Answer:
Question: A car (assumed to be a Ford Taurus) is traveling around a turn that is banked at 7 degrees. The turn has a radius of 29 m. The car has a mass of 1300 kg. The coefficient of static friction between the tires and the road is 0.68.
1. What is the "ideal speed?" That is, what speed would allow the car to make the turn without requiring friction?
2. What is the maximum speed the car can go around the turn without sliding?
No it does not have a timing belt
Pe=1/2Kx^2
Half times spring constant times distance squared over time
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Answer:
By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.
here weight of the child =21kgx9.8m/s2 = 205.8N
the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.
torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.
net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.
b) Ta = 2.5x151 = 377.5N-m
Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.
c)Ta = 2x151 = 302N-m
Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.
