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3 years ago

Identify the following formulas:

Engineering

1 answer:

gregori [183]3 years ago

4 0

1. In the image below
2. y=mx+b
3. Ax+By=C (I think)
4. y-y1=m(x-x1)

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To convert a whole number to a fraction the number can be written over a denominator of

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When you convert a whole number into a fraction, you put the whole number as the numerator and the denominator as 1.

Such as 12=12/1 or 5=5/1

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3 years ago

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3 years ago

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Define the Static Balancing.

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3 years ago

A non-inductive load takes a current of 15A at 125V. An inductor is then connected in series in order that the same current shal

Norma-Jean [14]

Answer:

The inductance of the inductor is 0.051H

Explanation:

From Ohm's law;

V = IR .................. 1

The inductor has its internal resistance referred to as the inductive reactance, X $_{L}$ , which is the resistance to the flow of current through the inductor.

From equation 1;

V = IX $_{L}$

X $_{L}$ = $\frac{V}{I}$ ................ 2

Given that; V = 240V, f = 50Hz, $\pi$ = $\frac{22}{7}$ , I = 15A, so that;

From equation 2,

X $_{L}$ = $\frac{240}{15}$

= 16Ω

To determine the inductance of the inductor,

X $_{L}$ = 2 $\pi$ fL

L = $\frac{X_{L} }{2 \pi f}$

= $\frac{16}{2*\frac{22}{7}*50 }$

= 0.05091

The inductance of the inductor is 0.051H.

4 0

4 years ago

A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a

kirill [66]

Answer:

105.70 mm

Explanation:

Poisson’s ratio, v is the ratio of lateral strain to axial strain.

E=2G(1+v) where E is Young’s modulus, v is poisson’s ratio and G is shear modulus

Since G is given as 25.4GPa, E is 65.5GPa, we substitute into our equation to obtain poisson’s ratio

$\begin{array}{l}\\65.5{\rm{ GPa}} = 2\left( {25.4{\rm{ GPa}}} \right)(1 + \upsilon )\\\\\upsilon = 0.2893\\\end{array}$

Original length $L_(i}$

$\upsilon = - \left( {\frac{{\left( {\frac{{{d_f} - {d_i}}}{{{d_i}}}} \right)}}{{\left( {\frac{{{L_f} - {L_i}}}{{{L_i}}}} \right)}}} \right)$

Where $d_{f}$ is final diameter, $d_{i}$ is original diameter, $L_{f}$ is final length and $L_{i}$ is original length.

$\begin{array}{l}\\0.2893 = - \left( {\frac{{\left( {\frac{{30.04{\rm{ mm}} - {\rm{30 mm}}}}{{{\rm{30 mm}}}}} \right)}}{{\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right)}}} \right)\\\\\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right) = - 4.6088 \times {10^{ - 3}}\\\end{array}$

$\begin{array}{l}\\105.2 - {L_i} = - \left( {4.6088 \times {{10}^{ - 3}}} \right){L_i}\\\\105.2 = 0.9953{L_i}\\\\{L_i} = 105.70{\rm{ mm}}\\\end{array}$

Therefore, the original length is 105.70 mm

7 0

4 years ago