The radius of the aorta in a test subject determined by MRI is 1.2 cm. At rest, his end diastolic volume is 120 mL and his end systolic volume is 50 mL. If we start the cardiac cycle at the opening of the mitral valve, the aortic valve in this person opens at 0.5 s and closes at 0.9 s. The entire cycle lasts 1.2 s in this person. The density of blood is 1.055 g mL-1 and its viscosity is 3.0 x 10^-3 Pa s. A pascal is one N m-2. Using this information: A. What is the stroke volume?B. What is the heart rate?C. What is the cardiac output?D. What is the period of ejection?E. What is the average velocity of blood in the aorta?F. What is the Reynolds number for blood during ejection?G. Is blood flow in the aorta laminar or turbulent?H. What is the kinetic energy of the ejected blood?I. W hat pressure-volume work is done during ejection?

A) SV = 70 mL

B) H.R = 49.8 bpm

C) H.R = 49.8 bpm

D) P_e = 0.4 s

E) v¯ = 38.68 cm/s

F) Re ≈ 3265

G) turbulent flow.

H) KE = 0.01105 J

I) Diastolic and Systolic pressure are not given so work can't be found

Explanation:

We are given;

Radius of aorta; r = 1.2 cm

Diastolic volume; DV = 120 mL

Systolic volume; SysV = 50 mL

Period that cycle lasts; t_cycl = 1.2 s

the aortic valve opens at; t_ao = 0.5 s aortic valve closes at; t_ac = 0.9 s

Density; ρ = 1.055 g/mL = 1.055 g/cm³

viscosity; μ = 3 × 10^(-3) Pa.s = 0.03 g/cm-s

A) Formula for stroke volume is;

SV = DV - SysV

SV = 120 - 50

SV = 70 mL

B) formula for heart rate is;

H.R = 1/t_cycl

H.R = 1/1.2

H.R = 0.83 beats per seconds

Converting to beats per minutes gives;

H.R = 0.83 × 60

H.R = 49.8 bpm

C) Formula for cardiac output is;

cardiac output = SV × HR

Let's convert SV to L to get;

SV = 0.07 L

Thus;

Cardiac output = 0.07 × 49.8

Cardiac output = 3.49 L/min

D) Period of ejection = t_ac - t_ao

Period of ejection = 0.9 - 0.5

Period of ejection; P_e = 0.4 s

E) average velocity; v¯ = SV/(A × P_e)

A = πr²

A = π × 1.2² cm²

SV = 70 mL = 70 cm³

Thus;

v¯ = 70/(π × 1.2² × 0.4)

v¯ = 38.68 cm/s

F) Reynolds number is calculated from;

Re = ρv¯D/μ

r = 1.2 and so diamter; D = 2r = 2 × 1.2 = 2.4 cm

Thus;

Re = 1.055 × 38.68 × 2.4/(0.03)

Re ≈ 3265

G) Reynolds number is greater than 2000 and thus it is a turbulent flow.

H) Formula for kinetic energy is;

KE = ½mv²

We know that; mass = volume × density

Thus; mass = 1.055 × 70 g

KE = ½ × 1.055 × 70 × 38.68²

KE = 110,490.12 g.cm²/s²

Converting to Joules gives;

KE = 0.01105 J

I) work volume relationship is given by;

W = ∫PdV

But we are not given the Diastolic and Systolic pressure, so it's not possible to calculate the work

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.

answera compound;