Air is compressed in a well insulated compressor from 95 kPa and 27 C to 600 kPa and 277 C. Use the air tables; assume negligible changes in kinetic and potential energy. Find the isentropic efficiency of the compressor. Find the exit temperature of the air if the compressor was reversible.

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.

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.

Answer:

sorry but I can't understand this Language.

Explanation:

unable to answer sorry

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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:

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).

Answer:

80⁰C

Explanation:

80°C.

Normal flat plate collectors can deliver heat at temperatures up to 80°C. Deficiency rates for normal flat plate collectors can be classified as visual losses, which produce with cumulative angles of the incident sunshine, and thermal losses, which upsurge fast with the working temperature intensities

Answer: all of the above

Explanation:

To drive safely in the station, the correct option is D. All of the above.

It should be noted that when a person is driving, it's essential for the person to drive carefully in a station.

One should not smoke in a station. Also, it's important that the driver should turn on their hazard lights

Furthermore, it's important for one to roll their windows down, turn off the radio and also maintain idle speed.

In conclusion, the correct option is D.

Read related link on:

https://brainly.com/question/18889428

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:

  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:

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

Explanation:

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

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.

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]

This question is incomplete, the complete question is;

Plexiglas, polymethylmethacrylate (PMMA), is often used as display windows and cases in art galleries and museums. If its flame imparts 32 kW/m^2 to the surface when it burns estimate the energy release rate of a 3 m^2 square sheet (one side). Assume its vaporization temperature is 350°C. Calculate with units, provide work.

a ⇒ [tex]q_{rad[/tex] = σT⁴, where σ = 5.67 × 10⁻¹¹ kW/m².K⁴

b ⇒ [tex]q"[/tex] = [tex]q_{flame[/tex] -  [tex]q_{rad[/tex]

c ⇒ [tex]m"[/tex] =  [tex]q"[/tex]/L, where L = 1.6 kJ/g

d ⇒ Q =  [tex]m"[/tex] × A × ΔHc, where ΔHc is 24.9 kJ/g

Answer:

a) [tex]q_{rad[/tex] = 8.54 kW/m²

b) [tex]q"[/tex] = 23.46 kW/m²

c) [tex]m"[/tex] = 14.6625 g/m²

d) Q = 1095.2888 kJ

Explanation:

Given the data in the question;

[tex]q_{flame[/tex] = 32 kW/m²

Area; A = 3m²

vaporization temperature; T = 350°C = ( 350 + 273 )K = 623 K

Now,

a) ⇒ [tex]q_{rad[/tex] = σT⁴, where σ = 5.67 × 10¹¹ kW/m².K⁴

we substitute

[tex]q_{rad[/tex] = ( 5.67 × 10⁻¹¹ kW/m².K⁴ ) × ( 623 K)⁴

[tex]q_{rad[/tex] = ( 5.67 × 10⁻¹¹ kW/m².K⁴ ) × 150644120641 K⁴

[tex]q_{rad[/tex] = 8.54 kW/m²

b) ⇒ [tex]q"[/tex] = [tex]q_{flame[/tex] - [tex]q_{rad[/tex]

we substitute

[tex]q"[/tex] = 32 kW/m² - 8.54 kW/m²

[tex]q"[/tex] = 23.46 kW/m²

c) ⇒ [tex]m"[/tex] =  [tex]q"[/tex]/L, where L = 1.6 kJ/g

we substitute

[tex]m"[/tex] =  23.46 / 1.6

[tex]m"[/tex] = 14.6625 g/m²

d) ⇒ Q =  [tex]m"[/tex] × A × ΔHc, where ΔHc is 24.9 kJ/g

we substitute

Q =  14.6625 × 3 × 24.9

Q = 1095.2888 kJ

Answer:

Data Rate Signaling

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.

Answer:

Voltage Regulator

Technician A is correct.

Explanation:

Technician B is not correct.  The voltage regulator is not installed between the output terminal of the alternator and the positive terminal of the battery as claimed by Technician B.  Technician A's opinion that the voltage regulator controls the strength of the rotor's magnetic field is correct.  The computer can also be used to control the output of the alternator by controlling the field current.

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.

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:

Force is a kinetic quantity. Force = m x a. Hence, the mass of the body has an effect on force calculation.

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]

Answer:

hola

Explanation:

Explanation:

so its an question?..........