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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.

Engineering
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 overhead 25m-long, uninsulated industrial steam pipe of 100-mm diameter, is routed through a building whose walls and air are
wariber [46]

Answer:

1) q=18414.93 W

2) C=12920$

Explanation:

Given data:

pipe length L=25m

pipe diameter D=100mm =0.1 m

air temperature T_{s1}=T_{\infty1} }=25 °C.....= 298.15k

pipe surface temp T_{s2}=150 °C.....=423.15k

surface emissivity e= 0.8

boiler efficiency η=0.90

natural gas price Cg=$0.02 per MJ

1) Total heat loss and radiation heat loss combined

          q=q_{conv} +q_{rad}

          q=A[h(T_{s2}-T_{s1})+eб(T_{s2}^4-T_{s1}^4)]....... (1)

б=5.67×10^-8 W/m^2K^4 (boltzmann constant)

area A =L.Dπ=25×0.1π=7.85 m^2

putting all these values in eq (1)

q=18414.93 W

2) suppose boiler is operating non stop annual energy loss will be

               E=q.t

                  =18414.93.3600.24.365

                  =5.81×10^11 J

   to find furnace energy consumption

               Ef =E/η

                  =6.46×10^5 MJ

   annual cost

                  C=Cg. Ef

                    =12920$

8 0
3 years ago
The rectangular frame is composed of four perimeter two-force members and two cables AC and BD which are incapable of supporting
Rama09 [41]

Answer:

Your question is lacking some information attached is the missing part and the solution

A) AB = AD = BD = 0, BC = LC

    AC = \frac{5L}{3}T, CD = \frac{4L}{3} C

B) AB = AD = BC = BD = 0

   AC = \frac{5L}{3} T, CD = \frac{4L}{3} C

Explanation:

A) Forces in all members due to the load L in position A

assuming that BD goes slack from an inspection of Joint B

AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C

B) steps to arrive to the answer is attached below

AB = AD = BC = BD = 0

AC = \frac{5L}{3} T,  CD = \frac{4L}{3}C

7 0
4 years ago
A sample of semiconductor has a cross sectional area of 1cm^2 and a thickness of 0.1cm.
pogonyaev

Answer:

a. 3.17*10¹⁹ b. 3.17*10¹⁴

Explanation:

a. Area A = 1cm²

Thickness h = 0.1cm

Energy of photon E = hf=hc/λ

Where f = frequency; c=speed of light; λ=wavelength = 6300Α=6300*10⁻¹⁰;

Planck's constant h = 6.626*10⁻³⁴ joule-seconds; speed of light c = 3*10⁸

Therefore E = (6.626*10⁻³⁴)*3*10⁸/6300*10⁻¹⁰=3.155*10¹⁹J

1Watt of light releases 3.17*10¹⁸ photons per second.

Volume of sample = Area * Thickness = 1*0.1=0.1cm³

Therefore, number of electron hole pairs that are generated per unit volume per unit time = 3.17*10¹⁸/0.1 = 3.17*10¹⁹photon/cm³-s

b. Steady state excess carrier concentration = 3.17*10¹⁹*10μs

=3.17*10¹⁹*10*10⁻⁶

=3.17*10¹⁴/cm³

6 0
3 years ago
5 Systems Modeling
Nesterboy [21]

Answer: None of the above

Explanation:

Business process modeling refers to the graphical representation of the business processes of a company, which is vital in the identification of potential improvements.

Business pticess modelling can be done through graphing methods, like data-flow diagram, flowchart etc. It is vital as business managers can effectively and quickly communicate their ideas.

It also enhances the customization of business processes, enhances the competitive advantage and enhances the process communication as well.

Therefore, the answer to the question will be "None of the above".

5 0
3 years ago
In a parallel one-dimensional flow in the positive x direction, the velocity varies linearly from zero at y = 0 to 32 m/s at y =
monitta

Answer:

Ψ = 10(y^2) + c

<em><u>y = 1.067m</u></em>

Explanation:

since the flow is one dimensional in positive X direction, the only velocity component is in X, which is denoted by u

while u is a function of y

we find the u in terms of y; u varies linearly wih y

we use similiraty to find the relation

32/1.6 =<em>u/y</em>

<em><u>u = 20y</u></em>

<em><u>Ψ = ∫20ydy</u></em>

<em><u>Ψ = 10(y^2) + c</u></em>

<em><u>(b)</u></em>

<em><u>the flow is half below y = 1.6*(2/3)=1.067 m</u></em>

<em><u>this is because at two third of the height of a triangle lies the centroid of triangle. since the velocity profile forms a right angled triangle , its height is 1.6 m . the flow is halved at y = 1.067m</u></em>

3 0
3 years ago
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