Respuesta :
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;
[tex]F_c = \dfrac{m \times v^2}{r}[/tex]
Where;
r = The radius of curvature of the hill = 200 m
Given that the weight of the occupants, W = m × g, we have;
[tex]F_c = 0.2 \times W = 0.2 \times m \times g[/tex]
[tex]\therefore 0.2 \times m \times g = \dfrac{m \times v^2}{r}[/tex]
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.
The velocity is 19.81 m/sec at which the occupants of the car appear to weigh 20% less than their normal weight and this can be determined by using the formula of centripetal force.
Given :
A car is going over the top of a hill whose curvature approximates a circle of radius 200 m.
The centripetal force acting on the car is given by the formula:
[tex]\rm F_c = \dfrac{mv^2}{r}[/tex] ---- (1)
where m is the mass of a car, v is the velocity of the car and r is the radius of curvature.
Now, it is given that the occupants of the car appear to weigh 20% less than their normal weight that means:
[tex]\rm F_c = 0.2\times W[/tex]
[tex]\rm F_c = 0.2 \times m \times g[/tex] --- (2)
Now, equate equation (1) with equation (2).
[tex]\rm \dfrac{mv^2}{r}=0.2\times m \times g}{}[/tex]
[tex]\rm v = \sqrt{0.2\times g \times r }[/tex]
[tex]\rm v = \sqrt{0.2\times 9.81 \times 200 }[/tex]
v = 19.81 m/sec
The velocity is 19.81 m/sec at which the occupants of the car appear to weigh 20% less than their normal weight.
For more information, refer to the link given below:
https://brainly.com/question/11324711
