Answer:
a) 0.167 μC/m^2
b) 1.887 * 10^4 V/m
Explanation:
Hello!
First let's find the surface charge density:
a)
Since thesatellite is metallic, the accumalted charge will be uniformly distribuited on its surface. Therefore the charge density σ will be:
σ = Q/A
Where A is the area of the satellite, which is:
A=4πr^2 = πd^2 = π(1.9m)^2
Therefore:
σ = (1.9)/(π (1.9)^2) μC/m^2 = 0.167 μC/m^2
Now let's calculate the electric field
b)
Just outside the surface of the satellite the elctric field will be:
E = σ/ε0
Where ε0=8.85×10^−12 C/Vm
Therefore:
E = (0.167*10^-6 C/m^2) / (8.85*10^-12 C/Vm) = 0.01887 *10^6 V/m
E = 1.887 * 10^4 V/m
This distance is known as the amplitude of the wave, and is the characteristic height of the wave, above or below the equilibrium position. Normally the symbol A is used to represent the amplitude of a wave. The SI unit of amplitude is the metre (m).
Answer:
Part a)
Part b)
Explanation:
Part a)
For force conditions of two blocks we will have
now from above equations we have
now we know that
now from above equation we have
Part b)
When heavier block is removed and F = 908 N is applied at the end of the string then we have
Answer:
Option B. Coulomb Repulsion
Explanation:
The reason for the heavy nucleus not being able to follow the trend or the N= Z line is because of the fact that as the atomic number, Z of an atom increases, the number of protons inside the atom also increases. Since the neutrons are charge less particles whereas the protons are positively charged particle and hence as these increases in number there is an increase in the repulsive force between the like charges, i.e., positively charged protons which is Coulomb repulsion.
