A 1100 kg car rounds a circular turn of radius 24.6 m. Of the road is flat and the coefficient of static friction between the tires and the road is 0.59, how fast can the car go without skidding

Velocity is 12.05m/s

Explanation: The Centripetal force in this case must be equal in magnitude to the frictional force in other for the car not to skidd.

But centripetal force F1 is equal to

F1 = Mass*(Velocity ²)/radius

That is,

F1 = M*V²/r

Also, the frictional force F2 is equal to

F2 = coefficient of friction* mass* acceleration due to gravity.

That is,

F2 = u*m*g

Recall that centripetal force must equal to the frictional force for the car not to skidd. Therefore,

M*V²/r = u*m*g

Making V² subject of the formula we have

r*u*m*g/M = V²

24.6*0.59*1100*10/1100 = V²

V² = 145.14

V=√{145.14}

V = 12.05m/s

i don’t know what the answer is i wish i

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000