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lord [1]
4 years ago
5

A second order system is to be subjected to inputs below 80Hz and is to operate with an amplitude response of ±5 percent. Calcul

ate the minimum value of natural frequency ωn, on to accomplish this goal.
Engineering
1 answer:
NeTakaya4 years ago
5 0

Answer:

So natural frequency will be 958.9616 rad/sec    

Explanation:

We have given amplitude response  \pm 5 %

Damped frequency f =80 Hz

So  \omega _d=2\pi\ f=2\times 3.14\times 80=502.4rad/sec

So e^{\frac{-\pi \zeta }{\sqrt{1-\zeta ^2}}}=0.05

We know that \zeta =cos\Theta

So \sqrt{1-\zeta ^2}=sin\Theta

So e^{-\pi cot\Theta }=0.05

{-\pi cot\Theta }=-3

cot\Theta =0.954

\Theta =46.34

cos\Theta =\zeta =0.690

Now we know that \omega _d=\omega _n\sqrt{1-\zeta ^2}

\omega _n=\frac{\omega _d}{\sqrt{1-\zeta ^2}}=\frac{502.4}{\sqrt{1-0.690^2}}=958.9616rad/sec

So natural frequency will be 958.9616 rad/sec

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An organization is struggling to differentiate threats from normal traffic and access to systems. A security engineer has been a
AnnZ [28]

Answer:

Web application Firewall (WAF)

Explanation:

The Web application Firewall (WAF) will be recommended. This will enormously help any organisation that is having a struggle of differentiating threats from normal traffic and access to systems.

WAF is also going to It aggregate data and provide metrics that will assist in identifying malicious actors. Another vital function of WAF will be to block unwanted web traffic from accessing your site. It will protect against hacks, brute force attacks, DDoS attacks, cross-site scripting, SQL injection, and zero-day exploits.

8 0
4 years ago
A rod is made from two segments: AB is steel and BC is brass. It is fixed at its ends and subjected to a torque of T = 680 N.m.
Ymorist [56]

Answer:

diameter of brass portion is 42.6 mm.

Explanation:

see the attached file

6 0
4 years ago
Which of the following is true about how the universe is expanding?
Law Incorporation [45]
B. The space between the galaxy is getting better
4 0
4 years ago
Read 2 more answers
I am trying to create a line of code to calculate distance between two points. (distance=[tex]\sqrt{ (x2-x1)^2+(y2-y1)^2}) My li
k0ka [10]

Answer:

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Explanation:

The distance formula is the difference of the x coordinates squared, plus the difference of the y coordinates squared, all square rooted.  For the general case, it appears you simply need to change how you have written the code.

point_dist = math.sqrt((math.pow(x2 - x1, 2) + math.pow(y2 - y1, 2))

Note, by moving the 2 inside of the pow function, you have provided the second argument that it is requesting.

You were close with your initial attempt, you just had a parenthesis after x1 and y1 when you should not have.

Cheers.

6 0
3 years ago
The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas
Lilit [14]

Answer:

A) 209.12 GPa

B) 105.41 GPa

Explanation:

We are given;

Modulus of elasticity of the metal; E_m = 67 GPa

Modulus of elasticity of the oxide; E_f = 390 GPa

Composition of oxide particles; V_f = 44% = 0.44

A) Formula for upper bound modulus of elasticity is given as;

E = E_m(1 - V_f) + (E_f × V_f)

Plugging in the relevant values gives;

E = (67(1 - 0.44)) + (390 × 0.44)

E = 209.12 GPa

B) Formula for upper bound modulus of elasticity is given as;

E = 1/[(V_f/E_f) + (1 - V_f)/E_m]

Plugging in the relevant values;

E = 1/((0.44/390) + ((1 - 0.44)/67))

E = 105.41 GPa

4 0
3 years ago
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