Question

One of the most studied objects in the night sky is the Crab nebula, the remains of a supernova explosion observed by the Chinese in 1054. In 1968 it was discovered that a pulsar-a rapidly rotating neutron star that emits a pulse of radio waves with each revolution-lies near the center of the Crab nebula. The period of this pulsar is 33 ms. What is the angular speed in rad/s of the Crab nebula pulsar?

Answer 1

The angular speed of the Crab nebula pulsar is 190.3 rad/s.

Step-by-step explanation:

Given that,

Time T= 33 ms = 0.033 s

The angular speed is equal to the 2π divided by time period.

We need to calculate the angular speed of the Crab nebula pulsar

Using formula of angular speed

`\omega=(2\pi)/(T)`

Where, T = time

`\omega` = angular speed

Put the value into the formula

`\omega=(2\pi)/(0.033)`

`\omega=190.3\ rad/s`

Hence, The angular speed of the Crab nebula pulsar is 190.3 rad/s.