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ANTONII [103]
3 years ago
7

GMA MIG weiding is a

Engineering
1 answer:
Dvinal [7]3 years ago
6 0

Gas metal arc welding (GMAW), sometimes referred to by its subtypes metal inert gas (MIG) welding or metal active gas (MAG) welding, is a welding process in which an electric arc forms between a consumable MIG wire electrode and the workpiece metal(s), which heats the workpiece metal(s), causing them to melt and join.

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Acquisition of resources from an external source is called?
densk [106]

Answer:

subcontracting

Explanation:

I hope this is right

6 0
3 years ago
When trying to solve a frame problem it will typically be necessary to draw many free body diagrams. a)-True b)-False
klasskru [66]

Answer:

True

Explanation:

When trying to solve a frame problem in Engineering or Physics, it will typically be necessary to draw more than one body diagram.

When we have several parts of the frame or a set of frames, we have the anchor point, as well as the intersections of frames. Besides that, usually, there is a particle or rigid body together with the frame system. In this sense, usually, it is required to analyze a body diagram for the particle or rigid body suspended, as well as the intersections of the frames. So, usually, it will be required a minimum of two body diagrams.

If the system is more complex, or there are many intersections points, it will be required more than two body diagrams.

Finally, indeed, it will typically be necessary to draw many-body diagrams. 

6 0
3 years ago
The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp
Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1

The Lower Surface Cp is given as

Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1

The difference of the Cp over the airfoil is given as

\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32

Now the Lift Coefficient is given as

C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192

Now the coefficient of moment about the leading edge is given as

C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0
3 years ago
In a cellular phone system, a mobile phone must be paged to receive a phone call. However, paging attempts don’t always succeed
Alina [70]

Answer:

The correct response will be "0.992". The further explanation to the following question is given below.

Explanation:

The probability that paging would be beneficial becomes 0.8  

Effective paging at the very first attempted is 0.8

On the second attempt the success probability will be:

⇒  0.2\times 0.8

⇒  0.16

On the third attempt the success probability will be:

⇒  0.2\times 0.2\times 0.8

⇒  0.032

So that the success probability will be:

⇒  0.8 + 0.16 + 0.032

⇒  0.992

7 0
3 years ago
Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW a
stealth61 [152]

Answer: a) 0.948 b) 117.5µf

Explanation:

Given the load, a total of 2.4kw and 0.8pf

V= 120V, 60 Hz

P= 2.4 kw, cos θ= 80

P= S sin θ - (p/cos θ) sin θ

= P tan θ(cos^-1 (0.8)

=2.4 tan(36.87)= 1.8KVAR

S= 2.4 + j1. 8KVA

1 load absorbs 1.5 kW at 0.707 pf lagging

P= 1.5 kW, cos θ= 0.707 and θ=45 degree

Q= Ptan θ= tan 45°

Q=P=1.5kw

S1= 1.5 +1.5j KVA

S1 + S2= S

2.4+j1.8= 1.5+1.5j + S2

S2= 0.9 + 0.3j KVA

S2= 0.949= 18.43 °

Pf= cos(18.43°) = 0.948

b.) pf to 0.9, a capacitor is needed.

Pf = 0.9

Cos θ= 0.9

θ= 25.84 °

(WC) V^2= P (tan θ1 - tan θ2)

C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2

f=60, π=22/7

C= 117.5µf

7 0
3 years ago
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