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Daniel [21]
3 years ago
6

A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw

een the tires and the road is μ = 0.800. If the car maintains a constant speed of 28 m/s, it will A. first veer toward the center for the first quarter-circle, then veer toward the outside for the second quarter circle. B. tend to veer toward the outside of the circle.
Physics
1 answer:
meriva3 years ago
8 0

Answer:

As  28m/s = 28m/s

Explanation:

r = the radius of the curve

m =  the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

a = \frac{v^{2}}{r}

therefore

\mu mg = m \frac{v^{2}}{r} \\\\ \mu =  \frac{v^{2}}{rg}

v = \sqrt{\mu rg}

\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s

As  28m/s = 28m/s

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