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3 years ago

What is "Engineering"?

1 answer:

5 0

Engineering is when both the application of science and math are used to solve problems. An engineer is a person who designs, builds, or maintains engines, machines, or public works. There are six large branches of engineering such as Mechanical, Chemical, Civil, Electrical, Management, and Geotechnical, they all have hundreds of subcategories of engineering under each different branch.

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Create a function (prob3_5) that will take inputs of vectors x and y in feet, scalar N, scalars L and W in feet and scalars T1 a

Shalnov [3]

Answer:

clear, clc

prob3_5([1,2,3],[6,5,7],12,11,22,55,76)

function T=prob3_5(x,y,N,L,W,T1,T2)

w=zeros(1,length(x));

for n=1:2:N

for i=1:length(x)

w(i)=w(i)+(2/pi)*(2/n)*sin(n*pi*x(i)/L).*sinh(n*pi*y(i)/L)/sinh(n*pi*W/L);

end

T=(T2-T1)*w+T1;

end

Explanation:

Please input the commands into MATLAB

3 0

3 years ago

liq [111]

Answer:

Explanation:

3 0

4 years ago

Which option distinguishes the step in the engineering design phase described in the following scenario?

rewona [7]

Answer:

reasearching the problem

Explanation:

EDG

7 0

3 years ago

While walking across campus one windy day, an engineering student speculates about using an umbrella as a "sail" to propel a bic

makvit [3.9K]

Answer:

$Given data:\\While walking across campus one windy day\\Frontal area, $A=0.3 m ^{2}$\\Wind speed $V=24 Km / hr$\\The drag coefficient $C_{D, b}=1.2$\\The combined mass $m=75 kg$\\Umbrella diameter, $D=1.22 m$\\Velocity of wind $V=24 \frac{ km }{ hr }$\\The rolling resistance $C_{R}=0.75 \%$$

Solution:

Note: Refer the diagram

$Basic equation:\\'s law of motion: $\sum F_{x}=m a_{x}$\\Lift coefficient, $C_{L}=\frac{F_{L}}{\frac{1}{2} \rho V^{2} A_{p}}$\\Drag coefficient, $C_{D}=\frac{F_{D}}{\frac{1}{2} \rho V^{2} A_{p}}$$

$From force balance equation:\\$\sum F_{x}=F_{D}-F_{R}=0$\\But $F_{D}=\left(C_{D, \alpha} A_{u}+C_{D, B} A_{b}\right) \frac{1}{2} \rho\left(V_{\nu}-V_{b}\right)^{2}$$F_{R}=C_{R} m g$\\Area of the Umbrella $A_{u}=\frac{\pi D_{u}^{2}}{4}$$A_{x}=\frac{\pi \times 1.22^{2}}{4} m ^{2}$$A_{v}=1.17 m ^{2}$$

Drag coefficient data for selected objects table at

Hemisphere (open end facing flow), $C_{D, x}=1.42$

Substituting all parameters,

$\begin{aligned}&F_{R}=0.0075 \times 75 \times 9.81\\&F_{R}=5.52 N\end{aligned}$

Then,

$\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}$

And the equation becomes,

$\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}$

Thus the floyds travels at $68.3^{\circ}$ wind speed.

7 0

4 years ago

Routers cannot be used to cut through material.<br><br> True<br> False

mario62 [17]

Answer:

Yes a router can be used to cut right through wood and sometimes it makes sense to do so. It leaves nice clean edges, can cut sharp curves and can follow a template

Explanation:

hope thats right

7 0

3 years ago

Read 2 more answers