1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kramer
2 years ago
9

Should you recommend changing fluids based on the color or smell of the fluid?

Engineering
1 answer:
iragen [17]2 years ago
3 0

Answer:

hi

Explanation:

Should you recommend changing fluids based on the color or smell of the fluid?

Only if the color is brown and the fluid smells burned

Yes, always.

No. Stick with the manufacturer's recommendations.

Only if the fluid is red and smells sweet

You might be interested in
¿Qué aditivo se debe incorporar a la masa de hormigón para aumentar su resistencia frente a los ciclos alternados de hielo-deshi
tamaranim1 [39]

Answer:

Los aditivos que deben incorporarse a la masa de concreto para aumentar su resistencia a los ciclos alternos de congelación y descongelación son;

1. Agentes de arrastre de aire (AEA) o

2. Materiales poliméricos súper absorbentes

Explanation:

La resistencia alterna de los ciclos de congelación y descongelación en el concreto puede aumentarse mediante la adición de agentes de arrastre de aire.(AEA) que es un surfactante, crea burbujas de aire muy pequeñas en el concreto resultante para mejorar la durabilidad y resistencia del cemento al ciclo repetido de congelación y descongelación o materiales poliméricos súper absorbentes

Ejemplos de agentes de arrastre de aire son;

Sulfonatos alcalinos

Acidos de resinas sulfonadas

Sales de ácidos grasos

Ejemplos de materiales poliméricos superabsorbentes son;

SAP0.26CT

SAP0.39PT.

6 0
4 years ago
an inclined manometer is connected to a pitor tube to measure the velocity at the center of a circular duct. If the inclined man
jekas [21]
57.5 m/s
I did 2.3/0.04
I’m not sure if it’s correct though
8 0
3 years ago
What are some aircraft aging problems? How can you as an Aviation Maintenance Manager monitor problems that relate to aircraft f
julia-pushkina [17]

Answer:

Answered

Explanation:

The two key processes that lead to aircraft ageing are fatigue and corrosion. These processes generally affect the aircraft structure, but can also affect wiring, flight controls, power plants, and other components. Fatigue and corrosion can work independently from one another, or they can interact. The interaction between fatigue and corrosion can increase the rate of ageing to a greater extent than that due to either process alone.

Fatigue predominately takes place in metal components, but it can also affect non-metallic materials. Fatigue occurs through cyclic loading patterns, where a component is repeatedly loaded. Bending a metal paper clip backwards and forwards is an example of fatigue; the paper clip will not break if only bent once, however, if it is repeatedly loaded, it will eventually break. Fatigue failures will often take place at loads much lower than the materials ultimate strength.

Generally, the initiation point for fatigue will be a microscopic crack that forms at a location of high stress, such as a hole, notch, or material imperfection. The crack will then grow as loads are repeatedly applied. If not detected and treated, the crack will eventually grow to a critical size and failure will occur at loads well below the original strength of the material.

The relationship between repetitive loading and fatigue crack growth, creates a link between fatigue related ageing, the number of flight cycles, and the number of flight hours that an aircraft has accumulated.

Aircraft components that are susceptible to fatigue include most structural components such as the wings, the fuselage, and the engine.

During initial manufacturing the research department is responsible for designing the aircraft to withstand fatigue. During operations the fatigue risk management of aviation maintenance will try to rectify the problems due to fatigue.

4 0
3 years ago
Sea B = 5.00 m a 60.0°. Sea C que tiene la misma magnitud que A y un ángulo de dirección mayor que el de A en 25.0°. Sea A ⦁ B =
uranmaximum [27]

Answer:

\| \vec A \| = 6.163\,m

Explanation:

Sean A, B y C vectores coplanares tal que:

\vec A = (\| \vec A \|\cdot \cos \theta_{A},\| \vec A \|\cdot \sin \theta_{A}), \vec B = (\| \vec B \|\cdot \cos \theta_{B},\| \vec B \|\cdot \sin \theta_{B}) y \vec C = (\| \vec C \|\cdot \cos \theta_{C},\| \vec C \|\cdot \sin \theta_{C})

Donde \| \vec A \|, \| \vec B \| y \| \vec C \| son las normas o magnitudes respectivas de los vectores A, B y C, mientras que \theta_{A}, \theta_{B} y \theta_{C} son las direcciones respectivas de aquellos vectores, medidas en grados sexagesimales.

