A 0.77 kg spike is hammered into a railroad tie. The initial speed of the spike is equal to 3.5 m/s. If

Question

A 0.77 kg spike is hammered into a railroad

tie. The initial speed of the spike is equal to

3.5 m/s.

If the tie and spike together absorb 88.3

percent of the spike’s initial kinetic energy

as internal energy, calculate the increase in

internal energy of the tie and spike.

Answer in units of J.

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Thanh Thu 2 months 2021-07-31T19:41:18+00:00 2 Answers 4 views 0

Answers ( )

    0
    2021-07-31T19:42:35+00:00

    Answer:

    ∆E = 4.16J

    Change in internal energy of the tie and spike is 4.16J

    Explanation:

    Given;

    Mass of spike m = 0.77kg

    Initial Speed of spike v = 3.5m/s

    Efficiency = 88.3%

    Kinetic energy of spike can be calculated as;

    K.E = 1/2 × mv^2

    Substituting the values;

    K.E = 1/2 × 0.77 × 3.5^2

    K.E = 4.71625J

    Change in internal energy of the tie and spike;

    ∆E = 88.3% of K.E

    ∆E = 0.883 × 4.71625J

    ∆E = 4.16J

    Change in internal energy of the tie and spike is 4.16J

    0
    2021-07-31T19:42:51+00:00

    Answer:

    4.168  J

    Explanation:

    Applying the formula of kinetic energy,

    Ek = 1/2mv²………………… Equation 1

    Where Ek = Kinetic energy of the spike, m = mass of the spike, v = initial velocity of the spike

    Given: m = 0.77 kg, v = 3.5 m/s.

    Substitute into equation 1

    Ek = 1/2(0.77)(3.5²)

    Ek = 4.72 J.

    From the question, if the tie and the spike together absorb, 88.3 percent of the spike’s initial kinetic energy.

    Ek’ = 88.3/100(4.72)

    Ek’ = 4.168  J

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