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?

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  1. 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. 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

    Friction force = Radial force(F)

    [tex]F=\dfrac{mv^2}{r}[/tex]

    By putting the values

    [tex]F=\dfrac{mv^2}{r}[/tex]

    [tex]F=\dfrac{1200\times 5.55^2}{25}[/tex]

    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.

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