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
The answer is C. You must divide your wavelength and your frequency to get your answer.
Answer:
v = 384km/min
Explanation:
In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.
You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:
You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:
hence, the speed of the Hubble is approximately 384km/min
It depends on what type of solid
Answer:
I would say A but am not sure
