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:
B) 12 m
Explanation:
Gravitational potential energy is:
PE = mgh
Given PE = 5997.6 J, and m = 51 kg:
5997.6 J = (51 kg) (9.8 m/s²) h
h = 12 m
Answer:
according to the new geometry lesson alcometry which will after 11 years in geometry
The power delivered to the coil is 0.5 Watts
<h3>Resistivity of a material</h3>
The formula for calculating the power delivered to it is expressed according to the equation
Power = I²R
where
I is the current
R is the resistance
Given
Current = 0.500A
Determine the resistance using the resistivity formula
R = ρL/A
Substitute the given parameters
R = 10^-6*(25)/0.002²
R = 2ohms
Substitute
Power = I²R
Power = (0.5)²(2)
Power = 0.5watts
Hence the power delivered to the coil is 0.5 Watts
Learn more on resistivity here: brainly.com/question/17010736
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