91影视

1. Mass of the meteorite $\mathrm{m}=4脳{10}^6\mathrm{kg}$.
2. Speed ${\mathrm{v}}_{\mathrm{i}}=15\mathrm{km}/\mathrm{s}=15脳{10}^3\mathrm{m}/\mathrm{s}$.
3. $1\mathrm{megaton}\mathrm{of}\mathrm{TNT}=4.2脳{10}^{15}\mathrm{J}$.
4. Energy associated with the atomic bomb explosion is data-custom-editor="chemistry" $13\mathrm{kilotons}\mathrm{of}\mathrm{TNT}$.

Using the formula of kinetic energy, we can find the meteorite鈥檚 loss of kinetic energy that would have been associated with the vertical impact. Using the energy loss and relation between 1 megaton of TNT and joule, we can find the energy loss as a multiple of multiple of the explosive energy of 1 megaton of TNT.

Formula:

$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$

We can calculate the kinetic energy with the expression,

$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$

Initial kinetic energy will be,

${\mathrm{K}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2$

Substitute the values in the above expression, and we get,

role="math" localid="1660904026139" $\begin{array}{rcl}{\mathrm{K}}_{\mathrm{i}}& =& \frac{1}{2}\left(4脳{10}^6\mathrm{kg}\right){\left(15脳{10}^3\frac{\mathrm{m}}{\mathrm{s}}\right)}^2\\ & =& 4.5脳{10}^{14}.\left(1\mathrm{kg}脳1\frac{{\mathrm{m}}^2}{{\mathrm{s}}^2}脳\frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}脳\frac{1\mathrm{N}}{1\mathrm{kg}.{\displaystyle \frac{\mathrm{m}}{{\mathrm{s}}^2}}}\right)\\ & =& 4.5脳{10}^{14}\mathrm{J}\end{array}$

Final kinetic energy will be,

${\mathrm{K}}_{\mathrm{f}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2$

Here v_f is the final speed, whose value is zero.

Substitute the values in the above expression, and we get,

${\mathrm{K}}_{\mathrm{f}}=0\mathrm{J}$

Therefore, the change in kinetic energy can be calculated as,

data-custom-editor="chemistry" $鈭�\mathrm{K}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}$

Substitute the values in the above expression, and we get,

data-custom-editor="chemistry" $$\begin{gathered} 鈭�\mathrm{K}=0\mathrm{J}-4.5脳{10}^{14}\mathrm{J} \\ 鈭�\mathrm{K}=-4.5脳{10}^{14}\mathrm{J} \end{gathered}$$

Therefore, the meteorite鈥檚 loss of kinetic energy that would have been associated with the vertical impact is $-4.5脳{10}^{14}\mathrm{J}$.

Given,

$1\mathrm{megaton}\mathrm{of}\mathrm{TNT}=4.2脳{10}^{15}\mathrm{J}$.

Therefore, the energy loss in terms of megatons of TNT can be calculated as,

$$\begin{gathered} 鈭�\mathrm{K}=-\left(4.5脳{10}^{14}\mathrm{J}\right)\left(\frac{1\mathrm{megaton}\mathrm{TNT}}{4.2脳{10}^{15}\mathrm{J}}\right) \\ 鈭�\mathrm{K}=-0.1\mathrm{megaton}\mathrm{TNT} \end{gathered}$$

Therefore, the energy loss as a multiple of the explosive energy of 1 megaton of TNT, which is $4.2脳{10}^{15}\mathrm{J}$, is $-0.1\mathrm{megaton}\mathrm{TNT}$.

Given,

$1\mathrm{Megaton}=1000\mathrm{kilotons}$.

The energy associated with the atomic bomb explosion is $13\mathrm{kilotons}\mathrm{of}\mathrm{TNT}$.

We can calculate the number of bombs as,

$\begin{array}{rcl}\mathrm{N}& =& \frac{0.1脳1000\mathrm{kilotons}\mathrm{of}\mathrm{TNT}}{13\mathrm{kilotons}\mathrm{TNT}}\\ & =& 7.69\\ & & ~8\end{array}$

Therefore, the number of Hiroshima bombs that would have been equivalent to the meteorite impact is 8.