11. Which of the following is the brake fluid most often used?

Engineering

2 answers:

Olenka [21]3 years ago

4 0

Dot 3 is mostly used in a lot of v4 and v6

Nadya [2.5K]3 years ago

4 0

Answer:

B. DOT 3

Explanation:

The braking system of a vehicle uses hydraulic power generated by a master cylinder to activate the four-wheel brake assemblies. The hydraulic power is transmitted throughout the brake system via brake fluid. Brake fluid is made from a silicone based or glycol based product. .The three classes of brake fluid commonly used in modern automobiles are as follows:

DOT 3, DOT 4, and DOT 5. DOT stands for

Department of Transportation specification grades; the increasing numbers corresponds to increasing boiling point. DOT 3 and DOT 4 are glycol based, and DOT 3 most commonly used in disc brake systems. The glycol base is hygroscopic means it attract water from the environment.

brake fluid should be disposed properly. it should not be exposed to fire as this mit pose a ue fire hazard. t shouldn't be thrown into septic tanks.

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A mass of 1.9 kg of air at 120 kPa and 24°C is contained in a gas-tight, frictionless piston–cylinder device. The air is now com

emmainna [20.7K]

Answer:

W=-260.66 kJ (negative answer means, that the work was done on the gas)

Explanation:

1) Convert temperature from C to K- T=24+273=297K- all temperature in the gas problems should be used in Kelvins;

2) We need to analyse type of the process- it is given, that the temperature is constant, so it is an Isothermal process, which means, that the equation of the process is: pV=const (constant);

3) Work, done on the system, should be calculated using the following equation: $W=\int\limits^{Vb}_{Va} {p} \, dV$

4) To calculate initical and final volumes (Va and Vb), we can use the following equation: pV=mRT, so V=mRT/p. Note, that the pressure is changing, thus we can calculate volumes for the both cases- initial and final, using initial (120kPa) and final (600kPa) pressures, in addition, we can find equation for the pressure, as function of the volume, which we need to use for the integration in step 3: p=mRT/V;

5) Now we can calculate the integral, given in the step 3: $W=mRT ln(\frac{Vb}{Va})$ . As we have pressure as a known values, we can re-write the equation, using pressures: $W=mRT ln(\frac{pa}{pb})=1.9*0.287*279*ln(\frac{120}{160})=-260.66 kJ$