Exercises 2 (Plotting in rectangular coordinates)
- The strain \(\varepsilon\) induced in a wire when subjected to a series of stress \(\sigma\) values produced the following results:
\(\sigma\) (M Pa) | \(10.8\) | \(21.6\) | \(33.3\) | \(37.8\) | \(45.9\) |
\(\varepsilon\) (\(\times 10^{-5}\)) | \(12\) | \(24\) | \(37\) | \(42\) | \(51\) |
Show that the stress is related to the strain by a law of the form \(\sigma=E\varepsilon\), where \(E\) is a constant. Determine the law for the wire under test.
- During a test on a simple lifting machine, the following results were obtained showing the applied force, \(F\), for the load, \(L\), lifted:
\(F\) (N) | \(19\) | \(37\) | \(50\) | \(93\) | \(125\) | \(149\) |
\(L\) (N) | \(40\) | \(120\) | \(230\) | \(410\) | \(540\) | \(680\) |
It is thought that the equation relating \(F\) and \(L\) is of the form \(F = kL + c\) where both \(k\) and \(c\) are constants. Assuming that the law holds true, find the force necessary to lift a load of \(1\,\mathrm{kN}\).
- The variation in pressure, \(p\), within a vessel at a temperature, \(T\), follows a law of the form \(p = aT + b\). Verify that the data below relates the data by this law and determine the law.
\(p\) (kPa) | \(248\) | \(253\) | \(257\) | \(262\) | \(266\) | \(270\) |
\(T\) (K) | \(273\) | \(278\) | \(283\) | \(288\) | \(293\) | \(298\) |
Determine graphically the solution to the simultaneous equations: \[\begin{align*} 2.5x + 0.45 - 3y &= 0\\ 1.6x + 0.8y - 0.8 &= 0 \end{align*}\]
(After Chapter 4: Quadratic equations) Plot graphs of:
- \(y = 2x^2\)
- \(y = 2x^2 - 4\)
- \(y = 2x^2 - 2x + 0.5\)
- \(y = 2x^2 + x - 6\)
(After Chapter 9: Long division and factorisation) Plot the graph of \(y = 4x^3 - 4x^2 - 15x + 18\) for values of \(x\) from \(-3\) to \(+3\) and using the graph, determine the roots of the polynomial.
(After Chapter 8: Exponential functions) On the same axes and to the same scale plot the equations \(y = 1.5e^{-1.18x}\) and \(y = 1.1(1 - e^{-2.3x})\). Determine the solution to the equations from your graph.