which the block is falling, and its mass is also known. Now as the block is under free fall, the potential energy will be converted into kinetic energy when the block falls to the ground. From this, we can find the kinetic energy of the block just before it landed.

We are given enough information to find the area of the body wave, and we know the power required to develop these waves. Using the formula for the intensity of the wave, we can find the intensity of the body wave.

A similar procedure has to be used to find the intensity of the surface waves.

Formula:

The intensity of a wave,

$\mathit{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{A}}$ 鈥�(颈)

The power exerted by the body,

$\mathit{P}\mathbf{=}\frac{\mathbf{Kinetic}\mathbf{}\mathbf{Energy}}{\mathbf{螖迟}}$ 鈥�(颈颈)

The potential energy of the body,

鈥�(颈颈颈)

$\mathit{P}\mathit{E}\mathbf{=}\mathit{m}\mathit{g}\mathit{h}$

As the block is falling from a certain height and no other forces are acting on it, according to conservation of energy, potential energy at that height will be converted into kinetic energy just before it comes to rest.

$PE=KE$

Hence, the kinetic energy of the block using equation (iii) is given as:

$KE=mgh$

We take,$m=7.3脳{10}^7\text{鈥塳驳}$

$\begin{array}{l}KE=(7.3脳{10}^7\text{鈥塳驳})脳9.8\text{鈥尘}/{\text{s}}^{\text{2}}脳500\text{鈥尘}\\ =3.6脳{10}^{11}\text{鈥塉}\end{array}$

We take,$m=1.7脳{10}^8\text{鈥塳驳}$

$\begin{array}{rcl}KE& =& (1.7脳{10}^8\text{鈥塳驳})脳9.8\text{鈥尘}/{\text{s}}^{\text{2}}脳500\text{鈥尘}\\ & =& 8.3脳{10}^{11}\text{鈥塉}\end{array}$

So, the range of the kinetic energy will be from$3.6脳{10}^{11}\text{鈥塉}$to.$8.3脳{10}^{11}\text{鈥塉}$

Now$20\%$of the kinetic energy is converted into the seismic wave.

When,$KE=3.6脳{10}^{11}\text{鈥塉}$

$\begin{array}{c}K{E}_{wave}=\frac{20}{100}脳(3.6脳{10}^{11}\text{鈥塉})\\ =0.72脳{10}^{11}\text{鈥塉}\end{array}$

The power exerted by the body using equation (ii) is given as:

$\begin{array}{c}P=\frac{0.72脳{10}^{11}\text{鈥塉}}{0.5\text{鈥塻}}\\ =1.44脳{10}^{11}\text{鈥塛}\end{array}$

The area of the surface of the body is given as:

$\begin{array}{l}{A}_{body}=\frac{1}{2}脳4蟺r^2\\ =2蟺脳{(200脳{10}^3\text{鈥尘})}^2\\ A_{body}=2.5脳{10}^{11}{\text{鈥尘}}^{\text{2}}\end{array}$

Intensity of the body wave using equation (i) can be given as:

$\begin{array}{c}{I}_{body}=\frac{1.44脳{10}^{11}\text{鈥塛}}{2.5脳{10}^{11}{\text{鈥尘}}^{\text{2}}}\\ I_{body}=0.58\text{鈥塛}/{\text{m}}^{\text{2}}\end{array}$

When,

$\begin{array}{l}K{E}_{wave}=\frac{20}{100}脳(8.3脳{10}^{11}\text{鈥塉})\\ K{E}_{wave}=1.66脳{10}^{10}\text{鈥塉}\end{array}$

The power exerted by the body using equation (ii) is given as:

$\begin{array}{c}P=\frac{1.66脳{10}^{11}\text{鈥塉}}{0.5\text{鈥塻}}\\ P=3.32脳{10}^{11}\text{鈥塛}\end{array}$

The area of the block is given as:

$A_{body}=2.5脳{10}^{11}{\text{鈥尘}}^{\text{2}}$

Intensity of the body wave using equation (i) can be given as:

$\begin{array}{l}{I}_{body}=\frac{3.32脳{10}^{11}\text{鈥塛}}{2.5脳{10}^{11}{\text{鈥尘}}^{\text{2}}}\\ I_{body}=1.33\text{鈥塛}/{\text{m}}^{\text{2}}\end{array}$

So, the intensity of the body wave ranges from$0.58\text{鈥塛}/{\text{m}}^{\text{2}}$ to.$1.33\text{鈥塛}/{\text{m}}^{\text{2}}$

Now$20\%$of the kinetic energy is converted into the seismic wave.

When$KE=3.6脳{10}^{11}\text{鈥塉}$

$\begin{array}{l}K{E}_{wave}=\frac{20}{100}脳(3.6脳{10}^{11}\text{鈥塉})\\ K{E}_{wave}=0.72脳{10}^{11}\text{鈥塉}\end{array}$

The power radiated by the body wave using equation (ii) is given as:

$\begin{array}{c}P=\frac{0.72脳{10}^{11}\text{鈥塉}}{0.5\text{鈥塻}}\\ =1.44脳{10}^{11}\text{鈥塛}\end{array}$

The area of the seismic surface is given as:

$\begin{array}{l}{A}_{surface}=2蟺rd\\ =2蟺脳(200脳{10}^3\text{鈥尘})脳5.0\text{鈥尘}\\ A_{surface}=6.28脳{10}^6{\text{鈥尘}}^{\text{2}}\end{array}$

Intensity of the surface wave using equation (i) can be given as:

$\begin{array}{l}{I}_{surface}=\frac{1.44脳{10}^{11}\text{鈥塛}}{6.28脳{10}^6\text{鈥�}{m}^2}\\ =22.92脳{10}^3\text{鈥塛}/{\text{m}}^{\text{2}}\\ I_{surface}鈮�23脳{10}^3\text{鈥塛}/{\text{m}}^{\text{2}}\end{array}$

When $KE=8.3脳{10}^{11}\text{鈥塉}$

$\begin{array}{l}K{E}_{wave}=\frac{20}{100}脳(8.3脳{10}^{11}\text{鈥塉})\\ K{E}_{wave}=1.66脳{10}^{10}\text{鈥塉}\end{array}$

The power radiated by the body wave using equation (ii) is given as:

$\begin{array}{c}P=\frac{1.66脳{10}^{11}\text{鈥塉}}{0.5\text{鈥塻}}\\ =3.32脳{10}^{11}\text{鈥塛}\end{array}$

The area of the seismic surface is given as:

$A_{surface}=6.28脳{10}^6{\text{鈥尘}}^{\text{2}}$

Intensity of the surface wave using equation (i) can be given as:

$\begin{array}{l}{I}_{surface}=\frac{3.32脳{10}^{11}\text{鈥塉}}{6.28脳{10}^6\text{鈥�}{m}^2}\\ =52.86脳{10}^3\text{鈥塛}/{\text{m}}^{\text{2}}\\ I_{surface}鈮�53脳{10}^3\text{鈥塛}/{\text{m}}^{\text{2}}\end{array}$

So, the intensity of the surface wave ranges from $23脳{10}^3\text{鈥塛}/{\text{m}}^{\text{2}}to53脳{10}^3\text{鈥塛}/{\text{m}}^{\text{2}}$.

Several factors are involved in determining which seismic waves are being detected by thedistant seismograph. In this situation, the waves which have higher intensity will be detected by the distant seismograph, and in this case, they are the surface waves.