Technician a says that the burning of the air-and-fuel mixture in the cylinder is called fuel activation. technician b says that the basic function of the upper-end components is to control the flow of air-and-fuel mixture into the engine and the flow of exhaust gases out of the engine after combustion. who is right? a. both technicians a and b b. technician a only c. neither technician a nor b d. technician b only

The answer is A. Both Technicians A and B. Technician B is correct because the basic function of the upper-end components of an air-fuel mixture is to control the flow of it into the mixture. And also the statement of technician A is correct also.

10 is the . . .

    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