1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
MArishka [77]
3 years ago
13

) If the blood viscosity is 2.7x10-3 Pa.s, length of the blood vessel is 1 m, radius of the blood vessel is 1 mm, calculate the

tubular resistance of the blood vessel (in GPa.S/m3 ). If the blood pressure at the inlet of the above vessel is 43 mm Hg and if the blood pressure at the outlet of the above vessel is 38 mm Hg. Calculate the flow rate (in ml/min).
Engineering
1 answer:
Illusion [34]3 years ago
5 0

Answer:

a) the tubular resistance of the blood vessel is 6.88 Gpa.s/m³.

b) the flow rate is 5.8 ml/min

Explanation:

Given the data in the question;

Length of blood vessel L = 1 m

radius r = 1 mm = 0.001 m

blood viscosity μ = 2.7 × 10⁻³ pa.s = 2.7 × 10⁻³ × 10⁻⁹ Gpa.s = 2.7 × 10⁻¹² Gpa.s

Now, we know that Resistance = 8μL / πr⁴

so we substitute

Resistance = [8 × (2.7 × 10⁻¹²) × 1] / [π(0.001)⁴]

Resistance = [2.16 × 10⁻¹¹] / [3.14159 × 10⁻¹²]

Resistance = 6.8755 ≈ 6.88 Gpa.s/m³

Therefore, the tubular resistance of the blood vessel is 6.88 Gpa.s/m³.

b)

blood pressure at the inlet of the vessel = 43 mm Hg

blood pressure at the outlet of the vessel = 38 mm Hg

flow rate = ?

we know that;

flow rate Q = ΔP / R

where ΔP is change in pressure and R is resistance.

ΔP = Inlet pressure - Outlet pressure = 43 - 38 = 5 mm Hg =  665 pa

R = 6.8755 Gpa.s/m³ = { 6.8755 × 10⁹ / 60 × 10⁶ } = 114.5916 pa.min.ml⁻¹

so we substitute

Q =  665 pa / 114.5916 pa.m.ml⁻¹

Q = 5.8 ml/min

Therefore, the flow rate is 5.8 ml/min

You might be interested in
I WILL GIVE 20 POINTS!!
Alex777 [14]

Answer:

Use a resume header

Explanation:

Create a Summary

Research industry, employer keywords

there are some hints okay

5 0
3 years ago
Read 2 more answers
There are 10 vehicles in a queue when an attendant opens a toll booth. Vehicles arrive at the booth at a rate of 4 per minute. T
dmitriy555 [2]

Answer:

as slated in your solution, if delay time is 2.30 mins, hence 9 vehicle will be on queue as the improved service commenced.

Explanation:

4 vehicle per min, in 2 mins of the delay time 8 vehicles while in 0.3 min average of 1 vehicle join the queue. making 9 vehicle maximum

3 0
3 years ago
Consider the string length equal to 7. This string is distorted by a function f (x) = 2 sin(2x) - 10sin(10x). What is the wave f
fenix001 [56]

Answer:

hello your question has a missing part below is the missing part

Consider the string length equal to \pi

answer : 2cos(2t) sin(2x) - 10cos(10t)sin(10x)

Explanation:

Given string length = \pi

distorted function f(x) = 2sin(2x) - 10sin(10x)

Determine the wave formed in the string

attached below is a detailed solution of the problem

8 0
3 years ago
Nanotechnology is a scientific area that deals with making or changing things that are incredibly _______________. *
Dafna1 [17]
B tiny nano means small!!
3 0
3 years ago
Read 2 more answers
A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please cal
Sphinxa [80]

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

\frac{d_{2} }{d_{1} } =\frac{1}{2} (\sqrt{1+8F^{2} } -1)

10*2=\sqrt{1+8F^{2} } -1

Solving for F:

F = 7.4162

v=F\sqrt{gd_{1} }

Here, g = gravity = 32.2 ft/s²

v=7.4162*\sqrt{32.2*1} =42.0833ft/s

b) The flow rate:

q=v*L*d_{1} =42.0833*80*1=3366.664ft^{3} /s

c) The Froude number:

v_{2} =\frac{q}{L*d_{2} } =\frac{3366.664}{80*10} =4.2083ft/s

F=\frac{v_{2}}{\sqrt{gd_{2} } } =\frac{4.2083}{\sqrt{32.2*10} } =0.2345

d) The flow energy dissipated:

E=\frac{(d_{2}-d_{1})^{3} }{4d_{1}d_{2}} =\frac{(10-1)^{3} }{4*1*10} =18.225ft

e) The critical depth:

d_{c} =(\frac{(\frac{q}{L})^{2}  }{g} )^{1/3} =(\frac{(\frac{3366.664}{80})^{2}  }{32.2} )^{1/3} =3.8030ft

4 0
4 years ago
Other questions:
  • the velocity of a particle is given by v=16t^2i+4t^3j+(5t+2k) m/s, where t is in seconds. if the particle is at the origin when
    10·1 answer
  • A 15 cm × 15 cm circuit board dissipating 20 W of power uniformly is cooled by air, which approached the circuit board at 20C w
    11·1 answer
  • I need answers for this sheet please.
    15·1 answer
  • What is the theoretical density in g/cm3 for Lead [Pb]?
    13·1 answer
  • SEICUL UC CULTELL allsvel.
    9·2 answers
  • Shear modulus is analogous to what material property that is determined in tensile testing? (a)- Percent reduction of area (b) Y
    11·1 answer
  • In the High Low Logic Index low levels are bearish and high levels are bullish, generally True False
    13·1 answer
  • Computer system analyst advantage​
    5·2 answers
  • The moisture content in air (humidity) is measured by weight and expressed in pounds or ____________________.
    14·1 answer
  • Of the cost reduction strategies for workers' compensation mentioned in the required readings, which one do you think would work
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!