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Brilliant_brown [7]
3 years ago
8

List three reasons for surfacing metals.

Engineering
2 answers:
Dmitrij [34]3 years ago
8 0
Chemical,fatigue,and metal transfer or adhesion.
Solnce55 [7]3 years ago
4 0

Answer:

JUST BECUASE

WHY NOT

LOOKS BETTER

Explanation:

You might be interested in
A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i
Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter d_{o} = 12.8 mm

Final diameter d_{f} = 10.7

Engineering stress  \alpha _{E} = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4(d_{o} )²  ; Ag = π/4(d_{f} )²

% = π/4 [ ( (d_{f} )² - (d_{o} )²) / ( π/4  (d_{o} )²) ]

= ( (d_{f} )² - (d_{o} )²) / (d_{o} )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress  \alpha _{T} = \alpha _{E} ( 1 +  E_{E} )

E_{E}  is engineering strain

E_{E}  = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49  

E_{E} = 0.431

so we substitute the value of E_{E}  into our initial equation;

True stress  \alpha _{T} = 460 ( 1 +  0.431)

True stress  \alpha _{T} = 460 (1.431)

True stress  \alpha _{T} = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

6 0
2 years ago
I need ideas for what to build because I have some spare wood.
Misha Larkins [42]

Answer:

small guitar with no strings?

Explanation:

it would be fun to make i think

6 0
3 years ago
The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature
Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

P_{1} = 5 Mpa

T_{1} = 1623°C

                       = 1896 K

V_{1} = 0.05 m^{3}

Also given \frac{V_{2}}{V_{1}} = 20

Therefore, V_{2} = 1  m^{3}

R = 0.27 kJ / kg-K

C_{V} = 0.8 kJ / kg-K

Also given : P_{1}V_{1}^{1.25}=C

   Therefore, P_{1}V_{1}^{1.25} = P_{2}V_{2}^{1.25}

                     5\times 0.05^{1.25}=P_{2}\times 1^{1.25}

                     P_{2} = 0.1182 MPa

a). Work transfer, δW = \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

                                  \left [\frac{5\times 0.05-0.1182\times 1}{1.25-1}  \right ]\times 10^{6}

                              = 527200 J

                             = 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

   = \frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

  =\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W

  =\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200

  = 197.7 kJ

6 0
3 years ago
What is an example of an innovation in logistics?
Scorpion4ik [409]

Answer:

geolocation technologies, drones, automated transportation vehicles

Explanation:

3 0
2 years ago
The brakes are being bled on a passenger vehicle with a disc/drum brake system. Technician A says that the drums should be remov
exis [7]

The brakes are being bled on a passenger vehicle with a disc/drum brake system is described in the following

Explanation:

1.Risk: Continued operation at or below Rotor Minimum Thickness can lead to Brake system failure. As the rotor reaches its minimum thickness, the braking distance increases, sometimes up to 4 meters. A brake system is designed to take kinetic energy and transfer it into heat energy.

2.Since the piston needs to be pushed back into the caliper in order to fit over the new pads, I do open the bleeder screw when pushing the piston back in. This does help prevent debris from traveling back through the system and contaminating the ABS sensors

3.There are three methods of bleeding brakes: Vacuum pumping. Pressure pumping. Pump and hold.

4,Brake drag is caused by the brake pads or shoes not releasing completely when the brake pedal is released. ... A worn or corroded master cylinder bore causes excess pedal effort resulting in dragging brakes. Brake Lines and Hoses: There may be pressure trapped in the brake line or hose after the pedal has been released.

4 0
3 years ago
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