Amolten salt reactor using thorium-232 as a blanket and uranium-233 as the fuel is being planned to run entirely off of the u-233 produced from the blanket. it is currently being fueled only by u-235. because the th-232 in the blanket can be pumped continuously from the blanket surrounding the core, pa-233 can be extracted before any radiative capture can occur, thereby removing any losses of pa-233 due to neutron absorption. the pa-233 is stored in a container and the u-233 is accumulated outside the core. a. write down the nuclear reactions that will lead to the production of u-233 from the neutron capture of th-232. why can you ignore th-233 for the rest of this problem? b. set up the rate equations you would use to analyze the pa-233 and u-233 content over time. c. how much th-232 would be needed produce 200 kg of u-233 fuel from the blanket in one year? you can keep your answers in terms of number of atoms. q = 1014 neutrons/cm² s oc of th-232 = 8 barns

50 meters per 5 seconds

Wooden , walls ,doors ,support

    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