A 1200 kg car travels at a constant speed of 15 m/s in a circular loop with a diameter of 45 m. What is the coefficient of friction on the car?

(a)

KE = m v^2 / 2 = (1200 kg)(20 m/s)^2 / 2 = 240,000 J

(b)

The energy is entirely dissipated by the force of friction in the brake system.

(c)

W = delta KE = KEf - KEi = (0 - 240,000) J = -240,000 J

(d)

Fd = delta KE

F = (delta KE) / d = (-240,000 J) / (50 m) = -4800 N

The magnitude of the friction force is 4800 N.

mark brainliest if correct

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.