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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 20 km (about the size of the San Francisco area). If a neutron star rotates once every second, what is the speed of a particle on the star's equator?

Respuesta :

Answer:

Speed,[tex]v=1.25\times 10^5\ m/s[/tex]

Explanation:

Given that,

Radius of the neutron stare, R = 20 km = 20,000 m

Time taken by the Neutron star to rotate, t = 1 s

We need to find the speed of a particle on the star's equator. The total distance covered divided by total time taken is called speed of an object. Here,

[tex]v=\dfrac{2\pi R}{t}[/tex]

[tex]v=\dfrac{2\pi \times 20000}{1}[/tex]

v = 125663.70 m/s

or

[tex]v=1.25\times 10^5\ m/s[/tex]

So, the speed of a particle on the star's equator is [tex]v=1.25\times 10^5\ m/s[/tex]. Hence, this is the required solution.

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