Me ayudas plis noce

For each function , sketch the Bode asymptotic magnitude and asymptotic phase plots.

horrorfan [7]

Answer:

attached below

Explanation:

a) G(s) = 1 / s( s+2)(s + 4 )

Bode asymptotic magnitude and asymptotic phase plots

attached below

b) G(s) = (s+5)/(s+2)(s+4)

phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4

attached below

c) G(s)= (s+3)(s+5)/s(s+2)(s+4)

solution attached below

5 0

3 years ago

Since the engineering design process may take the engineer back to its beginning, the process is considered ________

Valentin [98]

Answer:

Cyclical

Explanation:

I looked at the next question on edgenuity and it said it in the question.

7 0

3 years ago

Substances A and B have retention times of 16.63 and 17.63 min, respectively, on a 30 cm column. An unretained species passes th

Svet_ta [14]

Answer:

The time required to elute the two species is 53.3727 min

Explanation:

Given data:

tA = retention time of A=16.63 min

tB=retention time of B=17.63 min

WA=peak of A=1.11 min

WB=peak of B=1.21 min

The mathematical expression for the resolution is:

$Re_{s} =\frac{2(t_{B}-t_{A})}{W_{A}+W_{B} } =\frac{2*(17.63-16.63)}{1.11+1.21} =0.8621$

The mathematical expression for the time to elute the two species is:

$\frac{t_{2}}{t_{1}} =(\frac{Re_{B} }{Re_{s} } )^{2}$

Here

ReB = 1.5

$t_{2} =t_{1} *(\frac{Re_{B} }{Re_{s} } )^{2} =17.63*(\frac{1.5}{0.8621} )^{2} =53.3727min$

6 0

3 years ago

A thick steel slab (rho= 7800 kg/m3 , cp= 480 J/kg K, k= 50 W/m K) is initially at 300 °C and is cooled by water jets impinging

dimaraw [331]

Answer:

t = 2244.3 sec

Explanation:

calculate the thermal diffusivity

$\alpha = \frac{k}{\rho c}$

$= \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s$

Temperature at 28 mm distance after t time = = 50 degree C

we know that

$\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})$

$\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]$

$0.909 = erf{\frac{3.8245}{\sqrt{t}}}$

from gaussian error function table , similarity variable w calculated as

erf w = 0.909

it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have

w = 0.08073

$erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]$

$0.08073 = \frac{3.8245}{\sqrt{t}}$

solving fot t we get

t = 2244.3 sec

3 0

3 years ago

How can you drop two eggs the fewest amount of times, without them breaking?

qwelly [4]

You can use like a thickish paper but not to heavy and do it spiral and fall it with bubble paper and light things and then you spin it while you drop and it won’t crack

7 0

3 years ago

Me ayudas plis noce ​

Me ayudas plis noce