Answer:
The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".
Explanation:
Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.
Let's look at the speed of the car

We use movies and find lips

The moment is defined by

The moment change

Let's replace the speeds in this equation

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.
Answer:
If we are looking for evidence of something that exists outside of our visible Universe and leaves no trace within it, it seems that the idea of a Multiverse is fundamentally untestable. But there are all sorts of things that we cannot observe that we know must be true. Decades before we directly detected gravitational waves, we knew that they must exist, because we observed their effects.
Explanation:
Maybe helps lol
That is meters per second, same as velocity.
Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s
Your answer is "<span>surface of a sphere"
Hope this helps.</span>