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garri49 [273]
3 years ago
6

3. It's

Engineering
1 answer:
jeyben [28]3 years ago
6 0

Answer:

right answer is option no d

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on the same scale for stress, the tensile true stress-true strain curve is higher than the engineeringstress-engineering strain
Bess [88]

Answer:

The condition does not hold for a compression test

Explanation:

For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension.  The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.

<em>Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve.</em><em> </em>does not hold for compression test

5 0
3 years ago
A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h
hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

D_{1} = 150 mm

A_{1}= \frac{\pi }{4}\times 150^{2}

              = 17671.45 mm^{2}

D_{2} = 250 mm

A_{2}= \frac{\pi }{4}\times 250^{2}

              = 49087.78 mm^{2}

The centrifugal force acting on the flywheel is fiven by

F = M ( R_{2} - R_{1} ) x w^{2} ------------(1)

Here F = ( -UTS x A_{1} + UCS x A_{2} )

Since density, \rho = \frac{M}{V}

                        \rho = \frac{M}{A\times t}

                        M = \rho \times A\times tM = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t

                        M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37

                        M = 8252963901

∴ R_{2} - R_{1} = 50 mm

∴ F = 763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}

  F = 33618968.38 N --------(2)

Now comparing (1) and (2)

33618968.38 = 8252963901\times 50\times \omega ^{2}

∴ ω = 4036.61

We know

\omega = \frac{2\pi N}{60}

4036.61 = \frac{2\pi N}{60}

∴ N = 38546.82 rpm

7 0
4 years ago
What is the composition of the two phases that form when a stream of 40% A, 39% B, and 21% C separates into two phases? Label th
irga5000 [103]

Answer:

vapor fraction = 0.4 and 0.08

Explanation:

At reasonably high temperatures, a mixture will exist in the form of a sub cooled liquid. Between these extremes, the mixture exists in a two phrase region where it is a vapor liquid equilibrium. From a vapor-liquid phase diagram, a mixture of 40% A, 39% B, and 21% C separates to give the vapor compositions of 0.4 and 0.08.

8 0
3 years ago
Draw the schematics for the following two battery connections. Can you explain the value of the output voltage in the series con
matrenka [14]

if two 1.5 ya alt batteries are connected to head to the tail the voltage is 3.0 volt it is the because the battery is insidious reduce a voltage equal to number of battery is multiplied by the voltage of individual

5 0
2 years ago
Zachary finished an internship as a Software Quality Assurance Engineer. For which job is he best qualified?
astra-53 [7]

Answer:

B. A software development firm needs someone to find and fix bugs on multiple computer platforms.

Explanation:

A software quality assurance engineer is someone who monitors every phase of the software development process so as to ensure design quality, making sure that the software adheres to the standards set by the development company. Finding bugs would make this intern a amazing bug finder

6 0
3 years ago
Read 2 more answers
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