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Artemon [7]
3 years ago
9

Technician A says that the low level brake fluid switch on a master cylinder will turn on the brake warning light when the syste

m is low on fluid. Technician B says that the low level switch also monitors the condition of the fluid and will activate the warning light when the brake fluid needs to be replaced. Who is correct?
Engineering
1 answer:
igor_vitrenko [27]3 years ago
4 0

Answer:

Technician A

Explanation:

The brake fluid switch or the brake switch is a sensor provided in the master cylinder monitors the brake fluid level in the vehicle. It is found inside the brake fluid reservoir. When the brake fluid level is too low, it warns the ABS by sending a signal. This device keeps us and the vehicle safe.

Thus in the context, technician A is correct.

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A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m K), and the wire/sheath interface i
Semmy [17]

Answer:

maximum allowable electrical power=4.51W/m

critical radius of the insulation=13mm

Explanation:

Hello!

To solve this heat transfer problem we must initially draw the wire and interpret the whole problem (see attached image)

Subsequently, consider the heat transfer equation from the internal part of the tube to the external air, taking into account the resistance by convection, and  conduction as shown in the attached image

to find the critical insulation radius we must divide the conductivity of the material by the external convective coefficient

r=\frac{k}{h} =\frac{0.13}{10}=0.013m=13mm

3 0
3 years ago
The yield stress of a steel is 250Mpa. A steel rod used for implant in a femurneeds to withstand 29KN. What should the diameter
OleMash [197]

Answer:

r = 1.922 mm

Explanation:

We are given;

Yield stress; σ = 250 MPa = 250 N/mm²

Force; F = 29 KN = 29000 N

Now, formula for yield stress is;

σ = F/A

A = F/σ

Where A is area = πr²

Thus;

r² = 2900/250π

r² = 3.6924

r = √3.6924

r = 1.922 mm

3 0
3 years ago
Which is not one of the primary characteristic of unit testing:
Neporo4naja [7]

Answer:

Unit testing is a software development process in which the smallest testable parts of an application, called units, are individually and independently scrutinized for proper operation. This testing methodology is done during the development process by the software developers and sometimes QA staff. The main objective of unit testing is to isolate written code to test and determine if it works as intended.

Unit testing is an important step in the development process, because if done correctly, it can help detect early flaws in code which may be more difficult to find in later testing stages.

Unit testing is a component of test-driven development (TDD), a pragmatic methodology that takes a meticulous approach to building a product by means of continual testing and revision. This testing method is also the first level of software testing, which is performed before other testing methods such as integration testing. Unit tests are typically isolated to ensure a unit does not rely on any external code or functions. Testing can be done manually but is often automated. It might be helpful

6 0
2 years ago
All machines have three fundamental hazards: moving parts, point of operation, and?
OlgaM077 [116]

Answer:

All machines have three fundamental hazards: moving parts, point of operation, and the power transmission.

Explanation:

The unit that supplies power to the machine is a critical hazard due to high energy sources being potential fatal if proper protocols are not followed. This is why lockout tagout (LOTO) measures are put in place in order to protect people while they work on equipment.

3 0
2 years ago
As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial tempera
saveliy_v [14]

Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length  = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod \overline T = 548 \ K

\rho = 7900 \ kg/m^3

K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K

\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\  B_i = \dfrac{h(\rho/4)}{K} \\ \\  =0.657

Here, we can't apply the lumped capacitance method, since Bi > 0.1

\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\

0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\  0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o   \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81

t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\  t_f= 2162.5 \\ \\ t_f = 36 mins

However, on a single rod, the energy extracted is:

\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) )  \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\  \theta = 1.54 \times 10^7 \ J

Hence, for centerline temperature at 50 °C;

The surface temperature is:

T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}

5 0
3 years ago
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