physics

posted by .

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=

  • physics -

    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.

  • physics -

    THANK YOU

    HOW WOULD I WRITE THE f & r IN SCIENTIFIC NOTATION

  • physics -

    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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Physics

    A 1100 kg car is moving toward the north along a straight road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to rest in a distance of 140 m. (a) If the road is completely level, what is the constant force …
  2. Physics

    A 1100 kg car is moving toward the north along a straight road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to rest in a distance of 140 m. (a) If the road is completely level, what is the constant force …
  3. physics

    A 1000 kg car in neutral gear slow from 15 m/s to a stop in 5.0 s due to the force of friction. A. What was the acceleration of the car. B. What was the car’s initial momentum?
  4. Physics

    A 1000-kg car is driving toward the north along a straight road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to a rest in a distance of 140 m. What is the constant force applied to the car to bring it to …
  5. physics

    A force of -9000 N is used to stop a 1500kg car traveling at 20 m/s. What braking distance is needed to bring the car to a halt?
  6. Physics

    A car travels at 64 mi/h when the brakes are suddenly applied. Consider how the tires of a moving car come in contact with the road. When the car goes into a skid (with wheels locked up), the rubber of the tire is moving with respect …
  7. physics

    A 1,500 kg car starts from rest and accelerates to a speed of 12 m/s over a distance of 68 m a) calculate the total work on the car b) calculate the net force that caused the car to accelerate (constant force)
  8. Physics

    A 1550 kg car is traveling with a speed of 12.4 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 50.3 m?
  9. Physics

    Calculate the force needed to bring a 1100–kg car to rest from a speed of 83.0 km/h in a distance of 118 m (a fairly typical distance for a non-panic stop). Suppose instead the car hits a concrete abutment at full speed and is brought …
  10. Physics

    A force of -9000N is used to stop a 1500 kg car traveling at 20m/s. What braking distance is needed to bring the car to a halt?

More Similar Questions