Explanation:
We know that the sky appears to us like a sphere called as celestial sphere which appears to rotate around an imaginary axis because of Earth's rotation. Since the axis cuts the celestial sphere at celestial poles all the object seems to circle around the celestial poles.
Condition 1: The stars rise and set perpendicular to the horizon
The observer is at the equator
Condition 2: The stars circle the sky parallel to the horizon
The observer is at the Pole of the Earth
Condition 3: The celestial equator passes through the zenith
The observer is at the equator
Condition 4: In the course of a year, all stars are visible
The observer is at the equator
Condition 5: The Sun rises on March 21 and does not set until September 21 (ideally)
The observer is at North Pole
A spinning top is the answer
21.75 Miles Per Hour
I got this by multiplying 7.25(3) because I know 20 minutes is 1/3 of 1 he
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