PLEASE HELP 1. Find the distance to the image in the following curved mirror problems and state whether the image is virtual or real AND whether it is upright or inverted. (4 points each)
a. An object is placed 1.0 cm from a convex mirror with a focal length of 4.0 cm.
b. An object is placed 12 cm from a convex mirror with a focal length of 5.0 cm.
c. An object is placed 110 mm from a concave mirror with a focal length of 5.0 cm.
d. An object is placed 9.0 cm from a concave mirror with a focal length of 9.0 cm.
e. An object is placed 1.5 cm from a concave mirror with a focal length of 3.0 cm.

Gravity bc it’s gravity

No because scientist are still changing there mines today so they are not correct to this day

    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