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Juliette [100K]
4 years ago
12

A 70 kg person running at 4 m/s what is the kinetic energy

Physics
1 answer:
gladu [14]4 years ago
3 0

Let:

m = mass (kg)

v = velocity (m/s)

KE = 0.5m(v^2)

KE = 0.5(70)(4^2)

KE = 560 Joules.

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The value is  v  =  2.3359 *10^{4} \ m/s

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From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

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Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

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And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

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Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

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Generally from the law energy conservation we have that

        T__{E}} =  KE_p

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       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

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