When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about 18.0 km (about the size of a typical city). if a neutron star rotates once every 0.860 seconds, (a) what is the speed of a particle on the star's equator and (b) what is the magnitude of the particle's centripetal acceleration? (c) if the neutron star rotates faster, do the answers to (a) and (b) increase, decrease, or remain the same?

Explanation:

a ) radius of star  R = 18 x 10³ m

time period of rotation T = .86 s.

angular velocity ω = 2π / T

= 2 X 3.14 / .86

= 7.3 radian / s .

velocity required

= ω R

7.3 x 18 x 10³

= 131.4 m /s

b )

centripetal acceleration

= ω² R

= ω R ω

= 131.4 x  7.3

959.22 m/s²

c )

If the neutron star rotates faster , both velocity of the particle on the equator

and centripetal acceleration increases.

part 1: describe the mass and the surface temperature of dwarf stars.

part 2: explain using complete sentences where on the hr diagram you would expect to find a dwarf star based on its physical properties.

part 1: the mass of a dwarf star is very small and the surface temperature of dwarf stars are high and they have the lowest absolute brightness.

part 2: on the hr diagram you would expect to find a dwarf star based on its physical properties at an absolute brightness of -2 to -4

hope this