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: \(\boxed{13 \, \text{m/s}}\)
Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.