Answer:
The gravitational force between m₁ and m₂, is approximately 1.06789 × 10⁻⁶ N
Explanation:
The details of the given masses having gravitational attractive force between them are;
m₁ = 20 kg, r₁ = 10 cm = 0.1 m, m₂ = 50 kg, and r₂ = 15 cm = 0.15 m
The gravitational force between m₁ and m₂ is given by Newton's Law of gravitation as follows;
Where;
F = The gravitational force between m₁ and m₂
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
r₂ = 0.1 m + 0.15 m = 0.25 m
Therefore, we have;
The gravitational force between m₁ and m₂, F ≈ 1.06789 × 10⁻⁶ N
we know that center of mass is given as
r = (m₁ 
+ m₂ 
)/(m₁ + m₂)
taking derivative both side relative to "t"
dr/dt = (m₁ d/dt + m₂ d/dt)/(m₁ + m₂)
v = (m₁ 
+ m₂ 
)/(m₁ + m₂)
taking derivative again relative to "t" both side
dv/dt = (m₁ d/dt + m₂ d/dt)/(m₁ + m₂)
a= (m₁ 
+ m₂ 
)/(m₁ + m₂)
Answer:
A. I and V
Explanation:
According to Le Chatelier's Principle, increasing the product side will cause the equilibrium to shift back towards the reactant side, so I is true. By the same principle, II is false.
For gases, decreasing the pressure will cause the equilibrium to shift towards the side with higher number of moles. So V is true.
The reaction is endothermic, so increasing the temperature will shift the equilibrium to the products, so IV is false. And adding a catalyst has no effect on the equilibrium, so III is false.
