Since the cyclist travels along the circumference from point P to point S, which are opposite ends of the diameter of the circle, we can visualize the displacement as the straight-line distance between P and S.
1. Determine the Displacement:
Using the Pythagorean theorem, we find:
\[ \text{Displacement} = R\sqrt{2} = 2\sqrt{2} = \sqrt{8} \, \text{km}. \]
Answer: \(\sqrt{8} \, \text{km}\)
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).