A 12 kg box sliding on a horizontal floor has an initial speed of 4.0 m/s. The coefficient of friction between the box and the floor is 0.20. The box moves a distance of 4.0 m in 2.0 s. The magnitude of the change in momentum of the box during this time is most nearly

47.04 N*s or 47.04 kg*m/s

Explanation:

momentum is equal to mass times velocity the mass stays constant so we need to find the final and initial velocity. We are given the initial velocity of 4.0 m/s so plugging that into the momentum formula that is an initial momentum of 48 kg*m/s.

Now we can solve for  the final velocity by using the kinematic formulas. First we must find the net force which is the friction so friction is equal to coefficient * normal force with the normal force equaling the force of gravity. Therefore we get a frictional force  of 23.52 N

Now to find the acceleration we use newtons second a=f/m and get acceleration is equal to 1.96 m/s^2.

Now just find the Vfinal; Vfinal= Viinital +acceleration* time.

since the acceleration is negative we get 0.08m/s. Now just plug in to find final momentum of .96kgm/s. Then find the difference to be -47.04 kg*m/s.

Alternatively if you know that impulse is equal to change in momentum and equals force times time. We know the friction force is 23.52N in the negative direction applied for 2 seconds so we once again get  -47.04 N*s which is the same unit as kg*m/s

however since the question just asks for magnitude we take the absolute value and get 47.04 N*s

you have to draw the persons position over there

probably not the "magnets"in the magnet, so to speak, are thought to be electrons orbiting in shells. vurrent loops two sides to the loop