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(a) Find the useful power output (in W) of an elevator motor that lifts a 2800 kg load a height of 35.0 m in 12.0 s, if it al

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(2800).8) (3S) 12 2 二 186700W )cost860 kw) 8670KW)[ 0:09 00 A t 58 2211 О-20 2 1165.5 3/s

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