If he feels the chances of low, normal, and high precipitation are 30 percent, 20 percent, and 50 percent respectively, What is
1 answer:
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Answer:
The frequency of the oscillation is 2.45 Hz.
Explanation:
Given;
mass of the spring, m = 0.5 kg
total mechanical energy of the spring, E = 12 J
Determine the spring constant, k as follows;
E = ¹/₂kA²
kA² = 2E
k = (2E) / (A²)
k = (2 x 12) / (0.45²)
k = 118.519 N/m
Determine the angular frequency, ω;
Determine the frequency of the oscillation;
ω = 2πf
f = (ω) / (2π)
f = (15.396) / (2π)
f = 2.45 Hz
Therefore, the frequency of the oscillation is 2.45 Hz.
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.