Answer:
zero
Explanation:
Work done = - 400 J
In an isothermal process, the temperature remains constant. So, according to the first law of thermodynamics
dQ = dU + dW
As the temperature remains constant and the change in internal energy is the function of change in temperature, here change is temperature is zero so the change in internal energy is also zero.
Given that,
Mass of trackler, m₁ = 100 kg
Speed of trackler, u₁ = 2.6 m/s
Mass of halfback, m₂ = 92 kg
Speed of halfback, u₂ = -5 m/s (direction is opposite)
To find,
Mutual speed immediately after the collision.
Solution,
The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.
Answer:
a) P1=100kpa
V1=6m³
V2=?
P2=50kpa
rearranging mathematically the expression for Boyle's law
V2=(P1V1)/P2=(100×6)/50=12m³
b) same apartment as in (a) but only the value of P2 changes
=> V2=(100×6)/40=15m³
Explanation:
since temperature is not changing we use Boyle's law. mathematically expressed as P1V1=P2V2