In 39 s ; a pump delivers 0.619 dm^3 of oil
into barrels on a platform 39 m above the
pump intake pipe. The density of the oil is
0:832 g/cm3 :
The acceleration of gravity is 9.8 m/s2 :
Calculate the work done by the pump. An-
swer in units of kJ.
part2
Calculate the power produced by the pump.
Answer in units of kW.
2007-01-19 14:17:37 · 3 answers · asked by kavita 1 in Science & Mathematics ➔ Physics
The change in energy of the oil (hence the work of the pump) is equal to
mass * gravity * change in height
mass = density * volume = .832g/cm3 * .619dm3 * 1000cm3/dm3= 515 g
work = 515 g * 9.8 m/s2 * 39m * .001 kg/g = 197 J or 0.197 kJ.
Power is work/time, or 0.197/39
2007-01-19 14:30:06 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I dm^3 = 1000cc
.619dm^3 = 619cc
619cc x .832g/cc = 515.008g. = .515008 kg
.515008kg x 39m = 20.09 kg-m of work.
Power = work / time = 20.09 kg-m/39sec.
Power = .515 kg-m/sec
2007-01-19 22:34:05 · answer #2 · answered by Richard S 6 · 0⤊ 0⤋
That's 0.619x1000x0.832=515.008 grams=0.515 kg. So work done=0.515x39metresx9.8=196.8 joules=0.1968 kJ. So power output=196.8/39secs=5.05 watts=0.005 kW.
2007-01-19 22:32:44 · answer #3 · answered by zee_prime 6 · 0⤊ 0⤋