Suppose the roller-coaster car in fig.6–41 passes point 1 with a speed of if the average force of friction is equal to 0.23 of its weight, with what speed will it reach point 2? the distance traveled is 45.0 m.

The values in figure 1 shows h1 = 39 m, h2 = 13 m, h3 = 25 with a speed of 1.5m/s

The Initial speed u = 1.5m/s.

Vertical distance covered between point 1 and point 2 is (39-13) =26m

The expanse 45 m is equal to a height of (45/6) =7.5 (As frictional force is mg/6)

With the kinematics equation v^2 =u^2 + 2as

V^2 = 1.5^2 + 2*9.8 (26 + 7.5)

V = 25.7 m/s

answer: i think he can see forward ten seconds and move forward ten seconds but giorno can undo his actions with ger.

explanation:

where is the circuit diagram ?