Considering the deuterium-tritium fusion reaction with the tritium nucleus at rest: ¹₂H + ¹₃H → ²₄He + ⁰₁n the electric potential energy (in electron volts) at this distance is 17.58MeV
<h3>How is the electric potential energy of deuterium-tritium fusion reaction calculated?</h3>
The reaction is ¹₂H + 1₃H → ²₄He + ⁰₁n
Value of Q = (Mass of ¹₂H + Mass of ¹₃H - Mass of ²₄He- Mass of n) x 931 MeV
Mass of ¹₂H = 2.014102
Mass of ¹₃H = 3.016049
Mass of ²₄He = 4.002603
Mass of n = 1.00867
Therefore Value of Q = [2.014102+3.016049−4.002603−1.00867] × 931 MeV
Therefore Value of Q = 0.01887 × 931 MeV
= 17.58MeV
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Answer:
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Static frictional force = ƒs = (Cs) • (Fɴ)
2.26 = (Cs) • m • g
2.26 = (Cs) • (1.85) • (9.8)
Cs = 0.125
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
1.49 = (Cκ) • m • g
1.49 = (Cκ) • (1.85) • (9.8)
Cκ = 0.0822
Answer:
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Explanation:
