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# A medical technician is trying to determine what percentage of apatient's artery is blocked by plaque. To do this, she measures theblood pressure just

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A medical technician is trying to determine what percentage of apatient's artery is blocked by plaque. To do this, she measures theblood pressure just before the region of blockage and finds that itis $$\displaystyle{1.20}\times{10}^{{{4}}}{P}{a}$$, while in the region of blockage it is $$\displaystyle{1.15}\times{10}^{{{4}}}{P}{a}$$. Furthermore, she knows that blood flowingthrough the normal artery just before the point of blockage istraveling at 30.0 cm/s, and the specific gravity of this patient'sblood is 1.06. What percentage of the cross-sectional area of thepatient's artery is blocked by the plaque?

2021-04-16
We are given with
$$\displaystyle{P}_{{{1}}}={1.2}\times{10}^{{{4}}}{P}{a}$$
$$\displaystyle{P}_{{{2}}}={1.15}\times{10}^{{{4}}}{P}{a}$$
$$\displaystyle{v}_{{{1}}}={30}$$ cm/s
=0.3 m/s
specific gravity RD = 1.06
$$\displaystyle{P}_{{{b}}}={R}{D}\times$$ density of water
$$\displaystyle={1.06}\times{1000}{k}\frac{{g}}{{m}^{{{3}}}}$$
$$\displaystyle={1060}{k}\frac{{g}}{{m}^{{{3}}}}$$
as the heights are equal we can write
$$\displaystyle{h}_{{{1}}}={h}_{{{2}}}$$
and according to the bernoulis equation we get
$$\displaystyle{P}_{{{2}}}+{\left(\frac{{1}}{{2}}\right)}{p}_{{{b}}}{{v}_{{{2}}}^{{{2}}}}={P}_{{{1}}}+{\left(\frac{{1}}{{2}}\right)}{p}_{{{b}}}{{v}_{{{1}}}^{{{2}}}}$$
$$\displaystyle{\left(\frac{{1}}{{2}}\right)}{p}_{{{b}}}{{v}_{{{2}}}^{{{2}}}}={\left({P}_{{{1}}}-{P}_{{{2}}}\right)}+{\left(\frac{{1}}{{2}}\right)}{p}_{{{b}}}{{v}_{{{1}}}^{{{2}}}}$$
$$\displaystyle{v}_{{{2}}}=\sqrt{{{\left[{2}\frac{{{P}_{{{1}}}-{P}_{{{2}}}}}{{p}}{b}\right]}+{{v}_{{{1}}}^{{{2}}}}}}$$
=.......m/s
=.......cm/s
so according to the equation of continuity we get
$$\displaystyle{A}_{{{1}}}{v}_{{{1}}}={A}_{{{2}}}{v}_{{{2}}}$$
$$\displaystyle{\left(\frac{{A}_{{{2}}}}{{A}_{{{1}}}}\right)}=$$......%