Antimatter doesn’t really do anything by it’s own, but if we let 1 kg react with 1 kg of matter (non-anti-matter), we get E = mc^2 with m = 2 kg. So 1.810^17 J, or 1.810^11 MJ. If we assume that 10 MJ/kg is represented by about 1 cm, the bar would have to be 1.810^10 cm or about 1.810^8 m. A standard A4 piece of paper is about 30 cm tall, so 6.0*10^8 A4 papers are needed. I.e. 600 million papers.
So we definitely have enough paper, but it would be a very tall stack.
dQw4w9WgXcQ@lemm.ee 1 week ago
Antimatter doesn’t really do anything by it’s own, but if we let 1 kg react with 1 kg of matter (non-anti-matter), we get E = mc^2 with m = 2 kg. So 1.810^17 J, or 1.810^11 MJ. If we assume that 10 MJ/kg is represented by about 1 cm, the bar would have to be 1.810^10 cm or about 1.810^8 m. A standard A4 piece of paper is about 30 cm tall, so 6.0*10^8 A4 papers are needed. I.e. 600 million papers.
So we definitely have enough paper, but it would be a very tall stack.
Lv_InSaNe_vL@lemmy.world 1 week ago
That’s only about 180,000km (~112,000 miles) or just under half way to the moon.