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: The answer is C.) 25 m/s^2.
Explanation: If you input 5 as s, you would have to use the exponent 2. This means that you have to multiply 5 by 5. 5 x 5= 25.
Edit: Also, because the surface is frictionless, it will make the object go faster too. Nothing can really slow it down unless something blocks it.
Answer:
Both charges must have the same charge, Qt/2.
Explanation:
Let the two charges have charge Q1 and Q2, respectively.
Use Coulombs's Law to find an expression for the force between the two charges.
, where
Ke is Coulomb's contant and
r is the distance between the charges.
We know from the question that
Q1 + Q2 = Qt
So,
Q2 = Qt - Q1

Simplify to obtain,

In order to find the value of Q1 for which F is the maximum, we will use the optimization technique of calculus.
Differentiate F with respect to Q1,

Equate the differential to 0, to obtain the value of Q1 for which F is the maximum.

It follows that
.
Because there’s no elemental degradation, the mass of the compound EQUALS the sum of the masses.
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