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Citrus2011 [14]
3 years ago
5

Limestone (calcium carbonate) particles

Engineering
1 answer:
Aleksandr-060686 [28]3 years ago
3 0

Answer:

606506 026216n ytj dtyjjtj t

Explanation:

tyjtyjtyjtyj ty

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For the following circuit, V"#$=120∠30ºV.Redraw the circuit in your solution.a.(4) Calculate the total input impedance seen by t
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Explanation:

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List 5 characteristics of metal?​
RSB [31]

Answer:

- Son buenos conductores de electricidad.

- Excelentes conductores de calor.

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3 years ago
Technician A says that the connecting rod and main bearing caps should be marked before removing to ensure that they can be inst
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Technician A

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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
3 years ago
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