Aroller coaster car starts from rest at the top of a hill 15 m high and rolls down to ground level. from there it starts into a vertical loop. exactly at the top of the loop it is 10 m high and the radius of curvature of the track is 4.0 m. assuming negligible friction and air resistance, what is the magnitude of the normal force acting on a 65 kg passenger at the top of the loop?

955.5N

Explanation:

The normal force is given by the difference between the centripetal force and gravity at the top of the loop:

mass m = 65kg

radius of the loop r = 4m

velocity v = ?

g = 9.8 m/s²

To find the centripetal force, you need to find the velocity of the car at the top of the loop.

Use energy conservation:

At the top of the hill:

At the top of the loop:

Setting both energies equal and canceling the mass m gives:

Solving for v:

Using v in the first equation:

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.

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