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bearhunter [10]
3 years ago
12

Explain the prosses of welding

Engineering
2 answers:
klio [65]3 years ago
7 0

Answer:

Welding is a fabrication process whereby two or more parts are fused together by means of heat, pressure or both forming a join as the parts cool. Welding is usually used on metals and thermoplastics but can also be used on wood.

Explanation:

serious [3.7K]3 years ago
4 0
Welding is like a Fabrication Process ,which is where two or more parts are fused together by heat,Pressure or both forming or joining together as the parts cool.HOPE THIS HELPED!
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A piece of aluminum wire is 500 ft long and has a diameter of 0.03 inches. What is the resistance of the piece of wire?​
dexar [7]

Answer:

8.85 Ω

Explanation:

Resistance of a wire is:

R = ρL/A

where ρ is resistivity of the material,

L is the length of the wire,

and A is the cross sectional area.

For a round wire, A = πr² = ¼πd².

For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.

Given L = 500 ft and d = 0.03 in = 0.0025 ft:

R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)

R = 8.85 Ω

5 0
3 years ago
Whose job is it to ensure that a group stays focused and on schedule?
Sedbober [7]

Answer:

the leader or organizer. you should have assigned jobs, so one of them should be in charge. if you did not assign jobs, assign them now, and one person has a job, and they must be held accountable. if they cannot do their job, someone might have to take over, but then you tell your prof. that they could not do their part, so hopefully you will get the credit you deserve and they will not.

Explanation:

3 0
3 years ago
Read 2 more answers
Tensile testing provides engineers with the ability to verify and establish material properties related to a specific material.
Sedbober [7]

Answer:

True

Explanation:

Tensile testing which is also referred to as tension testing is a process which materials are subjected to so as to know how well it can be stretched before it reaches breaking point. Hence, the statement in the question is true

7 0
2 years ago
Convert 850 nm wavelength into frequency, eV, wavenumber, joules and ergs.
Sholpan [36]

Answer:

Frequency = 3.5294\times 10^{14}s^{-1}

Wavenumber = 1.1765\times 10^6m^{-1}

Energy = 2.3365\times 10^{-19}J

Energy = 1.4579 eV

Energy = 2.3365\times 10^{-12}erg

Explanation:

As we are given the wavelength = 850 nm

conversion used : (1nm=10^{-9}m)

So, wavelength is  850\times 10^{-9}m

The relation between frequency and wavelength is shown below as:

Frequency=\frac{c}{Wavelength}

Where, c is the speed of light having value = 3\times 10^8m/s

So, Frequency is:

Frequency=\frac{3\times 10^8m/s}{850\times 10^{-9}m}

Frequency=3.5294\times 10^{14}s^{-1}

Wavenumber is the reciprocal of wavelength.  

So,  

Wavenumber=\frac{1}{Wavelength}=\frac{1}{850\times 10^{-9}m}

Wavenumber=1.1765\times 10^6m^{-1}

Also,  

Energy=h\times frequency

where, h is Plank's constant having value as 6.62\times 10^{-34}J.s

So,  

Energy=(6.62\times 10^{-34}J.s)\times (3.5294\times 10^{14}s^{-1})

Energy=2.3365\times 10^{-19}J

Also,  

1J=6.24\times 10^{18}eV

So,  

Energy=(2.3365\times 10^{-19})\times (6.24\times 10^{18}eV)

Energy=1.4579eV

Also,  

1J=10^7erg

So,  

Energy=(2.3365\times 10^{-19})\times 10^7erg

Energy=2.3365\times 10^{-12}erg

5 0
3 years ago
A steady, incompressible, two-dimensional velocity field is given by the following components in the x-y plane: u=1.85+2.05x+0.6
amm1812

1) a_x=4.287+2.772x\\a_y=-5.579+2.772y

2) 8.418

Explanation:

1)

The two components of the velocity field in x and y for the field in this problem are:

u=1.85+2.05x+0.656y

v=0.754-2.18x-2.05y

The x-component and y-component of the acceleration field can be found using the following equations:

a_x=\frac{du}{dt}+u\frac{du}{dx}+v\frac{du}{dy}

a_y=\frac{dv}{dt}+u\frac{dv}{dx}+v\frac{dv}{dy}

The derivatives in this problem are:

\frac{du}{dt}=0

\frac{dv}{dt}=0

\frac{du}{dx}=2.05

\frac{du}{dy}=0.656

\frac{dv}{dx}=-2.18

\frac{dv}{dy}=-2.05

Substituting, we find:

a_x=0+(1.85+2.05x+0.656y)(2.05)+(0.754-2.18x-2.05y)(0.656)=\\a_x=4.287+2.772x

And

a_y=0+(1.85+2.05x+0.656y)(-2.18)+(0.754-2.18x-2.05y)(-2.05)=\\a_y=-5.579+2.772y

2)

In this part of the problem, we want to find the acceleration at the point

(x,y) = (-1,5)

So we have

x = -1

y = 5

First of all, we substitute these values of x and y into the expression for the components of the acceleration field:

a_x=4.287+2.772x\\a_y=-5.579+2.772y

And so we find:

a_x=4.287+2.772(-1)=1.515\\a_y=-5.579+2.772(5)=8.281

And finally, we find the magnitude of the acceleration simply by applying Pythagorean's theorem:

a=\sqrt{a_x^2+a_y^2}=\sqrt{1.515^2+8.281^2}=8.418

4 0
3 years ago
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