Physics, 02.12.2020 01:30 darianhaynes

1. *A car is going over the top of a hill whose curvature approximates a circle of radius 200 m. At what velocity will the occupants of the car appear to weigh 20% less than their normal weight (or
their normal weight times 0.8)?

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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:

$\rm F_c = \dfrac{mv^2}{r}$ ---- (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:

$\rm F_c = 0.2\times W$

$\rm F_c = 0.2 \times m \times g$ --- (2)

Now, equate equation (1) with equation (2).

$\rm \dfrac{mv^2}{r}=0.2\times m \times g}{}$

$\rm v = \sqrt{0.2\times g \times r }$

$\rm v = \sqrt{0.2\times 9.81 \times 200 }$

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.

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Answer from: aide1234564

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.

Answer from: Quest

it's when something is going so fast it makes a loud noise, breaking the sound barrier, for example: a jet makes a loud noise, not only cause of the jet but because they reached a point where they went so fast that you can year the noise from various parts, kinda like a police car, lets say the car was going so fast that you hear the siren of the cop car right next to you, but the car is probably a mile a way, it's cos the siren was next to you a few seconds if you understand me

Answer from: Quest

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.

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