A battery does work in moving charge around a circuit i. e. sustaining a current through the circuit. To illustrate this point, consider a resistor with a voltage V across it and a current I flowing through it. Focus on a single charge, q, passing through the resistor. Find the work W done on the charge by the electric field in the resistor. When thinking about an electric circuit, you usually focus not on the motion of individual charges, but rather on the continuous current (charge per unit time) flowing through the circuit. Thus, rather than considering the work done on a particular charge, it is useful to compute the work done per unit time on the charge flowing through the circuit, or in other words, the power. Find the electrical power P delivered to the resistor via the work done on the individual charges passing through it. (Again, this power ultimately appears in the form of heat).

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    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

Answerapplying a direct current to a non magnetic material in order to create a electromagnetic field; plz check first good luck;