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loris [4]
3 years ago
6

Disc brake rotors that are too thin cannot handle as much heat and will experience ___________.

Engineering
1 answer:
konstantin123 [22]3 years ago
6 0

Answer:

brake fade

Explanation:

Among the most frequent problems in the brake disc is their overheating. When there is overheating, the temperature of the disc rises to critical values, thus the brake pad starts to slide with respect to the disc, and the efficiency of the brake system decreases to a minimum.

Overheating of the brake disc is detected when inspecting the parts. Steel is the most common base material in the creation of brake discs. When heated, the material becomes its color. A disc of steel at critical temperature, turns bright orange and later on cooling becomes purple.

If a color change is seen on the brake discs when inspecting, a service station must be contacted without delay for an in-depth examination. After detecting the problem, the brake discs and pads will be changed in a mandatory way.

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Explain by Research how a basic generator works ? using diagram<br>​
natulia [17]
Correcto no se muy bien de que se trata el tema porque está en inglés.
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2 years ago
¿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

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Sales de ácidos grasos

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6 0
3 years ago
The time factor for a doubly drained clay layer
Margarita [4]

Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

T_v=\frac{\pi }{4}(\frac{U}{100})^2

Solving for 'U' we get

\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i)\frac{z}{H}=0.25=U=0.71 = 71% consolidation

ii)\frac{z}{H}=0.5=U=0.45 = 45% consolidation

iii)\frac{z}{H}=0.75U=0.3 = 30% consolidation

Part b)

The degree of consolidation is given by

\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm

Thus a settlement of 50.46 centimeters has occurred

For time factor 0.7, U is given by

T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59

thus consolidation of 85.59 % has occured if time factor is 0.7

The degree of consolidation is given by

\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm

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You did not put the following in
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