For motion on curved roads:
• Use the formula vmax = \(\sqrt{µgr}\) for safe turning.
• Ensure all values are in consistent units for accurate results.
13.4m/s
13m/s
22.4m/s
17m/s
1. Maximum Speed on a Curved Road: - The maximum speed of a vehicle on a curved road is given by:
\[v_\text{max} = \sqrt{\mu gr},\]
where \(\mu\) is the coefficient of friction, g is the acceleration due to gravity, and r is the radius of curvature.
2. Substitute the Values: - \(\mu = 0.34\), g = 10 m/s2, r = 50 m:
\[v_\text{max} = \sqrt{0.34 \times 10 \times 50} = \sqrt{170} \approx 13 \, \text{m/s}.\]
Final Answer: \(\boxed{13 \, \text{m/s}}\)
A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2 \(\Omega\) then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is _____ N.
Net gravitational force at the center of a square is found to be \( F_1 \) when four particles having masses \( M, 2M, 3M \) and \( 4M \) are placed at the four corners of the square as shown in figure, and it is \( F_2 \) when the positions of \( 3M \) and \( 4M \) are interchanged. The ratio \( \dfrac{F_1}{F_2} = \dfrac{\alpha}{\sqrt{5}} \). The value of \( \alpha \) is 

A block of mass 100 kg slides over a distance of 10 m on a horizontal surface. If the coefficient of friction between the surfaces is 0.4, then the work done against friction (in J) is:
Identify A in the following reaction. 