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Blizzard [7]
3 years ago
11

Decide whether or not the Final Value Theorem is applicable to the following functions. If not, indicate why you cannot apply it

. If so, apply the Final Value Theorem. Note: you do not need to perform the inverse Laplace transform.a. X(s)= s/3s+7b. X(s)= 10/3s^2+7s+4
Engineering
1 answer:
LUCKY_DIMON [66]3 years ago
3 0

Answer:

a) Yes applicable, x(∞) = 0

b) Yes applicable, x(∞) = 0

Explanation:

Final Value Theroem indicates that final value of x(t) can be found by Laplace Transform X(s) as follows:

x(\infty)= \lim\limits_{s \to0} sX(s)

a) X(s)= \frac{s}{3s+7}

Since the pole is at left side, FVT is applicable

x(\infty)= \lim\limits_{s \to0} s\frac{s}{3s+7}=0

b) X(s)= \frac{10}{3s^2+7s+4}

\frac{10}{3s^2+7s+4}= \frac{10}{(3s+4 )(s+1)}

Since the poles are at left side, FVT is applicable

x(\infty)= \lim\limits_{s \to0} s \frac{10}{(3s+4 )(s+1)}=0

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Describe ICP/OES in detail.
alex41 [277]

Answer:

ICP -OES stand for inductively coupled plasma optical emission spectroscopy

Explanation:

It is techniques that known as trace level technique which help to identify and quantify the element present in sample by using spectra emission.

The analysis process include desolvates, ionization and excitation of the sample. The sample is identify by analyzing the emission line from it and quantify by analyzing the intensity of same emission lines.

7 0
3 years ago
Read 2 more answers
The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-k
sammy [17]

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

(s_max)² = 0.7046

s_max = 0.8394m

8 0
3 years ago
(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1
olya-2409 [2.1K]

Answer:

The answer is "4.35 \times 10^{-3}\  mm and 157.5 MPa".

Explanation:

In point A:

The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point  

y =  yo + \frac{k}{\sqrt{x}}

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.  

Replacing the equation for each condition:  

y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k

People can get yo = 275 MPa with both equations and k= 15.5 Mpa mm^{\frac{1}{2}}.

To substitute the answer,  

310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\  mm

In point b:

The equation is \sigma y = \sigma 0 + k y d^{\frac{1}{2}}

equation is:

75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\

by putting the above value in the formula we get the \sigma y value that is= 157.5 MPa

5 0
3 years ago
Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia
nekit [7.7K]

Answer:

98°C

Explanation:

Total surface area of cylindrical fin = πr² + 2πrl , r = 0.015m; l= 0.1m; π =22/7

22/7*(0.015)² + 22/7*0.015*0.1 = 7.07 X 10∧-4 + 47.1 X 10∧-4 = (54.17 X 10∧-4)m²

Temperature change, t = (50 - 25)°C = 25°C = 298K

Hence, Temperature =  150 X (54.17 X 10∧-4) X 298/123 = 242.14/124 = 2.00K =

∴ Temperature change = 2.00K

But temperature, T= (373 - 2)K = 371 K

In °C = (371 - 273)K = 98°C         

7 0
3 years ago
A 0.40-m3 insulated piston-cylinder device initially contains 1.3 kg of air at 30°C. At this state, the piston is free to move.
Setler79 [48]

Answer:

(a) The Final Temperature is 315.25 K.

(b) The amount of mass that has entered  0.5742 Kg.

(c) The work done is 56.52 kJ.

(d) The entrophy generation is 0.0398 kJ/kgK.

Explanation:

Explanation is in the following attachments.

6 0
3 years ago
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