A satellite that is in a circular orbit 230 km above the surface of the planet Zeeman-474 has an orbital period of 89 min. The r
1 answer:
Answer:
Mass of the planet = 6.0 ×
Explanation:
Time period = 2π (R + h) / v
Orbital speed (v) = √GM / (R + h)
T² = 4π² (R + h)² / (GM/ (R + h))
= 4π² (R + h)³ / GM
making m the subject of the formula
m = 4π² (R + h)³ / GT²
= 4π² ( 6.38 × + 230 × 10³ )³ / ( 6.67 ×
) × (89 × 60)²
= 4π² ( 6610000)³ / ( 6.67 × ) × (89 × 60)²
= 5.99 ×
= 6.0 ×
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Answer:
The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.
Explanation:
In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.
Momentum is Mass X velocity.
5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.
The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.
Next possible options are in kinetic Energy
Initial Kinetic energy =
Final kinetic energy =
Change in kinetic energy = Final Kinetic Energy - Initial Kinetic Energy
Change in kinetic energy of 5kg ball =
Since the 5-kg ball has initial velocity (u), the magnitude of the change in velocity will be reduced.
Change in kinetic energy of 10kg ball:
the ball is initially at rest, therefore the initial velocity (u) will be zero (0)
Δ K.E =
From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.
Hence, The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball