Certain neutron stars (extremely dense stars) are believed to be rotating at about 1 rev/s. if such a star has a radius of 20 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation? hint: the equator of the star is where the maximum centripetal acceleration occurs.

M = 4.7 10²⁴ kg

Explanation:

For this exercise we must use the law of universal gravitation

         F = G m M / r²

Newton's second law with centripetal acceleration

         a = v² / r

         v = w r

         a = w² r

         F = m a

        G m M / r2 = m w2 r

        G M = w² r³

        M = w² r³ / G

Let's reduce the magnitudes to the SI system

        w = 1 rev / s (2π rad / 1 rev) = 2π rad / s

        r = 20 km (1000m / 1 km) = 2 10⁴ m

Let's calculate

       M = (2π)²2 (2 10⁴)³ / 6.67 10⁻¹¹

       M = 47.35 10²³ kg

       M = 4.7 10²⁴ kg

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

a cold ice-tea.

pour in room temperature or cold water. cover the pitcher and refrigerate. steep white or green tea for 6 to 8 hours; steep black or oolong tea for 8 to 12 hours.jun 28, 2015