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aleksandr82 [10.1K]
3 years ago
5

Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by

a minus sign. Leave no cells blank - be certain to enter "0" wherever required. Omit the "$" sign in your response.)

Engineering
1 answer:
vladimir2022 [97]3 years ago
4 0

Answer:

The answer for the question :

"Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by a minus sign. Leave no cells blank - be certain to enter "0" wherever required. Omit the "$" sign in your response.)"

is explained in the attachment.

Explanation:

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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
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