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a box has a momentum of 1.30 kg*m/s to the right. a 12.8 N force pushes it to the right for 0.218 s. what is the final momentum of the box. PLEASE HELP

Respuesta :

nkirotedoreen46

Answer:

4.09 kgm/s

Explanation:

Concept tested: Second Newton's law of motion

So what does the second Newton's law of motion state?

  • According to the second Newton's law of motion, the resultant force and the rate of change of momentum are directly proportional.
  • That is; [tex]F=\frac{(mVf-mVo)}{t}[/tex], where, mVf is the final momentum and mVo is the initial momentum.

In this case;

  • Force = 12.8 N
  • Initial momentum, mVo = 1.3 kgm/s
  • Time, t = 0.218 seconds

Therefore; replacing the values of the variables in the formula;

we get;

[tex]12.8N=\frac{(mVf-1.3kgm/s)}{0.218s}[/tex]

[tex](12.8N)(0.218s)=(mVf-1.3kgm/s)[/tex]

[tex]2.7904+1.3=mVf[/tex]

Solving for final momentum;

[tex]mVf =4.0904 kgm/s[/tex]

      = 4.09 kgm/s

Therefore, the final momentum of the box is 4.09 kgm/s

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