(03.07 MC) A 660 kg car is travelling around a circular track of radius 55.0 m at 11.5 m/s.
Calculate the force and its direction that provides the centripetal force acting on the car to keep it
in the circular path. (3 points)
a) 24 N toward the center of the circle
b) 24 N along the line tangent to the circle
c) 1.587 E3 N along the line tangent to the circle
d) 1.587 E3 N toward the center of the circle

D) 1.587 E3 N toward the center of the circle

This is the same as saying 1.587*10^3 newtons, or we can say 1587 newtons

Work Shown:

m = mass = 660 kilograms

v = velocity = 11.5 meters per second

r = radius of circle = 55.0 meters

F = centripetal force = inward pulling force toward center of circle

F = (m*v^2)/r

F = (660*11.5^2)/55

F = 1587

F = 1.587 * 10^3

F = 1.587 E 3

The "E 3" notation is what many calculators display scientific notation to be

The units for the force here are in newtons.

The force is pulling on the object toward the center of the circle, which helps keep the car on the circular path, rather than it going off on a tangent.

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

i already answered it by 34/s