Given:
- Radius of the circular track: \( R = 9 \, \text{m} \)
- Number of revolutions completed: 120 revolutions
- Time taken: 3 minutes
Step 1: Calculate the Angular Velocity
The angular velocity \( \omega \) is given by:
\[ \omega = \frac{\text{Total revolutions}}{\text{Time taken}} \times 2\pi \, \text{rad/s}. \]
Substituting the given values:
\[ \omega = \frac{120 \, \text{revolutions}}{3 \, \text{minutes}} \times \frac{2\pi \, \text{rad}}{1 \, \text{revolution}}. \]
Converting time to seconds:
\[ \omega = \frac{120 \times 2\pi}{3 \times 60} \, \text{rad/s} = \frac{4\pi}{3} \, \text{rad/s}. \]
Step 2: Calculate the Centripetal Acceleration
The centripetal acceleration \( a_{\text{centripetal}} \) is given by:
\[ a_{\text{centripetal}} = \omega^2 R. \]
Substituting the values of \( \omega \) and \( R \):
\[ a_{\text{centripetal}} = \left(\frac{4\pi}{3}\right)^2 \times 9. \]
Simplifying:
\[ a_{\text{centripetal}} = \frac{16\pi^2}{9} \times 9 = 16\pi^2 \, \text{m/s}^2. \]
Therefore, the magnitude of the centripetal acceleration of the monkey is \( 16\pi^2 \, \text{m/s}^2 \).
A sportsman runs around a circular track of radius $ r $ such that he traverses the path ABAB. The distance travelled and displacement, respectively, are:
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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).