what is the power output needed from a motor to lift in the absence of friction a mass of 1.5 ×10⁴ kg 25 m in 6.0 s at constant speed​

the power output needed is  61.25 × 10⁴

Explanation:

The computation of the power output needed is shown below;

Given that

m =  1.5 ×10⁴ kg

v = 25m  ÷ 6.0 s

And g = 9.8 m/sec^2

Now based on the above information

p = f × v

= mg × v

= 1.5 ×10⁴ × 9.8 × 25 ÷ 6

= 61.25 × 10⁴

Hence, the power output needed is  61.25 × 10⁴

- measure reactants precisely. ( first choice)

- stir to the reaction take place. ( second choice)

- wait patiently until the reaction completes. ( fourth choice)

- make sure to recover as much copper as possible. ( last choice)

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr