A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction between the road and the car's tires that will allow the car to travel atthis speed without sliding?

a. 1.23
b. 0.662
c. 0.816
d. 0.952
e. 0.736

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is,

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

As there is no vertical motion, therefore, . So,

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

Plug in the given values and solve for 'μ'. This gives,

Therefore, option (c) is correct.

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards