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 .

Explanation:

The given data is as follows.

       initial speed (u) = 4.0 m/s,     mass (m) = 12 kg

      Distance (s) = 4.0 m,    time (t) = 2.0 sec

First, we will calculate the acceleration as follows.

           s =

         4 =

         a = -2

Now, the final speed will be calculated as follows.

            v = u + at

               =

              = 0

Therefore, change in momentum will be calculated as follows.

            = m(v - u)

                           =

                           = -48 kg m/s

The negative sign indicates the change in momentum.

Thus, we can conclude that the change in momentum of the box during this time is most nearly 48 kg m/s.

  the unit of the measure =

explanation:

  the property that is quantified by the energy needed per of material to raise the temperature by one.

  si unit of energy = joule (j)

  si unit of temperature = kelvin (k)

  si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg