A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute the energy lost during the impact. Express your answer as an absolute value ΔE| and as a percentage n of the original system energy E.

Explanation:

The given data is as follows.

          Mass, m = 75 g

         Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

where,   = mass of the projectile

            = mass of block

              v = velocity after the impact

Now, putting the given values into the above formula as follows.

                                  =

                              v = 0.898 m/s

Now, equation for energy is as follows.

               E =

                  =

                  = 13500 J

Now, energy after the impact will be as follows.

             E' =

                 = 20.19 J

Therefore, energy lost will be calculated as follows.

            = E  E'

                       = (13500 - 20) J

                       = 13480 J

And,   n =

             =

             = 99.85

             = 99.9%

Thus, we can conclude that percentage n of the original system energy E is 99.9%.

hello, i am papaguy.

i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!