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3 years ago

15

A neutron star is the remnant left after certain supernovae (explosions of giant stars). Typically, neutron stars are about 18 k

m in diameter and have around the same mass as our sun. What is a typical neutron star density in g/cm^3?

1 answer:

Tresset [83]3 years ago

5 0

Answer:

$Density=6.51*10^{14}g/cm^{3$

Explanation:

Sun mass:

Ms=1.989 × 10^30 kg

Neutron star has the same mass.

Radius Neutron:

R=9Km (because diameter is 18Km)

R=9*10^3m

Density neutron star:

$D=Mn/Vol=Ms/(4/3*\pi*R^3)=1.989*10^{30} /(4/3*\pi*(9*10^{3})^{3})=6.51*10^{17}kg/m^{3}$

$D=6.51*10^{17}kg/m^{3}*(1000g/1Kg)*(1m^{3}/1000000cm^{3})=6.51*10^{14}g/cm^{3$

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3 years ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

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Al's mass is 102.92 kg

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So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

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$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

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$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

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Explanation:hope that helps

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