Aroller coaster car of mass m is on a track that forms a circular loop of radius r in the vertical plane. if the car is to just maintain contact with the track at the top of the loop, what is the minimum value for its speed at that point?

v = \sqrt{Rg}

Explanation:

As the car just maintain the contact at the top point of the circular track, it means the weight of the car is just balance by the centripetal force acting on the track.

Let v be the speed of the car.

M g = Mv^2 / R

g = v^2 / R

v^2 = R g

i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.

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