A mass of 0.079 kg hangs from a vertical spring in the lab room. You pull down on the mass and throw it vertically downward. The speed of the mass just after leaving your hand is 3.80 m/s.(a) While the mass moves downward a distance of 0.11 m, how much work was done on the mass by the Earth? Include the appropriate sign. Work done by Earth = J(b) At the instant in part (a) when the mass has moved downward a distance of 0.11 m, the speed of the mass has decreased to 1.52 m/s. How much work was done on the mass by the spring? Include the appropriate sign. Work done by spring = J

Answer (d) valerie rides her bike south at 8 kilometers per hour///owen jogs south at 8 kilometers per hour ;

    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

Ibelieve the answer is d. densityhope my answer has you : )

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