3. What is the minimum amount of energy required for a 70.0-kg climber carrying a 20.0-kg...

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Answer 1

Answer #1

$3.$ $E = \Delta U = mgh$

$E = (70+20) \times 9.8 \times 8850 = 7805.7kJ$

$4.$

$\frac{KE_1}{KE_2 } = \frac{m_1 v_1^2}{m_2v_2 ^2} = \frac{P_1 v_1 } {P_2v_2 } = \frac{P_1 }{2P_2}$

$P_1 = P_2$ By conservation of momnetum

$\frac{KE_1}{KE_2} = \frac{1}{2}$

$\frac{P_1}{P_2} = \frac{1}{1}$

Answer 2

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