For a train moving around a curve, the required banking angle θ is given by:
\[\tan \theta = \frac{v^2}{Rg}\]
where \(v = 12 \, \text{m/s}\), \(R = 400 \, \text{m}\), and \(g = 10 \, \text{m/s}^2\).
Substitute the values:
\[\tan \theta = \frac{12^2}{10 \times 400} = \frac{144}{4000} = \frac{h}{1.5}\]
where \(h\) is the height by which the outer rail should be raised over the inner rail, and the distance between the rails is 1.5 m.
Solving for \(h\):
\[h = \frac{144 \times 1.5}{4000} = 5.4 \, \text{cm}\]
Thus, the required height is 5.4 cm.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: