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3 years ago

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Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 8.0 mm. A tensile force of 1000 N produc

es an elastic reduction in diameter of 2.8 10-4 mm. Compute the modulus of elasticity for this alloy, given that Poisson's ratio is 0.30.

1 answer:

Dimas [21]3 years ago

3 0

Answer:

1.7 × 10^11 Pa

Explanation:

Please see the attachments below

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Answer:

J = 2.044x10⁶ A/m²

v = 1.50x10⁻⁴ m/s

Explanation:

The current density (J) of the copper wire is giving by:

$J = \frac {I}{A}$

<em>where I: electric current and A: cross-sectional area of the copper wire</em>

<u>The cross-sectional area of the copper wire can be calculated by:</u>

$A = \frac {\pi d^{2}}{4} = \frac {\pi (1.02 \cdot 10^{-3} m)^{2}}{4} = 8.17 \cdot 10^{-07} m^{2}$

<u>Substituting the calculated area in the equation (1) we have:</u>

$J = \frac {1.67 A}{8.17 \cdot 10^{-7} m^{2}} = 2.044 \cdot 10^{6} \frac {A}{m^{2}}$

Hence, the current density is 2.044x10⁶ A/m².

To find the drift speed (v), we need to use the next equation:

$v = \frac {J}{n q}$

<em>where n: the free-electron density, q: module of the charge of the electron </em>

$v = \frac {2.044 \cdot 10^{6} \frac {A}{m^{2}}}{(8.5 \cdot 10^{28} {m^{-3}}) (1.6 \cdot 10^{-19} C)}$

$v = 1.50 \cdot 10^{-04} \frac {m}{s}$

So, the drift speed is 1.50x10⁻⁴ m/s.

Have a nice day!

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