Answer:
500 kg
Explanation:
It is given that,
The mass of a open train car, M = 5000 kg
Speed of open train car, V = 22 m/s
A few minutes later, the car’s speed is 20 m/s
We need to find the mass of water collected in the car. It is based on the conservation of momentum as follows :
initial momentum = final momentum
Let m is final mass
MV=mv

Water collected = After mass of train - before mass of train
= 5500 - 5000
= 500 kg
So, 500 kg of water has collected in the car.
Answer: The statement "The charge cannot be created or destroyed describes the principle of the conservation of charge".
Explanation:
According to the conservation of charge, the charge can neither be created nor destroyed. It can be transferred from one system to another.
In an isolated system, the total electric charge remains constant. The net quantity of electric charge is always conserved in the universe.
Therefore, "the charge cannot be created or destroyed" describes the principle of the conservation of charge.
Answer:

Explanation:
Since the system is in international space station
so here we can say that net force on the system is zero here
so Force by the astronaut on the space station = Force due to space station on boy
so here we know that
mass of boy = 70 kg
acceleration of boy = 
now we know that


now for the space station will be same as above force




Answer:
a) ![(Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]](https://tex.z-dn.net/?f=%28Qa%2Ag%2AVb%29-%28Qh%2AVb%2Ag%29%3D%28Qh%2AVb%2Aa%29%5C%5Cwhere%20%5C%5Cg%3Dgravity%20%5Bm%2Fs%5E2%5D%5C%5Ca%3Dacceleration%20%5Bm%2Fs%5E2%5D)
b) a = 19.61[m/s^2]
Explanation:
The total mass of the balloon is:
![massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\](https://tex.z-dn.net/?f=massball%3Ddensityheli%2Avolumeheli%5C%5C%5C%5Cmassball%3D0.41%20%5Bkg%2Fm%5E3%5D%2A0.048%5Bm%5E3%5D%5C%5Cmassball%3D0.01968%5Bkg%5D%5C%5C%5C%5C)
The buoyancy force acting on the balloon is:
![Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]](https://tex.z-dn.net/?f=Fb%3Ddensityair%2Agravity%2Avolumeball%5C%5CFb%3D1.23%5Bkg%2Fm%5E3%5D%2A9.81%5Bm%2Fs%5E2%5D%2A0.048%5Bm%5E3%5D%5C%5CFb%3D0.579%5BN%5D)
Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.
In the attached image we can see the free body diagram and the equation deducted by Newton's second law
False
Balanced forces mean that there is no net force acting on the object. therefore, the object will not accelerate.