How does the temperature of water change when it is heated on a stove top and then begins to boil?
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Answer:
v=0.60 m/s
Explanation:
Given that
m ₁= 390 kg ,u ₁= 0.5 m/s
m₂ = 250 kg ,u₂ = 0.76 m/s
As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.
Pi = Pf
m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v
Now putting the values in the above equation
390 x 0.5 + 250 x 0.76 = (390 + 250 ) v
v=0.60 m/s
Therefore the velocity of the system will be 0.6 m/s.