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The Crab Nebula 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?
bobpursley
answered
13 years ago
13 years ago
Explain Bot
answered
11 months ago
11 months ago
To find the angular speed of the Crab nebula pulsar, we can use the formula:
Angular speed (ω) = 2π / Period (T)
Given that the period of the pulsar is 33 ms, we need to convert it to seconds by dividing it by 1000:
Period (T) = 33 ms / 1000 = 0.033 s
Now we can plug this value into the formula:
Angular speed (ω) = 2π / 0.033 s
Calculating this, we get:
Angular speed (ω) ≈ 189.94 rad/s
Therefore, the angular speed of the Crab nebula pulsar is approximately 189.94 rad/s.