Asked by Kobz
A car of mass 1550 kg is travelling at 75.0 km/h on a dry horizontal concrete surface when the driver applies the brakes licking the wheels. (a) How far does the car travel before coming to rest? (b) How long does it take the car to stop after the driver applies the brakes?
Coefficient of friction: static=0.948 kinetic=0.709
Coefficient of friction: static=0.948 kinetic=0.709
Answers
Answered by
drwls
Why would he lick the wheels? Do you mean that he LOCKS the wheels?
(a) Initial kinetic energy = (Friction force)*(length of skid)
M*V^2/2 = (0.709)*M*g*X
M cancels out. Solve for X
X = V^2/[(1.418*g]
(b) Initial momentum = (Friction force)*(Time of skid)
M*V = (0.709)M*g*T
T = V/(0.709g)
(a) Initial kinetic energy = (Friction force)*(length of skid)
M*V^2/2 = (0.709)*M*g*X
M cancels out. Solve for X
X = V^2/[(1.418*g]
(b) Initial momentum = (Friction force)*(Time of skid)
M*V = (0.709)M*g*T
T = V/(0.709g)
Answered by
umar hassan
m*155
t*55 u 36 155*55*36
t*55 u 36 155*55*36
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