An explosion at time t = 0 releases energy 𝐸 at the origin in a space filled with a gas of density σ . Subsequently, a hemispherical blast wave propagates radially outwards as shown in the figure.

Let R denote the radius of the front of the hemispherical blast wave. The radius R follows the relationship 𝑅 = 𝑘 𝑡 ^𝑎𝐸 ^𝑏𝜌 ^𝑐 , where k is a dimensionless constant. The value of exponent a is ___________. (Rounded off to one decimal place)

The correct answer is 0.4.

Step-by-step Explanation:

We are given that the radius of the hemispherical blast wave, $R$, depends on time $t$, energy $E$, and gas density $\rho$ (indicated in the diagram as $\rho$ and in the text as $\sigma$) according to the following relation:

$R=k{t}^a{E}^b{\rho }^c$

where $k$ is a dimensionless constant.

Let us write down the dimensions of each variable involved using mass ($M$), length ($L$), and time ($T$):

1. Radius $R$ is a length dimension:
$\left[R\right]=L$

2. Time $t$ is a time dimension:
$\left[t\right]=T$

3. Energy $E$ has dimensions of work or kinetic energy:
$\left[E\right]=M{L}^2{T}^{-2}$

4. Density $\rho$ is mass per unit volume:
$\left[\rho \right]=M{L}^{-3}$

Substituting these dimensions into the relationship, we get:

$M^0{L}^1{T}^0={\left(T\right)}^a{\left(M{L}^2{T}^{-2}\right)}^b{\left(M{L}^{-3}\right)}^c$

Simplifying the right-hand side by grouping the bases $M$, $L$, and $T$:

$M^0{L}^1{T}^0=M^{b+c}{L}^{2b-3c}{T}^{a-2b}$

By comparing the exponents of $M$, $L$, and $T$ on both sides, we get a system of three linear equations:

For $M$:
$b+c=0$

For $L$:
$2b-3c=1$

For $T$:
$a-2b=0$

Now, we solve these equations step-by-step:

From the mass equation:
$c=-b$

Substituting $c=-b$ into the length equation:
$2b-3(-b)=1$
$5b=1$
$b=0.2$

Now, substituting $b=0.2$ into the time equation to find $a$:
$a=2b$
$a=2\left(0.2\right)=0.4$

Thus, the value of the exponent $a$ is 0.4.