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Crazy boy [7]
3 years ago
15

A major artery with a 1.6 cm2 cross-sectional area branches into 18 smaller arteries, each with an average cross-sectional area

of 0.7 cm2. By what factor is the average velocity of the blood reduced when it passes into these branches?
Physics
1 answer:
Brums [2.3K]3 years ago
5 0

To solve this problem it is necessary to apply the continuity equations for which it is defined that the proportion of Area in the initial section is equal to the final section. In other words,

A_1 v_1 = A_2 v_2

Where,

A_i = Cross sectional area at each section

v_i =Velocities of fluid at each section

The total area of the branch is eighteen times of area of smaller artery. The average cross-sectional area of each artery is 0.7cm^2.

Therefore the Cross-sectional area at the end is

A_2= 18*0.7cm^2

A_2 = 12.6cm^2

Applying the previous equation we have then

A_1 v_1 = A_2 v_2

(1.7cm^2) v_1 = (12.6cm^2)v_2

The ratio of the velocities then is

\frac{v_1}{v_2} = \frac{1.7}{12.6}

\frac{v_1}{v_2} = 0.135

Therefore the factor by which the velocity of blood will reduce when it enters the smaller arteries is 0.1349

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4 0
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In order to model the motion of an extinct ape, scientists measure its hand and arm bones. From shoulder to wrist, the arm bones
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0.37 m

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The center of mass y = (m₁y₁ + m₂y₂)/(m₁ + m₂) where m₁ = mass of arm bones = 4.0 kg, y₁ = distance center of mass of arm bones  from shoulder = 0.30 m, m₂ = mass of hand bones = 1.0 kg and y₂ = distance of center of mass hand bones from shoulder = x₁ + distance of center of hand bones from wrist = 0.60 m + 0.05 m = 0.65 m

Substituting these into the equation for the center of mass, we have

y = (m₁y₁ + m₂y₂)/(m₁ + m₂)

y = (4.0 kg × 0.30 m + 1.0 kg × 0.65 m)/(4.0 kg + 1.0 kg)

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