Physics, 01.08.2020 03:01, sa12340

Part (a) Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing through the aorta (in grams of blood per second)? 94.2 Attempts Remain 33% Part (b) If all the blood that flows through the aorta then branches into the major arteries, what is the velocity of blood in the major arteries? Give your answer in cm/s X Attempts Remain v4.71 A 33%
Part (c) The blood flowing in the major arteries then branches into the capillaries. If the velocity of blood in the capillaries is measured to be 0.04 cm/s, what is the cross sectional area of the capillary system in cm2? Grade Summary Deductions Potential cm2 A = 0% 100% cos( tan acos Submissions sin ( 7 8 9 НОME л Attempts remaining: 10 (4% per attempt) detailed view cotan0 asin acotan A E 4 5 6 atan cosh sinh0 1 2 3 tanh) ODegrees cotanh 0 + END Radians BACKSPACE CLEAR DEL
The aorta (the main blood vessel coming out of the heart) has a radius of about 1.0 cm and the total cross section of the major arteries is about 20 cm2. The density of blood is about the same as water, 1 g/cm3.

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answered: sanfratello1081

This question involves the concepts of volumetric flow rate, continuity equation, and flow velocity.

a) Total current of the blood passing through the aorta is "94.2 g/s".

b) The velocity of blood in major arteries is "4.71 cm/s".

c) The cross-sectional area of the capillary system is "2356.2 cm²".

First, we will find the volumetric flow rate of the blood, using the continuity equation's formula:

Q=Av

where,

Q = volumetric flow rate = ?

A = cross-sectional area of aorta

A = $\pi(r)^2=\pi(1\ cm)^2= 3.14\ cm^2$

v = flow velocity = 30 cm/s

Therefore,

$Q=(3.14\ cm^2)(30\ cm/s)$

Q = 94.25 cm³/s

Now, the blood current will be given as:

I = Qρ

where,

I = current = ?

ρ = blood density = 1 g/cm³

Therefore,

I = (94.2 cm³/s)(1 g/cm³)

I = 94.2 g/s

Now, this volumetric flow rate will be constant in major arteries:

$Q = A_r v_r\\\\v_r=\frac{Q}{A_r}$

where,

Ar = cross-section area of major arteries = 20 cm²

vr = flow velocity of blood in major arteries = ?

Therefore,

$v_r=\frac{94.25\ cm^3/s}{20\ cm^2}$

vr = 4.71 cm/s

Now, this volumetric flow rate will be constant in capillaries:

$Q = A_c v_c\\\\A_c=\frac{Q}{v_c}$

where,

Ac = cross-section area of capillaries = ?

vc = flow velocity of blood in capillaries = 0.04 cm/s

Therefore,

$A_c=\frac{94.25\ cm^3/s}{0.04\ cm/s}$

Ac = 2356.2 cm²

Learn more about the continuity equation here:

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answered: suieemage

a) 94.26 g/s

b) 4.713 cm/s

c) 2356.5 cm^2

Explanation:

a) velocity of blood through the aorta = 30 cm/s

radius of aorta = 1 cm

density of blood = 1 g/cm^3

Area of the aorta = $\pi r^{2}$ = 3.142 x $1^{2}$ = 3.142 cm^2

Flow rate through the aorta Q = AV

where A is the area of aorta

V is the velocity of blood through the aorta

Q = 3.142 x 30 = 94.26 cm^3/s

Current of blood through aorta = Qρ

where ρ is the density of blood

= 94.26 x 1 = 94.26 g/s

b) Velocity of blood in the major aorta = 30 cm/s

Area of the aorta = 3.142 cm^2

Velocity of blood in the major arteries = ?

Area of major arteries = 20 cm^2

From continuity equation

$A_{ao} V_{ao} = A_{ar} V_{ar}$

where

$V_{ao}$ = velocity of blood in the major arteries

$A_{ao}$ = Area of the aorta

$V_{ar}$ = velocity of blood in the major arteries

$A_{ar}$ = Area of major arteries

substituting values, we have

3.142 x 30 = 20 $V_{ar}$

94.26 = 20 $V_{ar}$

$V_{ar}$ = 94.26/20 = 4.713 cm/s

c) From continuity equation

$A_{ar} V_{ar} = A_{c} V_{c}$

where

$A_{ar}$ = Area of major arteries = 20 cm/s

$V_{ar}$ = velocity of blood in the major arteries = 4.713 cm/s

$A_{c}$ = Area of the capillary system = ?

$V_{c}$ = velocity of blood in the capillary system = 0.04 cm/s

substituting values, we have

20 x 4.713 = $A_{c}$ x 0.04

94.26 = 0.04 $A_{c}$

$A_{c}$ = 94.26/0.04 = 2356.5 cm^2

answered: Guest

Rain is pulled by gravity which is a non contact force

answered: Guest

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from wikipedia