| ▲ | androiddrew an hour ago | |
I kind of want to know how they safely deal with its release (destruction) | ||
| ▲ | goda90 3 minutes ago | parent | next [-] | |
Ejecting the warp core is a common strategy. | ||
| ▲ | gus_massa 21 minutes ago | parent | prev | next [-] | |
As the sibling comments say, it's a single positron so the energy released is tiny. I tried to find positrons sources in common life, the best I could find was https://en.wikipedia.org/wiki/Positron_emission_tomography The relevant paragraph are > For a typical dose (245 MBq) of FDG, one of the most common radiotracers used for PET neuroimaging and cancer patient management,[48] the effective radiation dose is 4.7 mSv.[49] and > This limitation restricts clinical PET primarily to the use of tracers labelled with fluorine-18, which has a half-life of 110 minutes and can be transported a reasonable distance before use. Where "245 MBq" means 2*10^8 positrons per second for. And "110 minutes" means a half disappears every 2 hours, so the total sum is equivalent to 4 hours but I'm missing some small constant and they ae probably averaging the dose. I guess "3 hours" is a good compromise but multiply it by 2 or 4 if you prefer a smaller or bigger number. Anyway, 3 hours are 10^4 seconds, so the total numbers of positrons in a PET is approximately 2*10^8 * 10^4 = 2*10^12. | ||
| ▲ | SwellJoe an hour ago | parent | prev | next [-] | |
It's one particle. It's not a bomb. | ||
| ▲ | Retr0id an hour ago | parent | prev | next [-] | |
If it's a single particle I don't think there's much to worry about? | ||
| ▲ | ungovernableCat 41 minutes ago | parent | prev [-] | |
E=mc*2 but m is really really tiny | ||