This is the correct equation for centripetal acceleration—acceleration towards the center of motion. in this case it isn't necessary to introduce a numerical factor. such a factor is often displayed explicitly as a constant k in front of the right-hand side; for example, a = kv2/r. as it turns out, k = 1 gives the correct expression. a good technique sometimes introduced in calculus-based textbooks involves using unknown powers of the dimensions. this problem would then be set up as [a] = [v]b[r]c. writing out the dimensions and equating powers of each dimension on both sides of the equation would result in b = 2 and c = −1.

    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

Answershe should not have multiplied the sulfur atoms by the subscript 4