Question

1. *A car is going over the top of a hill whose curvature approximates a circle of radius 200 m. At

Answer 1

Answer:

The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight is approximately 19.81 m/s

Explanation:

The given parameters are;

The curvature of the hill, r = 200 m

Due to the velocity, v, the occupants weight = 20% less than the normal weight

The outward force of an object due to centripetal (motion) force is given by the following equation;

$F_c = \dfrac{m \times v^2}{r}$

Where;

r = The radius of curvature of the hill = 200 m

Given that the weight of the occupants, W = m × g, we have;

$F_c = 0.2 \times W = 0.2 \times m \times g$

$\therefore 0.2 \times m \times g = \dfrac{m \times v^2}{r}$

v = √(0.2 × g × r)

By substitution, we have;

v = √(0.2 × 9.81 × 200) ≈ 19.81

The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight ≈ 19.81 m/s.