6 points) Assume that 5.364x 1027MeV are needed to power the submarine for a 1-yearvoyage. Based on your result from Part A, how much 235U (in kilograms) isneeded for the submarine reactor to produce this much energy.

a person with a mass of 106kg has a weight of 1060n

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

this is a good question! i'll do some research and come back when i have the answer!

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j