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olga_2 [115]
3 years ago
13

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.85 m/s2

. The car makes it one quarter of the way around the circle before it skids off the track. From these data, determine the coefficient of static friction between the car and track.
Physics
1 answer:
Tasya [4]3 years ago
7 0

Answer:

Coefficient of friction = 0.836

Explanation:

If v be the speed after one quarter of the circular path

v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8

tangential acceleration = 5.8 m/s²

radial acceleration = v² /r = 5.8

total acceleration = √2 x 5.8

m x√2 x 5.8 = m x g xμ

μ = √2 x 5.8 / 9.8 = 0.836

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The Coulomb's force acting between two charges is
F=k_e  \frac{q_1 q_2}{r^2}
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Using q_1 = q_2 = -3 C, we can find the distance between the two charges when the force is F=19.2 N:
r=  \sqrt{k_e  \frac{q_1 q_2}{F} }=  \sqrt{ 8.99\cdot 10^9 \frac{(-3.0C)^2}{19.2 N} }=6.5 \cdot 10^4 m = 65 km
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