Remember Dark Energy is repulsive, the more the faster the universe expands (indeed accelerates), so the faster it would cool down and increase its size, not the other way. Now, no one knows what it is. For instance, if Dark Energy would rather be some other geometric effect, it could have a different ending than just if it is a cosmological constant term. So far, though, all experimental data do not favor anything else that is not compatible with just a cosmological constant ... We'll need to wait few years (~10 years) to get to know much better this with missions and ground experiments that are undergoing or planned to be launched.
Answer: ⇒ Answer is 3
<h2>Explanation
: momentum = mass × velocity</h2>
"A small force may produce a large change in momentum by acting on a very massive object".
THEY HAVEN'T GIVEN US THE TIME PERIOD NOR THE DISTANCE TRAVELED. THEREFORE WE CANNOT ACTUALLY DECIDE IF THE FORCE IS KEPT FOR A LONG TIME OR SHORT TIME. ANYWAYS SINCE THE MASS IS GIVEN AS MASSIVE , THE MOMENTUM SHOULD BE DEFINITELY HIGH.
WHY I SAY OTHERS ARE WRONG:
1) For a small force to give a large change in momentum, it should act for a long time interval.
2) By applying a large force for a short time interval, the change of momentum should be large.
3) Correct answer.
4) Acting over a short distance can be the same as acting over a short period of time.Therefore the distance should be large in order for a larger momentum.
I HOPE IT HELPS!
The answer correct is C:The cell is most likely prokaryotic ,but it could be eukaryotic since some unicellular are eukaryotic.
Answer:
10 km/hr/s
Explanation:
The acceleration of an object is given by

where
v is the final velocity
u is the initial velocity
t is the time
For the car in this problem:
u = 0

t = 6 s
Substituting in the equation,

Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ =
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m