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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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A displacement transducer has the following specifications: Linearity error ± 0.25% reading Drift ± 0.05%/○C reading Sensitivity
White raven [17]

Answer:

The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm

Explanation:

uncertainty in a nominal displacement

= (u^2 + v^2)^(1/2)

assume from specifications that k = 5v/5cm

                                                         = 1v/cm

u^2 = (0.0025*2)^(2) + (0.005*10*2)^2 + (0.0025*2)^2

      = 0.01225v

v = 2v * 0.001

  = 0.002v

uncertainty in a nominal displacement

= (u^2 + v^2)^(1/2)

= ((0.01225)^2 + (0.002)^2)^(1/2)

= 0.0124 cm

Therefore, The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm

8 0
3 years ago
determine the position d of the 6- kn load so that the average normal stress in each rod is the same.
Zinaida [17]

The load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

<h3>What is meant by torque?</h3>

The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.

Let the beam is of length L

Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero

$N_1 * x=N_2 *(L-x)$

also, we know that stress at both ends are same

$\frac{N_1}{12}=\frac{N_2}{8}$

$2 * N_1=3 * N_2$

Now from two equations we have

$\frac{3}{2} N_2 * x=N_2 *(L-x)

solving the above equation we have

$x=\frac{2}{5} L

so the load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

The complete question is:

47. the beam is supported by two rods ab and cd that have cross-sectional areas of $$12mm^2 and $$8mm^2, respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.

To learn more about torque refer to:

brainly.com/question/20691242

#SPJ4

7 0
2 years ago
You are in charge of ordering the concrete for a basement wall concrete pour. The wall forms are all set up and ready. The wall
choli [55]

Answer:

189.15cy

Explanation:

To understand this problem we need to understand as well the form.

It is clear that there is four wall, two short and two long.

The two long are \rightarrow 120ft5in+2(10ft)

The two long are \rightarrow 122ft1in=122.08ft

The two shors are \rightarrow 86ft4.5in = 86.375ft

The height and the thickness are 14ft and 0.83ft respectively.

So we only calculate the Quantity of concrete,

Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3

That in cubic yards is equal to 180.15 (1cy=27ft^3)

Hence, we need order 5% plus that represent with the quantity

Q_{ordered}=1.05*180.15=189.15cy

8 0
3 years ago
Tahir travel twice as far as ahmed, but onley one third as fast. Ahmed starts travel on tuesday at noon at point x to point z 30
shepuryov [24]

Answer:

6:00 pm the next day

Explanation:

Given that

Tahir traveled twice as far as Ahmed. We say,

Ahmed traveled a distance, D

Tahir would travel a distan, 2D

Tahir traveled 1/3 as fast as Ahmed, so we say

Ahmed traveled at a speed, S

Tahir would travel at a speed, S/3

If Ahmed starts travel on tuesday at noon at point x to point z 300km, by 9:00pm,

Time taken by Ahmed to travel is

9:00 pm - 12:00 pm = 9 hours

Ahmed, traveled 300 km in 9 hours, meaning he traveled at 33.3 km in an hour.

Speed, S that Ahmed traveled with is 33.3 km/h

Remember, we stated that Tahir travels at a speed of S/3, that is, The speed of Tahir is

33.3/3 = 11.1 km/h.

300 km would then be traveled in 300 km/11.1 km/h = 27 hours.

Tahir started traveling, 3 hours after Ahmed, that is 12:00 pm + 3:00 hrs = 3:00 pm, and if he's to spend 27 hours on the journey he would reach destination z at 6:00 pm the next day

7 0
2 years ago
I need a thesis statement about Engineers as Leaders.
algol [13]

Answer:

Engineers are a very beneficial contribution in which offers great solutions to national problems.

5 0
2 years ago
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