What is the surface temperature of a distant star having a peak wavelength of 475 nm?
1 answer:
You might be interested in
Answer:
<em>A) the moment of inertia of the system decreases and the angular speed increases. </em>
Explanation:
The complete question is
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, It is true to say that
A) the moment of inertia of the system decreases and the angular speed increases.
B) the moment of inertia of the system decreases and the angular speed decreases.
C) the moment of inertia of the system decreases and the angular speed remains the same.
D) the moment of inertia of the system increases and the angular speed increases.
E) the moment of inertia of the system increases and the angular speed decreases
In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as
....1
where and
are the initial and final moment of inertia respectively.
and and
are the initial and final angular speed respectively.
Also, we know that the moment of inertia of a rotating body is given as
....2
where is the mass of the rotating body,
and is the radius of the rotating body from its center.
We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.
From equation 1, we see that in order for the angular momentum to be conserved, the decrease from to
will cause the angular speed of the system to increase from
to
.
From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.
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