Every 2 seconds, a coal fired power plant produces enough electricity to do 9X10^8 J of work. What is the plant's power output? *

2 Answers
Douglas K.
May 10, 2018

#"Power" = 4.5xx10^8" W"#

Explanation:

The SI unit #"Joule"# can be decomposed into basic units in many ways; one way is:

#"Joule" = "Watts"xx "seconds"#

Therefore, we can divide the energy (in #"Joules"#) by a specified time interval (in #"seconds"#) and obtain the average power output (in #"Watts"#) for that time interval:

#"Power" = (9xx10^8" J")/(2" s")#

#"Power" = 4.5xx10^8" W"#

Nam D.
May 11, 2018

#4.5*10^8# watts

Explanation:

Power is given through the equation,

#P=W/t#

where:

  • #P# is the power in watts

  • #W# is the work done in joules

  • #t# is the time taken in seconds

So, we get:

#P=(9*10^8 \ "J")/(2 \ "s")#

#=4.5*10^8 \ "W"#

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