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Fiesta28 [93]
2 years ago
7

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 s

ystolic 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?
Physics
1 answer:
never [62]2 years ago
7 0

Answer:

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

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Anna [14]

Answer:

3) D: 31 m/s

4) D: 84.84 metres

Explanation:

3) Initial velocity along the x-axis is;

v_x = v_o•cos θ

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v_y = v_o•sin θ

Plugging in the relevant values, we have;

v_x = 31 cos 60

v_x = 31 × 0.5

v_x = 15.5 m/s

Similarly,

v_y = 31 sin 60

v_y = 31 × 0.8660

v_y = 26.85 m/s

Thus, magnitude of the initial velocity is;

v = √(15.5² + 26.85²)

v ≈ 31 m/s

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6 0
2 years ago
a stone is thrown by a person from the top of the building, which is 200m tall. at the same time, another stone is thrown with v
solniwko [45]

Answer:

The time after which the two stones meet is tₓ = 4 s

Explanation:

Given data,

The height of the building, h = 200 m

The velocity of the stone thrown from foot of the building, U = 50 m/s

Using the II equation of motion

                             S = ut + ½ gt²

Let tₓ be the time where the two stones  meet and x be the distance covered from the top of the building

The equation for the stone dropped from top of the building becomes

                            x = 0 + ½ gtₓ²

The equation for the stone thrown from the base becomes

                S - x = U tₓ - ½ gtₓ²  (∵ the motion of the stone is in opposite direction)

Adding these two equations,

                      x + (S - x) = U tₓ

                               S = U tₓ

                               200 = 50 tₓ

∴                                  tₓ = 4 s

Hence, the time after which the two stones meet is tₓ = 4 s   

6 0
3 years ago
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vampirchik [111]

Answer:

Explanation:

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Ultraviolet light from sun.

Heat from a stove burner.

X-ray from an x-ray machine.

Alpha particle emit from a radio active decay of uranium.

Sound waves from your stereo.

Microwave from micro oven.

ultraviolet light from a black light.

Gamma radiations from a supernova.

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8 0
3 years ago
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What happens when a light ray is parallel to the principal axis ?
Phantasy [73]

Answer:

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Hope this helps...

7 0
3 years ago
1. Determine the image distance in each of the following.
miskamm [114]

It is given that for the convex lens,

Case 1.

u=−40cm

f=+15cm

Using lens formula

v

1

−

u

1

=

f

1

v

1

−

40

1

=

15

1

v

1

=

15

1

−

40

1

v=+24.3cm

The image in formed in this case at a distance of 24.3cm in left of lens.

Case 2.

A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that

u=∞

f=15cm

Now, using mirror’s formula

v

1

+

u

1

=

f

1

v

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+

∞

1

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15

1

v=+15cm

The image is formed at a distance of 15cm in left of mirror

6 0
3 years ago
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