Answer:
weight!!!! Free fall is the motion of a body where its weight is the only force acting on an object.
Answer:
Fnet = 0
Explanation:
- Since the block slides across the floor at constant speed, this means that it's not accelerated.
- According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
- This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

- In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

⇒ 169 N + Fn = Fg = 216 N (3)
- This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
- Fn = 216 N - 169 N = 47 N (4)
Initial velocity u = 50 miles/hour
acceleration a = 10 miles/hour
Time t = 2 hours
Distance travelled S = ut + (at^2)/2
Substituting the values in the second equation of motion,
S = 50*2 + (10 * 2 *2)/2
S = 100 + 20
S = 120 miles
Therefore the distance travelled by the car in the next two hours is 120 miles
Here in the above situation when boy pushes the girl there will be equal and opposite force on boy as per Newton's III law.
Due to this action reaction law boy will also go back with some speed.
Since there is no external force on this girl + boy system so we can use momentum conservation principle here.
As per momentum conservation




So boy will go back with speed 0.2 m/s
Part b)
Since the boy and girl will always exert same force on each other by Newton's III law so it has no it matter whether the boy pushed the girl, the girl pushed the boy, or they put their hands together and pushed each other away.
As in all above cases the as per Newton's III law the force on them is always equal and opposite.
The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:

(1)
where
P is the power
A is the area
In our problem, the intensity is

. At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:

And so if we re-arrange (1) we find the power emitted by the source: