1. Add:
(i) 5xy, -2xy, -11xy, 8xy
(iv) 3a - 2b + c, 5a + 8b -70​

Answer:

[tex]T_2=315.69k[/tex]

Explanation:

Initial Temperature [tex]T_1=500K[/tex]

Initial Pressure [tex]P_1=1000kPa[/tex]

Final Pressure [tex]P_2=200kPa[/tex]

Generally the gas equation is mathematically given by

[tex]\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{n-1}{n}}[/tex]

Where

n for [tex]CO=1.4[/tex]

Therefore

[tex]\frac{T_2}{500}=\frac{200}{1000}^{\frac{1.4-1}{1.4}}[/tex]

[tex]T_2=315.69k[/tex]

Complete Question

Nitrogen (N2) flows through a pipe, entering at at 4 m/sec at 1000 kPa, 2270C. For a pipe inside diameter of 3 cm, find the volumetric flow rate (m3/sec) and the mass flow rate of the gas (kg/sec) assuming you have an ideal gas Then using your ideal gas mass flow rate find the rate at which enthalpy enters the pipe (kJ/sec) NO Cp, Cv, k permitted

Answer:

[tex]H=9.91kJ/sec[/tex]

Explanation:

From the question we are told that:

Velocity [tex]v=4 m/sec[/tex]

Pressure [tex]P=1000kPa[/tex]

Temperature [tex]T=227 \textdegree C[/tex]

Diameter [tex]d=3cm=>0.03m[/tex]

Generally the equation for volumetric Flow Rate is mathematically given by

[tex]V_r=(\frac{\pi*d^2}{4}v)[/tex]

[tex]V_r=(\frac{\pi*(0.03)^2}{4} *4)[/tex]

[tex]V_r=0.002827m^3/s[/tex]

Generally the equation for mass Flow Rate is mathematically given by

[tex]m_r=\frac{PV_r}{RT}[/tex]

[tex]m_r=\frac{1000*0.002827}{0.297*(227+273)}[/tex]

[tex]m_r=0.019kg/sec[/tex]

Generally the equation for mass Flow Rate is mathematically given by

Using gas Table for enthalpy Value

[tex]T=500K=>h=520.75kg[/tex]

Therefore

[tex]H=mh[/tex]

[tex]H=0.019*520.75[/tex]

[tex]H=9.91kJ/sec[/tex]

Answer:

Tradicionalmente, se han distinguido 5 tipos de mantenimiento, que se diferencian entre sí por el carácter de las tareas que incluyen:

Explanation:

Mantenimiento Correctivo: Es el conjunto de tareas destinadas a corregir los defectos que se van presentando en los distintos equipos y que son comunicados al departamento de mantenimiento por los usuarios de los mismos.

Mantenimiento Preventivo: Es el mantenimiento que tiene por misión mantener un nivel de servicio determinado en los equipos, programando las intervencions de sus puntos vulnerables en el momento más oportuno. Suele tener un carácter sistemático, es decir, se interviene aunque el equipo no haya dado ningún síntoma de tener un problema.

Mantenimiento Predictivo: Es el que persigue conocer e informar permanentemente del estado y operatividad de las instalaciones mediante el conocimiento de los valores de determinadas variables, representativas de tal estado y operatividad. Para aplicar este mantenimiento, es necesario identificar variables físicas (temperatura, vibración, consumo de energía, etc.) cuya variación sea indicativa de problemas que puedan estar apareciendo en el equipo. Es el tipo de mantenimiento más tecnológico, pues requiere de medios técnicos avanzados, y en ocasiones, de fuertes conocimientos matemáticos, físicos y/o técnicos.

Mantenimiento Cero Horas (Overhaul): Es el conjunto de tareas cuyo objetivo es revisar los equipos a intervalos programados bien antes de que aparezca ningún fallo, bien cuando la fiabilidad del equipo ha disminuido apreciablemente de manera que resulta arriesgado hacer previsiones sobre su capacidad productiva. Dicha revisión consiste en dejar el equipo a Cero horas de funcionamiento, es decir, como si el equipo fuera nuevo. En estas revisiones se sustituyen o se reparan todos los elementos sometidos a desgaste. Se pretende asegurar, con gran probabilidad un tiempo de buen funcionamiento fijado de antemano.

