Un objeto se suelta desde determinada altura y emplea un tiempo t en caer al suelo. Si se cuadruplica la altura desde la cual se
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1) First of all, since we know the radius of the wire (
), we can calculate its cross-sectional area
2) Then, we can calculate the current density J inside the wire. Since we know the current,
, and the area calculated at the previous step, we have
3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity
of the aluminium, the electric field is given by
1) First of all, since we know the radius of the wire (
2) Then, we can calculate the current density J inside the wire. Since we know the current,
3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity
Answer:
mass consumed by 235U each day = 2 kg
Explanation:
electrical power produced = 1 GW = 1 × 10⁹ × (6.24151 × 10¹⁸ ) eV
= 6.24151× 10²¹ MeV/s
thermal energy = 0.420 * 250 = 105 MeV
= 5.94 × 10¹⁹ fission/second
=5.94 × 10¹⁹× 24 × 60 ×60)
= 5.13 × 10²⁴ fission/day
mu = 235.04393 × 1.660× 10 ⁻²⁷ = 390.1729× 10⁻²⁷ Kg
M = mu ×5.13 × 10²⁴
= 390.1729× 10⁻²⁷ ×5.13 × 10²⁴
M = 2 kg(approx.)
mass consumed by 235U each day = 2 kg