PV System for the Smith Family answers
1 answer:
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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 angle of elevation of the rocket is increasing at a rate of 48.780º per second.
Explanation:
Geometrically speaking, the distance between the rocket and the observer (), measured in kilometers, can be represented by a right triangle:
(1)
Where:
- Horizontal distance between the rocket and the observer, measured in kilometers.
- Vertical distance between the rocket and the observer, measured in kilometers.
The angle of elevation of the rocket (), measured in sexagesimal degrees, is defined by the following trigonometric relation:
(2)
If we know that , then the expression is:
And the rate of change of this angle is determined by derivatives:
Where:
- Rate of change of the angle of elevation, measured in sexagesimal degrees.
- Vertical speed of the rocket, measured in kilometers per hour.
If we know that and
, then the rate of change of the angle of elevation is:
The angle of elevation of the rocket is increasing at a rate of 48.780º per second.