Consider a neutron star with a mass equal to the sun, a radius of 10 kmkm, and a rotation period of 1.0 ss. What is the speed of a point on the equator of the star

6.28 * 10⁴ m/s

Explanation:

Using the relation :

Given :

Radius, R = 10km

Period, T = 1.0 s

v = (2πR) / T

Where v = velocity of a point

R in meters ;

1km = 1000m = 10³m

10km = 10 * 10³m = 1 * 10^4 m

v = (2π*10^4m) / 1 s

v = 6.2831853 * 10⁴ m/s

v = 6.28 * 10⁴ m/s

as a general rule, scissors have blades less than 6 in (15 cm) long and usually have handles with finger holes of the same size. shears have blades scissors used to cut hair or fabric must be much sharper

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer