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 = / R² = mRω²
ω² = GM = / R³
ω = √(GM = / R³)
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
Answer:
ΔP = 14.5 Ns
I = 14.5 Ns
ΔF = 5.8 x 10³ N = 5.8 KN
Explanation:
The mass of the ball is given as 0.145 kg in the complete question. So, the change in momentum will be:
ΔP = mv₂ - mv₁
ΔP = m(v₂ - v₁)
where,
ΔP = Change in Momentum = ?
m = mass of ball = 0.145 kg
v₂ = velocity of batted ball = 55.5 m/s
v₁ = velocity of pitched ball = - 44.5 m/s (due to opposite direction)
Therefore,
ΔP = (0.145 kg)(55.5 m/s + 44.5 m/s)
<u>ΔP = 14.5 Ns</u>
The impulse applied to a body is equal to the change in its momentum. Therefore,
Impulse = I = ΔP
<u>I = 14.5 Ns</u>
the average force can be found as:
I = ΔF*t
ΔF = I/t
where,
ΔF = Average Force = ?
t = time of contact = 2.5 ms = 2.5 x 10⁻³ s
Therefore,
ΔF = 14.5 N.s/(2.5 x 10⁻³ s)
<u>ΔF = 5.8 x 10³ N = 5.8 KN</u>
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Power is the energy in a system per time. It will have units of Watts which is equal to joules per second. It can be expressed as:
P = E / t
where E = Force x distance
P = Fd / t
t = Fd / P
t = 8 (9.72) / 3.0
t = 25.92 s
metamorphic, sedimentary, igneous
