A 905 kg test car travels around a circular track with a circumference of 3.25 km. If the magnitude of the force that maintains the car's circular motion is 2140 N, what is the car's tangential speed?

from circumference, find radius.

force= mass*v^2/r

solve for v.

29kg

35 m/s

To find the car's tangential speed, we need to use the formula:

Tangential speed = Circumference / Time

First, let's calculate the time it takes for the car to complete one full revolution around the circular track.

The circumference is given as 3.25 km, which is equal to 3.25 x 1000 meters (since 1 km = 1000 meters). So, the circumference is 3250 meters.

The time taken to complete one full revolution is known as the period. We can calculate the period using the formula:

Period = 2π * sqrt(mass / force)

Where π is approximately equal to 3.14, sqrt refers to the square root, mass is the mass of the car, and force is the magnitude of the force maintaining the circular motion.

Let's substitute the given values into the formula:

Period = 2π * sqrt(905 kg / 2140 N)

Calculating the square root of (905 kg / 2140 N) gives us approximately 0.398.

Period = 2π * 0.398

Now, multiply 2π by 0.398:

Period ≈ 2 x 3.14 x 0.398 ≈ 2.5 seconds

Now that we have the period, we can find the tangential speed using the formula:

Tangential speed = Circumference / Period

Substituting the known values:

Tangential speed = 3250 m / 2.5 s

Calculating 3250 meters divided by 2.5 seconds gives us a tangential speed of approximately 1300 m/s.

Therefore, the car's tangential speed is approximately 1300 m/s.