A bucket of water of mass 14.8 kgkg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.330 mm with mass 11.8 kgkg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 11.0 mm to the water. You can ignore the weight of the rope. Part A What is the tension in the rope while the bucket is falling? Take the free fall acceleration to be ggg = 9.80 m/s2m/s2 . nothing N SubmitRequest Answer Part B With what speed does the bucket strike the water? Take the free fall acceleration to be ggg = 9.80 m/s2m/s2 . nothing m/s SubmitRequest Answer Part C What is the time of fall? Take the free fall acceleration to be ggg = 9.80 m/s2m/s2 . nothing s SubmitRequest Answer Part D While the bucket is falling, what is the force exerted on the cylinder by the axle? Take the free fall acceleration to be ggg = 9.80 m/s2m/s2 .

Multiply the 3 dimensions, assuming it is a rectangular cuboid, to get the volume in cm³ divide mass by volume to get density in g/cm³

Answer form flower;

    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

Ithink the answer is d