Areducing elbow in a horizontal pipe is used to deflect water flow by an angle θ = 45° from the flow direction while accelerating it. the elbow discharges water into the atmosphere. the cross-sectional area of the elbow is 150 cm2 at the inlet and 25 cm2 at the exit. the elevation difference between the centers of the exit and the inlet is 40 cm. the mass of the elbow and the water in it is 50 kg. determine the anchoring force needed to hold the elbow in place. take the momentum-flux correction factor to be 1.03 at both the inlet and the outlet. the density of water is 1000 kg/m3, and the flow rate of water is 30 kg/s.

The third hammer 8 pounds

    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

Since the pully is a simple machine so the mechanical advantage is a 1