By definition, a halo is a part of a galaxy wherein it mainly consists of scattered stars forming bulges having a significantly spherical structure. In addition, the galactic structure is commonly made up of old and metal-rich stars forming clouds of gas and dirt at the spirals of a galaxy.
Answer:
The force must he apply to the sled is of F= 764.4 N.
Explanation:
m= 60 kg
g= 9.8 m/s²
μ=0.3
W= m*g
W= 588 N
Fr= μ*W
Fr= 176.4 N
F= W + Fr
F= 764.4 N
Answer:
w₂ = 22.6 rad/s
Explanation:
This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.
Let's write the moment two moments
initial instant. Before releasing bricks
L₀ = I₁ w₁
final moment. After releasing the bricks
= I₂W₂
L₀ = L_{f}
I₁ w₁ = I₂ w₂
w₂ = I₁ / I₂ w₁
let's reduce the data to the SI system
w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s
let's calculate
w₂ = 6.0/2.0 7.54
w₂ = 22.6 rad/s
1) 0.0011 rad/s
2) 7667 m/s
Explanation:
1)
The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

where
is the angular displacement of the object
t is the time elapsed
is the angular velocity
In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is
rad
And the time taken is

Therefore, the angular velocity of the telescope is

2)
For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

where
v is the linear velocity
is the angular velocity
r is the radius of the circular orbit
In this problem:
is the angular velocity of the Hubble telescope
The telescope is at an altitude of
h = 600 km
over the Earth's surface, which has a radius of
R = 6370 km
So the actual radius of the Hubble's orbit is

Therefore, the linear velocity of the telescope is:
