The answer is B. Unbalanced force
<span>The angular momentum of a particle in orbit is
l = m v r
Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"
m_1 v_1 r_1 = m_2 v_2 r_2
Assuming that the mass did not change, conservation of angular momentum demands that
v_1 r_1 = v_2 r_2
or
v1 = v_2 (r_2/r_1)
Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have
v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s
Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.
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Answer:
The slope of a velocity graph represents the acceleration of the object. So, the value of the slope at a particular time represents the acceleration of the object at that instant.
decrease yeah free 100 percent
Explanation:
Given:
Final speed of mass A = Va
Final speed of mass B = Vb
Mass of A = Ma
Mass of B = Mb
Ma = 2 × Mb
By conservation of linear momentum,
0 = Ma × Va + Mb × Vb
0 = 2 × Mb × Va + Mb × Vb
Vb = - 2 × Va
Energy of the spring, U = 1/2 × k × x^2
1/2 k x² = 1/2 × Ma × Va² + 1/2 × Mb × Vb²
35 = 1/2 × Ma × Va² + 1/2 × Mb × Vb²
Ma × Va² + Mb × Vb² = 70
2 × Mb (-Vb/2)² + Mb × Vb² = 70
1/2 × Mb × Vb² + Mb × Vb² = 70
3/2 × Mb × Vb² = 70
Mb × Vb² = 140/3
= 46.7 J
Ma = 2 × Mb and Vb = - 2 × Va
Ma/2 × (4 × Va²) = 140/3
Ma × Va² = 70/3
Kinetic energy of mass A, KEa = 1/2 × Ma × Va² = 23.3 J
Kinetic energy of mass B = 1/2 × Mb × Vb² = 46.7 J