Mantenimiento En Uso: es el mantenimiento básico de un equipo realizado por los usuarios del mismo. Consiste en una serie de tareas elementales (tomas de datos, inspecciones visuales, limpieza, lubricación, reapriete de tornillos) para las que no es necesario una gran formación, sino tal solo un entrenamiento breve. Este tipo de mantenimiento es la base del TPM (Total Productive Maintenance, Mantenimiento Productivo Total).

Explanation:

Find a minimal set of cycles that covers all vertices and edges of the circuit graph. For each cycle, define a "mesh" current, and write the Kirchhoff's Voltage Law (KVL) equation with respect to each of the edges in the cycle. Where an edge is part of more than one cycle, all current(s) defined for the edge will contribute to the voltage there.

This will give as many equations as there are mesh currents. Solve the resulting system of equations. The (signed) sum of the mesh currents through any edge is the current in that circuit branch.

__

Example

Consider the attached circuit. It shows mesh currents I1, I2, and I3 in graph cycles with those numbers. The KVL equations are ...

  mesh 1: I1(R3 +R2 +R1) -I2·R1 -I3·R2 = Vi (the voltage across the current source)

  mesh 2: -I1·R1 +I2(r1 +1/(sC)) -I3(1/(sC)) = Vs

  mesh 3: -I1·R2 -I2(1/(sC)) +I3(R2 +sL +1/(sC)) = 0

You will note that the matrix of equation coefficients is symmetric.

__

In this example, you will end with I1 as a function of Vi. If I1 is a given source value, that relation can be used to find Vi.

Answer:

forward voltage triggering

temperature triggering

dv/dt triggering

light triggering

gate triggering

Then turning off;

Turn off is accomplished by a "negative voltage" pulse between the gate and cathode terminals

Explanation:

hope it helps

Answer:

Explanation:

Increased by the constant. Take a very simple case.

4  +  5 +  6 = 15

The mean is 5  (obtained by dividing the total (15) by the number of terms (3).

Now add a constant say 6

4 + 6 = 10

5 + 6 = 11

6 + 6 = 12

Total = 33/3 = 11

So the mean 5 is increased by the constant 6.

Now do the same thing more symbolically.

4 + c

5 + c

6 + c

Total = 15 + 3c

Divide by 3 you get 5 + c

If you want a more formal proof involving n terms, leave a note.

Try to be as relax as possible.

Provide names of all parties involved.

Provide vehicle information and identification details.

Provide full names, address, registration numbers and insurance company details.

Explanation:

After a collision one may be confused, afraid and have no attention about the details that what happened because all the collision event happens in a short interval of time. So the first thing one should do during information exchange is to sit back and relax and be calm so that one can remind the things at some extent. After that provide all the details about injured people and the involved vehicles.

Answer:

some of the professions and human resource related to engineering are:

Aerospace engineering involves the study, design and development of spacecrafts using Core science principles.

Electrical engineering involves the study and application of core science principles especially physics and mathematics into providing Electrical related solutions

Explanation:

Engineering is a major branch of applied science. In general Engineering is concerned with the design and building of engines ( i.e. application of scientific/science facts   )

some of the professions and human resource related to engineering are:

Aerospace engineering involves the study, design and development of spacecrafts using Core science principles.

Electrical engineering involves the study and application of core science principles especially physics and mathematics into providing Electrical related solutions

Answer:

X = 1 (1st digit in the code)

Y = 0 (2nd digit)

Z = 4 (3rd multiplier digit)

104 → 10 × 10^4 Ω

→ 10 × 10000Ω

→ 100 kΩ

resistors are marked 104, 105, 205, 751, and 754. The resistor marked with 104 should be 100kΩ (10x10^4), 105 would be 1MΩ (10x10^5), and 205 is 2MΩ (20x10^5). 751 is 750Ω (75x10^1), and 754 is 750kΩ (75x10^4).

Here we need to understand how a code in a resistor gives us information on the resistor. Here we will see that the code means that the resistance is 100,000 Ω.

When we use numbers, let's assume that we have 3 single-digit numbers abc.

So if the code in our resistor is abc, this will mean that the resistance of the resistor is:

ab×10^c Ω

Using this general rule we can see that if the code is 104, then the resistance will be:

r = 10×10^4 Ω

 = 100,000 Ω

Then we can conclude that the resistive value of this resistor is  100,000 Ω

If you want to learn more, you can read:

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Answer:

[tex]Q_2 = 32[/tex] mL/s

Explanation:

Given :

The flow is incompressible viscous flow.

The initial flow rate, [tex]Q_1[/tex] = 1 mL/s

Initial diameter, [tex]D_1= D_0[/tex]

Initial length, [tex]L_1=L_0[/tex]

The initial pressure difference to maintain the flow, [tex]P_1=P_0[/tex]

We know for a viscous flow,

[tex]$\Delta P = \frac{32 \mu V L}{D^2}$[/tex]

[tex]$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$[/tex]

[tex]$Q \propto \Delta P \times D^4$[/tex]

[tex]$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{1}{32}$[/tex]

∴ [tex]Q_2 = 32[/tex] mL/s

The flow rate if the pressure difference was increased to 2P0 and the diameter was increased to 2D0 is; Q2 = 32 mL/s

We are given;

Initial flow rate; Q1 = 1 mL/s

Initial uniform diameter; D0

Initial Length; L0

Initial Pressure difference; P0

Relationship between pressure, flow rate and diameter for vicious flow is given by;

Q1/Q2 = (P1/P2) × (D1/D2)⁴

Where;

Q1 is initial flow rate

Q2 is final flow rate

P1 is initial pressure difference

P2 is final pressure difference

D1 is initial diameter

D2 is final diameter

We are told that the pressure difference was increased to 2P0 and the diameter was increased to 2D0. Thus;

P2 = 2P0

D2 = 2D0

Thus;

1/Q2 = (P0/2P0) × (D0/2D0)⁴

>> 1/Q2 = ½ × (½)⁴

1/Q2 = 1/32

Q2 = 32 mL/s

Read more about vicious flow at; https://brainly.com/question/2684299

Answer:

sorry but I can't understand this Language.

Explanation:

unable to answer sorry

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

[tex]$J=\frac{\pi}{32}d^4$[/tex]

[tex]$J=\frac{\pi}{32}\times (46)^4$[/tex]

J = 207.6 [tex]mm^4[/tex]

So the shear stress at point  A is :

[tex]$\tau_A =\frac{Tc_A}{J}$[/tex]

[tex]$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$[/tex]

[tex]$\tau_A = 4913.29 \ MPa$[/tex]

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

Explanation:

conductor, flux, and movement of conductor in magnetic field are some of the factors that induce emf.

Answer:

i really font onow why tbh eot you

Answer:

maybe take a really common toy kids play with or often see, find the average height for the toy and do the math to see how many of those toys stacked ontop of eachother would make up 2000 feet. For example (this isn't accurate btw just an idea of what it would sound like but) "Have you ever seen a barbie doll? well if you stack 400 barbie dolls ontop of their head it would be equal to 2000 feet."

Explanation:

sometimes taking common or beloved objects children have into your examples makes them have a better image of how small or how big something is.

Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

A) Classifying the soil according to USCS system

 ( using 2nd image attached below )

description of sand :

The soil is a coarse sand since  ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Answer: The tracer lathe is a roughing operation for the output shaft on rear wheel drive transmissions.

Explanation:

The causes of low compression in an engine cylinder and leakage past the exhaust can be identified by performing mechanical tests on the engine

The correct technician is Tech B; Blown head gasket would generally leak past the exhaust valve

The reason for arriving at the above selection is as follows:

A cylinder leakage test is generally performed after there is an indication of low compression from one or more cylinders. The test allows the identification of the part of the cylinder that is leaking.

The cylinder that shows sign of low compression is often the location that the cylinder leakage test is performed, although, the test can be performed on every cylinder

A cylinder leakage test is valid only when

Therefore, the positioning of the piston in the bottom dead center during a leakage test as mentioned by Tech A  is not correct

The route of coolant in the engine are sealed by the cylinder head gasket, where there is a blown head gasket, the coolant will be allowed to leak into the cylinders and into the exhaust, causing white smoke and sweet odor of antifreeze

Therefore, Tech B is correct, coolant leakage into the exhaust is an indication of a blown head gasket

Learn more about engine mechanical testing here:

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Answer:

Explanation:

Minerals and rocks are the same. Aggregates of minerals form rocks. Minerals determine the texture of a rock. Most rocks are made of a single mineral type.

