Select the correct answer.

Engineering

2 answers:

nata0808 [166]3 years ago

7 0

The correct answer is d

valentina_108 [34]3 years ago

6 0

D-regulatory controls

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$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,

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