Una curva de radio R = 75 [m] tiene un peralte hecho de tal forma que un auto andando a v1 = 35 [m/s] no derrape incluso con el piso congelado (sin roce). ¿Cuál es el coeficiente de roce estático mínimo necesario entre las ruedas y el pavimento para que, en un día seco y soleado, el auto no derrape cuando avanza a v2 = 118 [m/s]?

053.

Explanation:

Given that the radius of curvature of the path, R = 75 m.

Speed of the car on that path ,

The centripetal force, acting on the body having mass, m, when it moves with the velocity v  on curved path having radiusR

Ris

Gravitational force, = mg.

Let tha angle of superelevation is

As the car does not skid even with zero friction, so

On sunny day, let the minimun static friction coefficient between the wheels and the pavement is  

As = 118 m/s is greater than   v_so the car tends to skid in upper direction and the frictional

force,f, will acts is downward direction.

As there is no skidding, so

where

Where N is the normal reaction can be determined by balancing the force in perpendicular direction of the plane.

From equation (ii)

=0.53

Hence, the minimum coefficient of friction is 0.53.

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg