Athin, diverging lens having a focal length of magnitude 45.0 cm has the same principal axis as a concave mirror with a radius of 60.0 cm. the center of the mirror is 20.0 cm from the lens, with the lens in front of the mirror. an object is placed 23.0 cm in front of the lens.
1) where is the final image due to the lens-mirror combination? enter the image distance with respect to the mirror. follow the sign convention. (express your answer to three significant figures.)
2) is the final image real or virtual?
3) is the final image upright or inverted?
4) suppose now that the concave mirror is replaced by a convex mirror of the same radius. where is the final image due to the lens-mirror combination? enter the image distance with respect to the mirror. follow the sign convention. (express your answer to three significant figures.)

    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