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:
the charge of the particle is 2.47 x 10⁻¹⁹ C
Explanation:
Given;
mass of the particle, m = 6.64 x 10⁻²⁷ kg
velocity of the particle, v = 8.7 x 10⁵ m/s
strength of the magnetic field, B = 1.3 T
radius of the circle, r = 18 mm = 1.8 x 10⁻³ m
The magnetic force experienced by the charge is calculated as;
F = ma = qvB
where;
q is the charge of the particle
a is the acceleration of the charge in the circular path
Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C
This is simply F=ma so 70N/30m/s^2 will give you the max mass which would be in kg, and the mass would be 2.333333kg a very light plane I might say
Answer:
d) What is the force if we doubled both the masses AND we doubled the distance
In the field of electromagnetism, when two charged plates that are situated opposite to each other by a certain distance, it forms an energy called the electric field. This energy is due to the difference in potential energy with respect to distance. Thus,
E = V/d
However, the voltage in volts is energy per coulomb. Thus,
V = (8x10-17 J/electron)*(1electron/1.60218x10^-19 C)
V = 499.32 volts
Therefore,
E = 499.32 volts /2.5 m
E = 199.73 N/C
The electric field that caused the change in potential energy is equal to 199.73 Newtons per Coulomb.
