Acar of mass 1200 kg drives around a curve with a radius of 25.0 m. if the driver maintains a speed

Acar of mass 1200 kg drives around a curve with a radius of 25.0 m. if the driver maintains a speed of 20.0 km/hr, what is the force of friction between the tires and the road? what is the minimum coefficient of static friction required to keep the car in this turn?

What is the potential energy of a 30kg rock that falls 15 meters

1. Expert says:

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation:

2. Expert says:

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3. friendsalwaysbae says:

F=1487.52 N

μ = 0.12

Explanation:

Given that

m= 1200 kg

r= 25 m

v= 20 km/hr = 20 x 5/18 m/s

v= 5.55 m/s

To maintain the speed 5.55 m/s ,friction force should be equal to the radial force

$F=\dfrac{mv^2}{r}$

By putting the values

$F=\dfrac{mv^2}{r}$

$F=\dfrac{1200\times 5.55^2}{25}$

F=1487.52 N

So friction force = 1478.52 N

We also know that

Friction force = μ m g

1478.52 N = μ  x 1200 x 10                         ( take g =10 m/s²)

μ = 0.12

This is  minimum coefficient of static friction required to keep the car in this turn.