Oxygen and carbon dioxide
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:
Particles in a water wave exchange kinetic energy for potential energy. When particles in water become part of a wave, they start to move up or down. This means that kinetic energy (energy of movement) has been transferred to them
Explanation:
hope this helps u ....
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Answer:0.27
Explanation:
Given
One worker Pushes with force 
other Pulls it with a rope of rope 
mass of crate 
both forces are horizontal and crate slides with a constant speed
Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

where
is the friction force



where
is the coefficient of static friction



Answer:
Imp = 25 [kg*m/s]
v₂= 20 [m/s]
Explanation:
In order to solve these problems, we must use the principle of conservation of linear momentum or momentum.
1)

where:
m₁ = mass of the object = 5 [kg]
v₁ = initial velocity = 0 (initially at rest)
F = force = 5 [N]
t = time = 5 [s]
v₂ = velocity after the momentum [m/s]
![(5*0) +(5*5) = (m_{1}*v_{2}) = Imp\\Imp = 25 [kg*m/s]](https://tex.z-dn.net/?f=%285%2A0%29%20%2B%285%2A5%29%20%3D%20%28m_%7B1%7D%2Av_%7B2%7D%29%20%3D%20Imp%5C%5CImp%20%3D%2025%20%5Bkg%2Am%2Fs%5D)
2)
![(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})\\(0.075*0)+(30*0.05)=(0.075*v_{2})\\v_{2}=20 [m/s]](https://tex.z-dn.net/?f=%28m_%7B1%7D%2Av_%7B1%7D%29%2B%28F%2At%29%3D%28m_%7B1%7D%2Av_%7B2%7D%29%5C%5C%280.075%2A0%29%2B%2830%2A0.05%29%3D%280.075%2Av_%7B2%7D%29%5C%5Cv_%7B2%7D%3D20%20%5Bm%2Fs%5D)