Complete question:

A steel component with a tensile strength of 800 MPa and fracture toughness Kic=20 MPa Nm is known to contain internal cracks (also called tunnel cracks) with the maximum length of 1.4 mm. This steel is being considered for use in a cyclic loading application for which the designed stresses vary from 0 to 420 MPa. Would you recommend using this steel in this application?

a. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.

b. Yes, this because the tensile strength of steel is much higher than the applied highest stress of 420 MPa.

c. Yes, this because the calculated critical stress to fracture for the cracks is higher than the highest applied stress of 420 MPa and the steel can withstand the stress of 420 MPa.

d. No. Although the calculated critical stress to fracture for the cracks is slightly higher than the highest applied stress of 420 MPa and the steel may withstand the static stress of 420 MPa, the cyclic loading may cause rapid fatigue fracture.

Answer:

A. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.

Explanation:

we are not sure if to recommend this alloy for this application given that this material has already been left to experience fatigue degradation. the cyclic load application brings about a growth in the crack. We know that cyclic loading is continuous loading that is useful for the testing of fatigue. Therefore the answer to this question is option a. We cannot make recommendations except fatigue testing has been carried out.

thank you!

Answer:

Data Rate Signaling

Answer:

  5.9 watts

Explanation:

The secondary voltage is the primary voltage multiplied by the turns ratio:

  (120 V)(41) = 4920 V

The power is the product of voltage and current:

  (4920 V)(1.2·10^-3 A) = (4.92)(1.2) W = 5.904 W

The power consumed is about 5.9 watts.

Answer:

No it is not permitted

Explanation:

It is not permitted  because as per NEC 410.21 policy no other conductor is allowed to be passed through integral junction box luminaries unless such conductor supply recessed luminaries.

The marking will show that the Luminaries is of the right construction or right installation to ensure that the the conductors ( in the outer boxes ) will not be exposed to temperatures greater than the conductor rating, hence the lack of marking makes it not to be permitted.

Answer:

Check it out here

Explanation:

Given :

Power, P = 20 kW

Speed, N = 430 rpm

Allowable shear stress, τ = 65 MPa

Torque in the shaft is given by :

[tex]$P=\frac{2 \pi NT}{60}$[/tex]

[tex]$T=\frac{60 \times 20 \times 10^3}{2 \pi \times 430}$[/tex]

T = 444.37 N.m

Diameter of the solid shaft is

[tex]$d=\sqrt[3]{\frac{16 T}{\pi \tau}}[/tex]

[tex]$d=\sqrt[3]{\frac{16 \times 444.37}{3.14 \times 65}}[/tex]

[tex]$d=\sqrt[3]{34.83} $[/tex]

d = 3.265 m

d = 326.5 mm

Internal diameter of the hollow shaft is :

[tex]$\frac{T}{\frac{\pi}{32} \left( d_0^4 - d_i^4 \right)}=\frac{\tau}{d_0/2}$[/tex]

[tex]$\frac{444.37}{\frac{3.14}{32} \left( 0.036^4 - d_i^4 \right)}=\frac{65 \times 10^6}{0.036/2}$[/tex]

[tex]$\frac{444.37}{0.09 \left( 1.6 \times 10^{-6} - d_i^4 \right)}=\frac{65 \times 10^6}{0.018}$[/tex]

[tex]$\frac{7.99}{ \left( 1.6 \times 10^{-6} - d_i^4 \right)}=5850000$[/tex]

[tex]$1.3\times 10^{-6} = 1.6 \times 10^{-6} - d_i^4 \right)}$[/tex]

[tex]$d_i^4=300000$[/tex]

[tex]$d_i = 23.40$[/tex] mm

Percentage savings in the weight is given by :

Percentage saving = [tex]$\frac{W_{solid}-W_{hollow}}{W_{solid}}\times100$[/tex]

                                 [tex]$=\frac{V_{solid}-V_{hollow}}{V_{solid}}\times100$[/tex]

                                 [tex]$=\frac{d^2 - (d_0^2 - d_i^2)}{d^2} \times 100$[/tex]

                               [tex]$=\frac{(326.5)^2 - (0.036^2 - (32.40)^2)}{(326.5)^2} \times 100$[/tex]

                                 [tex]$=\frac{106602 - \left(1.29 \times 10^{-3} - 1049.76 \right)}{106602} \times 100$[/tex]

                                  [tex]$=\frac{106602 - 1049 }{106602} \times 100$[/tex]

                                  [tex]$=\frac{105553 }{106602} \times 100$[/tex]

                                  = 99.01 %