since child is moving along with the wagon and we need to find the acceleration of child inside that wagon then in this case the system of interest must be child + wagon
System of interest will be the system that is used to find the force or acceleration using Newton's law
Here we have to assume that system on which if we will calculate the forces then the net value of force on that system will help to calculate the unknown quantities
So here our system will be boy + wagon
Answer:
The average velocity is
and
respectively.
Explanation:
Let's start writing the vertical position equation :

Where distance is measured in meters and time in seconds.
The average velocity is equal to the position variation divided by the time variation.
= Δx / Δt = 
For the first time interval :
t1 = 5 s → t2 = 8 s
The time variation is :

For the position variation we use the vertical position equation :

Δx = x2 - x1 = 1049 m - 251 m = 798 m
The average velocity for this interval is

For the second time interval :
t1 = 4 s → t2 = 9 s


Δx = x2 - x1 = 1495 m - 125 m = 1370 m
And the time variation is t2 - t1 = 9 s - 4 s = 5 s
The average velocity for this interval is :

Finally for the third time interval :
t1 = 1 s → t2 = 7 s
The time variation is t2 - t1 = 7 s - 1 s = 6 s
Then


The position variation is x2 - x1 = 701 m - (-1 m) = 702 m
The average velocity is

Answer:
Explanation:
Actual weight, Wo = 5 N
Apparent weight, W = 4.5 N
density of water = 1 g/cm^3 = 1000 kg/m^3
density of gold, = 19.32 g/cm^3 = 19.32 x 1000 kg/m^3
Buoyant force = Actual weight - Apparent weight
Volume x density of water x g = 5 - 4.5
V x 1000 x 9.8 = 0.5
V = 5.1 x 10^-6 m^3
Weight of gold = Volume of gold x density of gold x gravity
W' = 5.1 x 10^-6 x 19.32 x 1000 x 9.8 = 0.966 N
As W' is less than W so, it is not pure gold.