Ok I\'ve been working on these for like forever and I have no way tocheck if my

Question

Ok I've been working on these for like forever and I have no way tocheck if my work or answers are correct and would appreciate if Icould get verifcation...thanks

12. A car is safely negotiating an unbanked circular turn at aspeed of 21 m/s. The maximum static frictional force acts on thetires, Suddenly a wet patch in the road reduces the maximum staticfrictional force by a factor of three. If the car is to continuesafely around the curve, to what speed must the driver slow thecar?


15. A model airplane is being flown on a guideline that can sustainat most 180N of tension. The mass of the plane is 0.75 kg, and it'sspeed is 28 m/s. Assuming that the guideline is nearly parallel tothe ground, what is the radius of the smallest horizontal circle inwhich the plane can be flown?

19. A "swing" ride at a carnival consists of chairs that are swungin a circle by 12.0-m cables attached to a vertical rotating pole,as the drawing shows. Suppose the total mass of a chair and itsoccupant is 220 kg. (a) Determine the tension in the cable attachedto the chair. (b) Find the speed of the chair...angle is 65degrees.

40. A downhill skier, whose mass is 50.0 kg, attains a speed of21.0 m/s just as she reaches the point where a jump is necessary(point A). When she leaves the ground, her velocity is horizontal.In other words, point A is at the bottom of the circular arc AB(radius = 27.0 m). Determine the normal force acting on the skis atpoint A.

Explanation / Answer

Speed v = 21 m/s     [rg ] =21 m / s
The maximum static frictional force acts on the tires = we know v = [ rg ]             = v ^ 2 / rg               = 45 / r maximum static frictional force in wet path F =mg - mg / 3                                                                     = 2mg / 3 where mg =static frctional force on dry path Static frictional coefficient ' = 2 /3                                              speed of the car on wet path v ' = [ 'r g ]                                                 = { [ 2/ 3 ] rg }                                                 = [ 2rg / 3 ]                                                 = [ 2*21 / 3]                                                  = 3.74m / s

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