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3 years ago

Identify how slip affects the strength of a metal

Engineering

1 answer:

Fiesta28 [93]3 years ago

5 0

My answer says there is a link but not,there isn’t so here a photo of my answer.

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The ___________ section of the bid package might include information concerning provisions for water and other utilities, sanita

dexar [7]

Answer:

Procurement Process

Explanation:

Procurement Process describes the series of activities that an organization partakes in to get products or services in order to achieve their goals. The choice of the procurement process is very important for the success of a construction project.

So during a bidding process, the procurement process is section where the organization will need to get water and other utilities, sanitation equipment or storage needed for the success of a construction project.

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4 years ago

What is engineering?

ANTONII [103]

Answer:

the branch of science and technology concerned with the design, building, and use of engines, machines, and structures. Anything that involves engines, wires, etc. basically

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3 years ago

Copper spheres of 20-mm diameter are quenched by being dropped into a tank of water that is maintained at 280 K . The spheres ma

Ivenika [448]

Answer:

The height of the water is 1.25 m

Explanation:

copper properties are:

Kc=385 W/mK

D=20x10^-3 m

gc=8960 kg/m^3

Cp=385 J/kg*K

R=10x10^-3 m

Water properties at 280 K

pw=1000 kg/m^3

Kw=0.582

v=0.1247x10^-6 m^2/s

The drag force is:

$F_{D} =\frac{1}{2} Co*p_{w} A*V^{2}$

The bouyancy force is:

$F_{B} =V*p_{w} *g$

The weight is:

$W=V*p_{c} *g$

Laminar flow:

$v_{T} =\frac{p_{c}-p_{w}*g*D^{2} }{18*u} =\frac{(8960-1000)*9.8*(20x10^{-3})^{2} }{18*0.00143} =1213.48 m/s$

Reynold number:

$Re=\frac{1000*1213.48*20x10^{-3} }{0.00143} \\Re>>1$

Not flow region

For Newton flow region:

$v_{T} =1.75\sqrt{(\frac{p_{c}-p_{w} }{p_{w} })gD }=1.75\sqrt{(\frac{8960-1000}{1000} )*9.8*20x10^{-3} } =2.186m/s$

$Re=\frac{1000*2.186*20x10^{-3} }{0.00143} =30573.4$

$Pr=\frac{\frac{u}{p} }{\frac{K}{pC_{p} } } =\frac{u*C_{p} }{k} =\frac{0.0014394198}{0.582} =10.31$

$Nu=2+(0.4Re^{1/2} +0.06Re^{2/3} )Pr^{2/5} (u/us)^{1/4} \\Nu=2+(0.4*30573.4^{1/2}+0.06*30573.4^{2/3} )*10.31^{2/5} *(0.00143/0.00032)^{1/4} \\Nu=476.99$

$Nu=\frac{h*d}{K_{w} } \\h=\frac{476.99*0.582}{20x10^{-3} } =13880.44W/m^{2} K$

$\frac{T-T_{c} }{T_{w}-T_{c} } =e^{-t/T} \\T=\frac{m_{c}C_{p} }{hA_{c} } =\frac{8960*10x10^{-3}*385 }{13880.44*3} =0.828 s$

$e^{-t/0.828} =\frac{320-280}{360-280} \\t=0.573\\heightofthewater=2.186*0.573=1.25m$

8 0

3 years ago

For three control stations,there should be how many start buttons in parallel with the suxiliary contact

valentinak56 [21]

The three load contacts connected between the three-phase power line and the motor close to connect the motor to the line. The normally open auxiliary contact connected in parallel with the two Start buttons closes to maintain the circuit to M coil when the Start button is released.

6 0

3 years ago

When the 2.8-kg bob is given a horizontal speed of 1.5 m/s, it begins to rotate around the horizontal circular path A. The force

Rama09 [41]

Answer:

The speed is the same at 1.5 m/s while

The work done by the force F is 0.4335 J

Explanation:

Here we have angular acceleration α = v²/r

Force = ma = 2.8 × 1.5²/r₁

and ω₁ = v₁/r₁ = ω₂ = v₁/r₂

The distance moved by the force = 600 - 300 = 300 mm = 0.3 m

If the velocity is constant

The speed is 1.5 m/s while the work done is

2.8 × 1.5²1/(effective radius) ×0.3

r₁ = effective radius

2.8*9.81 = 2.8 × 1.5²/r₁

r₁ = 0.229

The work done by the force = 2.8 × 1.5²*1/r₁ *0.3 = 0.4335 J

4 0

3 years ago

Identify how slip affects the strength of a metal​

Identify how slip affects the strength of a metal