Answer:
If I understand correctly. Line B is parallel to the circle. Also, the angle is less than 90.
- The size of the circle determines.
- The diameter should not be fixed either.
Hey there!
When light changes speed, it REFRACTS.
Your answer is going to be option C.
Hope this helps you.
Have a great day!
D. March because it is just below the 1 million marker on the graph and it is the only one that low.
Answer:
x component 3.88 y- component 14.488
Explanation:
We have given a vector A which has a magnitude of 15 m/sec which is at 75° counter-clock wise ( anti-clock wise) from x -axis which is clearly shown in bellow figure
Now x-component will be 15 cos75°=3.8822 ( as it makes an angle of 75° with x-axis )
y- component will be 15 sin 75°=14.488
For verification the resultant of x and y component should be equal to 15
So 
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
