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An explosion produces a spherical shock wave whose radius \(R\) expands rapidly. The rate of expansion depends on the energy \(E\) of the explosion and the elapsed time \(t\) since the explosion. For many explosions, the relation is approximated closely by $$ R=4.16 E^{0.2} t^{0.4} . $$ Here \(R\) is the radius in centimeters, \(E\) is the energy in ergs, and \(t\) is the elapsed time in seconds. The relation is valid only for very brief periods of time, perhaps a second or so in duration. a. An explosion of 50 pounds of TNT produces an energy of about \(10^{15}\) ergs. See Figure \(2.85\). How long is required for the shock wave to reach a point 40 meters ( 4000 centimeters) away? b. A nuclear explosion releases much more energy than conventional explosions. A small nuclear device of yield 1 kiloton releases approximately \(9 \times 10^{20}\) ergs. How long would it take for the shock wave from such an explosion to reach a point 40 meters away? c. The shock wave from a certain explosion reaches a point 50 meters away in \(1.2\) seconds. How much energy was released by the explosion? The values of \(E\) in parts a and b may help you set an appropriate window. Note: In 1947 the government released film of the first nuclear explosion in 1945 , but the yield of the explosion remained classified. Sir Geoffrey Taylor used the film to determine the rate of expansion of the shock wave and so was able to publish a scientific paper \({ }^{26}\) concluding correctly that the yield was in the 20 -kiloton range.

Step by step solution

01

Understanding the Relationship

The formula given is \( R = 4.16 E^{0.2} t^{0.4} \), where \( R \) is in centimeters, \( E \) is in ergs, and \( t \) is in seconds. This means that to find the time \( t \), we need to solve for \( t \) given specific values for \( R \) and \( E \).

02

Solve Part (a)

Given \( E = 10^{15} \) ergs for 50 pounds of TNT and \( R = 4000 \) cm, substitute these values into the equation. You'll have: \[ 4000 = 4.16 (10^{15})^{0.2} t^{0.4} \]First, calculate \( E^{0.2} \):\( (10^{15})^{0.2} = 10^{3} = 1000 \). Substitute this into the equation:\[ 4000 = 4.16 \times 1000 \times t^{0.4} \]Simplify to find \( t^{0.4} \):\[ 4000 = 4160 \times t^{0.4} \]\[ \frac{4000}{4160} = t^{0.4} \]\[ t^{0.4} = 0.9615 \]Solve for \( t \):\( t = (0.9615)^{2.5} \approx 0.889 \) seconds.

03

Solve Part (b)

For a nuclear device with \( E = 9 \times 10^{20} \) ergs, use \( R = 4000 \) cm. Plug them into the formula:\[ 4000 = 4.16 (9 \times 10^{20})^{0.2} t^{0.4} \]Calculate \( (9 \times 10^{20})^{0.2} \):\[ (9)^{0.2} \times (10^{20})^{0.2} = 9^{0.2} \times 10^{4} \approx 1.52 \times 10000 \]Substituting back:\[ 4000 = 4.16 \times 15200 \times t^{0.4} \]Isolating \( t^{0.4} \):\[ \frac{4000}{63232} = t^{0.4} \]\[ t^{0.4} \approx 0.0632 \]\( t = (0.0632)^{2.5} \approx 0.025 \) seconds.

04

Solve Part (c)

For \( R = 5000 \) cm in \( t = 1.2 \) seconds, find \( E \):\[ 5000 = 4.16 E^{0.2} (1.2)^{0.4} \]Calculate \( 1.2^{0.4} \approx 1.083 \), then:\[ 5000 = 4.16 \times E^{0.2} \times 1.083 \]Simplifying:\[ \frac{5000}{4.50888} = E^{0.2} \]\[ E^{0.2} \approx 1108 \]Solve for \( E \):\( E = 1108^{5} \approx 1.68 \times 10^{15} \) ergs.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Explosion Energy

The energy from an explosion is released suddenly, resulting in a rapid expansion of air and creation of a shock wave. This energy, often measured in ergs or similar units, dictates the nature and power of the explosion. To put it into perspective, an explosion of 50 pounds of TNT releases about \(10^{15}\) ergs. On a larger scale, nuclear explosions release significantly more energy, like a 1 kiloton nuclear device releasing approximately \(9 \times 10^{20}\) ergs. Understanding the energy released can help predict the behavior of the shock wave it generates.

Spherical Shock Wave

A spherical shock wave is a powerful wave that travels outward in all directions from the center of an explosion. As it moves, it exerts a high pressure on its surroundings, diminishing in strength with distance. While the shock wave travels quickly initially, it slows over time due to energy dissipation and increased radius. Important parameters are the radius \( R \) and the speed at which this radius expands, both impacted by the explosion's energy and the time elapsed.

Mathematical Modeling

Mathematical modeling is essential in understanding and predicting the behavior of shock waves. Using the equation \( R = 4.16 E^{0.2} t^{0.4} \), we can relate the radius of the shock wave to the energy \( E \) of the explosion and the time \( t \) since the explosion occurred. Through such equations, we can predict the expansion rate and other details of the shock wave鈥檚 growth. Mathematical models are powerful tools in physics, allowing scientists and engineers to simulate explosions without the need for dangerous, real-world experiments.

Expansion Rate

The expansion rate of a shock wave is crucial in understanding how far and how fast a shock can travel. It is governed by both the energy of the explosion and the time elapsed since the explosion. The formula \( R = 4.16 E^{0.2} t^{0.4} \) shows that the radius grows as a function of energy raised to the power of 0.2 and time to the power of 0.4. This mathematical relationship helps predict how quickly the shock wave will reach certain distances, like in the exercise demonstrating how quickly such a wave can reach 40 meters under different energy scenarios.