Edwin Hubble and Albert Einstein.
Electrons would be found outside the nucleus protons and neutrons are on the inside of the nucleus.
Answer:
E = 4.72 * 10⁻⁶ Nm²
Explanation:
Parameters given:
Outer radius, R = 3.70cm = 0.037m
Inner radius, r = 3.15cm = 0.0315m
Permittivity of free space, ε₀ = 8.85 * 10⁻¹² C/Nm²
Charge density: 1.22 * 10⁻³ C/m³
The question requires that we solve using Gauss law which states that the net electric field through a closed surface is proportional to the enclosed electric charge.
Hence,
E = Q/Aε₀
Charge Q is given as
Q = ρπ(R² ⁻ r²)L
A = 2π(R - r)L
E = [ρπ(R² ⁻ r²)L]/[2π(R - r)ε₀L]
Using difference of two squares,
(R² ⁻ r²) = (R + r)(R - r)
E =[ρ(R + r)]/(2ε₀)
E = [1.22 * 10⁻³ *(0.0370 + 0.0315)]/(2 * 8.85 * 10⁻¹²)
E = 4.72 * 10⁻⁶ Nm²
Answer:
Velocity
Explanation:
The gradient= change in velocity
——————————
Change in time
45-4753that is your answer
