Answer:
[tex]5.25\cdot 10^{40} kg m^2/s[/tex]
Explanation:
The angular momentum of the pulsar is given by:
[tex]L=m\omega r^2[/tex]
where
[tex]m=2.8\cdot 10^{30} kg[/tex] is the mass of the pulsar
[tex]r = 10.0 km = 1\cdot 10^4 m[/tex] is the radius
[tex]\omega[/tex] is the angular speed
Given the period of the pulsar, [tex]T=33.5\cdot 10^{-3} s[/tex], the angular speed is given by
[tex]\omega=\frac{2\pi}{T}=\frac{2 \pi}{33.5\cdot 10^{-3}s}=187.5 rad/s[/tex]
And so, the angular momentum is
[tex]L=m\omega r^2=(2.8\cdot 10^{30}kg)(187.5 rad/s)(1\cdot 10^4 m)^2=5.25\cdot 10^{40} kg m^2/s[/tex]
