Weight of the carriage 
Normal force 
Frictional force 
Acceleration 
Explanation:
We have to look into the FBD of the carriage.
Horizontal forces and Vertical forces separately.
To calculate Weight we know that both the mass of the baby and the carriage will be added.
- So Weight(W)

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with
, force of
acting vertically downward.Both are downward and Normal is upward so Normal force 
- Normal force (N)

- Frictional force (f)

To calculate acceleration we will use Newtons second law.
That is Force is product of mass and acceleration.
We can see in the diagram that
and
component of forces.
So Fnet = Fy(Horizontal) - f(friction) 
- Acceleration (a) =

So we have the weight of the carriage, normal force,frictional force and acceleration.
Part a:
= 56
= 60
= 63
The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The
is the overall medium,
is the medium of the first half, and
is the medium of the second half.
-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.
[] See attached
Part b:
The range is 7.
The interquartile range is the range of numbers between
and
. In other words, it is 50% of the data, directly in the middle.
This becomes 63 - 56 = 7
Part c:
79 is an outlier.
It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.
-> 63 + (7 +
) ≤ 79
-> 63 + 10.5 ≤ 79
-> 73.5 ≤ 79
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.
- Heather
A soccer ball would keep moving forever without physics, because without force to act upon the soccer ball, it could, or will not be able to stop the acceleration. And force is a factor in physics.
Hi there!
This is an example of a recoil collision.
Using the conservation of momentum:

The initial momentum is 0 kgm/s (objects start from rest), so:

We are given that the 15 kg block has a velocity of 12 m/s to the left, so:

Solve for v2':
