Fidle if you give me food i will live if you give me watter i will die what im i

to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

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

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0

3 years ago

If 3.5 waves have a distance of 42M what is the wavelength of one wave?

docker41 [41]

Answer:

The answer is 12

Explanation:

Just divide 42/3.5

5 0

2 years ago

A 100 Newton force applied to a machine lifts a 400 N object. What is the actual mechanical advantage of

uranmaximum [27]

Answer:

m = 4

Explanation:

It is given that,

A 100 Newton force applied to a machine lifts a 400 N object.

We need to find the mechanical advantage of this machine.

The mechanical advantage of a machine is given by the ratio of output force to the input force.

Here, output force is 400 N and input force is 100 N

So, mechanical advantage becomes :

$m=\dfrac{400}{100}\\\\m=4$

So, the mechanical advantage of this machine is 4.

7 0

3 years ago

If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the c

Lemur [1.5K]

Answer:

$\mu_s \geq 0.27$

Explanation:

The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

$f_s=\frac{mv^{2}}{R}$

But we know that:

$f_s\leq \mu_s N$

And the normal force is given by the sum of the forces in the vertical direction:

$N-mg=0 \implies N=mg$

Finally, we have:

$f_s=\frac{mv^{2}}{R} \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR} \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27$

So, the minimum value for the coefficient of friction is 0.27.

4 0

3 years ago

NEED HELP FROM SCIENCE GENIUS ! POINTS WILL BE GIVEN AND DO GIVE ME A DECENT ANSWER !!!

Ray Of Light [21]

Answer:

IT IS HOSPITAL OR AMBULANCE

SOME THING LIKE THAT IS WRITTEN ZOOM THAT PHOTO MORE

3 0

2 years ago

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