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
Answer:
Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.
Explanation:
The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is
K= constant depends on the system of units used.
W= weight =485 g
D= density =7.9 g/cm³
A = exposed specimen area =100 in² =6.452 cm²
K=534 to give CPR in mpy
K=87.6 to give CPR in mm/yr
mpy
=37.4mpy
mm/yr
=0.952 mm/yr
Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.
It is an example of balanced force.
hope this helps. good luck
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