А1  ^?? Analyzing the dimensions, find $a$, $b$ and $c$.

\[[m] = [\text{kg} \cdot \text{m}^2/\text{s}^2]^a \cdot [\text{kg}/\text{m}^3]^b \cdot [\text{s}]^c \]
Each dimension gives us a linear equalitie:
\[1 = 2a - 3b\]\[0 = a + b\]\[0 = -2a + c\]In the end: \[a = \frac{1}{5}; \quad b = - \frac{1}{5} ; \quad c = \frac{2}{5} \]

A2  ^?? Above are photos of the first atomic explosion, taken milliseconds after detonation. At each photo the scale and time from the beginning of the explosion are indicated. Having linearized the dependence from point A1, find the energy $E_0$ that the atomic bomb released. The air density is $1.3~\text{kg}/\text{m}^3$. The coefficient $S$ for air is $S = 1.03$.

\[R(t) = S \]

$t,~\text{ms}$	$R,~\text{m}$
16	215
25	258
53	340
62	345
90	385

To check our simple theory we can plot $\ln R$ vs. $\ln t$.

The slope is $0.34$, which likely corresponds to $1/3$. However, it is also quite close to $2/5$.

The slope
\[k = S E_0^{1/5} \rho_0^{-1/5} =1.0 \cdot 10^3~\text{m} \cdot \text{s}^{-2/5},\]so the energy
\[E = \rho_0 \left( \frac{k}{S} \right)^5 = 50 ~\text{TJ}.\]