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.
A uniform circular disc of radius \( R \) and mass \( M \) is rotating about an axis perpendicular to its plane and passing through its center. A small circular part of radius \( R/2 \) is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32