physics

Calculate the force needed to bring a 1070–kg car to rest from a speed of 89.0 km/h in a distance of 110 m (a fairly typical distance for a non-panic stop).

Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.34 m. Calculate the force exerted on the car and compare it with the force found in part (a).
FORCE=
RATIO=

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asked by stephanie
  1. Vo = 89 km/h = 89,000m/3600s = 24.72 m/s.

    A. V^2 = Vo^2 + 2a*d.
    0 = 24.72^2 + 220a.
    a = -2.78 m/s^2.

    F = M*a = 1070 * (-2.78) = -2972 N. or 2972 N. of opposing force.

    B. V^2 = Vo^2 + 2a*d.
    0 = 24.72^2 + 4.68a.
    a = -130.6 m/s^2.

    F = M*a = 1070 * (-130.6) = -139,712 N.

    Ratio = 139,712/2972 = 47 times part A.

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    posted by Henry
  2. THANK YOU

    HOW WOULD I WRITE THE f & r IN SCIENTIFIC NOTATION

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    posted by stephanie
  3. In Scientific notation, the number on the left side of the decimal is always less than 10(a single digit number):

    F = 139,712. = 1.39712*10^5.
    If I had placed the decimal between 13 and 9, that would not be in scientific notation; because 13 is not a single digit number.

    After locating the decimal point, I count the number of decimal places to the right(5). So that is my exponent; it is positive because I counted to the right of the decimal point.

    Ratio = 47. = 4.7*10^1.
    After locating the decimal, I counted only 1 decimal place to the right. Therefore, the exponent is 1.

    For numbers less than 1, the exponent will always be negative:

    0.0058 = 5.8*10^-3.
    Now look at 0.0058 and count the number of decimal places to the LEFT of your new decimal location(3).
    The exponent is negative because we counted to the left of our new decimal point.

    I hope this helps.

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    posted by Henry

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