The Crab Pulsar has a period of length 33.085 ms.It is estimated to have an equatorial radius of 15 km,about average for a neutron star.What is the value of the centripetal acceleration of an object on the surface at the equator of the pulsar? A) 5.4 $\times$ 10⁸ m/s² B) 2.7 $\times$ 10⁸ m/s² C) 8.9 $\times$ 10⁸ m/s² D) 6.7 $\times$ 10⁸ m/s²

Question

Quizplus · Accepted Answer

The correct answer is A because the centripetal acceleration of an object on the surface at the equator of the pulsar is given by the formula a = v^2/r, where v is the tangential velocity and r is the radius of the pulsar. The tangential velocity is given by the formula v = 2πr/T, where T is the period of the pulsar. Substituting the given values into these formulas, we get a = (2π(15 km))/(33.085 ms))^2/(15 km) = 5.4 x 10^8 m/s^2.

The Crab Pulsar Has a Period of Length 33 ×\times× 108 M/s2 B) 2

The Crab Pulsar Has a Period of Length 33 $\times$ 10⁸ M/s²
B) 2