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2 years ago

The best penetration is achieved with DCEN current. True or false

Engineering

1 answer:

GenaCL600 [577]2 years ago

6 0

Answer:

True

Explanation:

The best penetration is achieved with dcen current. Mild steel is expensive and requires the most amount of cleaning. Copper-coated steel welding rods are not used for gas tungsten arc welding because they will continue the weld or electrode.

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Is a compass a analog or a digital sensor?

ollegr [7]

A compass is a analog sensor

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2 years ago

Problem 3: Soil Classification using the AASHTO and USCS Systems

nataly862011 [7]

Solution:

$Given\\ $\quad W=3000 Ib , \quad m=\frac{W}{g}=\frac{3000}{322} \ slug =93.1677 slug$\\K_{e q}=2160 lbs / wp =2100 \frac{ lbs }{10} \frac{ x 12}{1 ft }=(2160 \times 12) lb / ft$$$

a) The natural frequency

$\begin{aligned}&\left(\omega_{n}\right)=\sqrt{\frac{K_{e q}}{m}}\\&=\sqrt{\frac{2160 \times 12}{93.1677}}\\&\omega_{n}=16.68 \text { rad } | s\\&\omega_{n}=\frac{2 \pi}{T}\\&16.68=\frac{2 \pi}{T}\\&T=0.3767 s\end{aligned}$

$Given, $t=10 s , \quad y(t)=6 in = A$\\$y(t)=A \cos \left(\omega_{n} t+\phi\right) \rightarrow 0$\\$6=6 \cos (16.68 \times 10+\phi)$\\$1=\cos (166.8+\phi)$\\$166.8+\phi=0$\\\phi=-166.8\)\\At $t=0, \quad y(0)=6 \cos (16.68 \times 0-166.8)$ {y(0)}=-5.74 in$

5 0

2 years ago

Engine vacuum is being discussed. Technician A says that when the engine is operating under light loads, engine vacuum is low. T

sdas [7]

Neither of the two technicians (Technician A and Technician B) is correct.

<h3>What is an engine vacuum?</h3>

An engine vacuum can be defined as a type of engine which is designed and developed to derive its force from air pressure that's being pushed against one side of the piston of an automobile, while having a partial vacuum on the other side.

In this scenario, we can infer and logically conclude that neither of the two technicians (Technician A and Technician B) is correct because engine vacuum is high when the engine is operating under light loads and vice-versa.

Read more on engine vacuum here: brainly.com/question/14602340

#SPJ12

6 0

2 years ago

A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th

Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588 = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ

b) $\Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)$

Where:

$x_A$ and $x_B$ are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, $x_A$ = 248 /(248 + 42.8) = 0.83

$x_B$ = 42.8/(248 + 42.8) = 0.1472

∴ $\Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472)$ = 1066.0279 J/K

5 0

2 years ago

Determine the initial void ratio, the relative density and the unit weight (in pounds per cubic foot) of the specimens for each

Elina [12.6K]

The initial void ratio is the parameter which is used to show the structural foundations for each specimen of sand so that the method and speed of compression would be measured.

Relative density is the mass per unit volume of each specimen of sand which is measured and it has to do with the relative ratio of the density of the sand.

Unit weight is the the exact weight per cubic foot of the sand which is measured.

Please note that your question is incomplete so I gave you a general overview to help you better understand the concept