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geniusboy [140]

3 years ago

11

There are many interesting applications of our energy density model to the flow of blood in the human circulatory system. One in

teresting phenomenon is an aneurysm, a sudden abnormal enlargement of a section of an artery due to a weakening of the arterial wall. If the blood flow rate remains constant through the artery, how does the pressure in the enlarged section (the aneurysm) compare to the pressure in the rest of the artery? Neglect any affects of resistance to flow in your explanation.

2 answers:

IgorLugansk [536]3 years ago

7 0

Answer:

Pressure in enlarged portion is greater than remaining portion of artery.

Explanation:

The Bernoullis equation at front and back side of arterial wall is given by:

qaws [65]3 years ago

5 0

Answer:

Pressure increases due to enlargement

Explanation:

Energy density is just a fancy name for pressure

Pressure is same at the bottom of the cups (same level-Pascal's law)

thus, Air pressure 1 + h1d1g = Air pressure 2 + h2d1g

= Air pressure 3 + (h2-h1)d2g +h1d1g

from the first 2, we get that since h2>h1, AP2<AP1

from the next 2, we get that since d2<d1, AP3>AP2

from first and third, we get that AP1>AP3

thus, finally AP1>AP3>AP2

for fluids flowing in tubes (blood vessel in this case)

P+0.5dv^2 + gh is constant (also called the bernoulli equation

for the same blood vessel, the heights remain same i.e h1=h2

for same flow rate, inc in area decreases the speed at which the blood flows as vA must remain same

hence, P increases due to enlargement

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1. A pulley system has efficiency 60%. Calculate mechanical advantage to make its velocity ratio 6 .How much amount of effort is

lesya [120]

Efficiency = Mechanical advantage / Velocity Ratio:

0.60 = M.A./6

MA = 0.6 * 6 = 3.6

The mechanical advantage is 3.6

Mechanical advantage = Force out / Force in

3.6 = 3000/ Force in

Force in = 3000/3.6

Force in = 833.33 N

Answer: 833.33N

8 0

2 years ago

Would u rather/ be a super hero. or super villian.

antiseptic1488 [7]

Answer:

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4 0

3 years ago

Read 2 more answers

Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south.

Nina [5.8K]

Answer:

The resultant vector $\vec R = \vec A+\vec B$ is given by $\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$ .

Explanation:

Let $\vec A = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})$ and $\vec B = 4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j})$ , both measured in meters. The resultant vector $\vec R$ is calculated by sum of components. That is:

$\vec R = \vec A+\vec B$ (Eq. 1)

$\vec R = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})+4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j})$

$\vec R = (6\cdot \cos 30^{\circ}-4\cdot \sin 30^{\circ})\,\hat{i}+(6\cdot \sin 30^{\circ}-4\cdot \cos 30^{\circ})\,\hat{j}$

$\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$

The resultant vector $\vec R = \vec A+\vec B$ is given by $\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m]$ .

8 0

3 years ago

Two Earth satellites, A and B, each of mass m = 940 kg , are launched into circular orbits around the Earth's center. Satellite

-Dominant- [34]

Answer:

The required work done is $6.5\times10^{9}\ J$

Explanation:

Given that,

Mass of each satellites = 940 kg

Altitude of A = 4500 km

Altitude of B = 11100 km

We need to calculate the potential energy

Using formula of potential

$U_{A}=-\dfrac{Gm_{A}m_{E}}{r_{A}}$

Put the value into the formula

$U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}$

$U_{A}=-3.44\times10^{10}\ J$

We need to calculate the potential energy

Using formula of potential

$U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}$

Put the value into the formula

$U_{B}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+11.10\times10^{6}}$

$U_{B}=-2.14\times10^{10}\ J$

We need to calculate the value of $k_{A}$

Using formula of $k_{A}$

$k_{A}=-\dfrac{1}{2}U_{A}$

Put the value into the formula

$k_{A}=\dfrac{1}{2}\times3.44\times10^{10}$

$k_{A}=1.72\times10^{10}\ J$

We need to calculate the value of $k_{B}$

Using formula of $k_{B}$

$k_{B}=-\dfrac{1}{2}U_{B}$

Put the value into the formula

$k_{B}=\dfrac{1}{2}\times2.14\times10^{10}$

$k_{B}=1.07\times10^{10}\ J$

We need to calculate the work done

Using formula of work done

$W=\Delta K+\Delta U$

$W=(k_{B}-k_{A})+(U_{B}-U_{A})$

$W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})$

$W=\dfrac{1}{2}(U_{B}-U_{A})$

Put the value into the formula

$W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})$

$W=6.5\times10^{9}\ J$

Hence, The required work done is $6.5\times10^{9}\ J$

7 0

2 years ago

14 kilometers per hour to meters per second

Alina [70]

Answer:

About 3.88 meters per second

Explanation:

14km= 14,000m

14,000m per hour

so 14,000 ÷ 60= 233.33 meters per minute

233.33÷60=3.88 meters per second

3 0

3 years ago