An astronomical unit (A.U.) is 1 point A) a term for defining the luminosity of a star B) the average distance from the Earth to
1 answer:
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Answer:
<em>Speed of the electron is 2.46 x 10^8 m/s</em>
<em></em>
Explanation:
momentum of the electron before relativistic effect =
where is the rest mass of the electron
V is the velocity of the electron.
under relativistic effect, the mass increases.
under relativistic effect, the new mass M will be
M =
where
c is the speed of light = 3 x 10^8 m/s
V is the speed with which the electron travels.
The new momentum will therefore be
==>
It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have
1.75 =
the equation reduces to
1.75 =
square both sides of the equation, we have
3.0625 = 1/
3.0625 - 3.0625 = 1
2.0625 = 3.0625
= 0.67
β = 0.819
substitute for
V/c = 0.819
V = c x 0.819
V = 3 x 10^8 x 0.819 = <em>2.46 x 10^8 m/s</em>
Answer:
The required angular speed the neutron star is 10992.32 rad/s
Explanation:
Given the data in the question;
mass of the sun M = 1.99 × 10³⁰ kg
Mass of the neutron star
M = 2( M
)
M = 2( 1.99 × 10³⁰ kg )
M = ( 3.98 × 10³⁰ kg )
Radius of neutron star R = 13.0 km = 13 × 10³ m
Now, let mass of a small object on the neutron star be m
angular speed be ω.
During rotational motion, the gravitational force on the object supplies the necessary centripetal force.
GmM = / R
² = mR
ω
²
ω² = GM
= / R
³
ω = √(GM
= / R
³)
we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²
we substitute
ω = √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)
ω = √( 2.65466 × 10²⁰ / 2.197 × 10¹²
ω = √ 120831133.3636777
ω = 10992.32 rad/s
Therefore, The required angular speed the neutron star is 10992.32 rad/s