Por definición de producto escalar, se encuentra que:

\vec A \,\bullet\, \vec B = \|\vec A \| \| \vec B \| \cos \theta_{B}\cdot \cos \theta_{A} + \|\vec A \| \| \vec B \| \sin \theta_{B}\cdot \sin \theta_{A}

\vec B \,\bullet\, \vec C = \|\vec B \| \| \vec C \| \cos \theta_{B}\cdot \cos \theta_{C} + \|\vec B \| \| \vec C \| \sin \theta_{B}\cdot \sin \theta_{C}

Asimismo, se sabe que \| \vec B \| = 5\,m, \theta_{B} = 60^{\circ}, \vec A \,\bullet \,\vec B = 30\,m^{2}, \vec B\, \bullet\, \vec C = 35\,m^{2}, \|\vec A \| = \| \vec C \| y \theta_{C} = \theta_{A} + 25^{\circ}. Entonces, las ecuaciones quedan simplificadas como siguen:

30\,m^{2} = 5\|\vec A \| \cdot (\cos 60^{\circ}\cdot \cos \theta_{A} + \sin 60^{\circ}\cdot \sin \theta_{A})

35\,m^{2} = 5\|\vec A \| \cdot [\cos 60^{\circ}\cdot \cos (\theta_{A}+25^{\circ}) + \sin 60^{\circ}\cdot \sin (\theta_{A}+25^{\circ})]

Es decir,

30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})

35\,m^{2} = \| \vec A \| \cdot [2.5\cdot \cos (\theta_{A}+25^{\circ})+4.330\cdot \sin (\theta_{A}+25^{\circ}})]

Luego, se aplica las siguientes identidades trigonométricas para sumas de ángulos:

\cos (\theta_{A}+25^{\circ}) = \cos \theta_{A}\cdot \cos 25^{\circ} - \sin \theta_{A}\cdot \sin 25^{\circ}

\sin (\theta_{A}+25^{\circ}) = \sin \theta_{A}\cdot \cos 25^{\circ} + \cos \theta_{A} \cdot \sin 25^{\circ}

Es decir,

\cos (\theta_{A}+25^{\circ}) = 0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A}

\sin (\theta_{A}+25^{\circ}) = 0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A}

Las nuevas expresiones son las siguientes:

30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})

35\,m^{2} = \| \vec A \| \cdot [2.5\cdot (0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A})+4.330\cdot (0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A})]

Ahora se simplifican las expresiones, se elimina la norma de \vec A y se desarrolla y simplifica la ecuación resultante:

30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})

35\,m^{2} = \| \vec A \| \cdot (4.097\cdot \cos \theta_{A} +2.865\cdot \sin \theta_{A})

\frac{30\,m^{2}}{2.5\cdot \cos \theta_{A}+ 4.330\cdot \sin \theta_{A}} = \frac{35\,m^{2}}{4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}}

30\cdot (4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}) = 35\cdot (2.5\cdot \cos \theta_{A}+4.330\cdot \sin \theta_{A})

122.91\cdot \cos \theta_{A} + 85.95\cdot \sin \theta_{A} = 87.5\cdot \cos \theta_{A} + 151.55\cdot \sin \theta_{A}

35.41\cdot \cos \theta_{A} = 65.6\cdot \sin \theta_{A}

\tan \theta_{A} = \frac{35.41}{65.6}

\tan \theta_{A} = 0.540

Ahora se determina el ángulo de \vec A:

\theta_{A} = \tan^{-1} \left(0.540\right)

La función tangente es positiva en el primer y tercer cuadrantes y tiene un periodicidad de 180 grados, entonces existen al menos dos soluciones del ángulo citado:

\theta_{A, 1} \approx 28.369^{\circ} y \theta_{A, 2} \approx 208.369^{\circ}

Ahora, la magnitud de \vec A es:

\| \vec A \| = \frac{35\,m^{2}}{4.097\cdot \cos 28.369^{\circ} + 2.865\cdot \sin 28.369^{\circ}}

\| \vec A \| = 6.163\,m

8 0
3 years ago
Define centrifugal pump. Give the construction and working of centrifugal pump. ​
fenix001 [56]
Centrifugal pump is a hydraulic machine which converts mechanical energy into hydraulic energy by the use of centrifugal force acting on the fluid. These are the most popular and commonly used type of pumps for the transfer of fluids from low level to high level.
7 0
3 years ago
Other questions:
  • In C++ the declaration of floating point variables starts with the type name float or double, followed by the name of the variab
    14·1 answer
  • - You have a bin wrench turning a 1/2 13 UNC bolt. You overcome 1200 lbs of resistance when you
    5·1 answer
  • A four-lane freeway (two lanes in each direction) is located on rolling terrain and has 12-ft lanes, no lateral obstructions wit
    14·1 answer
  • How does emotion affect a persons driving
    15·1 answer
  • We can model a certain battery as a voltage source in series with a resistance. The open-circuit voltage of the battery is 10 V
    10·1 answer
  • Injector orifice patterms and size will affect propellant mixing and distribution. a)-True b)-False
    13·1 answer
  • Give me uses of a grinding machine in agriculture.
    15·1 answer
  • A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water. If the base is in the surface of water,
    13·1 answer
  • A glass tube is inserted into a flowing stream of water with one opening directed upstream and the other end vertical. If the wa
    9·1 answer
  • Please help me on this it’s due now
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!