Aballoon is released from a tall building. the total mass of the balloon including the enclosed gas is 2.0 kg. its volume is 5.0 m3. the density of air is 1.3 kg/m3. will the balloon rise, fall, or remain stationary; and why?
(a) the balloon will fall because the upward buoyant force is less than its weight.(b) the balloon will rise because the upward buoyant force is greater than its weight.(c) the balloon will fall because the downward buoyant force is greater than the upward buoyant force.(d) the balloon will fall because its density is greater than that of air.(e) the balloon will remain stationary because its density is less than that of air.

    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

No. rocks can follow various pathways.