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Hunter-Best [27]
3 years ago
11

Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's

numbers on one line, each number followed by one space. (2 pts)
Ex:Enter weight 1:236.0Enter weight 2:89.5Enter weight 3:142.0Enter weight 4:166.3Enter weight 5:93.0You entered: 236.000000 89.500000 142.000000 166.300000 93.000000
Engineering
1 answer:
atroni [7]3 years ago
6 0

Answer:

See below in the explanation section the Matlab script to solve the problem.

Explanation:

prompt='enter the first weight w1:  ';

w1=input(prompt);

wd1=double(w1);

prompt='enter the second weight w2:  ';

w2=input(prompt);

wd2=double(w2);

prompt='enter the third weight w3:  ';

w3=input(prompt);

wd3=double(w3);

prompt='enter the fourth weight w4:  ';

w4=input(prompt);

wd4=double(w4);

prompt='enter the first weight w5:  ';

w5=input(prompt);

wd5=double(w5);

x=[wd1 wd2 wd3 wd4 wd5]

format short

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zvonat [6]

Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

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3 years ago
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Answer:

The answer is copying a program to give to someone else to use

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Answer:

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Explanation:

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The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

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The length is:

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The resistance is:

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hope it helps

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