Answer:
33.65°
Explanation:
radius, r = 53.1 m
m = 2.9 Mg = 2.9 x 10^6 g = 2900 kg
v = 67 km/h
convert km/h into m/s
v = 18.61 m/s
Let the angle of banking of road is θ, without friction
[tex]tan\theta =\frac{v^{2}}{rg}[/tex]
[tex]tan\theta =\frac{18.61^{2}}{53.1\times 9.8}[/tex]
tan θ = 0.6655
θ = 33.65°
Thus, the angle of banking of road is 33.65°.