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Sagot :
Answer:
Explanation:
The work-energy theorem says that the change in kinetic energy of a system is equal to the work done.
[tex]\frac{1}{2}mv^2_f-\frac{1}{2}mv^2_i=\Delta W^{}[/tex][tex]\frac{1}{2}m(v^2_f-v^2_i)=\Delta W^{}[/tex]Now in our case
[tex]\begin{gathered} m=1079\operatorname{kg} \\ vf=\text{unknown} \\ v_i=3.00m/s \\ \Delta W=2.00\times10^5J \end{gathered}[/tex]Therefore, the above equation gives
[tex]\frac{1}{2}(1079)(v^2_f-(3.00)^2^{}_{})=2.00\times10^5J[/tex]Now we need to solve for v_f .
Mutlipying both sides by 2 gives
[tex](1079)(v^2_f-(3.00)^2_{})=2.00\times10^5\times2[/tex]dividing both sides by 1079 gives
[tex](v^2_f-(3.00)^2_{})=\frac{2.00\times10^5J}{1079}\times2[/tex]Finally, adding 3.00^2 to both sides gives
[tex]v^2_f=\frac{2.00\times10^5J}{1079}\times2+3.00^2[/tex]Finally, simplifying the right-hand side gives
[tex]v^2_f=379.71[/tex]taking the square root of both sides gives
[tex]\boxed{v_f=19.49m/s}[/tex]Hence, the final speed of the car is 19.49 m/s.
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