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what will the stopping distance be a a 3000-kg car if -3000N of force are applied when the car is traveling 10 m/s

Respuesta :

Answer:

50 m

Explanation:

Acceleration= force/mass

3000/3000=1m/s^-2

Applying equation of motion:

V^2=U^2+2as; V is final velocity, u is initial velocity, a is acceleration and s is the distance covered.

0=10^2 -2*1s;

Solve for s

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