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katovenus [111]

3 years ago

5

Here you will program the Newton-Raphson root finding method with details specified as fol-lows. Your code will continue perform

ing iterations of this root finding method until it hasfound the root within the given tolerance. The function should return when it has found aroot for whichf(rt)is withintolof 0 (inclusive), or aftermaxiteriterations have beenperformed.

1 answer:

babymother [125]3 years ago

3 0

Answer:

function [rt,n_iter]=newtonsMethod(f,df,x0,tol,max_iter)

rt=x0;

if(abs(f(rt))<tol)

n_iter=0;

return;

end

for i=1:max_iter

rt=x0-f(x0)/df(x0);

if(abs(f(rt))<tol)

break;

end

x0=rt;

end

n_iter=i;

end

Explanation:

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An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formatio

saveliy_v [14]

Answer:

The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

Explanation:

Given that,

Thickness of A = 8.0 m

Conductivity = 25.0 m/d

Thickness of B = 2.0 m

Conductivity = 142 m/d

Thickness of C = 34 m

Conductivity = 40 m/d

We need to calculate the horizontal conductivity

Using formula of horizontal conductivity

$K_{H}=\dfrac{H_{A}K_{A}+H_{A}K_{A}+H_{A}K_{A}}{H_{A}+H_{B}+H_{C}}$

Put the value into the formula

$K_{H}=\dfrac{8.0\times25+2,0\times142+34\times40}{8.0+2.0+34}$

$K_{H}=41.9\ m/d$

We need to calculate the vertical conductivity

Using formula of vertical conductivity

$K_{V}=\dfrac{H_{A}+H_{B}+H_{C}}{\dfrac{H_{A}}{K_{A}}+\dfrac{H_{B}}{K_{B}}+\dfrac{H_{C}}{K_{C}}}$

Put the value into the formula

$K_{V}=\dfrac{8.0+2.0+34}{\dfrac{8.0}{25}+\dfrac{2.0}{142}+\dfrac{34}{40}}$

$K_{V}=37.2\ m/d$

Hence, The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

3 0

3 years ago

For some transformation having kinetics that obey the Avrami equation (Equation 11.17), the parameter n is known to have a value

Nastasia [14]

Explanation:

Below is an attachment containing the solution.

5 0

3 years ago

Discuss with your neighbor your brain as a computer.

Misha Larkins [42]

Senors are a type of device that produce a amount of change to the output to a known input stimulus.
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6 0

4 years ago

A particular NMOS device has parameters VT N = 0.6 V, L = 0.8µm, tox = 200 Å, and µn = 600 cm2 /V–s. A drain current of ID = 1.2

NeTakaya

Answer:

$W= 3.22 \mu m$

Explanation:

the transistor In saturation drain current region is given by:

$i_D}=K_a(V_{GS}-V_{IN})^2$

Making $K_a$ the subject of the formula; we have:

$K_a=\frac {i_D} {(V_{GS} - V_{IN})^2}$

where;

$i_D = 1.2m$

$V_{GS}= 3.0V$

$V_{TN} = 0.6 V$

$K_a=\frac {1.2m} {(3.0 - 0.6)^2}$

$K_a = 208.3 \mu A/V^2$

Also;

$k'_n}=\frac{\mu n (\frac{cm^2}{V-s} ) \epsilon _{ox}(\frac{F}{cm} ) }{t_{ox}(cm)}$

where:

$\mu n (\frac{cm^2}{V-s} ) = 600$

$\epsilon _{ox}=3.9*8.85*10^{-14}$

${t_{ox}(cm)=200*10^{-8}$

substituting our values; we have:

$k'_n}=\frac{(600)(3.988.85*10^{-14})}{(200*10^{-8})}$

$k'_n}=103.545 \mu A/V^2$

Finally, the width can be calculated by using the formula:

$W= \frac{2LK_n}{k'n}$

where;

L = $0.8 \mu m$

$W= \frac{2*0.8 \mu m *208.3 \mu}{103.545 \mu}$

$W= 3.22 \mu m$

4 0

4 years ago

A 100-horsepower, three-phase squirrel-cage induction motor is connected to a 240-volt line. A dual-element time-delay fuse is t

antoniya [11.8K]

Based on the National Electrical Code (NEC), a 450-Ampere fuse should be used to protect this polyphase (three-phase) squirrel cage inductor motor.

<u>Given the following data:</u>

Power rating = 100-horsepower.

Voltage = 240 volt.

<h3>How to determine the correct fuse rating?</h3>

According to Table 430.52 of the National Electrical Code (NEC), a dual-element time delay fuse should be calculated at 175% (1.75) of the full-load current rating for an alternating current (AC) polyphase (three-phase) squirrel cage inductor motor.

In this scenario, the squirrel cage inductor motor didn't list a NEMA design code on its nameplate. Thus, we'll assume that the inductor motor is design B and its fuse rating is given by:

Fuse rating = 248 × 1.75 × 434

In conclusion, the nearest standard fuse size which is above the computed value listed in Section 240.6 of the National Electrical Code (NEC) is 450 amperes.

Read more on National Electrical Code here: brainly.com/question/10619436

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6 0

2 years ago