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DaniilM [7]
3 years ago
9

My computer has a mass of 0.031080997078386 slug the Earth's surface.

Engineering
1 answer:
poizon [28]3 years ago
4 0

Answer:

The answer is "0.187 lbm and 1 lbf".

Explanation:

The mass = 0.031080997078386\  slug

Calculating mass on Mars:

\to m=m_g\frac{g}{g_e}

        =0.031080997078386 \times \frac{32.2}{5.35}\\\\=0.187 \ lbm

\to W=mg_e

        =0.187 \times 5.35\\\\=1 \ lbf

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Answer:

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A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i
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Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter d_{o} = 12.8 mm

Final diameter d_{f} = 10.7

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% = π/4 [ ( (d_{f} )² - (d_{o} )²) / ( π/4  (d_{o} )²) ]

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= - 0.3012 × 100

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Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress  \alpha _{T} = \alpha _{E} ( 1 +  E_{E} )

E_{E}  is engineering strain

E_{E}  = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49  

E_{E} = 0.431

so we substitute the value of E_{E}  into our initial equation;

True stress  \alpha _{T} = 460 ( 1 +  0.431)

True stress  \alpha _{T} = 460 (1.431)

True stress  \alpha _{T} = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

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