Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2 
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
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.
Answer:
The weight limit of 300kg is the maximum amount the machine can handle so it can be dangerous to exceed the maximum load.
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