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MTALHAHASSAN2

  • one year ago

Need help!! A car skids to a stop on a road with a uk of 1.1. If the intial speed of car is 50 km/h how many meters long are the skid marks.

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  1. Michele_Laino
    • one year ago
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    here we have to keep in mind that the work done by the friction force has to be equal to the kinetic energy change, so we can write: \[\Large {\mu _k}mgd = \frac{1}{2}m{v^2}\] where m is the mass of your car, and d is the requested distance. Furthermore g is the earth gravity, and v is the velocity of our car

  2. MTALHAHASSAN2
    • one year ago
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    |dw:1435905880197:dw|

  3. MTALHAHASSAN2
    • one year ago
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    |dw:1435905970542:dw|

  4. MTALHAHASSAN2
    • one year ago
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    |dw:1435906093552:dw|

  5. MTALHAHASSAN2
    • one year ago
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    I just covert 50 km/h in m/s

  6. MTALHAHASSAN2
    • one year ago
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    @Michele_Laino

  7. MTALHAHASSAN2
    • one year ago
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    |dw:1435906235486:dw|

  8. MTALHAHASSAN2
    • one year ago
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    Is i do it right

  9. MTALHAHASSAN2
    • one year ago
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    plz plz help me

  10. IrishBoy123
    • one year ago
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    you dropped g out of your calculation so you are out by a factor of about 10

  11. MTALHAHASSAN2
    • one year ago
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    @IrishBoy123 from which step

  12. IrishBoy123
    • one year ago
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    |dw:1435912438381:dw|

  13. IrishBoy123
    • one year ago
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    also, does 89m sound realistic?!

  14. MTALHAHASSAN2
    • one year ago
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    Idk i just get that answer

  15. MTALHAHASSAN2
    • one year ago
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    ut the answer at the back of my book is 8.64 m [F]

  16. MTALHAHASSAN2
    • one year ago
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    But*

  17. IrishBoy123
    • one year ago
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    i am referring to your answer given above - ie 89m. |dw:1435912972594:dw| i get \(\frac{50,000^2}{3600^2 \times 2 \times 1.1 \times 9.8} = 8.95m\)

  18. MTALHAHASSAN2
    • one year ago
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    |dw:1435912949019:dw|

  19. IrishBoy123
    • one year ago
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    are you finished?

  20. MTALHAHASSAN2
    • one year ago
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    I still don't get the answer

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