A 2100 kg truck traveling north at 38 km/h turns east and accelerates to 55 km/h. (a) What is the change in the truck’s kinetic energy

Question

A 2100 kg truck traveling north at 38 km/h turns east and accelerates to 55 km/h. (a) What is the change in the truck’s kinetic energy

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Kiệt Gia 3 months 2021-08-07T20:23:25+00:00 1 Answers 4 views 0

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    2021-08-07T20:25:21+00:00

    Answer:

    Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J

    Explanation:

    Given:

    Mass of truck(m) = 2,100 kg

    Initial speed(v1) = 38 km/h = 38,000 / 3600 = 10.56 m/s

    Final speed(v2) = 55 km/h = 55,000 / 3600 = 15.28 m/s

    Find:

    Change in kinetic energy (ΔKE)

    Computation:

    Change in kinetic energy (ΔKE) = 1/2(m)[v2² – v1²]

    Change in kinetic energy (ΔKE) = 1/2(2100)[15.28² – 10.56²]

    Change in kinetic energy (ΔKE) = 1,050[233.4784 – 111.5136]

    Change in kinetic energy (ΔKE) = 1,050[121.9648]

    Change in kinetic energy (ΔKE) = 128063.04

    Change in kinetic energy (ΔKE) = 12.8 × 10⁴ J

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