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1 answer:
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Answer:
Zero
Explanation:
The work done by a force on an object is given by:
where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and the displacement of the object
In this situation, the force is the force of gravity acting on the satellite. This force always points towards the centre of the trajectory, so it is always perpendicular to the direction of motion of the satellite (since the orbit is circular), so and
. Therefore, the work done by gravity is also zero.
Answer:
If we square both sides we got:
We divide both sides by and we got:
Now we can apply log on both sides and we got:
And solving for n we got:
And replacing we got:
And since n needs to be an integer the correct answer would be n=2 for the filter order.
Explanation:
For this case we can use the formula for the Butterworth filter gain given by:
[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]
Where:
G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value
represent the corner frequency
represent the original frequency
n represent the filter order and that's the variable that we need to find
If we square both sides we got:
We divide both sides by and we got:
Now we can apply log on both sides and we got:
And solving for n we got:
And replacing we got:
And since n needs to be an integer the correct answer would be n=2 for the filter order.