The diagram shows the top view of a 65-kg student at point A on an amusement park ride. The ride spins the student in a horizontal circle with a radius of 2.5 m, at a constant speed of 8.6 m/s. The floor is lowered and the student remains against the wall without falling to the floor. The magnitude of the centripetal force acting on the student at point A is approximately

Question 4 options:

1398 N

47 N

223 N

1923 N

1923 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 65 Kg

Radius (r) = 2.5 m

Velocity (v) = 8.6 m/s

Centripetal force (F) =?

The centripetal force, F, can be obtained by using the following formula:

F = mv²/r

F = 65 × 8.6² / 2.5

F = 65 × 73.96 / 2.5

F = 4807.4 / 2.5

F = 1922.96 ≈ 1923 N

Thus, the magnitude of the centripetal's force acting on the student is approximately 1923 N

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.