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2 years ago

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Why is using the proper joining technique important? What could go wrong if the wrong joining technique is used?

1 answer:

Alex73 [517]2 years ago

5 0

Answer:

Explanation:

Using the proper technique is incredibly important because it prevents the materials being joined from breaking and/or causing an accident. If the wrong joining technique is used the materials may not hold in place and come apart easily instead. Also, some joining techniques are not meant for some materials and may instead cause the material to become weak and brittle causing it to break apart almost immediately.

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Kyla has obtained a bachelor’s degree in electronics engineering. In her search for a job, she comes across an advertisement tha

rodikova [14]

Answer:

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Explanation:

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2 years ago

Name two types of Transformers.

ipn [44]

Bumblebee and Starscream

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2 years ago

You are given a noninverting 741 op-amp with a dc-gain of 23.6 dB. The input signal to this amplifier is;Vin(t) = (0.18)∙cos(2π(

Vsevolod [243]

Answer:

Output voltage equation is $V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)$

Explanation:

Given:

dc gain $A = 23.6$ dB

Input signal $V_{in} (t) = 0.18 \cos (2\pi (57000)t +18.3)$

Now convert gain,

$A = 10^{\frac{23.6}{20} } = 15.13$

DC gain at frequency $f = 0$ is given by,

$A = \frac{V_{out} }{V_{in} }$

$V_{out} =AV_{in}$

$V_{out} = 15.13 \times 0.18 \cos (2\pi (57000)t +18.3)$

At zero frequency above equation is written as,

$V_{out} = 2.72 \times \cos 18.3$

$V_{out} = 2.72$

Now we write output voltage as input voltage,

$V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)$

Therefore, output voltage equation is $V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)$

7 0

2 years ago

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate is found to be 1% pe

LenaWriter [7]

Answer:

a) Q = 251.758 kJ/mol

b) creep rate is $= 1.751 \times 10^{-5} \% per hr$

Explanation:

we know Arrhenius expression is given as

$\dot \epsilon =Ce^{\frac{-Q}{RT}$

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is $\dot \epsilon = 5.5\times 10^{-2}$ % per hr

At 800 degree C creep rate is $\dot \epsilon = 1$ % per hr

activation energy for creep is $\frac{\epsilon_{800}}{\epsilon_{700}}$ = $= \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}$

$\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}$

$\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}]$

solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

$C = \epsilon e^{\frac{Q}{RT}}$

$=- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr$

$\epsilon_{500} = C e^{\frac{Q}{RT}}$

$= 1.804 \times 10^{12} e{\frac{251758}{8.314(500+273}}$

$= 1.751 \times 10^{-5} \% per hr$

4 0

2 years ago

A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas (both gases are monatomic) at 27

lidiya [134]

Answer:

Q = 62 ( since we are instructed not to include the units in the answer)

Explanation:

Given that:

$n_{HCl} = 3 \ kmol\\n_{Ar} = 7 \ k mol$

$T_1 = 27^0 \ C = ( 27+273)K = 300 K$

$P_1 = 200 \ kPa$

Q = ???

Now the gas expands at constant pressure until its volume doubles

i.e if $V_1 = x\\V_2 = 2V_1$

Using Charles Law; since pressure is constant

$V \alpha T$

$\frac{V_2}{V_1} =\frac{T_2}{T_1}$

$\frac{2V_1}{V_1} =\frac{T_2}{300}$

$T_2 = 300*2\\T_2 = 600$

mass of He =number of moles of He × molecular weight of He

mass of He = 3 kg × 4

mass of He = 12 kg

mass of Ar =number of moles of Ar × molecular weight of Ar

mass of He = 7 kg × 40

mass of He = 280 kg

Now; the amount of Heat Q transferred = $m_{He}Cp_{He} \delta T + m_{Ar}Cp_{Ar} \delta T$

From gas table

$Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar} = 0.5203 \ kJ/Kg/K$

∴ Q = $12*5.19*10^3(600-300)+280*0.5203*10^3(600-300)$

Q = $62.389 *10^6$

Q = 62 MJ

Q = 62 ( since we are instructed not to include the units in the answer)

5 0

3 years ago

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