A1 mol argon system in a 5m^3 container at 300k is well approximated as an ideal gas. the gas is adiabatically compressed to 4m^3, increasing the temperature to 350k.
i. determine the heat transfer, work done, and change in internal energy of the system during the compression.
ii. the system is instead taken from the original state, isothermally compressed to 4m3, then isochorically heated to 350k. what is the heat transfer, work done, and change in internal energy of this transformation?

Answer: fn = wcos50 = 38.89(9.81)(cos50) = 245.23 n

Ithink they do have the same properties just in less quantities i think

Sorry my friend i don't know the answer. i am just in 6th

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s