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zheka24 [161]
2 years ago
9

For a brass alloy, the following engineering stresses produce the corresponding plastic engineering strains prior to necking:

Engineering
1 answer:
tatuchka [14]2 years ago
4 0

Answer:

<h2><em> </em>The engineering stress necessary to produce an engineering strain of 0.28 is 339 Mpa</h2>

Explanation:

<em>On the basis of this information, we are going to compute the engineering stress necessary to produce an engineering strain of 0.28.</em>

<h2>Engineering           Engineering </h2><h2> Stress (MPa)             Strain</h2><h2>315                             0.105 </h2><h2>340                           0.220 </h2>

For clarity and you will find the solving in the attached file for your reference.

should you need further information do let me know.

​  

 

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The properties of the air in the inlet section with A1 = 0.25ab m2 in a converging-diverging channel are given as U1 = 25a,b m/s
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3 0
3 years ago
1. Calculate the battery life in years when a pacemaker has the following characteristics: Battery Ampere-hours = 1.5 Pulse volt
Wittaler [7]

Answer:

battery life in year = 9 years and 48 days

Explanation:

given data

Battery Ampere-hours = 1.5

Pulse voltage = 2 V

Pulse width = 1.5 m sec

Pulse time period = 1 sec

Electrode heart resistance = 150 Ω

Current drain on the battery = 1.25 µA

to find out

battery life in years

solution

we get first here duty cycle that is express as

duty cycle = \frac{width}{period}      ...............1

duty cycle = 1.5 × 10^{-3}

and applied voltage will be

applied voltage = duty energy × voltage    ...........2

applied voltage = 1.5 × 10^{-3} × 2

applied voltage = 3 mV

so current will be

current = \frac{applied\ voltage}{resistance}   ................3

current = \frac{3}{150}

current = 20 µA

so net current will be

net current = 20 - 1.25

net current = 18.75 µA

so battery life will be

battery life = \frac{1.5}{18.75*10^{-6}}

battery life = 80000 hours

battery life in year = \frac{80000}{8760}

battery life in year = 9.13 years

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4 0
3 years ago
Estimate the endurance strength, Se, of a 37.5-mm- diameter rod of AISI 1040 steel having a machined finish and heat-treated to
7nadin3 [17]

Answer:

endurance length is 236.64 MPa

Explanation:

data given:

d = 37.5 mm

Sut = 760MPa

endurance limit is

Se = 0.5 Sut

   = 0.5*760 = 380 MPa

surface factor is

Ka = a*Sut^b

where

Sut is ultimate strength

for AISI 1040 STEEL

a = 4.51, b = -0.265

Ka = 4.51*380^{-0.265}

Ka = 0.93

size factor is given as

Kb =1.29 d^{-0.17}

Kb = 0.669

Se = Sut *Ka*Kb

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Se = 236.64 MPa

therefore endurance length is 236.64 MPa

4 0
2 years ago
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