contestada

2. What is the power rating of an engine capable of lifting a 100 kg object 5 m vertically
in 4 seconds?

Respuesta :

nasser3
Work=f.d
Work=100*50 = 500
Power = work/time = 500/4
=125 watt
Cricetus

The power rating of an engine will be "125 watt".

Given values are:

Force,

  • f = 100

Time,

  • t = 4 seconds

Now,

→ [tex]Work \ done= f\times d[/tex]

                     [tex]= 100\times 50[/tex]

                     [tex]= 500[/tex]

hence,

The power will be:

= [tex]\frac{Work}{time}[/tex]

= [tex]\frac{500}{4}[/tex]

= [tex]125 \ watt[/tex]

Thus the response above is right.

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https://brainly.com/question/13534333

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