The Tevatron at Fermilab accelerates protons to 0.6 TeVkinetic energy. What's this in joules? How far would a 6 g mass have to fall near Earth's surface to gain this much energy?
given
KE = 0.6 TeV
we have 1 eV = 1.6 X 10-19 J
KE = 0.6 X 1012 X 1.6 X 10-19
KE = 9.6 X 10-8 J
now
PE = KE
m g d = 9.6 X 10-28
6 X 10-3 X 9.8 X d = 9.6 X 10-8
d = 9.6 X 10-28 / 6 X 10-3 X 9.8
d = 1.632 X 10-6 m
or
d = 1.632 m
so it should be far of d = 1.632 m
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