Suppose the person, whose mass is m, is being held up against the wall with an angular velocity of ω ′ = 2 ωmin. the magnitude of the frictional force between the person and the wall is 1. f = 1 5 m g . 2. f = 1 4 m g . 3. f = m g . 4. f = 1 2 m g . 5. f = 4 m g . 6. f = 2 m g . 7. f = 1 3 m g . 8. f = 5 m g . 9. f = 3 m g .

3. F = m g

Explanation:

The centripetal force the person is exerting on the wall is;

The maximum force due to  static friction is f = µ R ,

where R is  the normal force exerted by  the wall on the person. For the person to be held up against the wall, the maximal  friction force must be larger than the force of  gravity mg.

The actual force is now equal to or less than the maximum µR

                                                 f ≤ fmax = µR

When the angular velocity increase beyond the minimum angular velocity ωmin that is required  to hold the person against the wall,  the frictional force does to increase any more. Therefore, f still equals mg regardless of  the maximum frictional force.  

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

i believe the answer is electric potential.

explanation: