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
For the answer to the question above, first find out the gradient.
<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>
<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>
<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>
<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>
<span>y is the velocity of the car, x is the time.
</span>I hope this helps.
Answer:
The sun touches earth during daytime and the suns rays heat our earth giving us heat. The sun heating the earth is also considered radiation.
Explanation:
Answer:
B = 0.129 T
Explanation:
Given,
frequency, f = 60 Hz
maximum emf = 5200 V
Number of turns, N = 130
Area per turn = 0.82 m²
We know,
ω = 2 π f
ω = 2 π x 60 = 376.99 rad/s
now, Magnetic field calculation
B = 0.129 T
Hence, the magnetic field is equal to B = 0.129 T
