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.

jupiter

explanation:

it is so massive and has such a strong gravitational pull its likely prevented the formation of a planet between itself and mars in the asteroid belt

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stan bts

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.