To find the additional distance traveled by the tip of the second hand compared to the minute hand in 30 minutes, we need to calculate the respective circumferences traveled by each hand and then find their difference.
Thus, the value of \(x\) in meters is nearly 139.4 meters.
The distance traveled by the tip of the minute hand in one revolution is:
\( x_{\text{min}} = \pi \times r_{\text{min}} = \pi \times \frac{60}{100} \, \text{m}. \)
\( x_{\text{min}} = 3.14 \times 0.6 = 1.884 \, \text{m}. \)
The distance traveled by the tip of the second hand in 30 minutes is:
\( x_{\text{second}} = 30 \times 2\pi \times r_{\text{second}} \)
\( x_{\text{second}} = 30 \times 2 \times 3.14 \times \frac{75}{100} \, \text{m}. \)
\( x_{\text{second}} = 30 \times 4.71 = 141.3 \, \text{m}. \)
The difference in distance is:
\( x = x_{\text{second}} - x_{\text{min}} = 141.3 - 1.884 \, \text{m}. \)
\( x = 139.4 \, \text{m}. \)
Final Answer: 139.4 m.
A wheel of radius $ 0.2 \, \text{m} $ rotates freely about its center when a string that is wrapped over its rim is pulled by a force of $ 10 \, \text{N} $. The established torque produces an angular acceleration of $ 2 \, \text{rad/s}^2 $. Moment of inertia of the wheel is............. kg m².
A point particle of charge \( Q \) is located at \( P \) along the axis of an electric dipole 1 at a distance \( r \) as shown in the figure. The point \( P \) is also on the equatorial plane of a second electric dipole 2 at a distance \( r \). The dipoles are made of opposite charge \( q \) separated by a distance \( 2a \). For the charge particle at \( P \) not to experience any net force, which of the following correctly describes the situation?

