Why is Uranium-235 explosive?
\(v_{boom}=3.4354*\cfrac{19100}{92}=713.22\,ms^{-1}\)
Given sonic boom at \(v_{sonic}=342.53\,ms^{-1}\) from the post "Sonic Boom Updated" dated 30 Oct 2017.
\(\cfrac{v_{boom}}{v_{sonic}}=\cfrac{713.22}{342.53}=2.08\)
After the detonation, the explosive front moves off at sonic boom speed that is twice less than \(v_{boom}\) of Uranium-235, two head on collisions at \(342.53\,ms^{-1}\) can generate enough \(KE\) (\(2\times342.53\,ms^{-1}\)) and trigger nuclear disintegration of the Uranium. A-bomb is destructive because of this moving nuclear front at sonic boom speed. Otherwise a bomb is destructive after the release of heat and pressure built up just after detonation. The material does no continue to disintegrate as it is thrown from the explosion. The built up of heat and pressure is limited by the bomb's containment. Any weakness in the containment will reduce its destructiveness. The destructiveness of an A-bomb on the other hand, depends on it attaining \(v_{boom}\) that is half of the sonic boom. Prevent \(v_{boom}\) and the A-bomb is no more dangerous than boiling iron at \(3635.16\,K\). If the A-bomb is further contained and the nuclear material is prevented from attaining \(v_{boom}\), its destructiveness will be reduced.
Maybe... and thankfully \(^{235}U\) is difficult to obtain.
Which might lead us to water with \(v_{boom}=v_{sonic}\). A sonic boom is totally destructive, very loud (ie high pressure) and very hot, but not radioactive thereafter and not usually sustained.
Tabooo....
