A 0.32 m radius automobile tire starts to rotate from rest and it is acceleratin

Question

A 0.32 m radius automobile tire starts to rotate from rest and it is accelerating at a constant, angular acceleration of 1.8rad/s2. What is the centripetal acceleration of a point on the outer edge of the tire after 5.0 s? (Show all the steps of your work) A star has mass 5.0 Times 10kg. and it is compressed to a radius of 2.0 Times 106m. G = 6.67 Times 10-11 Nm2/kg2 is universal gravitational constant. What will be the gravitational force on a mass of 1.5 Times 10-3kg object lying on the surface of that star? (Show all the steps of your work ) A star has mass 2.5 Times 1030 kg. and it has a radius of 2.3 Times 108m. G = 6.67 Times 10-11 Nm2/kg2 is universal gravitational constant. What will be the gravitational force on a mass of 0.15kg object lying on the surface of that star? (Show all the steps of your work)

Explanation / Answer

18)

here ,

angular speed , wf = a*t

wf = 1.8 * 5

wf = 9 rad/s^2

Now, centripetal acceleration , ac = wf^2 * r

centripetal acceleration , ac = 9^2 * 0.32

centripetal acceleration , ac = 25.92 m/s^2

the centripetal acceleration is 25.92 m/s^2

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