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soldier1979 [14.2K]
3 years ago
15

Sea A una matriz 3x3 con la propiedad de que la transformada lineal x → Ax mapea R³ sobre R³.

Engineering
1 answer:
skelet666 [1.2K]3 years ago
7 0

Answer:

ax

Explanation:

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g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What
earnstyle [38]

Answer:

The friction factor is 0.303.

Explanation:

The flow velocity (v), measured in meters per second, is determined by the following expression:

v = \frac{4\cdot \dot V}{\pi \cdot D^{2}} (1)

Where:

\dot V - Flow rate, measured in cubic meters per second.

D - Diameter, measured in meters.

If we know that \dot V = 0.01\,\frac{m^{3}}{s} and D = 0.05\,m, then the flow velocity is:

v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}

v \approx 5.093\,\frac{m}{s}

The density and dinamic viscosity of the glycerin at 20 ºC are \rho = 1260\,\frac{kg}{m^{3}} and \mu = 1.5\,\frac{kg}{m\cdot s}, then the Reynolds number (Re), dimensionless, which is used to define the flow regime of the fluid, is used:

Re = \frac{\rho\cdot v \cdot D}{\mu} (2)

If we know that \rho = 1260\,\frac{kg}{m^{3}}, \mu = 1.519\,\frac{kg}{m\cdot s}, v \approx 5.093\,\frac{m}{s} and D = 0.05\,m, then the Reynolds number is:

Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }

Re = 211.230

A pipeline is in turbulent flow when Re > 4000, otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor (f), dimensionless, is determined by the following expression:

f = \frac{64}{Re}

If we get that  Re = 211.230, then the friction factor is:

f = \frac{64}{211.230}

f = 0.303

The friction factor is 0.303.

4 0
2 years ago
Which of the following would be addressed by an employer completing an EAP template?
zloy xaker [14]
I can help you, but what are the options that were given to you?
4 0
2 years ago
A room has a width of 14.1 feet, a length of 15.5 feet, and a ceiling height of 12.0 ft. The average flow rate for this room's a
prohojiy [21]

Answer:

Your question lacks the time required hence i will calculate the Average flow rate using a general concept and an assumed time value of 25 seconds  

ANSWER : 104.904 ft^3/sec

Explanation:

General concept : Average flow rate is the volume of fluid per unit time through an area

Hence the average flow rate of the air conditioning unit of this room

Volume of the room / time taken for the air to cycle the room = v / t

assuming the time taken = 25 seconds

volume of room = width * length * height

                          = 14.1 * 15.5 * 12 = 2622.6 ft^3

Average flow rate = V/ t

                              = 2622.6 / 25  = 104.904 ft^3/sec

8 0
3 years ago
What is the hardest part of engineering?
Vikki [24]

ANSWER:

Aerospace Engineering. ...

Chemical Engineering. ...

Biomedical Engineering.

EXPLANATION:

This is all i know but ... I hope this helps~

7 0
1 year ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in
Sonja [21]

The load is 17156 N.

<u>Explanation:</u>

First compute the flexural strength from:  

σ = FL / πR^{3}

   = 3000 \times (40 \times 10^-3) / π (5 \times 10^-3)^3

σ = 305 \times 10^6 N / m^2.

We can now determine the load using:

F = 2σd^3 / 3L

  = 2(305 \times 10^6) (15 \times 10^-3)^3 / 3(40 \times 10^-3)

F = 17156 N.  

7 0
3 years ago
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