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zheka24 [161]
3 years ago
6

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

pulse for each rotation of the star. The period T of rotation is found by measuring the time between pulses. Suppose a pulsar has a period of rotation of T = 0.0922 s that is increasing at the rate of 3.64 x 10-7 s/y. (a) What is the pulsar's angular acceleration
Physics
2 answers:
kow [346]3 years ago
7 0

Answer:

- 8.5 x 10^-12 rad/s²

Explanation:

Time period, T = 0.0922 s

dT / dt = 3.64 x 10^-7 s/year

As we know that

α = dω/dt

where, α is the angular acceleration and ω is the angular velocity

ω = 2π/T

\alpha =\frac{d\omega }{dt}

\alpha =\frac{d\left ( \frac{2\pi }{T} \right ) }{dt}

\alpha =-\frac{2\pi}{T^{2}}\times \frac{dT}{dt}

By substituting the values

\alpha =-\frac{2\times 3.14}{0.0922^{2}}\times \frac{3.64\times 10^{-7}}{365\times 24\times 3600}

α = - 8.5 x 10^-12 rad/s²

Elenna [48]3 years ago
6 0

Answer:

-8.54883\times 10^{-12}\ rad/s^2

Explanation:

T = Time period = 0.0922 s

\dfrac{dT}{dt}=3.64\times 10^{-7}\ s/y

Angular speed is given by

\omega=\dfrac{2\pi}{T}

Angular acceleration is given by

\alpha=\dfrac{d\omega}{dt}\\\Rightarrow \alpha=\dfrac{d\dfrac{2\pi}{T}}{dt}\\\Rightarrow \alpha=-\dfrac{2\pi}{T^2}\dfrac{dT}{dt}\\\Rightarrow \alpha=-\dfrac{2\pi}{0.0922^2}\times\dfrac{3.64\times 10^{-7}}{365.25\times 24\times 60\times 60}\\\Rightarrow \alpha=-8.52541\times 10^{-12}\ rad/s^2

The pulsar's angular acceleration is -8.52541\times 10^{-12}\ rad/s^2

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