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Sagot :
Answer:
[tex]F_A =5.625*10^1^6N[/tex]
Explanation:
From the question we are told that
50,000 years ago,
A giant 4.5 107-kg meteor
180-m-deep hole
20,000 m/s
Generally for this problem the energy change is given as
[tex]\triangle E=\frac{1}{2} mv^2 +mgd[/tex]
Having the potential and kinetic energy in place
Mathematically solving for Average force[tex]F_A[/tex]
[tex]\triangle E=F_a*d[/tex]
[tex]F_A =\frac{1/2* 4.5*10^7(20,000)^2-kg+4.5*10^7*9.81*160}{160}[/tex]
Therefore Average force [tex]F_A[/tex] is given by
[tex]F_A =5.625*10^1^6N[/tex]
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