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lutik1710 [3]
3 years ago
14

For a bronze alloy, the stress at which plastic deformation begins is 275 MPa, and the modulus of elasticity is 115 GPa. (a) Wha

t is the maximum load that may be applied to a specimen with a cross-sectional area of 325 mm2 without plastic deformation
Engineering
1 answer:
Anton [14]3 years ago
3 0

Answer:

89375 N

Explanation:

Rearrange the formula for normal stress for F:

\sigma=\frac{F}{A}

F=\sigma*A

Convert given values to base units:

275 MPa = 275*10^{6} Pa

325 mm^{2} = 0.000325 m^{2}

Substituting in given values:

F = (275*10^{6})*(0.000325)=89375 N

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The Employee

Explanation:

Because it is there responsibility

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What would happen to a plane if the weight force becomes greater than the lift force?
mrs_skeptik [129]

Answer:

When the lift is greater than the weight, the aircraft gains altitude. ... Drag must be overcome for the aircraft to move, and movement is essential to obtain lift. To overcome drag and move the aircraft forward, another force is essential. This force is thrust.

Explanation:

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8 0
3 years ago
Read 2 more answers
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
GenaCL600 [577]

Answer:

62.14\ \text{miles}

6213727.37\ \text{miles}

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = 10^5\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}

1\ \text{mile}=1609.34\ \text{m}

\dfrac{10^5}{1609.34}=62.14\ \text{miles}

The chain would extend 62.14\ \text{miles}

Dislocation density = 10^{10}\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}

\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}

The chain would extend 6213727.37\ \text{miles}

3 0
3 years ago
11. Which of these is NOT true when dealing with refrigerants?
Alexus [3.1K]
Answer is an increase in pressure will cause an decrease in the pressure
4 0
3 years ago
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
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