Answer:
F = 2.30 10⁴ N
Explanation:
The force required to link two gates must be equal to or greater than the electrostatic force of repulsion, because the protons have equal charges.
F = k q₁ q₂ / r²
Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C²
In this case the proton charge is 1.6 10⁻¹⁹ C and the distance between them is approximately the diameter of the core r = 10⁻¹⁵ m
Let's calculate
F = 8.99 10⁹ (1.6 10⁻¹⁹)² / (10⁻¹⁵)²
F = 2.30 10⁴ N
The bond strength must be equal to or greater than this value
Answer:
Orbital period, T = 1.00074 years
Explanation:
It is given that,
Orbital radius of a solar system planet,
The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :
M is the mass of the sun
T = 31559467.6761 s
T = 1.00074 years
So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.
Answer:
Explanation:
Initial kinetic energy of M = 1/2 M vi²
let final velocity be vf
v² = u² + 2a s
vf² = vi² + 2 (F / M) x D
Kinetic energy
= 1/2 Mvf²
= 1/2 M ( vi² + 2 (F / M) x D
1/2 M vi² + FD
Ratio with initial value
1/2 M vi² + FD) / 1/2 M vi²
RK = 1 + FD / 2 M vi²
Answer:
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