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4. A 1300 kg car makes a turn with a 45 m radius. Coefficient of static friction between car tires and the road is 0.28. At which spee
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Answer: v = 11 m/s
Explanation:
If we ASSUME that the road is horizontal, not banked
The maximum friction force available is μmg
This will need to supply all of the required centripetal force
mv²/R = μmg
v = √(μgR) = √(0.28(9.8)(45)) = 11.1121…≈ 11 m/s