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Ganezh [65]
3 years ago
8

Write a function separatethem that will receive one input argument which is a structure containing fields named length and width

, and will return the two values separately. Here is an example of calling the function >> myrectangle = struct('length',33,'width',2); You just created a structure myrectangle with the above line of code. Now, if you send myrectangle into separatethem function, it should access the length and width fields and store it in output variables l and w (Again, no tips or tricks! do not overthink on this one. Just access the fields length and width from myrectangle input argument and store it in output variables l and w >> [l w] = separatethem(myrectangle) l = 33 w = 2
Engineering
1 answer:
SpyIntel [72]3 years ago
6 0

Answer:

%seperatethem.m

%Creation and definition of functions

function [l w] = separatethem(myrectangle)

%Call the structure field length.

l = myrectangle.length(1);

%Call the structure field length.

w = myrectangle.width(1);

%Ending call function.

end

%trial of program

%Create a structure with two fields length and width.

myrectangle = struct('length',33,'width',2);

%Call the function separate them.

[l w] = separatethem(myrectangle)

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Answer:

True, <em>Regeneration is the only process where increases the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than working fluid leaving the compressor</em>.

Option: A

<u>Explanation: </u>

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3 years ago
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A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

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load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

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7 0
3 years ago
A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f
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Answer:

Head loss is 1.64

Explanation:

Given data:

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Discharge (Q) = 0.16 m3/s

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We know, head\ loss  = \frac{f L V^2}{( 2 g D)}

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

V = \frac{Q}{A}

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obtained Darcy friction factor  

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Re = \frac{\rho V D}{\mu}

where,\rho = density of water

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Answer:

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