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rodikova [14]
3 years ago
7

A star with the mass (M=2.0×1030kg) and size (R=3.5×108m) of our sun rotates once every 30.0 days. After undergoing gravitationa

l collapse, the star forms a pulsar that is observed by astronomers to emit radio pulses every 0.200 s . By treating the neutron star as a solid sphere, deduce its radius.
Physics
1 answer:
Gre4nikov [31]3 years ago
8 0

Answer:

r=97.22x10^{3} m

Explanation:

Using the angular formulas can determine the radius using both values neutron star and the the knowing star so

L=I*w

L_{1}=I_{1}*w_{1}=L_{2}=I_{2}*w_{2}

I_{1}*w_{1}=I_{2}*w_{2}

I=Inertia of the star

w=angular velocity

I=\frac{2*m*r^{2}}{5}

w=\frac{2\pi}{t}

Notice the angular velocity determinate by the time and the Inertia have the radius value so

\frac{2}{5}*m*r_{sn}^{2}*\frac{2\pi }{t_{1}}=\frac{2}{5}*m*r_{s}^{2}*\frac{2\pi }{t_{2}}

r_{sn}^{2}*\frac{1}{t_{1}}=r_{s}^{2}*\frac{1}{t_{2}}

r_{sn}^{2}=r_{s}^{2}*\frac{t_{1}}{t_{2}}

t_{1}=0.2s\\t_{2}=30day*\frac{24hr}{1day}*\frac{60minute}{1hr}*\frac{60seg}{1minute}=2.592x10^{6}s

r_{sn}=3.5x10^{8}m*\sqrt{\frac{0.2s}{2.592x^{6}s}}

r_{sn}=97.22x10^{3} m

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Explanation:

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2. To algebraically add vectors, split each vector into x and y components.

Aₓ = 5.0 cos 45 = 3.5

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Answer:

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where V = voltage / potential difference, I = current, and R = resistance

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