When radio waves try to pass through a city, they encounter thinvertical slits: the separations between the buildings. this causesthe radio waves to diffract. in this problem, you will see howdifferent wavelengths refract as they pass through a city andrelate this to reception for radios and cell phones. you will usethe angle from the center of the central intensity maximum to thefirst intensity minimum as a measure of the width of the centralmaximum (where nearly all of the diffracted energy isfound). consider radio waves of wavelength λ entering a city where the buildings have an average separation of a. part a find the angle θ to the first minimum from the center of the central maximum. express your answer in terms of λ and a

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    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