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16 February, 01:10

A 2200 kg car doubles its speed from 50 km/hr to 100 km/hr. By how many times does the kinetic energy from the car's forward motion increase?

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Answers (2)
  1. 16 February, 01:40
    0
    Answer: 4

    Explanation:

    Given that:

    Mass of car M = 2200 kg

    Initial speed Vi = 50 km/hr

    Final speed Vf = 100 km/hr

    Kinetic energy is the energy possessed by a moving object. It is measured in joules, and depends on the mass (m) of the object and the speed (v) by which it moves

    i. e K. E = 1/2MV^2

    So, when traveling at 50 km/h

    KE = 1/2x 2200kg x (50km/h) ^2

    KE = 0.5 x 2200 x 2500

    KE1 = 2750000J

    So, when traveling at 100 km/h

    KE = 1/2x 2200 x (100 km/h) ^2

    KE = 0.5 x 2200 x 10000

    KE2 = 11000000J

    Thus, the number of times kinetic energy increases is obtained by dividing KE2 by KE1

    i. e 11000000J / 2750000J

    = 4

    Thus, the kinetic energy from the car's forward motion increase 4 times
  2. 16 February, 02:52
    0
    Answer:4

    Explanation:m=2200, u=50, v=100

    The unit for speed is in m/s we have to conver to m/s

    For ke1=50km/h to m/s 50*1000/3600=13.89

    So keep=1/2mv^2

    1/2 (2200) 13.89^2 = 212225.31

    Ke2=1/2mv^2=100*1000/3600=27.78

    1/2mv^2 = 1/2 (2200) 27.78^2

    =848901.24 ... soke2/ke1 = 848901.24/212225.31=4

    That d increase in ke
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