The answer is B.
The planet cannot be too hot or too cold it has to be the right distance from its sun to maintain life.
The force is gravitational because when something is falling is call gravitational
(a) The object moves with uniform velocity from A to B.
(b) The object moves with constant velocity from B to C.
(c) The object moves with increasing velocity from C to D.
<h3>
Velocity of the object from point A to B</h3>
V(A to B) = (6 - 0)/(4 - 0) = 1.5 m/s
<h3>
Velocity of the object from point B to C</h3>
V(B to C) = (6 - 6)/(11 - 4) = 0 m/s
<h3>
Velocity of the object from point C to D</h3>
V(C to D) = (7 - 6)/(12 - 11) = 1 m/s
final velocity = 1 + 1.5 m/s = 2.5 m/s
Thus, we can conclude the following;
The object moves with uniform velocity from A to B.
The object moves with constant velocity from B to C.
The object moves with increasing velocity from C to D.
Learn more about velocity here: brainly.com/question/6504879
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The answer is false. The speed of the astronaut cancels out the force of gravity, causing a 'stationary freefall'. While under these effects, it is not required for an astronaut to 'strengthen' his body.
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.