If this 1300 kg car is driven at 27 m/s through the bottom of a circular dip in the road that has a radius of 600 m, by how much do these springs compress compared to when the car is driven on a flat road

The springs compress by an extra 12% when driven through the bottom of a circular dip compared to being driven on a flat road.

Explanation:

On a flat road, the springs just support the car's weight.

The weight of the car = force on the springs

And from Hooke's law, F = kx

On a flat road,

k x = mg

x(flat) = 1300 × 9.8/k = 12740/k

Through the bottom of the circular dip, the springs not only balance the weight of the car, they also balance the force that keeps it in circular motion.

kx = (mv²/r) + mg

kx = (1300 × 27²/600) + (1300×9.8) = 14319.5N

x(circular dip) = 14319.5/k

If the value of the springs' spring-constant is available, we'd get straight values for This, but in their absence,

x(circular dip)/x(flat road) = 14319.5/12740 = 1.12

The springs compress by an extra 12% when driven through the bottom of a circular dip compared to being driven on a flat road.

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .