Answer:
<em>The comoving distance and the proper distance scale</em>
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Explanation:
The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster. Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.
Vas happenin!
The third one makes no since because the clouds carry the rain. It isn’t always cold when it’s going to rain
The fourth one is a good one
The second one again it’s not always cold when it’s raining
The first one could be it also
Hmmm I would go with the last one
Sorry if it’s wrong
Answer: 459.14 N
Explanation:
from the question, we have
diameter = 10 m
radius (r) = 5 m
weight (Fw) = 670 N
time (t) = 8 seconds
Circular motion has centripetal force and acceleration pointing perpendicular and inwards of the path, therefore we apply the equation below
∑ F = F c = F w − Fn ..............equation 1
Fn = Fw − Fc = mg − (mv^2 / r) ...................equation 2
substituting the value of v as (2πr / T) we now have
Fn = mg − (m(2πr / T )^2) / r
Fn= mg − (4(π^2)mr / T^2) ..........equation 3
Fw (mass of the person) = mg
therefore m = Fw / g
m = 670 / 9.8 = 68.367 kg
now substituting our values into equation 3
Fn = 670 - ( (4 x (π^2) x 68.367 x 5 ) / 8^2)
Fn = 670 - 210.86
Fn = 459.14 N
Answer:
h’ = 1/9 h
Explanation:
This exercise must be solved in parts:
* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy
starting point. Higher
Em₀ = U = m g h
final point. Lower, just before the crash
Em_f = K = ½ m 
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v_b = 
* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2m 0 + m v_b
final instant. Right after the crash
p_f = (2m + m) v
the moment is preserved
p₀ = p_f
m v_b = 3m v
v = v_b / 3
v = ⅓
* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together
Starting point. Lower
Em₀ = K = ½ 3m v²
Final point. Higher
Em_f = U = (3m) g h'
Em₀ = Em_f
½ 3m v² = 3m g h’
we substitute
h’= 
h’ = 
h’ = 1/9 h
Answer:
v=9.6 km/s
Explanation:
Given that
The mass of the car = m
The mass of the truck = 4 m
The velocity of the truck ,u= 12 km/s
The final velocity when they stick = v
If there is no any external force on the system then the total linear momentum of the system will be conserve.
Pi = Pf
m x 0 + 4 m x 12 = (m + 4 m) x v
0 + 48 m = 5 m v
5 v = 48
v=9.6 km/s
Therefore the final velocity will be 9.6 km/s.
