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physics

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What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 95 km/h?

i got the answer to be 13062.5 N. i don't think that's right...

  • physics - ,

    Impluse= change in momentum
    Force*time= 1100*26.4m/s

    I don't get your numbers, about one fourth of that. Did your conversion of km/hr to m/s same as mine?

    Here is a neat trick. Put in the google search window...

    1100kg*95km/hr* 1/(8seconds) in newtons

  • physics - ,

    First convert kg/h to m/s
    95 km/h = (95000 m/h)/(3600) s/h
    = 26.4 m/s

    Force x Time = momentum change
    = 1100*26.4 = 2.90*10^4 kg m/s

    Divide that by the time interval (8.0s) to get the average force in Newtons

  • physics - ,

    Change velocity from km/h to m/s
    95 km/h*(1000 m/km)*(1 h/3600 s)= 26.4 m/s
    a=ΔV/Δt=26.4/8=3.3 m/s^2
    ΣF=m*a=1100kg*3.3 m/s^2=3630 N

    Tada!

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