The Crab Pulsar has a period that is currently of length 33.085 ms.It is estimated to have an equatorial radius of 15 km,about average for a neutron star.The pulsar is slowing in its rotation so that it is expected to come to rest 9.5 $\times$ 10¹⁰ s in the future.What is the value of the tangential acceleration of an object on the neutron star's equator? A) 3.0 $\times$ 10^-5 m/s² B) 2.0 $\times$ 10^-9 m/s² C) 3.2 $\times$ 10^-10 m/s² D) 7.6 $\times$ 10^-7 m/s² E) 4.8 $\times$ 10^-6 m/s²

Question

Quizplus · Accepted Answer

The tangential acceleration of an object on the neutron star's equator is given by:$$a_t = \frac{v^2}{r} = \frac{(2\pi r / T)^2}{r} = \frac{4\pi^2 r}{T^2}$$where:* v is the tangential velocity* r is the equatorial radius* T is the periodSubstituting the given values, we get:$$a_t = \frac{4\pi^2 (15 	imes 10^3 	ext{ m})}{(33.085 	imes 10^{-3} 	ext{ s})^2} = 3.0 	imes 10^{-5} 	ext{ m/s}^2$$Therefore, the correct choice is A.

The Crab Pulsar Has a Period That Is Currently of Length