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A roller coaster at an amusement park has a dip that bottoms out in a vertical circle of radius 11.0 m. A passenger feels the seat of the car pushing upward on her with a force equal to three times her weight at the bottom of the dip. What is her speed as she goes through the dip

Respuesta :

ogorwyne

Answer:

14.69

Explanation:

The circle of radius R = 11.0m

Force f = 3 times her Weight

= 3W = 3mg

Then the net force acting on her

F = W + F(centripetal)

= mg + mv²/r

= Mg + mv²/r = 3mg

Then her speed is going to be

V = √2gr

Where g =9.81m/s²

R = 11.0

When we put these values into the formula we get:

V = √2x9.81x11.0

V = √215.82

V = 14.69