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

x = 10.1

step-by-step explanation:

all of the angles of a tirangle add up to 180º

180 - 65 = 115 (since we already know the value of one angle)

(2x + 7) + (8x + 7) = 115

(combins like terms) 10x + 14 = 115

(isolate the variable subtract 14 from both sides): 10x = 101

(find the variable divide both sides by 10): x = 101/10

x = 10.1

answer: kljdcdhjbdlchjad oac apoich a

step-by-step explanation: