Adouble-slit experiment is set up using red light (l = 701 nm). a first order bright fringe is seen ata given location on a screen. what wavelength of visible light(between 380 nm and 750 nm) would produce a dark fringe at theidentical location on the screen?

b) a new experiment is created with the screen at a distance of 2.1m from the slits (with spacing 0.12 mm). what is the distancebetween the second order bright fringe of light with l = 696 nm and the third order bright fringe oflight with l = 407 nm? (give theabsolute value of the smallest possible distance between these twofringes: the distance between bright fringes on the same side ofthe central bright fringe.)

Domain eukarya; kingdom plantae should be the answer.

    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

490 n  f=125n a=2.5m/s2