Math Group
Question:
lala2pretty
Student
College - Sophomore
A rain cloud contain 5.37 x 10 to the seventh power kg of water vapor.
The acceleration of gravity is 9.81 m/s^2
How long would it take for a 1.53kW pump to raise the same amount of water to the cloud's altitude of 1.54 km? Answer in units of s.
Answers:
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krishna-agrawala Teacher
Graduate SchooleNotes Editor
Posted by krishna-agrawala on Wednesday November 4, 2009 at 6:37 AMGiven:
mass of water = m = 5.37 x 10^7 kg
Displacement of water required = s = 1.54 km = 1540 m
Acceleration of gravity = g = 9.81 m/s^2
Power of pump = 1.53 kW = 1530 W
Total work or energy required to raise the water to required height is given by formula:
Work = m*g*s* = (5.37 x 10^7)*9.81*1540 = 811.26738 x 10^12
Time take to do this work is given by the formula:
Time = Work/(Power of pump) = (811.26738 x 10^12)/1530
= 530.2401176 x 10^9 s
Answer:It will take 530.240 x 10^9 s for the pump.
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neela Teacher
Graduate SchooleNotes Editor
Posted by neela on Wednesday November 4, 2009 at 7:46 AMThe mass of water to be be pumped=5.37*10^7 kg.
The energy required to raise the mass m=5.37*10^7 kg to the height of h = 1.54km=1540 meter is = mgh, where g is the acceleration due togravitaty =9.981m/s^2
mgh =5.37*10^7*9.81*1540 J
The energy generation by the 1.53 KW =1.53KW=1530 watt per second
Threfore, the time in seconds the pump could produce the energy to lift the mass of 5.37*10^ 7 kg of water to a height of 15400 metr = 5.37*10^7*9.81*1540/ (1530) seconds.
=530240117.6 seconds.
