Answer :
Answer:24.26m/s
Explanation:coefficient of static friction is = v^2/rg
1.0=v^2/60*9.81
1.0=v^2=588.6
V^2=588.6
V=24.26m/s
The maximum speed with which a 1400 kg rubber-tired car can take this curve without sliding will be "24.26 m/s".
Friction force
According to the question,
Radius = 60.0 m
Angle = 19.0°
Mass = 1400 kg
Acceleration due to gravity, g = 9.8
We know the formula,
The coefficient of static friction will be:
1.0 = [tex]\frac{v^2}{rg}[/tex]
By substituting the values,
1.0 = [tex]\frac{v^2}{60\times 9.8}[/tex]
1.0 = [tex]\frac{v^2}{588}[/tex]
By applying cross-multiplication, we get
v² = 1.0 × 588
v² = 588
v = √588
= 24.26 m/s
Thus the approach above is correct.
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