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ehidna [41]
3 years ago
8

The pressure difference, , across a partial blockage in an artery (called a stenosis) is approximated by the equation where is t

he blood velocity, the blood viscosity , the blood density , the artery diameter, the area of the unobstructed artery, and the area of the stenosis. Determine the dimensions of the constants and . Would this equation be valid in any system of units
Physics
1 answer:
Strike441 [17]3 years ago
5 0

The question is incomplete. The complete question is  :

The pressure difference, Δp, acK_uross a partial blockage in an artery (called a stenosis) is approximated by the equation :

$\Delta p=K_v\frac{\mu V}{D}+K_u\left(\frac{A_0}{A_1}-1\right)^2 \rho V^2$

Where V is the blood velocity, μ the blood viscosity {FT/L2}, ρ the blood density {M/L3}, D the  artery diameter, A_0 the area of the unobstructed artery, and A1 the area of the stenosis.  Determine the dimensions of the constants K_v and K_u. Would this equation be valid in any  system of units?

Solution :

From the dimension homogeneity, we require :

$\Delta p=K_v\frac{\mu V}{D}+K_u\left(\frac{A_0}{A_1}-1\right)^2 \rho V^2$

Here, x means dimension of x. i.e.

$[ML^{-1}T^{-2}]=\frac{[K_v][ML^{-1}T^{-1}][LT^{-1}]}{[L]}+[K_u][1][ML^{-3}][L^2T^{-2}]$

                    $=[K_v][ML^{-1}T^{-2}]+[K_u][ML^{-1}T^{-2}]$

So, $[K_u]=[K_v]=[1 ]=$ dimensionless

So, K_u and K_v  are dimensionless constants.

This equation will be working in any system of units. The constants K_u and K_v will be different for different system of units.

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This picture shows human red blood cells. The human body contains about 5 million red blood cells per cubic milliliter. This mea
Elanso [62]

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

                  Hello!

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

❖ The correct answer choice is B) is multicellular. When something is multicellular, it consists of two or more cells. When something is unicellular, it consists of only one cell and in this case we have 5 million red blood cells so that wouldn't make sense.

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

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4 0
3 years ago
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
A horizontal force acts on an object on a fric- tionless horizontal surface. If the force is halved and the mass of the object i
guapka [62]

(a)  If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b)  if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

The force on an object is determined by applying Newton's second law of motion;

F = ma

\frac{F_1}{m_1a_1} = \frac{F_2}{m_2a_2}

(a)

when the force is halved, F₂ = 0.5F₁,

mass is doubled, m₂ = 2m₁

\frac{F_1}{m_1a_1} = \frac{0.5F_1}{2m_1a_2} \\\\2m_1a_2F_1 = 0.5F_1 m_1a_1\\\\2a_2 = 0.5a_1\\\\a_2 = \frac{0.5a_1}{2} = \frac{a_1}{2 \times 2} = \frac{a_1}{4} \\\\a_2 = \frac{1}{4} (a_1)

Thus, If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b)

when the force is doubled, F₂ = 2F₁,

mass is halved, m₂ = 0.5m₁

\frac{F_1}{m_1 a_1} = \frac{2F_1}{0.5m_1 a_2} \\\\0.5m_1a_2 F_1 = 2F_1m_1a_1\\\\0.5a_2 = 2a_1\\\\a_2 = \frac{2a_1}{0.5} \\\\a_2 = 4(a_1)

Thus, if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

Learn more here:brainly.com/question/19887955

3 0
2 years ago
What are the characteristics of an ideal transformer ​
Vsevolod [243]

It is an imaginary transformer which has no core loss, no ohmic resistance and no leakage flux. The ideal transformer has the following important characteristic. The resistance of their primary and secondary winding becomes zero. The core of the ideal transformer has infinite permeability.

4 0
3 years ago
A radio wave has a frequency of 5.5 × 104 hertz and travels at a speed of 3.0 × 108 meters/second. What is its wavelength
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Use v=fλ
3x10^8=5.5x 10^4 λ
λ=5.45x10^3m
4 0
3 years ago
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