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Paraphin [41]
3 years ago
6

What is the kinetic energy of a 74.0 kg skydiver falling at a terminal velocity of 52.0m/s?

Physics
1 answer:
skad [1K]3 years ago
8 0

Answer:

100,048

Explanation:

K.E = 1/2 m (v)^2

K.E = 1^/2 * 74 * (52)^2

K.E = 100,048J =100.048kJ

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The initial kinetic energy of the two blocks together, just before entering the rough path, is
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\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
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