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Zigmanuir [339]
3 years ago
10

The human circulatory system consists of a complex branching pipe network ranging in diameter from

Engineering
1 answer:
Stels [109]3 years ago
8 0

Answer: the average velocity decreases

Explanation:

From the provided data we have:

Vessel    avg. diameter[mm] number

Aorta                 25.0                   1

Arteries             4.0                    159

Arteioles           0.06                 1.4*10^7

Capillaries         0.012               2.9*10^9

from the information, let \hat{m} be the mass flow rate, \rho is density, n number of vessels, and A is the cross-section area for each vessel

the flow rate is constant so it is equal for all vessels,

The average velocity is related to the flow rate by:

\hat{m} = v* \rho * A * n

we clear the side where v is in:

v = \frac{\hat{m}}{\rho A n}

area is π*R^2 where R is the average radius of the vessel (diameter/2)

we get:

v = \frac{\hat{m}}{\rho \pi R^2 n}

you can directly see in the last equation that if we go from the aorta to the capillaries, the number of vessels is going to increase ( n will increase and R is going to decrease ) . From the table, R is significantly smaller in magnitude orders than n, therefore, it wont impact the results as much as n. On the other hand, n will change from 1 to 2.9 giga vessels which will dramatically reduce the average blood velocity

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Determine the depreciation expense for 2018 and 2019 using the following​ methods: (a)​ Straight-line (SL),​ (b) Units of produc
prohojiy [21]

Answer:

Check Explanation.

Explanation:

(1). The straight-line method: the general clue with this method is that in the two years, depreciation is the same. The formula for Calculating depreciation is given below;

straight-line method = (cost - Residual value)/ useful life in years.

From the question we know that the cost of acquisition is $30,000,000, the residual value of the asset is $4,000,000 and useful life is 7 years. Therefore;

straight-line method = ($30,000,000 - $4,000,000)/ 7.

= $3, 714,285.71 Per year.

That is $3, 714,285.71 for 2018 and 2019.

(2).Units of production​ (UOP) = (cost - Residual value)/ useful life in units.

= ($30,000,000 - $4,000,000)/ 4,375, 000.

Units of production​ (UOP) = $6 per mile.

Hence, the depreciation in 2018 = Depreciation per unit × 2018 year usage.

= 6 × 1,100,000 mile.

= $6,600,000.

depreciation in 2019 = Depreciation per unit × 2019 year usage.

= 6 × 1,200,000.

= $7,200,000.

Double-declining-balance​ (DDB)= (cost - accumulated depreciation) × 2 × 1/(useful life years).

Double-declining-balance​ (DDB) = (30,000,000 - 0)× 2 × (1/7).

= $8,571,428.57 depreciation in 2018.

= $8,571,428.6 depreciation in 2018

Double-declining-balance​ (DDB) = (30,000,000 - 8,571,428.57) × 2 × 1/7.

= $6,122,449.00 depreciation in 2019.

====================================================================

Total depreciation for straight-line method(2018 and 2019) = $7,428,571.42.

Total depreciation for Units of production​ (UOP)(2018 and 2019) = $13,800,000.

Total depreciation for Double-declining-balance (DDB)= $ 14,693,877.6.

5 0
3 years ago
Joe, a technician, is attempting to connect two hubs to add a new segment to his local network. He uses one of his CAT5 patch ca
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Answer:

Option D. is correct

Explanation:

Joe uses one of his CAT5 patch cables to connect two hubs to add a new segment to his local network. As he can only connect to it from a workstation within that segment,  he is not able to reach the new network segment from his workstation.

The most problem is that the technician used a straight-through cable.

Option D. is correct.

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3 years ago
Singularity is an important property of a square matrix. This is also known as degenerate. What is the value of the determinant
DENIUS [597]

Answer:

For a Singular matrix, the determinant must be equivalent to 0.

Explanation:

A matrix is a rectangular array in which elements are arranged in rows and columns.

Each square matrix has a determinant. The determinant is a numerical idea that has a fundamental function in finding the arrangement just as investigation of direct conditions. For a Singular matrix, the determinant must be equivalent to 0.

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3 years ago
Please help i give brainliest​
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A mock-up

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It is made of cheap and easy to access parts.

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3 years ago
A soil is at a void ratio e = 0.90 with a specific gravity of the solid particles Gs = 2.70.
Alexus [3.1K]

Answer:

The correct answers are:

a. % w = 33.3%

b. mass of water = 45g

Explanation:

First, let us define the parameters in the question:

void ratio e  = \frac{V_v}{V_s} =  \frac{\left\begin{array}{ccc}volume&of&void\end{array}\right}{\left\begin{array}{ccc}volume&of&solid\end{array}\right}------ (1)

Specific gravity G_{s} = \frac{P_s}{P_w} =  \frac{\left\begin{array}{ccc}density&of&soil\end{array}\right}{\left\begin{array}{ccc}density&of&water\end{array}\right}------ (2)

% Saturation S = \frac{V_w}{Vv} × \frac{100}{1} =  \frac{\left\begin{array}{ccc}volume&of&water\end{array}\right}{\left\begin{array}{ccc}volume&of&void\end{array}\right} × \frac{100}{1}--------(3)

water content w =  \frac{M_w}{M_s} = \frac{\left\begin{array}{ccc}mass&of&water\end{array}\right}{\left\begin{array}{ccc}mass&of&solid\end{array}\right} ------(4)

a) To calculate the lower and upper limits of water content:

when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.

when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.

Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.

To get the relationship between water content and saturation, we will manipulate the equations above;

w =  \frac{M_w}{Ms}

Recall; mass = Density × volume

w = \frac{V_wP_w}{V_sP_s} ------(5)

From eqn. (2)  G_{s} = \frac{P_s}{P_w}

∴ \frac{1}{G_s} = \frac{P_w}{P_s} ------(6)

putting eqn. (6) into (5)

w = \frac{V_w}{V_sG_s} -----(7)

Again, from eqn (1)

V_s = \frac{V_v}{e}

substituting into eqn. (7)

w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v}  = S

∴ w = \frac{Se}{G_s} -----(8)

With eqn. (7), we can calculate

upper limit of water content

when S = 100% = 1

Given, G_{s} = 2.7, e= 0.9

∴w= \frac{0.9*1}{2.7} = 0.333

∴ %w = 33.3%

Lower limit of water content

when S = 1% = 0.01

w= \frac{0.01*0.9 }{2.7} = 0.0033

∴ % w = 0.33%

b) Calculating mass of water in 100 cm³ sample of soil (P_w=\frac{1_g}{cm^{3} } )

Given, V_{s} = 100 cm^{3 }, S = 50% = 0.5

%S = \frac{V_w}{V_v} × \frac{100}{1} = \frac{V_w}{eV_s} × \frac{100}{1}

0.50 = \frac{V_w}{0.9* 100}  = 45cm^{3}

mass of water = P_wV_w= 1 * 45 = 45_{g}

7 0
4 years ago
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