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 tube of length 1m is filled completely with an ideal liquid of mass 2M, and closed at both ends. The tube is rotated uniformly in horizontal plane about one of its ends. If the force exerted by the liquid at the other end is \( F \) and the angular velocity of the tube is \( \omega \), then the value of \( \alpha \) is ______ in SI units.
For \( \alpha, \beta, \gamma \in \mathbb{R} \), if \[ \lim_{x \to 0} \frac{x^2 \sin(\alpha x) + (\gamma - 1)e^{x^2}}{\sin(2x - \beta x)} = 3, \] then \( \beta + \gamma - \alpha \) is equal to:

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: