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dimulka [17.4K]

4 years ago

5

At what distance should the warning triangle be placed if a vehicle breaks down on the autobahn?

1 answer:

igor_vitrenko [27]4 years ago

4 0

Answer:

The answer is 200 Meters

Explanation:

At what distance should the warning triangle be placed if a vehicle breaks down on the autobahn?

200 Meters

The Autobahn is the federal controlled-access highway system in Germany. The official German term is Bundesautobahn, which translates as "federal motorway". The literal meaning of the word Bundesautobahn is "Federal Auto Track".

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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

Malia is working with aluminum wire. while working with this type of wire, she must remember to a. always use pressure-type term

Y_Kistochka [10]

Answer:

B

Explanation:

Aluminium doesn't rust unless exposed to copper for a duration of time.

4 0

3 years ago

Knowing that the central portion of the link BD has a uniform cross-sectional area of 800 2, determine the magnitude of the load

IceJOKER [234]

Answer: 50

Explanation:

7 0

3 years ago

Which of the following are made up of electrical probes and connectors?

Mariulka [41]

Uhm is there a multiple choice?

6 0

3 years ago

What will happen in a wire drawing operation when the cross-sectional area has a reduction of 60% in a single pass?

Fofino [41]

Answer:

DRAWING LOAD IS $3.67 A_{O}\sigma$

Explanation:

wire drawing is a method of obtaining wire of bigger large diameter from iron rod . it is cold process which need die to obtain wire

drawing load for wire drawing is given as P = $A_{F}*\sigma*ln(\frac{A_{O}}{A_{F}})$

Where A f is initial area, Ao is original area, σ is yield stress

as given in question sectional area reduce 60%, therefore

$A_{f} = A_{O}- 0.6A_{O}$

= $0.4 A_{O}$

Due to change in area ,drawing load p is

p = $0.4A_{O}*\sigma*ln(\frac{A_{O}}{0.4A_{O}})$

p = $3.67 A_{O}\sigma$

4 0

3 years ago