Apinhole camera is just a rectangular box with a tiny hole in one face. the film is on the face opposite this hole, and that is where the image is formed. the camera forms an image without a lens. (a) make a clear ray diagram to show how a pinhole camera can form an image on the film without using a lens. (hint: put an object outside the hole, and then draw rays passing through the hole to the opposite side of the box.) (b) a certain pinhole camera is a box that is 25 cm square and 20.0 cm deep, with the hole in the middle of one of the 25 cm x 25 cm faces. if this camera is used to photograph a canada goose that is 18 cm high and 1.5 m in front of the camera, how large is the image of this bird on the film? what is the lateral magnification of this camera?

Pe = 1/2k(x)^2  and .05n/m = .05x10^-9m = x 

    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

That big glowy thing in the sky is called the sun